[{"Name":"Rate of Reaction","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Rate of Reaction","Duration":"5m 37s","ChapterTopicVideoID":28813,"CourseChapterTopicPlaylistID":292227,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:05.110","Text":"In this video, we\u0027ll begin our study of Kinetics."},{"Start":"00:05.110 ","End":"00:09.300","Text":"We\u0027re going to talk about chemical kinetics."},{"Start":"00:09.300 ","End":"00:14.415","Text":"Thermodynamics tells us whether a reaction can take place,"},{"Start":"00:14.415 ","End":"00:17.755","Text":"but it doesn\u0027t tell us how fast that will happen."},{"Start":"00:17.755 ","End":"00:20.715","Text":"That\u0027s why we need kinetics."},{"Start":"00:20.715 ","End":"00:25.680","Text":"Chemical kinetics tells us about the rate at which a reaction will take"},{"Start":"00:25.680 ","End":"00:32.340","Text":"place and gives us information about the route taken from reactants to products."},{"Start":"00:32.340 ","End":"00:36.660","Text":"We need to define the rate of reaction."},{"Start":"00:36.660 ","End":"00:39.045","Text":"Supposing we have a reaction,"},{"Start":"00:39.045 ","End":"00:42.230","Text":"a moles of A plus b moles of B,"},{"Start":"00:42.230 ","End":"00:47.870","Text":"and so on to give g moles of G plus h moles of H and so on."},{"Start":"00:47.870 ","End":"00:52.519","Text":"Then the average rate of reaction is defined as follows."},{"Start":"00:52.519 ","End":"00:59.510","Text":"The average rate is written as minus 1 over a is the coefficient here,"},{"Start":"00:59.510 ","End":"01:07.830","Text":"times Delta concentration of A divided by Delta t. That\u0027s also equal to minus 1"},{"Start":"01:07.830 ","End":"01:16.185","Text":"over b Delta concentration of B divided by Delta t. Then that\u0027s equal to 1 over g,"},{"Start":"01:16.185 ","End":"01:22.650","Text":"Delta concentration of G divided by Delta t. That\u0027s equal to 1 over"},{"Start":"01:22.650 ","End":"01:26.490","Text":"h times Delta the concentration of H divided by"},{"Start":"01:26.490 ","End":"01:31.815","Text":"Delta t. What are these Delta concentrations?"},{"Start":"01:31.815 ","End":"01:34.265","Text":"If we have a substance X,"},{"Start":"01:34.265 ","End":"01:38.735","Text":"delta of the concentration of X is the concentration of X"},{"Start":"01:38.735 ","End":"01:44.035","Text":"at time t_2 minus the concentration at time t_1."},{"Start":"01:44.035 ","End":"01:50.295","Text":"Delta t is the time that t_2 minus the time t_1."},{"Start":"01:50.295 ","End":"01:54.315","Text":"Now, the rate itself varies with time."},{"Start":"01:54.315 ","End":"02:00.905","Text":"The rate of reaction is actually the average rate of reaction over a period of time."},{"Start":"02:00.905 ","End":"02:08.090","Text":"Now, why do we need the negative sign before Delta A?"},{"Start":"02:08.090 ","End":"02:13.930","Text":"Here is a plot of the concentration versus time."},{"Start":"02:13.930 ","End":"02:17.410","Text":"If we look at A or B,"},{"Start":"02:17.410 ","End":"02:22.235","Text":"we see that the concentration decreases over time."},{"Start":"02:22.235 ","End":"02:29.160","Text":"If we take this time as t_1 and this time as t_2,"},{"Start":"02:29.160 ","End":"02:35.495","Text":"then the concentration at t_2 will be smaller than the concentration at t_1."},{"Start":"02:35.495 ","End":"02:40.585","Text":"Delta of the concentration will be negative."},{"Start":"02:40.585 ","End":"02:44.735","Text":"That\u0027s written here as the reactants A and B react,"},{"Start":"02:44.735 ","End":"02:48.800","Text":"their concentrations decrease with time so that"},{"Start":"02:48.800 ","End":"02:54.155","Text":"the concentration at time t_2 is less than the concentration of time t_1."},{"Start":"02:54.155 ","End":"02:59.930","Text":"Delta concentration of A will be negative, same for B."},{"Start":"02:59.930 ","End":"03:03.190","Text":"Now, the rate needs to be positive."},{"Start":"03:03.190 ","End":"03:07.670","Text":"We put a negative sign before Delta A and Delta B."},{"Start":"03:07.670 ","End":"03:10.205","Text":"That\u0027s the reason for the negative sign."},{"Start":"03:10.205 ","End":"03:12.965","Text":"Now, what about the products?"},{"Start":"03:12.965 ","End":"03:17.660","Text":"Here\u0027s a graph of concentration of the products versus time."},{"Start":"03:17.660 ","End":"03:25.310","Text":"We see that the concentration at t_2 here is greater than the concentration at t_1,"},{"Start":"03:25.310 ","End":"03:29.830","Text":"so Delta of the concentration is positive."},{"Start":"03:29.830 ","End":"03:33.920","Text":"That\u0027s written here as the products G and H are formed,"},{"Start":"03:33.920 ","End":"03:37.565","Text":"their concentrations increase with time so that"},{"Start":"03:37.565 ","End":"03:42.080","Text":"the concentration of G at t_2 is greater than the concentration at t_1,"},{"Start":"03:42.080 ","End":"03:46.320","Text":"and Delta of the concentration is positive."},{"Start":"03:46.320 ","End":"03:53.490","Text":"The same for H. We\u0027ll automatically get a positive rate of reaction."},{"Start":"03:53.720 ","End":"04:00.920","Text":"We divide by the stoichiometric coefficients so that all the rate expressions are equal."},{"Start":"04:00.920 ","End":"04:09.620","Text":"Because for example, a moles of A reacts at the same time as b moles of B."},{"Start":"04:09.620 ","End":"04:11.584","Text":"Here\u0027s an example."},{"Start":"04:11.584 ","End":"04:16.190","Text":"Consider the decomposition of hydrogen peroxide given by this equation."},{"Start":"04:16.190 ","End":"04:21.575","Text":"H_2O_2 decomposes to give us water and oxygen."},{"Start":"04:21.575 ","End":"04:25.925","Text":"Initially, the concentration of H_2O_2 is 3 molar."},{"Start":"04:25.925 ","End":"04:30.740","Text":"After 200 seconds, it goes down to 2.59."},{"Start":"04:30.740 ","End":"04:33.380","Text":"Calculate the average rate."},{"Start":"04:33.380 ","End":"04:37.430","Text":"The average rate is minus Delta,"},{"Start":"04:37.430 ","End":"04:43.100","Text":"the concentration of each tool to divide by Delta t. At 200 seconds,"},{"Start":"04:43.100 ","End":"04:45.985","Text":"we have a concentration of 2.59."},{"Start":"04:45.985 ","End":"04:50.690","Text":"That\u0027s minus the concentration at t equals 0."},{"Start":"04:50.690 ","End":"04:52.685","Text":"That\u0027s 0."},{"Start":"04:52.685 ","End":"04:56.460","Text":"That\u0027s 200 seconds."},{"Start":"04:56.810 ","End":"05:02.730","Text":"We have 2.59 minus 3 and that\u0027s divided by Delta t,"},{"Start":"05:02.730 ","End":"05:06.970","Text":"which is 200 minus 0, which is 200."},{"Start":"05:08.300 ","End":"05:18.405","Text":"This is minus, 0.41 turns into plus 0.41 divided by 200 seconds."},{"Start":"05:18.405 ","End":"05:22.220","Text":"We divide that out, we get 2.1 times 10 to the power minus"},{"Start":"05:22.220 ","End":"05:25.993","Text":"3 molar concentration per second,"},{"Start":"05:25.993 ","End":"05:29.265","Text":"so that is our rate."},{"Start":"05:29.265 ","End":"05:31.650","Text":"It\u0027s an average rate."},{"Start":"05:31.650 ","End":"05:37.470","Text":"In this video, we learned about the average rate of reaction."}],"ID":30310},{"Watched":false,"Name":"Measuring Rate of Reaction","Duration":"6m 43s","ChapterTopicVideoID":28814,"CourseChapterTopicPlaylistID":292227,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.740","Text":"In the previous video,"},{"Start":"00:01.740 ","End":"00:03.600","Text":"we defined the rate of reaction."},{"Start":"00:03.600 ","End":"00:07.000","Text":"In this video, we\u0027ll learn how to measure it."},{"Start":"00:07.160 ","End":"00:09.705","Text":"We\u0027re going to say a few words about"},{"Start":"00:09.705 ","End":"00:13.275","Text":"experimental methods for measuring the rate of reaction."},{"Start":"00:13.275 ","End":"00:16.890","Text":"Now reactions can be extremely slow, for example,"},{"Start":"00:16.890 ","End":"00:19.545","Text":"the decay of radioactive uranium,"},{"Start":"00:19.545 ","End":"00:21.600","Text":"or extremely fast, for example,"},{"Start":"00:21.600 ","End":"00:24.355","Text":"combustion of natural gas."},{"Start":"00:24.355 ","End":"00:30.004","Text":"The method used to measure the rate depends on how fast the reaction occurs."},{"Start":"00:30.004 ","End":"00:32.870","Text":"There are simple methods for slow reactions and"},{"Start":"00:32.870 ","End":"00:35.825","Text":"very sophisticated ones for fast reactions."},{"Start":"00:35.825 ","End":"00:41.180","Text":"For fast reactions, spectroscopic methods are usually used."},{"Start":"00:41.180 ","End":"00:45.525","Text":"Nowadays, even reactions that take attoseconds,"},{"Start":"00:45.525 ","End":"00:50.225","Text":"that\u0027s 10 to the power minus 18 seconds, can be investigated."},{"Start":"00:50.225 ","End":"00:53.660","Text":"In this video, we\u0027re going to talk about a simple method"},{"Start":"00:53.660 ","End":"00:57.320","Text":"where we measure the concentration as a function of time."},{"Start":"00:57.320 ","End":"01:02.870","Text":"We\u0027re going to take as an example the decomposition of hydrogen peroxide."},{"Start":"01:02.870 ","End":"01:08.900","Text":"Now we\u0027ve seen in a previous video that the chemical equation is H_2O_2,"},{"Start":"01:08.900 ","End":"01:11.308","Text":"that\u0027s hydrogen peroxide,"},{"Start":"01:11.308 ","End":"01:12.740","Text":"decomposing to give water,"},{"Start":"01:12.740 ","End":"01:15.070","Text":"H_2O, and 1/2 O_2."},{"Start":"01:15.070 ","End":"01:18.930","Text":"It decomposes to give us water and oxygen."},{"Start":"01:20.420 ","End":"01:25.700","Text":"Now, we can think of 2 simple ways of following the reaction."},{"Start":"01:25.700 ","End":"01:30.170","Text":"We can follow the volume of oxygen released as a function of time and"},{"Start":"01:30.170 ","End":"01:34.895","Text":"calculate its concentration using the ideal gas equation."},{"Start":"01:34.895 ","End":"01:41.870","Text":"Or we can remove small samples of the reaction mixture at various times and"},{"Start":"01:41.870 ","End":"01:44.960","Text":"test for the concentration of hydrogen peroxide by"},{"Start":"01:44.960 ","End":"01:49.430","Text":"titrating it with potassium permanganate in acidic solution."},{"Start":"01:49.430 ","End":"01:52.495","Text":"Here\u0027s the equation, it\u0027s a redox reaction."},{"Start":"01:52.495 ","End":"01:57.139","Text":"2 moles of permanganate plus 5 moles of hydrogen peroxide"},{"Start":"01:57.139 ","End":"02:01.865","Text":"plus 6 moles of H plus to give 2 moles of Mn^2 plus,"},{"Start":"02:01.865 ","End":"02:05.680","Text":"5 moles of oxygen and 8 moles of water."},{"Start":"02:05.680 ","End":"02:11.175","Text":"Now this is colorless all the time the H_2O_2 is still present."},{"Start":"02:11.175 ","End":"02:13.610","Text":"But when all the peroxide is reacted,"},{"Start":"02:13.610 ","End":"02:15.290","Text":"the color turns to purple,"},{"Start":"02:15.290 ","End":"02:18.260","Text":"that\u0027s the color of the permanganate."},{"Start":"02:18.260 ","End":"02:22.880","Text":"Now we can write that the average rate is minus Delta,"},{"Start":"02:22.880 ","End":"02:28.040","Text":"the concentration of H_2O_2 divided by Delta t. The units,"},{"Start":"02:28.040 ","End":"02:31.055","Text":"of course, are molar concentration per second."},{"Start":"02:31.055 ","End":"02:35.000","Text":"Now, that\u0027s minus the concentration of"},{"Start":"02:35.000 ","End":"02:40.435","Text":"H_2O_2 at t_2 minus the concentration of H_2O_2 at t_1."},{"Start":"02:40.435 ","End":"02:45.260","Text":"That\u0027s divided by the time interval t_2 minus t_1."},{"Start":"02:45.260 ","End":"02:47.750","Text":"Now because there\u0027s a minus sign here,"},{"Start":"02:47.750 ","End":"02:58.780","Text":"we can turn this around and get H_2O_2 t_1 minus H_2O_2 t_2 divided by t_2 minus t_1."},{"Start":"02:58.780 ","End":"03:02.810","Text":"Now here\u0027s some typical data for this reaction."},{"Start":"03:02.810 ","End":"03:09.080","Text":"We\u0027re starting with 3 molar H_2O_2 at time 0,"},{"Start":"03:09.080 ","End":"03:13.825","Text":"and we\u0027re measuring every 200 seconds."},{"Start":"03:13.825 ","End":"03:18.420","Text":"Delta t is 200 minus 0,"},{"Start":"03:18.420 ","End":"03:21.180","Text":"that\u0027s 200, 400 minus 200,"},{"Start":"03:21.180 ","End":"03:24.225","Text":"200, and it\u0027s always 200 here."},{"Start":"03:24.225 ","End":"03:27.955","Text":"Then, after 200 seconds,"},{"Start":"03:27.955 ","End":"03:32.660","Text":"the concentration goes from 3 to 2.59."},{"Start":"03:32.670 ","End":"03:40.935","Text":"The difference between t_1 and t_2 is 0.41."},{"Start":"03:40.935 ","End":"03:46.945","Text":"Then we can continue calculating the difference in concentration every 200 seconds,"},{"Start":"03:46.945 ","End":"03:50.050","Text":"and we see this, it decreases, 0.41,"},{"Start":"03:50.050 ","End":"03:53.790","Text":"0.35, 0.30 and so on."},{"Start":"03:53.790 ","End":"03:56.190","Text":"We can calculate the rate,"},{"Start":"03:56.190 ","End":"04:02.040","Text":"that\u0027s the Delta of the concentration of H_2O_2 divided by Delta t. Here\u0027s the rate,"},{"Start":"04:02.040 ","End":"04:04.620","Text":"2.1 times 10 to power minus 3,"},{"Start":"04:04.620 ","End":"04:08.580","Text":"decreasing to 1.8 times 10 to the minus 3,"},{"Start":"04:08.580 ","End":"04:11.220","Text":"and continuing to decrease."},{"Start":"04:11.220 ","End":"04:14.185","Text":"The rate decreases."},{"Start":"04:14.185 ","End":"04:18.625","Text":"Now we can plot the concentration versus time."},{"Start":"04:18.625 ","End":"04:21.990","Text":"I\u0027ve gone on to 3,000 seconds."},{"Start":"04:21.990 ","End":"04:26.060","Text":"This blue curve is the concentration versus time."},{"Start":"04:26.060 ","End":"04:33.680","Text":"We see that the concentration changes more rapidly at the beginning and then slows down."},{"Start":"04:33.680 ","End":"04:35.945","Text":"The change becomes smaller."},{"Start":"04:35.945 ","End":"04:38.315","Text":"The rate becomes smaller."},{"Start":"04:38.315 ","End":"04:44.920","Text":"We can summarize and say that the rate generally decreases as the reaction proceeds."},{"Start":"04:44.920 ","End":"04:51.364","Text":"Here\u0027s an example for t_1=400 seconds and t_2=600 seconds."},{"Start":"04:51.364 ","End":"04:54.710","Text":"We can calculate that the average rate at 500 seconds,"},{"Start":"04:54.710 ","End":"04:57.745","Text":"intermediate between 400 and 600,"},{"Start":"04:57.745 ","End":"05:00.880","Text":"is 1.5 times 10 to the minus 3."},{"Start":"05:00.880 ","End":"05:02.800","Text":"We saw that on the table,"},{"Start":"05:02.800 ","End":"05:06.425","Text":"we can also see that on this graph here."},{"Start":"05:06.425 ","End":"05:10.640","Text":"Now, as the time between 2 measurements get shorter,"},{"Start":"05:10.640 ","End":"05:13.070","Text":"the rate becomes more accurate."},{"Start":"05:13.070 ","End":"05:18.620","Text":"The smallest portion of time gives the most accurate rate."},{"Start":"05:18.620 ","End":"05:24.084","Text":"Now I\u0027m going to say a few words about the instantaneous rate of reaction."},{"Start":"05:24.084 ","End":"05:29.600","Text":"The instantaneous rate at a point in time is equal to the negative of"},{"Start":"05:29.600 ","End":"05:35.240","Text":"the slope of the tangent to the concentration versus time plot, at that point."},{"Start":"05:35.240 ","End":"05:36.590","Text":"That\u0027s a bit of a mouthful,"},{"Start":"05:36.590 ","End":"05:40.340","Text":"but you\u0027ll see what I mean by looking at the graph."},{"Start":"05:40.340 ","End":"05:44.360","Text":"I\u0027ve drawn the tangent to this curve of"},{"Start":"05:44.360 ","End":"05:49.790","Text":"the concentration versus time at 500 seconds here."},{"Start":"05:49.790 ","End":"05:53.825","Text":"This red curve, that\u0027s a tangent."},{"Start":"05:53.825 ","End":"05:59.695","Text":"Now, here\u0027s the mathematical definition of the instantaneous rate of reaction."},{"Start":"05:59.695 ","End":"06:06.950","Text":"Instead of Delta, we have d. The instantaneous rate is minus 1/a dA by dt,"},{"Start":"06:06.950 ","End":"06:08.540","Text":"I mean, the concentration of A,"},{"Start":"06:08.540 ","End":"06:11.045","Text":"I\u0027m not going to say concentration every time,"},{"Start":"06:11.045 ","End":"06:13.670","Text":"equal to minus 1/b dB by dt,"},{"Start":"06:13.670 ","End":"06:17.348","Text":"and that\u0027s equal to 1/g dG by dt,"},{"Start":"06:17.348 ","End":"06:20.070","Text":"and that\u0027s equal to 1/h dH by dt."},{"Start":"06:20.070 ","End":"06:28.850","Text":"We can write that the instantaneous rate is proportional to the derivative of"},{"Start":"06:28.850 ","End":"06:38.655","Text":"the concentration with respect to time."},{"Start":"06:38.655 ","End":"06:43.350","Text":"In this video, we learned how to find the rate of reaction."}],"ID":30311},{"Watched":false,"Name":"Exercise 1","Duration":"4m 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21s","ChapterTopicVideoID":30399,"CourseChapterTopicPlaylistID":292227,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":32435}],"Thumbnail":null,"ID":292227},{"Name":"The Rate Laws","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"The Rate Equation","Duration":"3m 29s","ChapterTopicVideoID":28815,"CourseChapterTopicPlaylistID":292228,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.510 ","End":"00:02.965","Text":"In the previous video,"},{"Start":"00:02.965 ","End":"00:08.090","Text":"we saw that the rate of reaction changes as the reaction proceeds."},{"Start":"00:08.090 ","End":"00:13.100","Text":"In this video we\u0027ll discuss the rate equation or rate law."},{"Start":"00:13.100 ","End":"00:18.400","Text":"We\u0027re going to talk about the rate equation or the rate law."},{"Start":"00:18.400 ","End":"00:25.165","Text":"Now for the reaction a moles of A plus b moles of B giving g moles of G and h moles of H,"},{"Start":"00:25.165 ","End":"00:27.745","Text":"we can often write not always"},{"Start":"00:27.745 ","End":"00:34.580","Text":"a rate equation or law of the form rate of reaction is equal to K,"},{"Start":"00:34.580 ","End":"00:37.135","Text":"it\u0027s a constant we\u0027ll talk about it,"},{"Start":"00:37.135 ","End":"00:43.345","Text":"times the concentration of A^m times the concentration of B^n."},{"Start":"00:43.345 ","End":"00:49.415","Text":"Now, A and B are the molar concentrations and m,"},{"Start":"00:49.415 ","End":"00:53.240","Text":"n are usually small positive whole numbers."},{"Start":"00:53.240 ","End":"00:57.230","Text":"But sometimes there\u0027s 0 or fractions or negative numbers,"},{"Start":"00:57.230 ","End":"01:00.324","Text":"but on the whole they\u0027re small positive numbers."},{"Start":"01:00.324 ","End":"01:04.325","Text":"It\u0027s important to note that m and n are not"},{"Start":"01:04.325 ","End":"01:08.440","Text":"necessarily equal to the stoichiometric coefficients,"},{"Start":"01:08.440 ","End":"01:11.870","Text":"and not necessarily equal to a and b."},{"Start":"01:11.870 ","End":"01:16.475","Text":"Now, this rate equation is determined experimentally."},{"Start":"01:16.475 ","End":"01:19.325","Text":"Now, what\u0027s the order of reaction?"},{"Start":"01:19.325 ","End":"01:26.880","Text":"Now, m that appears here is the order of the reaction respect to A and n,"},{"Start":"01:26.880 ","End":"01:31.525","Text":"it appears here is the order of the reaction with respect to B."},{"Start":"01:31.525 ","End":"01:34.010","Text":"If m is equal to 1, for example,"},{"Start":"01:34.010 ","End":"01:37.085","Text":"we\u0027d say the reaction is first-order in A."},{"Start":"01:37.085 ","End":"01:39.725","Text":"If for example, n is equal to 2,"},{"Start":"01:39.725 ","End":"01:42.955","Text":"the reaction is second order in B."},{"Start":"01:42.955 ","End":"01:45.270","Text":"For 1, it\u0027s first-order,"},{"Start":"01:45.270 ","End":"01:48.260","Text":"for 2 it\u0027s second-order, and so on."},{"Start":"01:48.260 ","End":"01:54.545","Text":"Now the total or overall order of reaction is the sum of m and n."},{"Start":"01:54.545 ","End":"02:01.250","Text":"I\u0027ve just written this q is equal to the sum of m and n. Now in a rage equation,"},{"Start":"02:01.250 ","End":"02:07.055","Text":"we also had a k. Now that\u0027s called the rate constant."},{"Start":"02:07.055 ","End":"02:09.440","Text":"k is the rate constant,"},{"Start":"02:09.440 ","End":"02:13.550","Text":"is different for each reaction and it depends on the temperature,"},{"Start":"02:13.550 ","End":"02:17.365","Text":"it\u0027s something that\u0027s measured experimentally."},{"Start":"02:17.365 ","End":"02:19.580","Text":"The larger the rate constant,"},{"Start":"02:19.580 ","End":"02:22.145","Text":"the faster the reaction occurs."},{"Start":"02:22.145 ","End":"02:25.580","Text":"Now, what are the units of the rate constant?"},{"Start":"02:25.580 ","End":"02:29.200","Text":"Now, the units will depend on the order of the reaction."},{"Start":"02:29.200 ","End":"02:35.390","Text":"The units of the rate constant will be the units of the rate divided by"},{"Start":"02:35.390 ","End":"02:42.350","Text":"the units of concentration to the power m plus n. Now,"},{"Start":"02:42.350 ","End":"02:44.840","Text":"units of the rate is molar concentration per"},{"Start":"02:44.840 ","End":"02:50.270","Text":"second divided by the units of concentration to the power m plus n,"},{"Start":"02:50.270 ","End":"02:54.290","Text":"that\u0027s molar concentration to the power m plus n. Of course,"},{"Start":"02:54.290 ","End":"02:55.910","Text":"1 didn\u0027t necessarily use seconds."},{"Start":"02:55.910 ","End":"03:00.020","Text":"Sometimes it\u0027s minutes or hours, or even years."},{"Start":"03:00.020 ","End":"03:02.015","Text":"When we divide these 2,"},{"Start":"03:02.015 ","End":"03:07.535","Text":"we get molar concentration to the power 1 from here,"},{"Start":"03:07.535 ","End":"03:09.935","Text":"minus m minus n,"},{"Start":"03:09.935 ","End":"03:14.150","Text":"and units of time seconds to the power minus 1."},{"Start":"03:14.150 ","End":"03:19.040","Text":"We can see that if m plus n is 2, for example,"},{"Start":"03:19.040 ","End":"03:22.415","Text":"this will be m to the power minus 1,"},{"Start":"03:22.415 ","End":"03:25.200","Text":"seconds to the power minus 1."},{"Start":"03:25.200 ","End":"03:29.040","Text":"In this video we learnt about the rate equation."}],"ID":30312},{"Watched":false,"Name":"Method of Initial Rates","Duration":"5m 39s","ChapterTopicVideoID":28816,"CourseChapterTopicPlaylistID":292228,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.500 ","End":"00:04.619","Text":"In the previous video we learned about the rate equation."},{"Start":"00:04.619 ","End":"00:09.195","Text":"In this video we\u0027ll learn about the method of initial rates."},{"Start":"00:09.195 ","End":"00:12.195","Text":"What\u0027s the initial rate?"},{"Start":"00:12.195 ","End":"00:19.140","Text":"The initial rate is the rate of reaction as near as possible to t=0."},{"Start":"00:19.140 ","End":"00:25.385","Text":"It can also be found by measuring the tangent to the reaction curve at time t = 0."},{"Start":"00:25.385 ","End":"00:27.755","Text":"Here\u0027s the method."},{"Start":"00:27.755 ","End":"00:30.080","Text":"You perform the reaction for"},{"Start":"00:30.080 ","End":"00:35.105","Text":"different initial concentrations and find the initial rates of reaction."},{"Start":"00:35.105 ","End":"00:40.730","Text":"For example, if you have p reactants you need at least p plus 1 experiments."},{"Start":"00:40.730 ","End":"00:42.320","Text":"If you have 2 reactants,"},{"Start":"00:42.320 ","End":"00:44.585","Text":"you need at least 3 experiments."},{"Start":"00:44.585 ","End":"00:47.860","Text":"If 3 you need 4 experiments."},{"Start":"00:47.860 ","End":"00:51.560","Text":"You can use the initial rates to find the orders of"},{"Start":"00:51.560 ","End":"00:54.605","Text":"the reaction with respect to the reactants,"},{"Start":"00:54.605 ","End":"00:56.795","Text":"the total order of the reaction,"},{"Start":"00:56.795 ","End":"00:59.060","Text":"and the rate constant."},{"Start":"00:59.060 ","End":"01:01.195","Text":"Here\u0027s an example."},{"Start":"01:01.195 ","End":"01:05.804","Text":"Consider the reaction A plus 2B to give products."},{"Start":"01:05.804 ","End":"01:07.550","Text":"From the following data,"},{"Start":"01:07.550 ","End":"01:11.180","Text":"find the order with respect to A and B,"},{"Start":"01:11.180 ","End":"01:13.204","Text":"and then the total order,"},{"Start":"01:13.204 ","End":"01:16.030","Text":"and then the rate constant."},{"Start":"01:16.030 ","End":"01:20.310","Text":"Here we have 3 experiments, 1, 2, 3."},{"Start":"01:20.310 ","End":"01:23.390","Text":"Here is the initial concentration of A,"},{"Start":"01:23.390 ","End":"01:26.675","Text":"0.8, 0.2, and 0.8."},{"Start":"01:26.675 ","End":"01:30.935","Text":"Note that we\u0027ve got 2 reactions in which the concentrations is the same."},{"Start":"01:30.935 ","End":"01:32.725","Text":"Then we have B,"},{"Start":"01:32.725 ","End":"01:35.790","Text":"0.4, 0.4, 0.2."},{"Start":"01:35.790 ","End":"01:41.230","Text":"Again, we have 2 reactions in which the concentration is the same."},{"Start":"01:42.500 ","End":"01:45.670","Text":"This one and this one for A,"},{"Start":"01:45.670 ","End":"01:48.970","Text":"and this one and this one for B."},{"Start":"01:48.970 ","End":"01:51.225","Text":"Here are the initial rates,"},{"Start":"01:51.225 ","End":"01:54.610","Text":"12.8, 3.2, 3.2."},{"Start":"01:54.610 ","End":"01:58.730","Text":"Now we can write that the initial rate is k,"},{"Start":"01:58.730 ","End":"02:00.905","Text":"is rate constant,"},{"Start":"02:00.905 ","End":"02:04.760","Text":"times the concentration of A at time t=0,"},{"Start":"02:04.760 ","End":"02:13.250","Text":"the power m times the concentration of B at time t=0 to the power n. We want to find k,"},{"Start":"02:13.250 ","End":"02:19.995","Text":"m, and n. Let\u0027s compare experiments 1 and 2."},{"Start":"02:19.995 ","End":"02:25.430","Text":"The initial rate of the first experiment divide by the initial rate"},{"Start":"02:25.430 ","End":"02:30.950","Text":"of the second is 12.8 divided by 3.2, and that\u0027s 4."},{"Start":"02:30.950 ","End":"02:35.119","Text":"Now let\u0027s take the ratio of this expression."},{"Start":"02:35.119 ","End":"02:37.320","Text":"Now it\u0027s k in both cases,"},{"Start":"02:37.320 ","End":"02:39.140","Text":"so the k cancels,"},{"Start":"02:39.140 ","End":"02:42.815","Text":"and we can see where the A or B cancels."},{"Start":"02:42.815 ","End":"02:46.055","Text":"We\u0027re taking experiments 1 and 2,"},{"Start":"02:46.055 ","End":"02:51.710","Text":"we can see that we have the same number for B,"},{"Start":"02:51.710 ","End":"02:54.410","Text":"0.4 for the experiment 1,"},{"Start":"02:54.410 ","End":"02:56.645","Text":"and 0.4 experiment 2,"},{"Start":"02:56.645 ","End":"02:59.725","Text":"so the ratio of these 2 will be 1."},{"Start":"02:59.725 ","End":"03:05.720","Text":"We only need the ratio of the initial concentrations of A."},{"Start":"03:05.720 ","End":"03:10.340","Text":"That\u0027s 0.80 for the first experiment,"},{"Start":"03:10.340 ","End":"03:13.894","Text":"0.20 for the second experiment."},{"Start":"03:13.894 ","End":"03:18.760","Text":"We have 0.80 divided by 0.20 to the power m,"},{"Start":"03:18.760 ","End":"03:25.969","Text":"and that gives us 4 to the power m. Obviously if 4=4 to the m,"},{"Start":"03:25.969 ","End":"03:28.040","Text":"m must be equal to 1."},{"Start":"03:28.040 ","End":"03:32.180","Text":"That\u0027s the order of reaction with respect to A."},{"Start":"03:32.180 ","End":"03:35.155","Text":"Now let\u0027s compare experiments 1 and 3."},{"Start":"03:35.155 ","End":"03:39.610","Text":"With the initial rate of the first compared to the initial rate of the third,"},{"Start":"03:39.610 ","End":"03:43.245","Text":"that\u0027s 12.8 divided by 3.2,"},{"Start":"03:43.245 ","End":"03:44.935","Text":"and again that\u0027s 4."},{"Start":"03:44.935 ","End":"03:51.500","Text":"Then we have the initial concentration in the first experiment compared to the third,"},{"Start":"03:51.500 ","End":"03:53.305","Text":"so it\u0027s the same for A,"},{"Start":"03:53.305 ","End":"03:55.125","Text":"that will give us just 1."},{"Start":"03:55.125 ","End":"03:59.850","Text":"But for B it\u0027s 0.40 divided by 0.20,"},{"Start":"03:59.850 ","End":"04:04.260","Text":"so it\u0027s 0.40 divided by 0.20 to"},{"Start":"04:04.260 ","End":"04:10.955","Text":"the power n. That\u0027s equal to 2 to the power n. If 4 is equal to 2 to the power n,"},{"Start":"04:10.955 ","End":"04:13.055","Text":"then n must be equal to 2."},{"Start":"04:13.055 ","End":"04:16.310","Text":"That\u0027s the order with respect to B."},{"Start":"04:16.310 ","End":"04:19.010","Text":"The total order must be m plus n,"},{"Start":"04:19.010 ","End":"04:20.210","Text":"that\u0027s 1 plus 2,"},{"Start":"04:20.210 ","End":"04:21.985","Text":"that\u0027s equal to 3."},{"Start":"04:21.985 ","End":"04:25.475","Text":"Now we know that the initial rate is equal to k"},{"Start":"04:25.475 ","End":"04:29.510","Text":"times the initial concentration of A to the power m times"},{"Start":"04:29.510 ","End":"04:32.960","Text":"initial concentration of B to the power n. We can"},{"Start":"04:32.960 ","End":"04:37.370","Text":"see that k is equal to the initial rate divided,"},{"Start":"04:37.370 ","End":"04:39.560","Text":"in this case we know the values of m and n,"},{"Start":"04:39.560 ","End":"04:43.915","Text":"so it\u0027s A to the power 1 times B to the power 2."},{"Start":"04:43.915 ","End":"04:46.350","Text":"We can just calculate that."},{"Start":"04:46.350 ","End":"04:49.235","Text":"If we look at the first experiment, for example,"},{"Start":"04:49.235 ","End":"04:53.795","Text":"k is equal to 12.8 divided by 0.80,"},{"Start":"04:53.795 ","End":"04:55.730","Text":"it\u0027s the concentration of A,"},{"Start":"04:55.730 ","End":"04:59.315","Text":"times 0.40, it\u0027s the concentration of B,"},{"Start":"04:59.315 ","End":"05:01.130","Text":"and for B it\u0027s squared."},{"Start":"05:01.130 ","End":"05:07.670","Text":"If we multiply all that out we get 100M to the power minus 2 times seconds,"},{"Start":"05:07.670 ","End":"05:09.040","Text":"the power of minus 1."},{"Start":"05:09.040 ","End":"05:13.025","Text":"Here we have M on the top divide by M,"},{"Start":"05:13.025 ","End":"05:16.280","Text":"and here\u0027s our M to the power minus 2."},{"Start":"05:16.280 ","End":"05:22.700","Text":"Now we have the value of the rate constant."},{"Start":"05:22.700 ","End":"05:29.520","Text":"Using this method we have found the order of reaction respect to A and B,"},{"Start":"05:29.520 ","End":"05:31.140","Text":"and the total order,"},{"Start":"05:31.140 ","End":"05:33.660","Text":"and also the rate constant."},{"Start":"05:33.660 ","End":"05:40.050","Text":"In this video we used the method of initial rates to determine the rate equation."}],"ID":30313},{"Watched":false,"Name":"Exercise 1 Part 1","Duration":"5m 39s","ChapterTopicVideoID":30387,"CourseChapterTopicPlaylistID":292228,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":32423},{"Watched":false,"Name":"Exercise 1 Part 2","Duration":"3m 48s","ChapterTopicVideoID":30388,"CourseChapterTopicPlaylistID":292228,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":32424},{"Watched":false,"Name":"Exercise 2","Duration":"5m 5s","ChapterTopicVideoID":30389,"CourseChapterTopicPlaylistID":292228,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":32425},{"Watched":false,"Name":"Exercise 3","Duration":"2m 38s","ChapterTopicVideoID":30390,"CourseChapterTopicPlaylistID":292228,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":32426},{"Watched":false,"Name":"Exercise 4","Duration":"4m 59s","ChapterTopicVideoID":30391,"CourseChapterTopicPlaylistID":292228,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":32427}],"Thumbnail":null,"ID":292228},{"Name":"Zero Order Reactions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Zero-Order Reactions","Duration":"3m 10s","ChapterTopicVideoID":28817,"CourseChapterTopicPlaylistID":292229,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.845","Text":"In a previous video,"},{"Start":"00:01.845 ","End":"00:03.840","Text":"we learned about the rate equation,"},{"Start":"00:03.840 ","End":"00:07.950","Text":"and in this video, we\u0027ll learn about zero-order reactions."},{"Start":"00:07.950 ","End":"00:12.375","Text":"We\u0027re going to write the rate equation for zero-order reactions."},{"Start":"00:12.375 ","End":"00:17.250","Text":"The fact that it\u0027s zero order means that the total order is 0."},{"Start":"00:17.250 ","End":"00:22.470","Text":"If we have a reaction where some reactant A turns into products,"},{"Start":"00:22.470 ","End":"00:26.430","Text":"the rate which we write as minus dA by dt."},{"Start":"00:26.430 ","End":"00:31.455","Text":"I\u0027ll just say A instead of concentration of A to make it less cumbersome."},{"Start":"00:31.455 ","End":"00:36.230","Text":"Minus the dA by dt is equal to k to the power of 0."},{"Start":"00:36.230 ","End":"00:41.450","Text":"That\u0027s the order 0. Now, A to the power 0 is just equal to 1."},{"Start":"00:41.450 ","End":"00:46.430","Text":"That means that the rate is equal to k. The rate is constant."},{"Start":"00:46.430 ","End":"00:49.745","Text":"Now I\u0027m going to integrate this rate equation."},{"Start":"00:49.745 ","End":"00:51.949","Text":"If you don\u0027t need the calculus,"},{"Start":"00:51.949 ","End":"00:55.675","Text":"you shouldn\u0027t just concentrate on the final result."},{"Start":"00:55.675 ","End":"01:05.490","Text":"We have minus dA by dt is equal to k. We can write that as dA equal to minus k dt."},{"Start":"01:05.490 ","End":"01:09.709","Text":"Now, if we integrate both sides of this equation,"},{"Start":"01:09.709 ","End":"01:16.665","Text":"so we have the integral from A at time 0 to at time t of dA,"},{"Start":"01:16.665 ","End":"01:19.760","Text":"and that\u0027s equal to minus k,"},{"Start":"01:19.760 ","End":"01:24.170","Text":"the integral from time 0 to time t dt."},{"Start":"01:24.170 ","End":"01:30.905","Text":"From the left-hand side we have A time t minus A at time 0."},{"Start":"01:30.905 ","End":"01:34.225","Text":"From the right-hand side we have minus kt."},{"Start":"01:34.225 ","End":"01:36.375","Text":"Here\u0027s our equation,"},{"Start":"01:36.375 ","End":"01:39.690","Text":"At minus A0 is equal to minus kt."},{"Start":"01:39.690 ","End":"01:46.870","Text":"We can write that A at time t is equal to A times 0 minus kt."},{"Start":"01:47.150 ","End":"01:51.090","Text":"For those of you don\u0027t need the calculus,"},{"Start":"01:51.090 ","End":"01:52.984","Text":"here\u0027s our final result."},{"Start":"01:52.984 ","End":"01:55.775","Text":"Now, what are the units of k?"},{"Start":"01:55.775 ","End":"01:58.700","Text":"We see that the rate is equal to k,"},{"Start":"01:58.700 ","End":"02:02.380","Text":"so k has the same units as the rate."},{"Start":"02:02.380 ","End":"02:05.360","Text":"If time is in units of seconds,"},{"Start":"02:05.360 ","End":"02:12.095","Text":"the units of k will be molar concentration per second."},{"Start":"02:12.095 ","End":"02:14.615","Text":"Same as the rate."},{"Start":"02:14.615 ","End":"02:20.525","Text":"Now we\u0027re going to plot the concentration versus the time."},{"Start":"02:20.525 ","End":"02:28.516","Text":"We can see that it\u0027s a straight line because we have A times t equal to A at time 0,"},{"Start":"02:28.516 ","End":"02:34.335","Text":"that\u0027s the intercept here, minus kt."},{"Start":"02:34.335 ","End":"02:40.310","Text":"We have a straight line with slope minus k. It\u0027s like"},{"Start":"02:40.310 ","End":"02:46.045","Text":"y is equal to mx plus c. Here\u0027s the c,"},{"Start":"02:46.045 ","End":"02:54.950","Text":"and here\u0027s the m is minus k and x is t. When we plot the concentration of A versus time,"},{"Start":"02:54.950 ","End":"02:59.780","Text":"we get a straight line with a negative slope and the slope is"},{"Start":"02:59.780 ","End":"03:04.940","Text":"minus k. That\u0027s how we can determine k. In this video,"},{"Start":"03:04.940 ","End":"03:08.940","Text":"we learned about zero-order reactions."}],"ID":30314},{"Watched":false,"Name":"Exercise 1","Duration":"2m 18s","ChapterTopicVideoID":31667,"CourseChapterTopicPlaylistID":292229,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33902},{"Watched":false,"Name":"Exercise 2","Duration":"4m 6s","ChapterTopicVideoID":31665,"CourseChapterTopicPlaylistID":292229,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33903},{"Watched":false,"Name":"Exercise 3","Duration":"1m 40s","ChapterTopicVideoID":31666,"CourseChapterTopicPlaylistID":292229,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33904}],"Thumbnail":null,"ID":292229},{"Name":"First Order Reactions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"First Order Reactions","Duration":"5m 54s","ChapterTopicVideoID":28818,"CourseChapterTopicPlaylistID":292230,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.830","Text":"In the previous video,"},{"Start":"00:01.830 ","End":"00:04.355","Text":"we learned about zero-order reactions."},{"Start":"00:04.355 ","End":"00:08.325","Text":"In this video, we\u0027ll talk about first order reactions."},{"Start":"00:08.325 ","End":"00:12.615","Text":"We\u0027re going to write the rate equation for first-order reactions."},{"Start":"00:12.615 ","End":"00:16.230","Text":"First-order means that the total order is one."},{"Start":"00:16.230 ","End":"00:19.874","Text":"If you have some reaction A to given products,"},{"Start":"00:19.874 ","End":"00:22.560","Text":"the rate which is minus d(A) by dt,"},{"Start":"00:22.560 ","End":"00:27.750","Text":"I\u0027m just going to say A instead of concentration of A minus"},{"Start":"00:27.750 ","End":"00:34.155","Text":"d(A) by dt is equal to k(A) to the power 1 and A to the power 1 is just A."},{"Start":"00:34.155 ","End":"00:37.045","Text":"We have the rate is equal to k(A)."},{"Start":"00:37.045 ","End":"00:40.110","Text":"Now we\u0027re going to integrate this equation."},{"Start":"00:40.110 ","End":"00:44.705","Text":"If you don\u0027t need to know how to integrate for this course,"},{"Start":"00:44.705 ","End":"00:47.720","Text":"then you can just look at the final expression."},{"Start":"00:47.720 ","End":"00:52.700","Text":"We have minus d(A) by dt is equal to k(A) and now if"},{"Start":"00:52.700 ","End":"00:57.455","Text":"we take the A to the other side and divide d(A) by A,"},{"Start":"00:57.455 ","End":"01:02.390","Text":"we get d(A) divided by A equal to minus kdt."},{"Start":"01:02.390 ","End":"01:06.170","Text":"Now let\u0027s integrate both sides of this equation."},{"Start":"01:06.170 ","End":"01:13.545","Text":"We have the integral of d(A) divided by A from A at time 0 to"},{"Start":"01:13.545 ","End":"01:21.960","Text":"A at time t and that\u0027s equal to minus kdt from time 0 to time t. Now,"},{"Start":"01:21.960 ","End":"01:27.120","Text":"the integral of 1 over A is ln A."},{"Start":"01:27.120 ","End":"01:30.416","Text":"On the left-hand side we have In of A time t,"},{"Start":"01:30.416 ","End":"01:34.380","Text":"that\u0027s the upper limit minus In at A at time 0,"},{"Start":"01:34.380 ","End":"01:39.040","Text":"the lower limit and the right hand side integral is minus kt."},{"Start":"01:39.040 ","End":"01:46.920","Text":"We have In A at time t minus In A at time 0 equal to minus kt and we can take"},{"Start":"01:46.920 ","End":"01:50.295","Text":"the minus In A_0 to the other side and write In"},{"Start":"01:50.295 ","End":"01:56.410","Text":"A at time t is equal to In A at time 0 minus kt."},{"Start":"01:56.510 ","End":"02:00.315","Text":"That\u0027s our first expression."},{"Start":"02:00.315 ","End":"02:04.845","Text":"Now, cause In of"},{"Start":"02:04.845 ","End":"02:12.305","Text":"a minus In of b is equal to In of a divided by b."},{"Start":"02:12.305 ","End":"02:17.540","Text":"We can take the ln of a at time 0 to the left-hand side and write ln"},{"Start":"02:17.540 ","End":"02:23.900","Text":"of A at time t divided by A times 0 equal to minus kt."},{"Start":"02:23.900 ","End":"02:27.770","Text":"That\u0027s another useful expression."},{"Start":"02:27.770 ","End":"02:32.160","Text":"Then we can take the exponential of each side,"},{"Start":"02:32.160 ","End":"02:37.025","Text":"so we have A time t divided by A times 0 on the left hand side,"},{"Start":"02:37.025 ","End":"02:43.640","Text":"equal to the e to minus kt on the right-hand side and then we can multiply both sides"},{"Start":"02:43.640 ","End":"02:51.305","Text":"by A times 0 and get A times t is equal to A times 0, exponential minus kt."},{"Start":"02:51.305 ","End":"02:54.965","Text":"That\u0027s the third expression we\u0027re going to use."},{"Start":"02:54.965 ","End":"02:57.574","Text":"Now, what are the units of k?"},{"Start":"02:57.574 ","End":"03:01.130","Text":"Now we know that logs are dimensionless,"},{"Start":"03:01.130 ","End":"03:04.730","Text":"so that if T is in units of seconds,"},{"Start":"03:04.730 ","End":"03:09.280","Text":"then k needs to be in units of seconds to the power minus 1."},{"Start":"03:09.280 ","End":"03:14.210","Text":"Also exponentials without units so kt is unitless,"},{"Start":"03:14.210 ","End":"03:17.620","Text":"so k has units of seconds to power of minus 1."},{"Start":"03:17.620 ","End":"03:19.460","Text":"We\u0027re going to take as an example,"},{"Start":"03:19.460 ","End":"03:23.060","Text":"the decomposition of hydrogen peroxide and we\u0027ve talked about this in"},{"Start":"03:23.060 ","End":"03:32.000","Text":"a previous video and we said that the chemical equation with H_2O_2 decomposing to water,"},{"Start":"03:32.000 ","End":"03:35.570","Text":"H_2O plus 1/2 O2."},{"Start":"03:35.570 ","End":"03:39.515","Text":"Hydrogen peroxide decomposes to water and oxygen."},{"Start":"03:39.515 ","End":"03:42.650","Text":"Now this is a first-order reaction,"},{"Start":"03:42.650 ","End":"03:48.560","Text":"and we\u0027ll see that it has indeed exponential decay as we saw in a previous video."},{"Start":"03:48.560 ","End":"03:50.270","Text":"According to this equation,"},{"Start":"03:50.270 ","End":"03:53.525","Text":"A time t is equal to A times 0,"},{"Start":"03:53.525 ","End":"03:57.730","Text":"e to the power minus kt exponential decay."},{"Start":"03:57.730 ","End":"04:00.665","Text":"Here\u0027s the graph we had in a previous video,"},{"Start":"04:00.665 ","End":"04:08.975","Text":"concentration starting at 3m and decaying exponentially."},{"Start":"04:08.975 ","End":"04:12.785","Text":"Now this is not very useful for finding the value of k,"},{"Start":"04:12.785 ","End":"04:16.010","Text":"but we can use one of the other expressions we found."},{"Start":"04:16.010 ","End":"04:18.935","Text":"Now we can also plot the In of"},{"Start":"04:18.935 ","End":"04:25.170","Text":"the concentration versus time in seconds and that\u0027s using the expression we"},{"Start":"04:25.170 ","End":"04:35.565","Text":"found that In of A time t is equal to In at A time 0 minus kt,"},{"Start":"04:35.565 ","End":"04:38.085","Text":"so that\u0027s a straight line."},{"Start":"04:38.085 ","End":"04:42.060","Text":"Here\u0027s y, In of A of t is y,"},{"Start":"04:42.060 ","End":"04:47.325","Text":"ln of A is 0 is c. Here it is."},{"Start":"04:47.325 ","End":"04:55.560","Text":"T is x, so t is x and m is minus k. The slope is minus"},{"Start":"04:55.560 ","End":"05:03.835","Text":"k. The slope of the graph is minus k. Let\u0027s solve a problem at time t 0,"},{"Start":"05:03.835 ","End":"05:13.025","Text":"the In of H_2O_2 is the In of 3 and that\u0027s 1.1 and at time t equal to 3 thousand seconds,"},{"Start":"05:13.025 ","End":"05:19.125","Text":"the In of H_2O_2 is equal to minus 1.09."},{"Start":"05:19.125 ","End":"05:21.150","Text":"We can work out k,"},{"Start":"05:21.150 ","End":"05:30.090","Text":"so k is equal to minus the In at 3,000 sessions minus In at 0,"},{"Start":"05:30.090 ","End":"05:32.732","Text":"divided by 3,000 seconds,"},{"Start":"05:32.732 ","End":"05:37.040","Text":"that\u0027s our Delta t and that is equal to"},{"Start":"05:37.040 ","End":"05:43.455","Text":"7.3 times 10 to the power minus 3 seconds to the power minus 1."},{"Start":"05:43.455 ","End":"05:46.550","Text":"That\u0027s our value of k. Now we could have"},{"Start":"05:46.550 ","End":"05:49.760","Text":"found it from any two points on the graph, of course."},{"Start":"05:49.760 ","End":"05:55.080","Text":"In this video, we\u0027ll learnt about first-order reactions."}],"ID":30315},{"Watched":false,"Name":"Half Life of First Order Reaction","Duration":"4m 46s","ChapterTopicVideoID":28819,"CourseChapterTopicPlaylistID":292230,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:04.860","Text":"In the previous video we learned about first-order reactions,"},{"Start":"00:04.860 ","End":"00:10.065","Text":"and in this video we\u0027ll talk about the half-life of first-order reactions."},{"Start":"00:10.065 ","End":"00:14.280","Text":"We\u0027re going to talk about the half-life of first-order reaction."},{"Start":"00:14.280 ","End":"00:18.570","Text":"Now the half-life which we\u0027ll call t_1/2 of a reaction is the time"},{"Start":"00:18.570 ","End":"00:23.280","Text":"taken for the concentration to reach half of its initial value."},{"Start":"00:23.280 ","End":"00:30.135","Text":"From the exact expression we had in the previous video the ln of A at time t;"},{"Start":"00:30.135 ","End":"00:39.203","Text":"now we\u0027re going to write it t=t_1/2 divided by A at time 0 is equal minus kt_1/2,"},{"Start":"00:39.203 ","End":"00:43.035","Text":"t_1/2 being of course the half-life,"},{"Start":"00:43.035 ","End":"00:52.890","Text":"so we have the ln of 1/2A_0 divided by A_0 = minus kt_1/2."},{"Start":"00:52.890 ","End":"00:58.050","Text":"The A_0 cancels and we get the ln of 1/2 and"},{"Start":"00:58.050 ","End":"01:04.155","Text":"that\u0027s equal to minus ln of 2 which is equal to minus 0.693,"},{"Start":"01:04.155 ","End":"01:07.320","Text":"and that\u0027s equal to minus kt_1/2."},{"Start":"01:07.320 ","End":"01:13.820","Text":"We can write that kt_1/2 is equal to 0.693 so"},{"Start":"01:13.820 ","End":"01:20.855","Text":"that t_1/2 is equal to 0.693 divided by k. Of course if we know t_1/2 half,"},{"Start":"01:20.855 ","End":"01:28.200","Text":"we can write that k is equal to 0.693 divided by t_1/2,"},{"Start":"01:28.200 ","End":"01:34.295","Text":"and that way we can find k. We can see that since k is a constant,"},{"Start":"01:34.295 ","End":"01:35.780","Text":"t_1/2 is also constant."},{"Start":"01:35.780 ","End":"01:37.969","Text":"The half-life is constant,"},{"Start":"01:37.969 ","End":"01:43.674","Text":"that means it\u0027s independent of the initial concentration for a first-order reaction."},{"Start":"01:43.674 ","End":"01:50.090","Text":"We can also use the half-life as a test to see whether a reaction is first-order."},{"Start":"01:50.090 ","End":"01:51.680","Text":"The half-life is constant,"},{"Start":"01:51.680 ","End":"01:54.365","Text":"we know the reaction is first-order."},{"Start":"01:54.365 ","End":"01:56.345","Text":"Here\u0027s an example."},{"Start":"01:56.345 ","End":"02:01.160","Text":"Now we found in the previous video that for the decomposition of hydrogen peroxide,"},{"Start":"02:01.160 ","End":"02:06.960","Text":"k is equal 7.3 times 10^minus 3 seconds per minus 1."},{"Start":"02:08.360 ","End":"02:13.020","Text":"We can calculate the half-life of this reaction and"},{"Start":"02:13.020 ","End":"02:18.470","Text":"then second part if the initial concentration is 3 molar,"},{"Start":"02:18.470 ","End":"02:24.835","Text":"how long will it take for the concentration to drop to 0.375 molar?"},{"Start":"02:24.835 ","End":"02:28.335","Text":"Let\u0027s begin by calculating t_1/2."},{"Start":"02:28.335 ","End":"02:33.530","Text":"K=7.3 times 10^minus 3 per second,"},{"Start":"02:33.530 ","End":"02:41.775","Text":"so t_1/2 is 0.693 divided by 7.3 times 10^minus 3 per second."},{"Start":"02:41.775 ","End":"02:43.560","Text":"If we divide these 2 numbers,"},{"Start":"02:43.560 ","End":"02:48.870","Text":"we get 949 and 1 over second^minus 1 is seconds."},{"Start":"02:48.870 ","End":"02:55.550","Text":"We see that the half-life for this reaction is 949 seconds."},{"Start":"02:55.550 ","End":"03:04.601","Text":"That means that after 949 seconds the concentration drops from 3 to 1.5,"},{"Start":"03:04.601 ","End":"03:09.690","Text":"and if we take 2 half-lives it drops from 1.5 to 0.75."},{"Start":"03:10.000 ","End":"03:15.825","Text":"So going from 3 to 1.5 to 0.75."},{"Start":"03:15.825 ","End":"03:17.900","Text":"If we wait another half-life,"},{"Start":"03:17.900 ","End":"03:22.855","Text":"it will drop to 0.375."},{"Start":"03:22.855 ","End":"03:25.685","Text":"After adding another half-life;"},{"Start":"03:25.685 ","End":"03:30.880","Text":"that\u0027s 3 half-lives, is 2,847 seconds."},{"Start":"03:30.880 ","End":"03:39.230","Text":"After 2,847 seconds the concentration will be 0.375 molar."},{"Start":"03:39.230 ","End":"03:43.864","Text":"Let\u0027s look for a moment at the half-life of a zero-order reaction."},{"Start":"03:43.864 ","End":"03:49.235","Text":"We can calculate the half-life for a zero-order reaction is t_1/2 equal to"},{"Start":"03:49.235 ","End":"03:55.033","Text":"A at at time 0 the initial concentration divided by 2k,"},{"Start":"03:55.033 ","End":"04:00.020","Text":"so we see that here the half-life is dependent on the initial concentration."},{"Start":"04:00.020 ","End":"04:01.495","Text":"Now, here\u0027s our proof."},{"Start":"04:01.495 ","End":"04:08.705","Text":"We saw that 80 minus A_0 is equal to minus kt for a zero-order reaction."},{"Start":"04:08.705 ","End":"04:13.660","Text":"Now at t_1/2 A will have dropped to 1/2A_0,"},{"Start":"04:13.660 ","End":"04:19.365","Text":"and we still have the minus A_0=minus kt_1/2."},{"Start":"04:19.365 ","End":"04:23.615","Text":"1/2 minus 1 is minus 1/2,"},{"Start":"04:23.615 ","End":"04:28.865","Text":"and changing the signs we get 1/2A_0 equal to kt_1/2."},{"Start":"04:28.865 ","End":"04:32.905","Text":"T_1/2 is equal to A_0 divided by 2k."},{"Start":"04:32.905 ","End":"04:37.250","Text":"We see that t_1/2 depends on the initial concentration,"},{"Start":"04:37.250 ","End":"04:39.180","Text":"it\u0027s not a constant."},{"Start":"04:39.180 ","End":"04:46.080","Text":"In this video, we talked about the half-life of a first-order and zero-order reaction."}],"ID":30316}],"Thumbnail":null,"ID":292230},{"Name":"Second Order Reactions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Second-Order Reactions","Duration":"5m 36s","ChapterTopicVideoID":28820,"CourseChapterTopicPlaylistID":292231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.770","Text":"In the previous video,"},{"Start":"00:01.770 ","End":"00:04.440","Text":"we learned about first-order reactions."},{"Start":"00:04.440 ","End":"00:05.610","Text":"In this video,"},{"Start":"00:05.610 ","End":"00:08.240","Text":"we\u0027ll talk about second-order reactions."},{"Start":"00:08.240 ","End":"00:12.870","Text":"We\u0027re going to talk about the rate equation for second-order reactions."},{"Start":"00:12.870 ","End":"00:16.620","Text":"Now for second-order, means the total order is 2."},{"Start":"00:16.620 ","End":"00:19.830","Text":"For action which A turns into products,"},{"Start":"00:19.830 ","End":"00:23.498","Text":"the rate is minus dA by dt and the concentration of A,"},{"Start":"00:23.498 ","End":"00:26.550","Text":"but I\u0027m just going to say A for short."},{"Start":"00:26.550 ","End":"00:30.900","Text":"It\u0027s minus dA by dt equal to KA^2."},{"Start":"00:30.900 ","End":"00:33.755","Text":"Now we\u0027re going to integrate the equation."},{"Start":"00:33.755 ","End":"00:37.625","Text":"If you don\u0027t need the actual integration process,"},{"Start":"00:37.625 ","End":"00:40.130","Text":"you can look at the final result."},{"Start":"00:40.130 ","End":"00:43.380","Text":"Minus dA by dt equal to KA^2."},{"Start":"00:43.540 ","End":"00:49.410","Text":"We\u0027re going to change the sign and divide both sides by A^2."},{"Start":"00:49.410 ","End":"00:53.560","Text":"We have dA divided by A^2 equal to minus kdt."},{"Start":"00:53.560 ","End":"00:57.245","Text":"Now if we integrate both sides of this equation,"},{"Start":"00:57.245 ","End":"01:02.670","Text":"we get the integral from A0 to A0t of dA divided"},{"Start":"01:02.670 ","End":"01:08.805","Text":"by A^2 equal to minus k the integral from 0 to t dt."},{"Start":"01:08.805 ","End":"01:18.510","Text":"Now the integral of 1 over A^2 dA is equal to minus 1 over A."},{"Start":"01:18.510 ","End":"01:24.330","Text":"We can write this as equal to minus 1 over At,"},{"Start":"01:24.330 ","End":"01:26.265","Text":"that\u0027s on the top limit,"},{"Start":"01:26.265 ","End":"01:31.590","Text":"plus 1 over A0 from the bottom limit equal to minus kt."},{"Start":"01:31.590 ","End":"01:39.825","Text":"Then we can rearrange it to get 1 over A at time t equal to 1 over A at time 0 plus kt."},{"Start":"01:39.825 ","End":"01:42.770","Text":"As we\u0027ll see, that\u0027s a straight line."},{"Start":"01:42.770 ","End":"01:45.680","Text":"First, let\u0027s just say a few words about the units of"},{"Start":"01:45.680 ","End":"01:48.805","Text":"k. Now if time is in units of seconds,"},{"Start":"01:48.805 ","End":"01:55.190","Text":"units of k will be molar concentration power minus 1 seconds to the power of minus 1."},{"Start":"01:55.190 ","End":"01:56.615","Text":"How did I get to that?"},{"Start":"01:56.615 ","End":"02:02.315","Text":"We can see from this equation here that k must have the same units"},{"Start":"02:02.315 ","End":"02:10.500","Text":"as 1 over the concentration times the time."},{"Start":"02:10.690 ","End":"02:20.750","Text":"That\u0027s 1 over molar concentration times 1 over seconds and here\u0027s our units."},{"Start":"02:20.750 ","End":"02:25.855","Text":"Now we\u0027re going to plot the inverse concentration versus time."},{"Start":"02:25.855 ","End":"02:30.560","Text":"Here\u0027s the inverse concentration versus time,"},{"Start":"02:30.560 ","End":"02:33.575","Text":"and here\u0027s the intercept, 1 over A0."},{"Start":"02:33.575 ","End":"02:41.140","Text":"Remember 1 over A at time t is equal to 1 over A times 0 plus kt."},{"Start":"02:41.140 ","End":"02:48.270","Text":"Here\u0027s our slope k. This is like Y is 1 over At,"},{"Start":"02:48.270 ","End":"02:52.425","Text":"the intercept is 1 over is 0."},{"Start":"02:52.425 ","End":"02:54.735","Text":"This is like mx,"},{"Start":"02:54.735 ","End":"02:58.230","Text":"where t is x."},{"Start":"02:58.230 ","End":"03:02.030","Text":"We have a slope of k. The slope of the graph is"},{"Start":"03:02.030 ","End":"03:07.170","Text":"k. Let\u0027s calculate what the slope is at time t equal to 0,"},{"Start":"03:07.170 ","End":"03:16.875","Text":"I\u0027ve taken A to be 3M so 1 over 3 is 0.333M to the power minus 1"},{"Start":"03:16.875 ","End":"03:20.820","Text":"and I\u0027ve taken A at time 3,000 seconds to be"},{"Start":"03:20.820 ","End":"03:28.104","Text":"0.396 so 1 over A is going to be 2.523."},{"Start":"03:28.104 ","End":"03:35.070","Text":"This is 0,1,2, and 3,"},{"Start":"03:35.070 ","End":"03:39.260","Text":"so k will be the value at time t,"},{"Start":"03:39.260 ","End":"03:48.940","Text":"2.523 minus the value at time 0.333 divided by the time interval, 3,000 seconds."},{"Start":"03:48.940 ","End":"03:54.645","Text":"We divide that, we get 7.3 times 10^-3,"},{"Start":"03:54.645 ","End":"03:56.495","Text":"and here the correct units,"},{"Start":"03:56.495 ","End":"04:04.160","Text":"7.3 times 10^-3 per second per molar concentration."},{"Start":"04:04.160 ","End":"04:08.345","Text":"Now a few words about the half-life of the second-order reaction."},{"Start":"04:08.345 ","End":"04:15.390","Text":"Now, 1 over the half-life concentration is 1 over a half A0,"},{"Start":"04:15.390 ","End":"04:17.550","Text":"1/2 the initial concentration."},{"Start":"04:17.550 ","End":"04:21.915","Text":"We know that\u0027s equal to 1 over A0 plus kt half."},{"Start":"04:21.915 ","End":"04:29.280","Text":"We can write 1 over a half A0 as 2 over A0 and then we"},{"Start":"04:29.280 ","End":"04:37.290","Text":"have 2 minus 1 over A0 that gives us 1 over A0 equals kt half."},{"Start":"04:37.290 ","End":"04:43.000","Text":"We can see that t half is equal to 1 over kA0."},{"Start":"04:43.070 ","End":"04:45.710","Text":"We can see that the half-life of"},{"Start":"04:45.710 ","End":"04:50.245","Text":"a second-order reaction increases as the reaction proceeds."},{"Start":"04:50.245 ","End":"04:52.530","Text":"As the reaction proceeds,"},{"Start":"04:52.530 ","End":"04:57.264","Text":"each half-life A0 is smaller,"},{"Start":"04:57.264 ","End":"05:00.995","Text":"so t-half will be longer because it\u0027s inversely proportional to"},{"Start":"05:00.995 ","End":"05:06.485","Text":"A0 and each half-life is twice as long as the previous one."},{"Start":"05:06.485 ","End":"05:11.225","Text":"We can summarize and say that for first-order,"},{"Start":"05:11.225 ","End":"05:13.835","Text":"the half-life was constant."},{"Start":"05:13.835 ","End":"05:18.330","Text":"For zero-order it decreases as"},{"Start":"05:18.330 ","End":"05:25.545","Text":"the reaction proceeds and for second-order it increases."},{"Start":"05:25.545 ","End":"05:31.270","Text":"Effectively, only for first order is the half-life a constant."},{"Start":"05:31.270 ","End":"05:36.960","Text":"In this video, we talked about second-order reactions."}],"ID":30317},{"Watched":false,"Name":"Pseudo First Order Reactions","Duration":"2m 35s","ChapterTopicVideoID":28821,"CourseChapterTopicPlaylistID":292231,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.080 ","End":"00:02.325","Text":"In the previous video,"},{"Start":"00:02.325 ","End":"00:04.710","Text":"we learned about second-order reactions."},{"Start":"00:04.710 ","End":"00:07.200","Text":"In this video, we\u0027ll learn that sometimes"},{"Start":"00:07.200 ","End":"00:11.985","Text":"second-order reactions can be treated as if they are first order."},{"Start":"00:11.985 ","End":"00:17.010","Text":"We\u0027re going to talk about pseudo first-order reactions."},{"Start":"00:17.010 ","End":"00:22.125","Text":"Sometimes second-order reactions can be treated as if they\u0027re first-order."},{"Start":"00:22.125 ","End":"00:25.530","Text":"We call this pseudo first-order reactions."},{"Start":"00:25.530 ","End":"00:29.295","Text":"The first example is the hydrolysis of ethyl acetate."},{"Start":"00:29.295 ","End":"00:37.985","Text":"Ethyl acetate reacts with water to give us acetic acid and ethanol."},{"Start":"00:37.985 ","End":"00:43.625","Text":"This is a second-order reaction with rate equation rate is equal to k,"},{"Start":"00:43.625 ","End":"00:49.565","Text":"the concentration of ethyl acetate times the concentration of water."},{"Start":"00:49.565 ","End":"00:52.670","Text":"If the solution is sufficiently dilute,"},{"Start":"00:52.670 ","End":"00:57.235","Text":"so that the concentration of water hardly changes during the reaction,"},{"Start":"00:57.235 ","End":"01:01.085","Text":"we can assume the concentration of water is constant."},{"Start":"01:01.085 ","End":"01:04.565","Text":"We can write the rate is equal to k prime."},{"Start":"01:04.565 ","End":"01:10.170","Text":"The concentration of ethyl acetate with"},{"Start":"01:10.170 ","End":"01:15.875","Text":"k prime equal to k times the concentration of water."},{"Start":"01:15.875 ","End":"01:21.025","Text":"We\u0027ve written it now as a first-order reaction."},{"Start":"01:21.025 ","End":"01:23.265","Text":"Here\u0027s another example."},{"Start":"01:23.265 ","End":"01:29.145","Text":"This is a redox reaction of persulfate ions with iodide ions."},{"Start":"01:29.145 ","End":"01:34.820","Text":"Here\u0027s persulfate ions reacting with 3 moles of"},{"Start":"01:34.820 ","End":"01:42.290","Text":"iodide to give us 2 moles of sulfate and 1 mole of tri-iodide."},{"Start":"01:42.290 ","End":"01:47.915","Text":"This is a second-order reaction with rate equation rate is equal to k,"},{"Start":"01:47.915 ","End":"01:54.340","Text":"the concentration of persulfate times the concentration of iodide."},{"Start":"01:54.340 ","End":"01:59.930","Text":"If we take a very high concentration of the persulfate compared to the iodide,"},{"Start":"01:59.930 ","End":"02:03.560","Text":"we can assume it stays constant throughout the reaction."},{"Start":"02:03.560 ","End":"02:09.380","Text":"The rate will now be k prime times the concentration of iodide with k"},{"Start":"02:09.380 ","End":"02:16.300","Text":"prime equal to k times the concentration of the persulfate."},{"Start":"02:16.300 ","End":"02:18.930","Text":"If we know k prime,"},{"Start":"02:18.930 ","End":"02:27.565","Text":"we can calculate k because we started with an initial concentration of the persulfate."},{"Start":"02:27.565 ","End":"02:33.780","Text":"In this video we learned about pseudo first-order reactions."}],"ID":30318},{"Watched":false,"Name":"Exercise 1","Duration":"5m 45s","ChapterTopicVideoID":31669,"CourseChapterTopicPlaylistID":292231,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33905},{"Watched":false,"Name":"Exercise 2","Duration":"3m 43s","ChapterTopicVideoID":31668,"CourseChapterTopicPlaylistID":292231,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33906}],"Thumbnail":null,"ID":292231},{"Name":"Reaction Mechanisms","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Elementary Reactions and Reaction Mechanisms","Duration":"4m 56s","ChapterTopicVideoID":28823,"CourseChapterTopicPlaylistID":293018,"HasSubtitles":true,"ThumbnailPath":"https://www.proprep.uk/Images/Videos_Thumbnails/28823.jpeg","UploadDate":"2022-05-04T08:00:41.8870000","DurationForVideoObject":"PT4M56S","Description":null,"MetaTitle":"Elementary Reactions and Reaction Mechanisms: Video + Workbook | Proprep","MetaDescription":"Chemical Kinetics - Reaction Mechanisms. Watch the video made by an expert in the field. Download the workbook and maximize your learning.","Canonical":"https://www.proprep.uk/general-modules/all/general-chemistry/chemical-kinetics/reaction-mechanisms/vid30437","VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:03.975","Text":"The previous videos, we learned about rate equations."},{"Start":"00:03.975 ","End":"00:05.940","Text":"In the next few videos,"},{"Start":"00:05.940 ","End":"00:08.830","Text":"we\u0027ll learn about reaction mechanisms."},{"Start":"00:08.830 ","End":"00:12.465","Text":"We\u0027re going to talk about reaction mechanisms."},{"Start":"00:12.465 ","End":"00:15.450","Text":"Now, reaction mechanisms or sequence of"},{"Start":"00:15.450 ","End":"00:17.940","Text":"elementary reactions that leads from"},{"Start":"00:17.940 ","End":"00:21.300","Text":"the reactants to the products in a chemical reaction."},{"Start":"00:21.300 ","End":"00:25.710","Text":"Now, what cannot prove that a proposed mechanism is the correct one,"},{"Start":"00:25.710 ","End":"00:29.505","Text":"only that it is reasonable or plausible."},{"Start":"00:29.505 ","End":"00:31.830","Text":"The reaction mechanism must lead to"},{"Start":"00:31.830 ","End":"00:36.645","Text":"the same rate equation as that measured experimentally."},{"Start":"00:36.645 ","End":"00:41.420","Text":"It must also agree with the stoichiometry of the overall reaction."},{"Start":"00:41.420 ","End":"00:45.725","Text":"Now, what\u0027s an elementary reaction?"},{"Start":"00:45.725 ","End":"00:48.470","Text":"Now, an elementary reaction describes"},{"Start":"00:48.470 ","End":"00:53.050","Text":"a distinct event such as the collision between particles."},{"Start":"00:53.050 ","End":"00:58.100","Text":"Elementary reactions or processes are reversible,"},{"Start":"00:58.100 ","End":"01:00.265","Text":"they can go in the opposite direction."},{"Start":"01:00.265 ","End":"01:04.550","Text":"Sometimes equilibrium could be obtained if the rate of"},{"Start":"01:04.550 ","End":"01:08.750","Text":"the forward reaction is equal to that of the reverse reaction."},{"Start":"01:08.750 ","End":"01:11.855","Text":"We\u0027ll hear more about that in future videos."},{"Start":"01:11.855 ","End":"01:15.500","Text":"Now we\u0027re going to define molecularity."},{"Start":"01:15.500 ","End":"01:19.610","Text":"Now, the molecularity of an elementary reaction is the number"},{"Start":"01:19.610 ","End":"01:23.710","Text":"of particles that take part in the reaction."},{"Start":"01:23.710 ","End":"01:27.740","Text":"If only 1 particle takes plot, for example,"},{"Start":"01:27.740 ","End":"01:32.870","Text":"if it decomposes, the reaction is unimolecular."},{"Start":"01:32.870 ","End":"01:36.865","Text":"If 2 particles steep part of the reaction is bimolecular."},{"Start":"01:36.865 ","End":"01:39.810","Text":"If the 3 particles take part,"},{"Start":"01:39.810 ","End":"01:43.865","Text":"the reaction\u0027s called trimolecular or termolecular."},{"Start":"01:43.865 ","End":"01:47.924","Text":"3 body collision is a rare event."},{"Start":"01:47.924 ","End":"01:50.450","Text":"Even in crashes between cars."},{"Start":"01:50.450 ","End":"01:53.645","Text":"It\u0027s unusual for 3 cars to crash simultaneously,"},{"Start":"01:53.645 ","End":"01:55.925","Text":"usually 2 crash into each other,"},{"Start":"01:55.925 ","End":"01:59.030","Text":"and then a third crashes into one of them."},{"Start":"01:59.030 ","End":"02:03.810","Text":"Now, what are the rate equations for elementary reactions?"},{"Start":"02:04.900 ","End":"02:09.875","Text":"Now the order of an elementary reaction with respect to a particular reactant"},{"Start":"02:09.875 ","End":"02:14.720","Text":"is the same as the stoichiometric coefficient and the chemical equation."},{"Start":"02:14.720 ","End":"02:16.985","Text":"This is very important."},{"Start":"02:16.985 ","End":"02:18.890","Text":"The order is the same as"},{"Start":"02:18.890 ","End":"02:21.440","Text":"the stoichiometric coefficient because"},{"Start":"02:21.440 ","End":"02:24.970","Text":"that tells us the number of particles that have collided."},{"Start":"02:24.970 ","End":"02:31.625","Text":"The total order of an elementary reaction is equal to its molecularity."},{"Start":"02:31.625 ","End":"02:35.500","Text":"For example, unimolecular reaction,"},{"Start":"02:35.500 ","End":"02:37.340","Text":"A turning into products,"},{"Start":"02:37.340 ","End":"02:45.805","Text":"the rate is equal to k times the concentration of A to the power 1 and unit."},{"Start":"02:45.805 ","End":"02:48.725","Text":"If we have a bimolecular reaction,"},{"Start":"02:48.725 ","End":"02:51.440","Text":"A plus A turning into products,"},{"Start":"02:51.440 ","End":"02:55.345","Text":"the rate will be equal to k A squared."},{"Start":"02:55.345 ","End":"03:00.655","Text":"The order is equal to the molecularity."},{"Start":"03:00.655 ","End":"03:03.360","Text":"Or we could have A plus B,"},{"Start":"03:03.360 ","End":"03:06.545","Text":"to different species giving products."},{"Start":"03:06.545 ","End":"03:10.820","Text":"Then the rate is equal to k A times B,"},{"Start":"03:10.820 ","End":"03:14.680","Text":"concentration of A times the concentration of B."},{"Start":"03:14.680 ","End":"03:19.885","Text":"Each one of these is 1 and the total order is 2."},{"Start":"03:19.885 ","End":"03:23.255","Text":"The same as the molecularity."},{"Start":"03:23.255 ","End":"03:25.880","Text":"If it\u0027s trimolecular,"},{"Start":"03:25.880 ","End":"03:29.510","Text":"we can have 3 possibilities A plus A plus A to"},{"Start":"03:29.510 ","End":"03:34.105","Text":"products and the rate is equal to k A to the power 3,"},{"Start":"03:34.105 ","End":"03:37.425","Text":"A plus A plus B to products."},{"Start":"03:37.425 ","End":"03:40.500","Text":"The rates k A squared times B,"},{"Start":"03:40.500 ","End":"03:42.780","Text":"or A plus B plus C to products."},{"Start":"03:42.780 ","End":"03:44.085","Text":"The rate is k,"},{"Start":"03:44.085 ","End":"03:49.580","Text":"concentration of A times concentration of B times concentration of C. In all these,"},{"Start":"03:49.580 ","End":"03:52.710","Text":"the total order is 3."},{"Start":"03:53.060 ","End":"03:58.170","Text":"Here with respect to A is 3 with respect to A is 2,"},{"Start":"03:58.170 ","End":"04:00.090","Text":"and 1 with respect to B,"},{"Start":"04:00.090 ","End":"04:02.550","Text":"and here\u0027s 1,1,1 with respect to A,"},{"Start":"04:02.550 ","End":"04:06.130","Text":"B, and C, but the total is 3."},{"Start":"04:06.130 ","End":"04:10.955","Text":"Now a few words about intermediate species."},{"Start":"04:10.955 ","End":"04:17.390","Text":"An intermediate species is produced in 1 elementary process and consumed in another."},{"Start":"04:17.390 ","End":"04:23.160","Text":"They do not appear in the overall chemical equation or rate equation."},{"Start":"04:24.280 ","End":"04:29.780","Text":"Now a very important concept is the rate determining step."},{"Start":"04:29.780 ","End":"04:35.570","Text":"Sometimes the particular elementary reaction is much slower than all the rest."},{"Start":"04:35.570 ","End":"04:42.380","Text":"It acts as a bottleneck and determines the overall rate of reaction."},{"Start":"04:42.380 ","End":"04:47.795","Text":"And we\u0027ll see several examples of this in the next few videos."},{"Start":"04:47.795 ","End":"04:55.170","Text":"In this video, we introduced reaction mechanisms and discussed some of the concepts."}],"ID":30437},{"Watched":false,"Name":"Slow Step Followed by Fast Step","Duration":"2m 35s","ChapterTopicVideoID":28825,"CourseChapterTopicPlaylistID":293018,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.050 ","End":"00:02.400","Text":"In the previous video,"},{"Start":"00:02.400 ","End":"00:05.505","Text":"we began a discussion of reaction mechanisms."},{"Start":"00:05.505 ","End":"00:07.350","Text":"In this video, we\u0027ll talk about"},{"Start":"00:07.350 ","End":"00:11.910","Text":"a reaction mechanism involving a slow step followed by a fast one."},{"Start":"00:11.910 ","End":"00:15.825","Text":"We\u0027re going to talk about a slow step followed by a fast one."},{"Start":"00:15.825 ","End":"00:21.870","Text":"Here\u0027s an example, the reaction of hydrogen with iodine monochloride."},{"Start":"00:21.870 ","End":"00:23.594","Text":"Here\u0027s the equation."},{"Start":"00:23.594 ","End":"00:26.310","Text":"Hydrogen, H_2, the gas phase,"},{"Start":"00:26.310 ","End":"00:30.840","Text":"plus 2 ICl in the gas phase, that\u0027s iodine monochloride."},{"Start":"00:30.840 ","End":"00:34.335","Text":"To give us iodine gas plus 2 HCl,"},{"Start":"00:34.335 ","End":"00:37.215","Text":"HCl is hydrogen chloride."},{"Start":"00:37.215 ","End":"00:42.510","Text":"The rate equation found experimentally is"},{"Start":"00:42.510 ","End":"00:49.205","Text":"that rate equals k times the concentration of H_2 times the concentration of ICl."},{"Start":"00:49.205 ","End":"00:52.324","Text":"Now here\u0027s the proposed mechanism."},{"Start":"00:52.324 ","End":"00:57.440","Text":"H_2 reacting with ICl to give us HI plus HCl."},{"Start":"00:57.440 ","End":"01:00.220","Text":"That\u0027s supposed to be a slow step."},{"Start":"01:00.220 ","End":"01:03.020","Text":"Then HI, which has an intermediate,"},{"Start":"01:03.020 ","End":"01:07.695","Text":"is formed in the first step and reacts in the second step,"},{"Start":"01:07.695 ","End":"01:12.195","Text":"HI plus ICl to give I_2 plus HCl."},{"Start":"01:12.195 ","End":"01:16.270","Text":"That\u0027s proposed to be the fast step."},{"Start":"01:16.370 ","End":"01:20.145","Text":"Let\u0027s look at the overall reaction."},{"Start":"01:20.145 ","End":"01:24.090","Text":"We have H_2 plus 2 ICl,"},{"Start":"01:24.090 ","End":"01:29.490","Text":"HI cancels to give us I_2 plus 2 HCl."},{"Start":"01:29.490 ","End":"01:31.880","Text":"The overall reaction is correct."},{"Start":"01:31.880 ","End":"01:33.590","Text":"Now we\u0027re asking the question,"},{"Start":"01:33.590 ","End":"01:36.325","Text":"is the reaction mechanism plausible?"},{"Start":"01:36.325 ","End":"01:40.924","Text":"First point is that the overall reaction is correct, that\u0027s essential."},{"Start":"01:40.924 ","End":"01:45.275","Text":"The second is that the elementary reactions are bimolecular,"},{"Start":"01:45.275 ","End":"01:49.685","Text":"not trimolecular, which is somewhat improbable,"},{"Start":"01:49.685 ","End":"01:52.805","Text":"and we have an intermediate HI."},{"Start":"01:52.805 ","End":"01:57.700","Text":"Now, the first slow step is the rate-determining step."},{"Start":"01:57.700 ","End":"02:00.045","Text":"Let\u0027s look at its rate equation."},{"Start":"02:00.045 ","End":"02:02.250","Text":"Its rate is k_1,"},{"Start":"02:02.250 ","End":"02:05.235","Text":"if the k_1 is rated the first reaction."},{"Start":"02:05.235 ","End":"02:10.705","Text":"The concentration of H_2 times the concentration of ICl."},{"Start":"02:10.705 ","End":"02:14.655","Text":"That\u0027s an agreement with experiment."},{"Start":"02:14.655 ","End":"02:17.480","Text":"Because it\u0027s an agreement with experiment,"},{"Start":"02:17.480 ","End":"02:19.760","Text":"the mechanism is plausible."},{"Start":"02:19.760 ","End":"02:22.130","Text":"That means the mechanism is reasonable,"},{"Start":"02:22.130 ","End":"02:26.575","Text":"not that it\u0027s necessarily the correct mechanism."},{"Start":"02:26.575 ","End":"02:31.620","Text":"In this video, we learnt about a simple reaction mechanism."}],"ID":30438},{"Watched":false,"Name":"Fast Reversible Step Followed by Slow Step","Duration":"5m 27s","ChapterTopicVideoID":28824,"CourseChapterTopicPlaylistID":293018,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.275","Text":"In the previous video,"},{"Start":"00:01.275 ","End":"00:06.200","Text":"we talked about a reaction mechanism which are slow steps followed by a fast one,"},{"Start":"00:06.200 ","End":"00:07.245","Text":"and in this video,"},{"Start":"00:07.245 ","End":"00:10.950","Text":"a rapid equilibrium will be followed by a slow step."},{"Start":"00:10.950 ","End":"00:14.670","Text":"We\u0027re going to talk about a reaction mechanism in which"},{"Start":"00:14.670 ","End":"00:18.600","Text":"a fast reversible step is followed by a slow step."},{"Start":"00:18.600 ","End":"00:23.750","Text":"For example, the reaction of nitrogen monoxide with oxygen."},{"Start":"00:23.750 ","End":"00:31.260","Text":"The chemical equation is 2 NO plus O_2 to give 2 NO_2 all in the gas phase."},{"Start":"00:31.260 ","End":"00:40.880","Text":"The rate equation is rate= k[NO]^2 concentration of NO^2 times the concentration of O_2."},{"Start":"00:40.880 ","End":"00:45.035","Text":"Now you can see that it\u0027s second order with respect to NO,"},{"Start":"00:45.035 ","End":"00:48.215","Text":"and first order with respect to O_2."},{"Start":"00:48.215 ","End":"00:50.785","Text":"In total, it\u0027s third-order."},{"Start":"00:50.785 ","End":"00:56.240","Text":"So it\u0027s unlikely to occur in one step since that would involve a 3-body collision."},{"Start":"00:56.240 ","End":"00:58.490","Text":"Now here\u0027s the proposed mechanism."},{"Start":"00:58.490 ","End":"01:02.245","Text":"First of all, 2 NO in equilibrium,"},{"Start":"01:02.245 ","End":"01:04.945","Text":"often called pre-equilibrium,"},{"Start":"01:04.945 ","End":"01:07.550","Text":"because it\u0027s before another step."},{"Start":"01:07.550 ","End":"01:11.065","Text":"2 NO in equilibrium with N_2O_2,"},{"Start":"01:11.065 ","End":"01:17.435","Text":"k_1 is going to be the rate constant for left to right and k_minus 1 from right to left."},{"Start":"01:17.435 ","End":"01:19.940","Text":"Now this is a fast step."},{"Start":"01:19.940 ","End":"01:23.945","Text":"Then the slow step is N_2O_2,"},{"Start":"01:23.945 ","End":"01:26.000","Text":"which is, of course,"},{"Start":"01:26.000 ","End":"01:32.347","Text":"an intermediate because it\u0027s formed in the first step and then reacts in the second step."},{"Start":"01:32.347 ","End":"01:35.120","Text":"So N_2O_2 plus O_2 to give"},{"Start":"01:35.120 ","End":"01:41.275","Text":"2 NO_2 and that\u0027s going to have the rate constant k_2 and that\u0027s slow step."},{"Start":"01:41.275 ","End":"01:44.480","Text":"First of all, let\u0027s see what we have overall."},{"Start":"01:44.480 ","End":"01:50.420","Text":"We have 2 NO plus O_2 giving 2 NO_2."},{"Start":"01:50.420 ","End":"01:55.800","Text":"And N_2O_2 is an intermediate so it drops out."},{"Start":"01:55.880 ","End":"01:59.175","Text":"The overall reaction is correct."},{"Start":"01:59.175 ","End":"02:02.705","Text":"We\u0027re going to ask, is the reaction mechanism plausible?"},{"Start":"02:02.705 ","End":"02:06.140","Text":"We saw that the overall chemical equation is correct and"},{"Start":"02:06.140 ","End":"02:12.085","Text":"the elementary reactions proposed here are bimolecular, not trimolecular."},{"Start":"02:12.085 ","End":"02:15.020","Text":"N_2O_2 is an intermediate."},{"Start":"02:15.020 ","End":"02:18.770","Text":"It\u0027s both formed and consumed in the mechanism."},{"Start":"02:18.770 ","End":"02:23.000","Text":"Now the second slow step is the rate-determining step."},{"Start":"02:23.000 ","End":"02:24.875","Text":"What\u0027s its rate equation?"},{"Start":"02:24.875 ","End":"02:31.530","Text":"Its rate is equal to k_2[N_2O_2] times O_2."},{"Start":"02:31.530 ","End":"02:36.320","Text":"No, the problem with this is the N_2O_2 appears in"},{"Start":"02:36.320 ","End":"02:42.230","Text":"the rate equation and N_2O_2 shouldn\u0027t appear as it\u0027s an intermediate."},{"Start":"02:42.230 ","End":"02:44.123","Text":"So let\u0027s find the rate equation."},{"Start":"02:44.123 ","End":"02:46.685","Text":"Now from the first step at equilibrium,"},{"Start":"02:46.685 ","End":"02:52.070","Text":"we have from left to right k_1 times a concentration of NO^2,"},{"Start":"02:52.070 ","End":"02:53.755","Text":"that\u0027s going from left to right,"},{"Start":"02:53.755 ","End":"02:59.190","Text":"and from right to left we have k_minus 1 times the concentration of N_2O_2."},{"Start":"02:59.190 ","End":"03:00.510","Text":"Here\u0027s our equilibrium,"},{"Start":"03:00.510 ","End":"03:05.670","Text":"we see 2 NO to N_2O_2 going left to right,"},{"Start":"03:05.670 ","End":"03:10.305","Text":"and N_2O_2 going to 2 NO, right to left."},{"Start":"03:10.305 ","End":"03:12.045","Text":"If there\u0027s an equilibrium,"},{"Start":"03:12.045 ","End":"03:13.730","Text":"the 2 rates are equal,"},{"Start":"03:13.730 ","End":"03:16.184","Text":"left to right and right to left."},{"Start":"03:16.184 ","End":"03:22.865","Text":"And now we can say the concentration of N_2O_2 is equal to k_1 NO^2,"},{"Start":"03:22.865 ","End":"03:25.325","Text":"divided by k_minus 1."},{"Start":"03:25.325 ","End":"03:27.800","Text":"Here\u0027s the expression for N_2O_2."},{"Start":"03:27.800 ","End":"03:35.660","Text":"Now if we substitute this expression for N_2O_2 into the rate equation, here we have it,"},{"Start":"03:35.660 ","End":"03:41.750","Text":"rate is equal to k_2 times the concentration N_2O_2 times the concentration of O_2,"},{"Start":"03:41.750 ","End":"03:46.460","Text":"then if we substitute the expression of N_2O_2 into this expression,"},{"Start":"03:46.460 ","End":"03:51.830","Text":"we\u0027ll get the rate is equal to k_1k_2 divided by k_minus 1,"},{"Start":"03:51.830 ","End":"03:54.485","Text":"times the concentration of NO^2,"},{"Start":"03:54.485 ","End":"03:57.005","Text":"times the concentration of O_2."},{"Start":"03:57.005 ","End":"04:05.570","Text":"Now this is the same as the experimental rate equation provided that this is k,"},{"Start":"04:05.570 ","End":"04:10.520","Text":"that this collection of rate constants is equal to k. I\u0027ve written it here."},{"Start":"04:10.520 ","End":"04:13.730","Text":"The rate agrees with experiment provided k is equal to"},{"Start":"04:13.730 ","End":"04:17.935","Text":"k_1 times k_2 divided by k_minus 1."},{"Start":"04:17.935 ","End":"04:20.960","Text":"Let\u0027s say a few words about the connection between"},{"Start":"04:20.960 ","End":"04:24.440","Text":"the equilibrium constant and the rate constants."},{"Start":"04:24.440 ","End":"04:26.600","Text":"Now from the equilibrium step,"},{"Start":"04:26.600 ","End":"04:28.331","Text":"we saw that left to right,"},{"Start":"04:28.331 ","End":"04:30.680","Text":"k times concentration of NO^2,"},{"Start":"04:30.680 ","End":"04:33.755","Text":"k_minus 1, the concentration of N_2O_2."},{"Start":"04:33.755 ","End":"04:38.420","Text":"We saw that these were equal and then we can write that k_1 divided by"},{"Start":"04:38.420 ","End":"04:45.670","Text":"k_minus 1 is equal to concentration of N_2O_2 divided by the concentration of NO^2."},{"Start":"04:45.670 ","End":"04:51.300","Text":"Now, you\u0027ll recall that that\u0027s just the equilibrium constant."},{"Start":"04:51.300 ","End":"05:00.610","Text":"We can see that the equilibrium constant is k is equal to k_1 divided by k_minus 1."},{"Start":"05:00.610 ","End":"05:04.700","Text":"Now we have a connection between equilibrium constant,"},{"Start":"05:04.700 ","End":"05:10.775","Text":"which is thermodynamics, and the rate constants which come from kinetics."},{"Start":"05:10.775 ","End":"05:14.750","Text":"Now this is a general result for reactions in equilibrium."},{"Start":"05:14.750 ","End":"05:20.450","Text":"It\u0027s a very important equation because it connects between kinetics and thermodynamics."},{"Start":"05:20.450 ","End":"05:23.690","Text":"In this video, we talked about a reaction mechanism in which"},{"Start":"05:23.690 ","End":"05:28.080","Text":"a fast reversible step is followed by a slow one."}],"ID":30439},{"Watched":false,"Name":"Steady State Approximation","Duration":"4m 58s","ChapterTopicVideoID":28826,"CourseChapterTopicPlaylistID":293018,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.415","Text":"In the previous video,"},{"Start":"00:02.415 ","End":"00:06.900","Text":"a rapid equilibrium was followed by a slow step and this video,"},{"Start":"00:06.900 ","End":"00:09.030","Text":"we will consider the reaction mechanisms,"},{"Start":"00:09.030 ","End":"00:13.560","Text":"the same reaction without assuming equilibrium."},{"Start":"00:14.600 ","End":"00:19.355","Text":"Here\u0027s the example we talked about in the previous video."},{"Start":"00:19.355 ","End":"00:21.860","Text":"The reaction of nitrogen monoxide with"},{"Start":"00:21.860 ","End":"00:29.185","Text":"oxygen and the chemical equation is 2 NO plus O_2 to give 2 NO_2."},{"Start":"00:29.185 ","End":"00:34.535","Text":"Now here\u0027s the proposed mechanism is the same as the 1 in the previous video,"},{"Start":"00:34.535 ","End":"00:39.544","Text":"except we don\u0027t assume equilibrium between the first 2 steps."},{"Start":"00:39.544 ","End":"00:42.680","Text":"We have first step 2 NO to N_2O_2,"},{"Start":"00:42.680 ","End":"00:46.325","Text":"the second step N_2O_2 to 2 NO,"},{"Start":"00:46.325 ","End":"00:49.100","Text":"that\u0027s the reverse of the first step."},{"Start":"00:49.100 ","End":"00:54.265","Text":"Then N_2O_2 plus oxygen to give 2 NO_2."},{"Start":"00:54.265 ","End":"00:57.130","Text":"Let\u0027s find the rate equation."},{"Start":"00:57.130 ","End":"00:59.120","Text":"Now from the last step,"},{"Start":"00:59.120 ","End":"01:00.935","Text":"we get that the rate is equal to"},{"Start":"01:00.935 ","End":"01:07.520","Text":"k_2 times the concentration of N_2O_2 times the concentration of O_2."},{"Start":"01:07.520 ","End":"01:11.345","Text":"However, N_2O_2 is an intermediate"},{"Start":"01:11.345 ","End":"01:16.020","Text":"and shouldn\u0027t appear in the rate equation so we have to get rid of N_2O_2."},{"Start":"01:16.030 ","End":"01:20.270","Text":"Now if we assume a steady-state for N_2O_2,"},{"Start":"01:20.270 ","End":"01:25.085","Text":"that means we assume the concentration is constant,"},{"Start":"01:25.085 ","End":"01:30.045","Text":"then its derivative or Delta"},{"Start":"01:30.045 ","End":"01:35.710","Text":"N_2O_2 divided by Delta t is equal to 0."},{"Start":"01:35.710 ","End":"01:41.400","Text":"We could write this just as a derivative d N_2O_2 dt."},{"Start":"01:41.960 ","End":"01:48.080","Text":"That\u0027s 0 because the derivative of a constant is 0."},{"Start":"01:48.080 ","End":"01:52.565","Text":"That means it appears and disappears at the same rate."},{"Start":"01:52.565 ","End":"01:59.985","Text":"Now this rate of appearance is k_1 times the concentration of NO^2."},{"Start":"01:59.985 ","End":"02:07.295","Text":"That\u0027s on the first equation and its rate of disappearance is k_minus 1 N_2O_2."},{"Start":"02:07.295 ","End":"02:11.142","Text":"That\u0027s on the second equation and"},{"Start":"02:11.142 ","End":"02:18.225","Text":"k_2 times N_2O_2 times O_2 that\u0027s on the third."},{"Start":"02:18.225 ","End":"02:22.580","Text":"Now if we equate the rate of appearance,"},{"Start":"02:22.580 ","End":"02:25.165","Text":"That\u0027s k_1, NO^2."},{"Start":"02:25.165 ","End":"02:26.560","Text":"I\u0027m just saying NO^2,"},{"Start":"02:26.560 ","End":"02:29.680","Text":"I mean the concentration of NO^2,"},{"Start":"02:29.680 ","End":"02:34.250","Text":"so k_1 NO^2 is equal to the rate of disappearance,"},{"Start":"02:34.250 ","End":"02:40.810","Text":"k_minus 1 N_2O_2 plus k_2 N_2O_2 times O_2."},{"Start":"02:40.810 ","End":"02:48.930","Text":"Then we can take the N_2O_2 as a common factor and find that N_2O_ 2 is equal to"},{"Start":"02:48.930 ","End":"02:58.090","Text":"k_1 times NO^2 divided by k _minus 1 plus k_2O_2."},{"Start":"02:58.180 ","End":"03:05.150","Text":"Now if we substitute that expression for N_2O_2 in the rate equation we"},{"Start":"03:05.150 ","End":"03:11.890","Text":"found before then we get that the rate equal to k_2 N_2O_2 O_2."},{"Start":"03:11.890 ","End":"03:13.845","Text":"When we substitute the N_2O_2,"},{"Start":"03:13.845 ","End":"03:15.090","Text":"we get k_1,"},{"Start":"03:15.090 ","End":"03:23.890","Text":"k_2 times NO^2 times O_2 divided by k_minus 1 plus k_2 O_2."},{"Start":"03:23.890 ","End":"03:28.820","Text":"This is a more complicated expression that we usually have and it will be"},{"Start":"03:28.820 ","End":"03:34.855","Text":"useful in a later video because it\u0027s useful in enzyme catalysis."},{"Start":"03:34.855 ","End":"03:41.560","Text":"Now the second step is faster than the third step. Here we have it."},{"Start":"03:41.560 ","End":"03:50.160","Text":"If the second step, k_minus 1 N_2O_2 is larger than k_2, N_2O_2 O_2."},{"Start":"03:50.160 ","End":"03:54.830","Text":"Then k_minus 1 will be larger than"},{"Start":"03:54.830 ","End":"04:01.130","Text":"k_2O_2 and then we can say that the rate is equal to k_1,"},{"Start":"04:01.130 ","End":"04:08.255","Text":"k_2 divided by k_minus 1, NO^2 times O_2."},{"Start":"04:08.255 ","End":"04:13.220","Text":"In other words, we\u0027re ignoring this k_2O_2."},{"Start":"04:13.220 ","End":"04:15.830","Text":"Now, this is the same result as we got"},{"Start":"04:15.830 ","End":"04:20.675","Text":"the assuming equilibrium and it agrees with experiment."},{"Start":"04:20.675 ","End":"04:25.430","Text":"Now if we assumed that the third step was fast and the second step,"},{"Start":"04:25.430 ","End":"04:30.010","Text":"that means k_2O_2 is greater than k_minus 1."},{"Start":"04:30.010 ","End":"04:34.315","Text":"The rate would be k_1 times NO^2."},{"Start":"04:34.315 ","End":"04:37.400","Text":"In other words, would be assuming this 1 is a small"},{"Start":"04:37.400 ","End":"04:43.825","Text":"1 and then O_2 cancels and k_2 cancels."},{"Start":"04:43.825 ","End":"04:46.985","Text":"Now, it\u0027s not useful in this reaction,"},{"Start":"04:46.985 ","End":"04:52.110","Text":"but this will be very useful when we get to the enzyme catalysis."},{"Start":"04:52.520 ","End":"04:57.990","Text":"In this video, we learned about the steady-state approximation."}],"ID":30440},{"Watched":false,"Name":"Chain Reactions","Duration":"3m 56s","ChapterTopicVideoID":28822,"CourseChapterTopicPlaylistID":293018,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.680 ","End":"00:02.940","Text":"In the previous videos,"},{"Start":"00:02.940 ","End":"00:05.505","Text":"we talked about various reaction mechanisms."},{"Start":"00:05.505 ","End":"00:08.610","Text":"In this video, we\u0027ll learn about chain reactions."},{"Start":"00:08.610 ","End":"00:11.490","Text":"We\u0027re going to talk about chain reactions."},{"Start":"00:11.490 ","End":"00:13.179","Text":"Now in a chain reaction,"},{"Start":"00:13.179 ","End":"00:15.330","Text":"a highly reactive intermediate,"},{"Start":"00:15.330 ","End":"00:17.595","Text":"which is often called a chain carrier,"},{"Start":"00:17.595 ","End":"00:21.090","Text":"reacts to form 1 or more reactive intermediates,"},{"Start":"00:21.090 ","End":"00:25.065","Text":"each of which reacts to form 1 or more intermediates and so on."},{"Start":"00:25.065 ","End":"00:28.110","Text":"An example is of formation of polymer chains,"},{"Start":"00:28.110 ","End":"00:32.130","Text":"nuclear fission, and many explosive reactions."},{"Start":"00:32.130 ","End":"00:33.705","Text":"Here\u0027s an illustration."},{"Start":"00:33.705 ","End":"00:37.245","Text":"Here\u0027s 1 intermediate reacting to give 2,"},{"Start":"00:37.245 ","End":"00:40.890","Text":"and each 1 of them reacting to give 2."},{"Start":"00:40.890 ","End":"00:44.695","Text":"Very rapidly, we have a lot of intermediates."},{"Start":"00:44.695 ","End":"00:46.265","Text":"If this goes on too long,"},{"Start":"00:46.265 ","End":"00:47.780","Text":"we can have an explosion."},{"Start":"00:47.780 ","End":"00:50.675","Text":"It can happen very rapidly."},{"Start":"00:50.675 ","End":"00:56.380","Text":"Let\u0027s take an example of the formation of HBr from hydrogen and bromine."},{"Start":"00:56.380 ","End":"01:01.455","Text":"The chemical reactions H_2 plus Br_2 to give 2HBr."},{"Start":"01:01.455 ","End":"01:03.525","Text":"Here\u0027s a mechanism."},{"Start":"01:03.525 ","End":"01:09.560","Text":"Now, chain reactions always begin with an initiation step."},{"Start":"01:09.560 ","End":"01:15.245","Text":"In this case, heat or light decomposes Br_2 into atoms."},{"Start":"01:15.245 ","End":"01:17.180","Text":"Now these atoms are radicals."},{"Start":"01:17.180 ","End":"01:21.620","Text":"Remember we have 1 unpaired electron in Br atoms."},{"Start":"01:21.620 ","End":"01:24.085","Text":"Then we have propagation."},{"Start":"01:24.085 ","End":"01:32.165","Text":"The Br formed in the initiation step reacts with hydrogen to give HBr and hydrogen,"},{"Start":"01:32.165 ","End":"01:34.729","Text":"and hydrogen atom is also radical."},{"Start":"01:34.729 ","End":"01:37.090","Text":"That\u0027s just 1 electron."},{"Start":"01:37.090 ","End":"01:44.895","Text":"Then the H could react with Br_2 to give HBr another Br atom,"},{"Start":"01:44.895 ","End":"01:47.325","Text":"which of course is a radical."},{"Start":"01:47.325 ","End":"01:50.805","Text":"Then that can react with H_2 and go on."},{"Start":"01:50.805 ","End":"01:57.195","Text":"These steps can go on and on propagating until something stops them."},{"Start":"01:57.195 ","End":"02:00.560","Text":"What stops them is called the terminations."},{"Start":"02:00.560 ","End":"02:04.290","Text":"We have initiation, propagation, and termination."},{"Start":"02:04.370 ","End":"02:11.145","Text":"Now the chain is terminated when a Br reacts with another Br to give Br_2,"},{"Start":"02:11.145 ","End":"02:15.000","Text":"or hydrogen reacts with Br to give HBr."},{"Start":"02:15.000 ","End":"02:18.215","Text":"We have 3 processes,"},{"Start":"02:18.215 ","End":"02:21.440","Text":"initiation, propagation, and termination."},{"Start":"02:21.440 ","End":"02:25.085","Text":"Let\u0027s look at a branching chain reaction."},{"Start":"02:25.085 ","End":"02:27.800","Text":"Now, in a branching chain reaction,"},{"Start":"02:27.800 ","End":"02:32.015","Text":"more than 1 chain carriers formed at each propagation step,"},{"Start":"02:32.015 ","End":"02:34.190","Text":"often leading to explosions."},{"Start":"02:34.190 ","End":"02:39.200","Text":"That\u0027s the same as I showed you in the diagram before."},{"Start":"02:39.200 ","End":"02:43.880","Text":"We had 2 intermediates formed in every step."},{"Start":"02:43.880 ","End":"02:45.650","Text":"Now here\u0027s an example,"},{"Start":"02:45.650 ","End":"02:49.480","Text":"the condition of a mixture of hydrogen and oxygen."},{"Start":"02:49.480 ","End":"02:51.870","Text":"Here\u0027s the mechanism."},{"Start":"02:51.870 ","End":"02:59.150","Text":"The initiation is when hydrogen is dissociated into 2 hydrogen atoms."},{"Start":"02:59.150 ","End":"03:00.319","Text":"Now these are radicals,"},{"Start":"03:00.319 ","End":"03:02.245","Text":"each have 1 electron."},{"Start":"03:02.245 ","End":"03:06.395","Text":"Once again, we have an initiation step."},{"Start":"03:06.395 ","End":"03:09.305","Text":"Now we have branching."},{"Start":"03:09.305 ","End":"03:13.820","Text":"The hydrogen formed in the initiation process can react with"},{"Start":"03:13.820 ","End":"03:18.965","Text":"oxygen to form HO and O, an O atom."},{"Start":"03:18.965 ","End":"03:22.840","Text":"An O atom has 2 unpaired electrons."},{"Start":"03:22.840 ","End":"03:27.770","Text":"We started with 1 intermediate and we formed 2."},{"Start":"03:27.770 ","End":"03:32.690","Text":"Then the O can react with H_2 to give us HO"},{"Start":"03:32.690 ","End":"03:38.090","Text":"plus H. We have the H formed again and it can react again."},{"Start":"03:38.090 ","End":"03:40.400","Text":"These are branching reactions."},{"Start":"03:40.400 ","End":"03:46.035","Text":"Every time 1 radical gives us 2 radicals."},{"Start":"03:46.035 ","End":"03:51.260","Text":"The number of radicals increases rapidly and an explosion can occur."},{"Start":"03:51.260 ","End":"03:56.460","Text":"In this video, we talked about chain reactions."}],"ID":30441},{"Watched":false,"Name":"Exercise 1","Duration":"3m 27s","ChapterTopicVideoID":31672,"CourseChapterTopicPlaylistID":293018,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33907},{"Watched":false,"Name":"Exercise 2","Duration":"4m 44s","ChapterTopicVideoID":31670,"CourseChapterTopicPlaylistID":293018,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33908},{"Watched":false,"Name":"Exercise 3","Duration":"3m 20s","ChapterTopicVideoID":31671,"CourseChapterTopicPlaylistID":293018,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33909}],"Thumbnail":null,"ID":293018},{"Name":"Effect of Temperature","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Arrhenius Equation 1","Duration":"6m 24s","ChapterTopicVideoID":28827,"CourseChapterTopicPlaylistID":292232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.410","Text":"In previous videos,"},{"Start":"00:01.410 ","End":"00:03.390","Text":"we spoke about rates of reactions."},{"Start":"00:03.390 ","End":"00:08.160","Text":"In this video we\u0027ll learn about the effect of temperature on these rates."},{"Start":"00:08.160 ","End":"00:12.600","Text":"We\u0027re going to talk about the effect of temperature on rates of reactions."},{"Start":"00:12.600 ","End":"00:16.080","Text":"Now, rates of reactions depend on the temperature"},{"Start":"00:16.080 ","End":"00:19.305","Text":"and most reactions go faster when the temperature is"},{"Start":"00:19.305 ","End":"00:23.160","Text":"raised and that\u0027s the reason we refrigerate food to"},{"Start":"00:23.160 ","End":"00:27.660","Text":"slow down its decomposition and heat food to cook it."},{"Start":"00:27.660 ","End":"00:30.884","Text":"Now we\u0027re going to talk about the important equation,"},{"Start":"00:30.884 ","End":"00:33.765","Text":"the Arrhenius equation for the rate constant."},{"Start":"00:33.765 ","End":"00:36.860","Text":"Now, in 1889 Arrhenius discovered"},{"Start":"00:36.860 ","End":"00:43.235","Text":"experimentally the plotting lnk versus 1/T often gives a straight line."},{"Start":"00:43.235 ","End":"00:47.930","Text":"Here\u0027s a plot of lnk against 1/T."},{"Start":"00:47.930 ","End":"00:55.320","Text":"Nowadays we write lnk=lnA minus E_a divided by RT."},{"Start":"00:55.320 ","End":"00:57.643","Text":"The intercept is lnA,"},{"Start":"00:57.643 ","End":"00:59.280","Text":"here we have it,"},{"Start":"00:59.280 ","End":"01:02.550","Text":"and the slope is minus E_a/R."},{"Start":"01:02.550 ","End":"01:08.870","Text":"We can compare this to y=mx plus c. Now,"},{"Start":"01:08.870 ","End":"01:11.825","Text":"A is called the pre-exponential factor."},{"Start":"01:11.825 ","End":"01:14.030","Text":"You\u0027ll see in a few moments why it\u0027s called"},{"Start":"01:14.030 ","End":"01:19.235","Text":"pre-exponential and E_a is called the activation energy."},{"Start":"01:19.235 ","End":"01:22.775","Text":"Together they\u0027re called the Arrhenius parameters."},{"Start":"01:22.775 ","End":"01:25.340","Text":"They themselves are almost independent of"},{"Start":"01:25.340 ","End":"01:29.165","Text":"temperature but they vary from reaction to reaction."},{"Start":"01:29.165 ","End":"01:33.245","Text":"Now, if the reaction gives a straight line experimentally,"},{"Start":"01:33.245 ","End":"01:35.855","Text":"we say it shows Arrhenius behavior."},{"Start":"01:35.855 ","End":"01:38.600","Text":"Not all reactions give a straight line,"},{"Start":"01:38.600 ","End":"01:41.195","Text":"not all show Arrhenius behavior."},{"Start":"01:41.195 ","End":"01:44.645","Text":"Now, another way of writing lnK is equal to lnA"},{"Start":"01:44.645 ","End":"01:50.045","Text":"minus E_a divide by RT is to take the exponential of both sides."},{"Start":"01:50.045 ","End":"01:58.035","Text":"Then we get k=Ae to the power of minus E_a divided by RT."},{"Start":"01:58.035 ","End":"02:04.030","Text":"Now you can see why a is called pre-exponential because it\u0027s before the exponential."},{"Start":"02:04.030 ","End":"02:07.250","Text":"This is called the Arrhenius equation."},{"Start":"02:07.250 ","End":"02:12.640","Text":"If we look at the Arrhenius equation we can see that k decreases"},{"Start":"02:12.640 ","End":"02:19.825","Text":"as 1/T increases or the k increases as the temperature increases."},{"Start":"02:19.825 ","End":"02:21.505","Text":"Let\u0027s see a plot of that."},{"Start":"02:21.505 ","End":"02:29.445","Text":"Here\u0027s k plotted against 1/T and you can see it decreases exponentially."},{"Start":"02:29.445 ","End":"02:34.800","Text":"As t increases, 1/T decreases."},{"Start":"02:34.800 ","End":"02:36.900","Text":"This is the hot side,"},{"Start":"02:36.900 ","End":"02:41.540","Text":"this is high temperatures and this is the cold side."},{"Start":"02:41.540 ","End":"02:45.460","Text":"We can see that k goes down,"},{"Start":"02:45.460 ","End":"02:48.400","Text":"it decreases as we go from"},{"Start":"02:48.400 ","End":"02:54.355","Text":"a hot temperatures to cold temperatures as the temperature decreases."},{"Start":"02:54.355 ","End":"03:01.525","Text":"Or we can say that k increases as t increases."},{"Start":"03:01.525 ","End":"03:06.910","Text":"T is going from right to left is increasing and k is also increasing."},{"Start":"03:06.910 ","End":"03:08.765","Text":"Here\u0027s an example."},{"Start":"03:08.765 ","End":"03:13.675","Text":"For the second-order reaction of ethyl bromide with hydroxide,"},{"Start":"03:13.675 ","End":"03:21.285","Text":"calculate the rate constant at 300 K and 400 K given the value of A,"},{"Start":"03:21.285 ","End":"03:27.730","Text":"4.3 times 10^11 liters per mole per second and E_a,"},{"Start":"03:27.730 ","End":"03:32.645","Text":"the activation energy as 90 kilojoules per mole."},{"Start":"03:32.645 ","End":"03:34.630","Text":"Here\u0027s the chemical equation,"},{"Start":"03:34.630 ","End":"03:39.755","Text":"C_2H_5Br, that\u0027s ethyl bromide,"},{"Start":"03:39.755 ","End":"03:42.995","Text":"plus OH minus, that\u0027s hydroxide,"},{"Start":"03:42.995 ","End":"03:50.074","Text":"giving us C_2H_5OH which is ethanol plus Br minus."},{"Start":"03:50.074 ","End":"03:55.530","Text":"Now, the intercept is lnA, that\u0027s 26.79,"},{"Start":"03:55.530 ","End":"03:58.980","Text":"taking the ln of 4.3 times 10^11,"},{"Start":"03:58.980 ","End":"04:03.180","Text":"and the slope should be minus E_a/R."},{"Start":"04:03.180 ","End":"04:09.615","Text":"E_a is 9 times 10^4 joules per mole,"},{"Start":"04:09.615 ","End":"04:12.515","Text":"often given E_a in kilojoules."},{"Start":"04:12.515 ","End":"04:18.420","Text":"We have to convert it to joules because R is in joules."},{"Start":"04:18.420 ","End":"04:26.840","Text":"We\u0027re dividing by R which is 8.3145 joules per kelvin per mole."},{"Start":"04:26.840 ","End":"04:31.610","Text":"Then if we divide these 2 of course the joules goes and the mole,"},{"Start":"04:31.610 ","End":"04:33.320","Text":"the power of minus 1 goes,"},{"Start":"04:33.320 ","End":"04:39.350","Text":"we\u0027re left with 1/k to the power minus 1 which is k. The arithmetic gives us minus"},{"Start":"04:39.350 ","End":"04:46.740","Text":"1.082 times 10^4 K. Now we can calculate,"},{"Start":"04:46.740 ","End":"04:51.650","Text":"lnk at 300 Kelvin is 26.79."},{"Start":"04:51.650 ","End":"04:54.805","Text":"That\u0027s the intercept minus,"},{"Start":"04:54.805 ","End":"05:02.090","Text":"and now we have minus E_a over R and that has to be multiplied by 1/T."},{"Start":"05:02.090 ","End":"05:08.983","Text":"We have minus 1.08 times 10^4 times 1 divided by 300,"},{"Start":"05:08.983 ","End":"05:17.600","Text":"that\u0027s 300 K. If we multiply that out we get minus 9.21,"},{"Start":"05:17.600 ","End":"05:21.680","Text":"so lnk is minus 9.21 and k is the"},{"Start":"05:21.680 ","End":"05:27.520","Text":"exponential of minus 9.21 and that\u0027s 1 times 10 to the power of minus 4."},{"Start":"05:27.520 ","End":"05:32.885","Text":"We can write the units because we\u0027re told it\u0027s a second order reaction."},{"Start":"05:32.885 ","End":"05:37.880","Text":"This is the same as m to the power of minus 1 seconds to minus 1."},{"Start":"05:37.880 ","End":"05:40.639","Text":"Now, at 400 Kelvin,"},{"Start":"05:40.639 ","End":"05:46.995","Text":"we have the same thing lnA minus E_a over R times 1/T,"},{"Start":"05:46.995 ","End":"05:53.880","Text":"and now it\u0027s 400 instead of 300 and that gives us minus 0.21."},{"Start":"05:53.880 ","End":"06:00.945","Text":"K is exponential of minus 0.21 and that 0.81."},{"Start":"06:00.945 ","End":"06:07.020","Text":"We can see this enormous increase in K,"},{"Start":"06:07.020 ","End":"06:15.970","Text":"in the value of K by increasing the temperature by 100 degrees enormous change."},{"Start":"06:15.970 ","End":"06:19.580","Text":"In this video we\u0027ll learnt about the Arrhenius equation and"},{"Start":"06:19.580 ","End":"06:24.150","Text":"in the next video we\u0027ll use it to solve a problem."}],"ID":30319},{"Watched":false,"Name":"Arrhenius Equation 2","Duration":"5m 14s","ChapterTopicVideoID":28828,"CourseChapterTopicPlaylistID":292232,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.605","Text":"In the previous video,"},{"Start":"00:01.605 ","End":"00:03.780","Text":"we learned about the Arrhenius equation."},{"Start":"00:03.780 ","End":"00:08.190","Text":"In this video we\u0027ll show an example of how to use it."},{"Start":"00:08.190 ","End":"00:11.820","Text":"First we\u0027re going to write an equation that"},{"Start":"00:11.820 ","End":"00:15.765","Text":"compares the rate constants at 2 different temperatures."},{"Start":"00:15.765 ","End":"00:22.020","Text":"We\u0027re going to write the lnk_1 is equal to InA minus E_a divided"},{"Start":"00:22.020 ","End":"00:28.390","Text":"by RT_1 where k_1 is the rate constant at temperature T_1."},{"Start":"00:28.390 ","End":"00:31.985","Text":"We\u0027re assuming the A and E_a are constant."},{"Start":"00:31.985 ","End":"00:35.765","Text":"We can write the same equation at a second, temperature,"},{"Start":"00:35.765 ","End":"00:40.820","Text":"lnk_2 is equal to lnA minus E_a divided by RT_2."},{"Start":"00:40.820 ","End":"00:46.370","Text":"So k2 is the rate constant at temperature T_2."},{"Start":"00:46.370 ","End":"00:53.315","Text":"We\u0027re going to subtract the first equation from the second equation."},{"Start":"00:53.315 ","End":"00:56.240","Text":"lnk2 minus Ink1,"},{"Start":"00:56.240 ","End":"01:00.250","Text":"that\u0027s equal to lnk2/k1."},{"Start":"01:00.250 ","End":"01:03.750","Text":"On the left hand side and on the right-hand side,"},{"Start":"01:03.750 ","End":"01:11.115","Text":"lnA cancels and we get minus Ea divided by RT_2."},{"Start":"01:11.115 ","End":"01:17.610","Text":"I\u0027ve taken out the Ea/R left with minus 1/T_2."},{"Start":"01:17.610 ","End":"01:25.695","Text":"Then that\u0027s minus minus Ea over RT_1, that\u0027s plus Ea/RT_1."},{"Start":"01:25.695 ","End":"01:29.460","Text":"Here we have EA over R times 1/T_1."},{"Start":"01:29.460 ","End":"01:31.800","Text":"Here\u0027s our final equation."},{"Start":"01:31.800 ","End":"01:36.185","Text":"The ln of the ratio between k2 and k1 is equal to"},{"Start":"01:36.185 ","End":"01:42.510","Text":"Ea/R times 1/T_1 minus 1/T_2."},{"Start":"01:42.510 ","End":"01:44.220","Text":"Let\u0027s look at this equation."},{"Start":"01:44.220 ","End":"01:46.935","Text":"When T_2 is greater than T_1,"},{"Start":"01:46.935 ","End":"01:50.745","Text":"1/T_1 minus 1/T_2 will be positive."},{"Start":"01:50.745 ","End":"01:54.900","Text":"That means lnk_2/k_1 will also be"},{"Start":"01:54.900 ","End":"01:59.590","Text":"positive and then that means that k_2 is greater than k_1."},{"Start":"02:00.440 ","End":"02:04.355","Text":"That just confirms what we\u0027ve always said,"},{"Start":"02:04.355 ","End":"02:07.175","Text":"that as the temperature gets higher,"},{"Start":"02:07.175 ","End":"02:09.935","Text":"k will also get higher."},{"Start":"02:09.935 ","End":"02:17.549","Text":"Now if we look at a particular value of 1 over T_1 minus1/T_2 in this equation,"},{"Start":"02:17.549 ","End":"02:26.295","Text":"we see that lnk2/k_1 gets larger as Ea over R gets larger."},{"Start":"02:26.295 ","End":"02:30.390","Text":"That\u0027s also true for k_2/k_1."},{"Start":"02:30.390 ","End":"02:34.695","Text":"Its increases as EA over r increases."},{"Start":"02:34.695 ","End":"02:38.090","Text":"That means that the higher the activation energy,"},{"Start":"02:38.090 ","End":"02:43.190","Text":"the greater the effect of the rate constant of changing the temperature."},{"Start":"02:43.190 ","End":"02:45.045","Text":"Here\u0027s an example."},{"Start":"02:45.045 ","End":"02:48.560","Text":"For the reaction of ethyl bromide with hydroxide."},{"Start":"02:48.560 ","End":"02:51.755","Text":"That\u0027s the same reaction as we spoke about before."},{"Start":"02:51.755 ","End":"02:59.385","Text":"Calculate the rate constant k_2 at T_2=320 K. Given that"},{"Start":"02:59.385 ","End":"03:08.205","Text":"k_1 is equal to 1.00 times 10^ minus 4 to the power minus 4 at T_1=300k."},{"Start":"03:08.205 ","End":"03:12.805","Text":"We\u0027re given the values of A and the value of Ea/R,"},{"Start":"03:12.805 ","End":"03:16.120","Text":"and they are the same as we had in the previous video."},{"Start":"03:16.120 ","End":"03:17.580","Text":"Here\u0027s our equation."},{"Start":"03:17.580 ","End":"03:22.880","Text":"Ink_2/k_1=EA /R times 1/T_1 minus"},{"Start":"03:22.880 ","End":"03:29.765","Text":"1/T_2./ We know that Ea/R is 1.082 times 10^4."},{"Start":"03:29.765 ","End":"03:35.070","Text":"Here it is. We can work out the value"},{"Start":"03:35.070 ","End":"03:43.590","Text":"of 1/300T_1 minus 1/320 and 320 is T_2."},{"Start":"03:43.590 ","End":"03:48.660","Text":"If we work that out, we get 2.254."},{"Start":"03:48.660 ","End":"03:54.450","Text":"Since we\u0027re talking about a logarithm, it\u0027s dimensionless."},{"Start":"03:54.760 ","End":"04:01.130","Text":"If we now take the exponential of 2.254,"},{"Start":"04:01.130 ","End":"04:05.180","Text":"we get k_2 divided by k_1, the exponentials."},{"Start":"04:05.180 ","End":"04:15.085","Text":"The log is just k_2/k_1 is equal to e to the power 2.254, and that\u0027s 9.526."},{"Start":"04:15.085 ","End":"04:22.215","Text":"We can conclude that k_2 is equal to k_1 times 9.526,"},{"Start":"04:22.215 ","End":"04:30.555","Text":"k_1 is 1.00 times 10^ minus 4 and we\u0027re going to multiply that by 9.526."},{"Start":"04:30.555 ","End":"04:35.580","Text":"So that\u0027s 9.526 times 10^ minus 4."},{"Start":"04:35.580 ","End":"04:37.880","Text":"Units are, of course,"},{"Start":"04:37.880 ","End":"04:42.935","Text":"liter per mole per second is a second-order reaction."},{"Start":"04:42.935 ","End":"04:48.635","Text":"We can see that the rate constant increases by a factor of 9.5 of"},{"Start":"04:48.635 ","End":"04:54.170","Text":"almost 10 for an increase in temperatures of only 20k,"},{"Start":"04:54.170 ","End":"04:56.905","Text":"only 20 degrees Celsius."},{"Start":"04:56.905 ","End":"04:59.550","Text":"This is a very dramatic effect."},{"Start":"04:59.550 ","End":"05:04.265","Text":"That\u0027s because for this case, EA was large."},{"Start":"05:04.265 ","End":"05:07.135","Text":"We had a high activation energy."},{"Start":"05:07.135 ","End":"05:11.999","Text":"In this video, we continue to learn about the Arrhenius equation."}],"ID":30320},{"Watched":false,"Name":"Exercise 1","Duration":"5m 50s","ChapterTopicVideoID":31674,"CourseChapterTopicPlaylistID":292232,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33910},{"Watched":false,"Name":"Exercise 2","Duration":"5m 44s","ChapterTopicVideoID":31673,"CourseChapterTopicPlaylistID":292232,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33911}],"Thumbnail":null,"ID":292232},{"Name":"Theoretical Models for Reactions","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Collision Theory","Duration":"7m 53s","ChapterTopicVideoID":28829,"CourseChapterTopicPlaylistID":292233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.890 ","End":"00:03.135","Text":"In the previous video,"},{"Start":"00:03.135 ","End":"00:04.986","Text":"we talked about the radius collision."},{"Start":"00:04.986 ","End":"00:06.574","Text":"In this video,"},{"Start":"00:06.574 ","End":"00:11.520","Text":"we\u0027ll talk about collision theory that explains the equation at the molecular level."},{"Start":"00:11.520 ","End":"00:15.010","Text":"We\u0027re going to talk about collision theory."},{"Start":"00:15.920 ","End":"00:19.770","Text":"Now, collision theory is a theoretical model of"},{"Start":"00:19.770 ","End":"00:23.130","Text":"reactions between molecules in the gas phase."},{"Start":"00:23.130 ","End":"00:28.770","Text":"Now reactions can only occur between 2 molecules if there\u0027s a collision between them."},{"Start":"00:28.770 ","End":"00:33.015","Text":"But however, not all collisions lead to a reaction."},{"Start":"00:33.015 ","End":"00:40.470","Text":"Only collisions with a relative kinetic energy greater than E_min lead to reaction."},{"Start":"00:40.470 ","End":"00:42.620","Text":"Then only collisions with"},{"Start":"00:42.620 ","End":"00:47.650","Text":"suitable relative orientation of the molecules lead to a reaction."},{"Start":"00:47.650 ","End":"00:52.100","Text":"We\u0027re going to start by talking about the collision frequency."},{"Start":"00:52.100 ","End":"00:56.980","Text":"Now the collision frequency is the number of collisions per second."},{"Start":"00:56.980 ","End":"01:00.920","Text":"It can be calculated from the kinetic theory of gases,"},{"Start":"01:00.920 ","End":"01:03.905","Text":"which we met when we talked about gases."},{"Start":"01:03.905 ","End":"01:07.850","Text":"Now the collision frequency is equal to Sigma,"},{"Start":"01:07.850 ","End":"01:13.280","Text":"the mean relative speed of the 2 molecules N_A^2,"},{"Start":"01:13.280 ","End":"01:16.475","Text":"N_A is Avogadro\u0027s number,"},{"Start":"01:16.475 ","End":"01:20.810","Text":"times the concentration of A times the concentration of B."},{"Start":"01:20.810 ","End":"01:28.415","Text":"Now this sigma is equal to Pi times r_A plus r_B^2,"},{"Start":"01:28.415 ","End":"01:34.900","Text":"where r_A is the radius of molecule A and r_B is the radius of molecule B."},{"Start":"01:34.900 ","End":"01:37.860","Text":"This is called the collision cross section."},{"Start":"01:37.860 ","End":"01:41.130","Text":"It\u0027s the area around a particle which the center of"},{"Start":"01:41.130 ","End":"01:45.305","Text":"another particle must lie in order for a collision to occur."},{"Start":"01:45.305 ","End":"01:51.960","Text":"Here\u0027s a picture. Here\u0027s r_A, r_B."},{"Start":"01:51.960 ","End":"01:55.565","Text":"The center of the blue molecule B,"},{"Start":"01:55.565 ","End":"02:02.705","Text":"must lie within this area in order for A and B to touch each other,"},{"Start":"02:02.705 ","End":"02:05.020","Text":"in order for A and B to collide."},{"Start":"02:05.020 ","End":"02:08.570","Text":"This is called the collision cross section."},{"Start":"02:08.570 ","End":"02:11.390","Text":"This is white area."},{"Start":"02:11.390 ","End":"02:20.110","Text":"Now the mean relative speed is equal to the square root of 8RT divided by Pi Mu."},{"Start":"02:20.110 ","End":"02:23.090","Text":"Mu is called the reduced mass and it\u0027s equal"},{"Start":"02:23.090 ","End":"02:26.150","Text":"to the molar mass of A times the molar mass of B,"},{"Start":"02:26.150 ","End":"02:30.875","Text":"divided by the sum of the molar masses of A and B."},{"Start":"02:30.875 ","End":"02:35.720","Text":"This is the mean relative speed between 2 colliding molecules."},{"Start":"02:35.720 ","End":"02:41.535","Text":"This u in a gas and Mu is called the reduced mass."},{"Start":"02:41.535 ","End":"02:48.020","Text":"We can see that the mean relative speed is proportional to T,"},{"Start":"02:48.020 ","End":"02:52.315","Text":"the temperature, square root of the temperature."},{"Start":"02:52.315 ","End":"02:57.590","Text":"It\u0027s proportional to the square root of the temperature."},{"Start":"02:57.590 ","End":"03:05.135","Text":"We can calculate this mean from the Maxwell-Boltzmann distribution of speeds."},{"Start":"03:05.135 ","End":"03:11.465","Text":"Now we can see that as u is proportional to the square root of temperature,"},{"Start":"03:11.465 ","End":"03:14.515","Text":"the collision frequency increases with temperature."},{"Start":"03:14.515 ","End":"03:16.610","Text":"This is not enough to explain"},{"Start":"03:16.610 ","End":"03:22.210","Text":"the high dependence of the rate constant on the temperature."},{"Start":"03:22.210 ","End":"03:31.190","Text":"In addition, the collision frequency gives far too high a rate constant."},{"Start":"03:31.190 ","End":"03:34.445","Text":"The theory isn\u0027t sufficient yet."},{"Start":"03:34.445 ","End":"03:37.670","Text":"The first correction to the theory is the fraction of"},{"Start":"03:37.670 ","End":"03:41.750","Text":"molecules with energy greater than E_min."},{"Start":"03:41.750 ","End":"03:45.050","Text":"We need to know this fraction in order to"},{"Start":"03:45.050 ","End":"03:50.435","Text":"have a more realistic expression for the rate constant."},{"Start":"03:50.435 ","End":"03:55.250","Text":"Now, here\u0027s the relative number of molecules plotted against the speed."},{"Start":"03:55.250 ","End":"03:59.750","Text":"Equally well, I could have done a plot against the kinetic energy."},{"Start":"03:59.750 ","End":"04:03.860","Text":"The kinetic energy is proportional to the speed squared."},{"Start":"04:03.860 ","End":"04:06.545","Text":"It\u0027s the same general principle."},{"Start":"04:06.545 ","End":"04:09.770","Text":"Let\u0027s take a particular energy, E_min."},{"Start":"04:09.770 ","End":"04:12.950","Text":"Now as we raise the temperature,"},{"Start":"04:12.950 ","End":"04:21.425","Text":"we see that more and more molecules have kinetic energy greater than E_min."},{"Start":"04:21.425 ","End":"04:25.130","Text":"We\u0027re talking about 100 Kelvin,"},{"Start":"04:25.130 ","End":"04:28.340","Text":"it\u0027s a very small quantity."},{"Start":"04:28.340 ","End":"04:33.170","Text":"At 300 Kelvin, it\u0027s a larger quantity,"},{"Start":"04:33.170 ","End":"04:36.320","Text":"and at 1,000 Kelvin even larger,"},{"Start":"04:36.320 ","End":"04:39.510","Text":"the area under this red curve."},{"Start":"04:39.590 ","End":"04:42.110","Text":"As the temperature increases,"},{"Start":"04:42.110 ","End":"04:47.545","Text":"the fraction of molecules with energy greater than E_min increases considerably."},{"Start":"04:47.545 ","End":"04:55.375","Text":"This fraction is equal to the exponential of minus E_min divided by RT."},{"Start":"04:55.375 ","End":"04:59.810","Text":"We can also calculate this from the Maxwell-Boltzmann distribution."},{"Start":"04:59.810 ","End":"05:03.230","Text":"This is the maximum Boltzmann distribution of speeds,"},{"Start":"05:03.230 ","End":"05:05.540","Text":"what we have in this picture."},{"Start":"05:05.540 ","End":"05:10.355","Text":"It rises quickly and then has a long tail,"},{"Start":"05:10.355 ","End":"05:13.075","Text":"rises and has a long tail."},{"Start":"05:13.075 ","End":"05:18.575","Text":"Now we\u0027re going to try to find an expression for the rate to constant."},{"Start":"05:18.575 ","End":"05:20.990","Text":"Now the rate of a reaction is equal to"},{"Start":"05:20.990 ","End":"05:25.165","Text":"the collision frequency times the fraction with enough energy."},{"Start":"05:25.165 ","End":"05:29.060","Text":"It\u0027s equal to this expression we had for the collision frequency"},{"Start":"05:29.060 ","End":"05:34.370","Text":"times the expression we had for the fraction with enough energy,"},{"Start":"05:34.370 ","End":"05:38.530","Text":"e^ minus E_min divided by RT."},{"Start":"05:38.530 ","End":"05:42.410","Text":"We can write that as equal to k times A,"},{"Start":"05:42.410 ","End":"05:45.230","Text":"concentration of A times the concentration of B,"},{"Start":"05:45.230 ","End":"05:49.130","Text":"so that k itself is equal to Sigma,"},{"Start":"05:49.130 ","End":"05:51.485","Text":"the mean relative speed,"},{"Start":"05:51.485 ","End":"05:57.510","Text":"N_A^2 e^ minus E_min divided by RT."},{"Start":"05:57.710 ","End":"06:02.155","Text":"Now if we compare this expression with the Arrhenius equation,"},{"Start":"06:02.155 ","End":"06:04.750","Text":"which is k equal to A,"},{"Start":"06:04.750 ","End":"06:07.710","Text":"e^ minus E_a divided by RT,"},{"Start":"06:07.710 ","End":"06:09.620","Text":"we can see that A,"},{"Start":"06:09.620 ","End":"06:11.560","Text":"the pre-exponential factor,"},{"Start":"06:11.560 ","End":"06:13.270","Text":"is equal to this expression,"},{"Start":"06:13.270 ","End":"06:17.935","Text":"Sigma times mean relative speed times N_A squared."},{"Start":"06:17.935 ","End":"06:22.765","Text":"E_a, the activation energy is equal to E_min."},{"Start":"06:22.765 ","End":"06:27.995","Text":"The pre-exponential factor comes from the rate of collisions."},{"Start":"06:27.995 ","End":"06:33.865","Text":"The activation energy is the minimum kinetic energy required for a reaction."},{"Start":"06:33.865 ","End":"06:41.180","Text":"However, this expression for the rate constant gives values that are too high."},{"Start":"06:41.180 ","End":"06:47.480","Text":"In order to decrease the expression for the rate constant,"},{"Start":"06:47.480 ","End":"06:49.355","Text":"in order to get smaller values,"},{"Start":"06:49.355 ","End":"06:52.690","Text":"we introduced the steric factor."},{"Start":"06:52.690 ","End":"06:57.935","Text":"This arises from the fact that for a reaction to occur,"},{"Start":"06:57.935 ","End":"07:03.250","Text":"the molecules must approach each other with the correct relative geometry."},{"Start":"07:03.250 ","End":"07:06.230","Text":"That reduces the rate constant."},{"Start":"07:06.230 ","End":"07:08.165","Text":"If only say, half,"},{"Start":"07:08.165 ","End":"07:11.015","Text":"the molecules have the right geometry,"},{"Start":"07:11.015 ","End":"07:14.665","Text":"that will certainly reduce the rate constant by half."},{"Start":"07:14.665 ","End":"07:16.460","Text":"In order to do this,"},{"Start":"07:16.460 ","End":"07:18.560","Text":"we introduced this P,"},{"Start":"07:18.560 ","End":"07:20.645","Text":"which has a value of less than 1,"},{"Start":"07:20.645 ","End":"07:23.225","Text":"and that\u0027s called the steric factor."},{"Start":"07:23.225 ","End":"07:24.620","Text":"Here\u0027s an example."},{"Start":"07:24.620 ","End":"07:28.480","Text":"Supposing we have the reaction, HI plus Cl."},{"Start":"07:28.480 ","End":"07:31.620","Text":"Now the Cl has to attach to the H,"},{"Start":"07:31.620 ","End":"07:39.515","Text":"so the Cl has to come towards the HI from the side where the H is."},{"Start":"07:39.515 ","End":"07:41.870","Text":"If it comes from the side where the I is,"},{"Start":"07:41.870 ","End":"07:43.595","Text":"there\u0027ll be no reaction."},{"Start":"07:43.595 ","End":"07:49.925","Text":"This is a steric factor that\u0027s going to reduce the rate constant."},{"Start":"07:49.925 ","End":"07:53.970","Text":"In this video, we learned about collision theory."}],"ID":30321},{"Watched":false,"Name":"Effect of Temperasture on Equilibrium Constants","Duration":"5m 27s","ChapterTopicVideoID":28830,"CourseChapterTopicPlaylistID":292233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.605","Text":"In a previous video,"},{"Start":"00:01.605 ","End":"00:07.185","Text":"we explained the effect of temperature on the equilibrium constant using thermodynamics."},{"Start":"00:07.185 ","End":"00:10.245","Text":"In this video, we explain it using kinetics."},{"Start":"00:10.245 ","End":"00:14.820","Text":"The important equation in some dynamics for"},{"Start":"00:14.820 ","End":"00:21.940","Text":"the temperature effect on the equilibrium constant is the Van\u0027t Hoff equation."},{"Start":"00:21.940 ","End":"00:26.400","Text":"It says that ln of K_2 divided by K_1,"},{"Start":"00:26.400 ","End":"00:34.789","Text":"where K is the equilibrium constant is equal to Delta H of the reaction, 0."},{"Start":"00:34.789 ","End":"00:38.540","Text":"That\u0027s the standard enthalpy of reaction,"},{"Start":"00:38.540 ","End":"00:41.134","Text":"divided by R, the gas constant,"},{"Start":"00:41.134 ","End":"00:44.330","Text":"times 1/T_1 minus 1/T_2,"},{"Start":"00:44.330 ","End":"00:48.850","Text":"where T is the temperature in Kelvin."},{"Start":"00:48.850 ","End":"00:55.580","Text":"The conclusions we drew from this was that for an endothermic reaction,"},{"Start":"00:55.580 ","End":"01:00.845","Text":"increase in temperature leads to an increase in K in the equilibrium constant."},{"Start":"01:00.845 ","End":"01:05.110","Text":"That favors the forward reaction, which is endothermic."},{"Start":"01:05.110 ","End":"01:09.380","Text":"K increases as the temperature increases."},{"Start":"01:09.380 ","End":"01:11.930","Text":"Now for an exothermic reaction,"},{"Start":"01:11.930 ","End":"01:15.770","Text":"it\u0027s increasing the temperature leads to a decrease in"},{"Start":"01:15.770 ","End":"01:20.075","Text":"K. K decreases as the temperature increases."},{"Start":"01:20.075 ","End":"01:25.225","Text":"That favors the reverse reaction, which is endothermic."},{"Start":"01:25.225 ","End":"01:33.188","Text":"In both cases, the endothermic reaction is favored by increasing the temperature,"},{"Start":"01:33.188 ","End":"01:35.735","Text":"and all this agrees with Le Chatelier."},{"Start":"01:35.735 ","End":"01:39.245","Text":"Now let\u0027s recall what we learned a few videos ago"},{"Start":"01:39.245 ","End":"01:44.120","Text":"about the relation between equilibrium constants and rate constants."},{"Start":"01:44.120 ","End":"01:47.720","Text":"We saw that the equilibrium constant is equal to the rate constant for"},{"Start":"01:47.720 ","End":"01:53.270","Text":"the forward reaction divided by the rate constant of the reverse reaction."},{"Start":"01:53.270 ","End":"01:57.470","Text":"Forward divided by reverse."},{"Start":"01:57.470 ","End":"02:00.905","Text":"Now, what about the Arrhenius equation,"},{"Start":"02:00.905 ","End":"02:02.150","Text":"which is from kinetics?"},{"Start":"02:02.150 ","End":"02:03.950","Text":"What does that tell us?"},{"Start":"02:03.950 ","End":"02:05.960","Text":"Just to recall what it is,"},{"Start":"02:05.960 ","End":"02:08.360","Text":"it\u0027s a ln of K_2 divided by K_1,"},{"Start":"02:08.360 ","End":"02:11.525","Text":"which is, K is the rate constant,"},{"Start":"02:11.525 ","End":"02:15.845","Text":"equal to E_a, which is the activation energy divided by R,"},{"Start":"02:15.845 ","End":"02:20.995","Text":"the gas constant, multiplied by 1/T1 minus 1/T2."},{"Start":"02:20.995 ","End":"02:25.100","Text":"It looks quite similar to the Van\u0027t Hoff equation."},{"Start":"02:25.100 ","End":"02:29.780","Text":"Now, increasing the temperature leads to an increase in rate constant."},{"Start":"02:29.780 ","End":"02:33.455","Text":"K goes up as the temperature increases."},{"Start":"02:33.455 ","End":"02:36.170","Text":"We also saw that reactions with"},{"Start":"02:36.170 ","End":"02:43.025","Text":"higher activation energy are more favored than those with lower activation energy."},{"Start":"02:43.025 ","End":"02:45.950","Text":"In other words, raising the temperature has"},{"Start":"02:45.950 ","End":"02:50.885","Text":"a greater effect on those with high activation energies."},{"Start":"02:50.885 ","End":"02:53.645","Text":"Now, what is a reaction profile?"},{"Start":"02:53.645 ","End":"02:57.970","Text":"We need to use this in order to explain what happens is the kinetics."},{"Start":"02:57.970 ","End":"03:01.550","Text":"A reaction profile is a diagram that shows"},{"Start":"03:01.550 ","End":"03:05.675","Text":"the energy changes that occur as a reaction proceeds."},{"Start":"03:05.675 ","End":"03:11.645","Text":"Here, the energy profiles for an endothermic reaction and an exothermic reaction."},{"Start":"03:11.645 ","End":"03:14.915","Text":"Let\u0027s look at the endothermic reaction first."},{"Start":"03:14.915 ","End":"03:17.930","Text":"The activation energy for"},{"Start":"03:17.930 ","End":"03:24.770","Text":"the forward reaction is greater than the activation energy for the reverse reaction."},{"Start":"03:24.770 ","End":"03:28.880","Text":"We need more energy to go over the barrier for"},{"Start":"03:28.880 ","End":"03:33.320","Text":"the forward reaction that we do from the reverse reaction."},{"Start":"03:33.320 ","End":"03:36.545","Text":"On the other hand, an exothermic reaction,"},{"Start":"03:36.545 ","End":"03:45.635","Text":"the activation energy for the reverse reaction is greater than for the forward reaction."},{"Start":"03:45.635 ","End":"03:50.270","Text":"Now, let\u0027s consider again the endothermic reaction."},{"Start":"03:50.270 ","End":"03:54.680","Text":"We saw that the activation energy is greater for the forward reaction,"},{"Start":"03:54.680 ","End":"03:56.495","Text":"which is endothermic,"},{"Start":"03:56.495 ","End":"03:59.600","Text":"which means that the forward reaction is more"},{"Start":"03:59.600 ","End":"04:03.965","Text":"sensitive to an increase in temperature than the reverse reaction."},{"Start":"04:03.965 ","End":"04:09.200","Text":"K_1 for the forward reaction changes more than K minus 1,"},{"Start":"04:09.200 ","End":"04:14.210","Text":"and since the equilibrium constant is the ratio of these 2,"},{"Start":"04:14.210 ","End":"04:19.435","Text":"it means K will increase and favoring the forward reaction."},{"Start":"04:19.435 ","End":"04:27.140","Text":"K will increase as the temperature increases and let\u0027s consider the exothermic reaction."},{"Start":"04:27.140 ","End":"04:36.500","Text":"Here, the activation energy is greater for the reverse process than the forward process."},{"Start":"04:36.500 ","End":"04:41.030","Text":"The activation energy is greater for the reverse reaction,"},{"Start":"04:41.030 ","End":"04:45.775","Text":"which is endothermic because the forward reaction is exothermic."},{"Start":"04:45.775 ","End":"04:49.160","Text":"It means that the reverse reaction is more"},{"Start":"04:49.160 ","End":"04:52.915","Text":"centered to increase in temperature than the forward reaction."},{"Start":"04:52.915 ","End":"04:56.178","Text":"K minus 1 changes more than K_1,"},{"Start":"04:56.178 ","End":"05:01.145","Text":"so the K decreases favoring the reverse direction."},{"Start":"05:01.145 ","End":"05:06.130","Text":"K decreases as the temperature increases."},{"Start":"05:06.130 ","End":"05:08.465","Text":"Let\u0027s summarize this."},{"Start":"05:08.465 ","End":"05:14.645","Text":"Thermodynamics, kinetics, and Le Chatelier\u0027s principle all give the same result;"},{"Start":"05:14.645 ","End":"05:20.020","Text":"raising the temperature favors the endothermic reaction."},{"Start":"05:20.020 ","End":"05:23.405","Text":"In this video, we explained the effect of temperature"},{"Start":"05:23.405 ","End":"05:27.660","Text":"on the equilibrium constant using kinetics."}],"ID":30322},{"Watched":false,"Name":"Transition-State Theory","Duration":"4m 55s","ChapterTopicVideoID":28831,"CourseChapterTopicPlaylistID":292233,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.290 ","End":"00:02.520","Text":"In the previous video,"},{"Start":"00:02.520 ","End":"00:04.620","Text":"we talked about collision theory."},{"Start":"00:04.620 ","End":"00:08.775","Text":"In this video, we\u0027ll learn about transition state theory."},{"Start":"00:08.775 ","End":"00:15.180","Text":"We\u0027re going to talk about transition state theory or activated-complex theory,"},{"Start":"00:15.180 ","End":"00:19.095","Text":"it\u0027s the same thing with 2 different names."},{"Start":"00:19.095 ","End":"00:25.635","Text":"Now collision theory relates only to gases and treats molecules as hard spheres."},{"Start":"00:25.635 ","End":"00:27.930","Text":"The transition state theory,"},{"Start":"00:27.930 ","End":"00:30.870","Text":"sometimes called activated-complex theory."},{"Start":"00:30.870 ","End":"00:36.390","Text":"Either [inaudible] or ACT is more general than collision theory."},{"Start":"00:36.390 ","End":"00:40.440","Text":"It applies to gases and solutions and takes account"},{"Start":"00:40.440 ","End":"00:44.630","Text":"of changes in molecular geometry as the reaction proceeds."},{"Start":"00:44.630 ","End":"00:49.790","Text":"Now it was first proposed by Henry Eyring and others in 1935,"},{"Start":"00:49.790 ","End":"00:53.440","Text":"but it has been improved many times since then."},{"Start":"00:53.440 ","End":"00:56.055","Text":"Now in transition state theory,"},{"Start":"00:56.055 ","End":"01:02.725","Text":"2 molecules approach each other and as they do so they form a transition state."},{"Start":"01:02.725 ","End":"01:05.070","Text":"This is not an intermediate,"},{"Start":"01:05.070 ","End":"01:09.245","Text":"it\u0027s something you can\u0027t isolate and measure."},{"Start":"01:09.245 ","End":"01:13.655","Text":"The former transition state or an activated complex,"},{"Start":"01:13.655 ","End":"01:19.665","Text":"that can either dissociate to form the initial reactants or proceed to form products,"},{"Start":"01:19.665 ","End":"01:22.730","Text":"it can either go backwards or forwards."},{"Start":"01:22.730 ","End":"01:24.605","Text":"Here\u0027s our diagram."},{"Start":"01:24.605 ","End":"01:31.400","Text":"We plot the potential energy as a function of the progress of the reaction."},{"Start":"01:31.400 ","End":"01:34.645","Text":"The reaction is progressing from left to right."},{"Start":"01:34.645 ","End":"01:38.855","Text":"Here\u0027s the activation energy for the forward reaction,"},{"Start":"01:38.855 ","End":"01:43.445","Text":"and this is the activation energy for the reverse reaction."},{"Start":"01:43.445 ","End":"01:47.990","Text":"At the top of the barrier we have the transition state,"},{"Start":"01:47.990 ","End":"01:50.845","Text":"highest energy in the barrier."},{"Start":"01:50.845 ","End":"01:52.500","Text":"The potential energy,"},{"Start":"01:52.500 ","End":"01:55.340","Text":"and that\u0027s a function of the positions of the atoms,"},{"Start":"01:55.340 ","End":"01:58.460","Text":"is plotted as the reaction progresses."},{"Start":"01:58.460 ","End":"02:03.205","Text":"For example, if we had an atom reacting with a diatomic molecule,"},{"Start":"02:03.205 ","End":"02:12.015","Text":"atom A reacting with a diatomic molecule B-C. Then A will come closer,"},{"Start":"02:12.015 ","End":"02:13.890","Text":"and as it becomes closer,"},{"Start":"02:13.890 ","End":"02:18.545","Text":"B-C will become further apart."},{"Start":"02:18.545 ","End":"02:23.540","Text":"This bond will weaken and lengthen as A comes closer."},{"Start":"02:23.540 ","End":"02:27.230","Text":"The transition state will be A, B, C,"},{"Start":"02:27.230 ","End":"02:34.740","Text":"very weak bonds between A and B and between B and C. As the reaction proceeds,"},{"Start":"02:34.740 ","End":"02:38.555","Text":"A will get closer to B forming a bond,"},{"Start":"02:38.555 ","End":"02:40.985","Text":"and C will get further away."},{"Start":"02:40.985 ","End":"02:44.255","Text":"Eventually, we have A-B molecule,"},{"Start":"02:44.255 ","End":"02:48.200","Text":"diatomic molecule and C a single atom."},{"Start":"02:48.200 ","End":"02:51.140","Text":"This is called the transition state,"},{"Start":"02:51.140 ","End":"02:53.405","Text":"and that\u0027s the top of the barrier."},{"Start":"02:53.405 ","End":"02:55.910","Text":"Here it\u0027s written as A comes closer,"},{"Start":"02:55.910 ","End":"02:58.690","Text":"the B-C bond gets weaker and longer."},{"Start":"02:58.690 ","End":"03:02.085","Text":"The transition state has 2 weak bonds,"},{"Start":"03:02.085 ","End":"03:05.885","Text":"and it has the highest potential energy, top of the barrier."},{"Start":"03:05.885 ","End":"03:09.185","Text":"Now, if the relative kinetic energy"},{"Start":"03:09.185 ","End":"03:15.440","Text":"between the molecule and the atom is less than the activation energy."},{"Start":"03:15.440 ","End":"03:18.425","Text":"If it\u0027s lower than the activation energy,"},{"Start":"03:18.425 ","End":"03:24.365","Text":"then the reaction will not proceed and the reactants will be re-formed,"},{"Start":"03:24.365 ","End":"03:27.065","Text":"it will go back to the reactants."},{"Start":"03:27.065 ","End":"03:31.040","Text":"They won\u0027t have enough energy to cross the activation barrier."},{"Start":"03:31.040 ","End":"03:33.560","Text":"However, if the relative kinetic energy of"},{"Start":"03:33.560 ","End":"03:36.770","Text":"collision is at least as high as the activation energy,"},{"Start":"03:36.770 ","End":"03:39.620","Text":"the activation barrier will be crossed,"},{"Start":"03:39.620 ","End":"03:42.055","Text":"and products will be formed."},{"Start":"03:42.055 ","End":"03:46.375","Text":"We\u0027ll cross barrier and get to the other side,"},{"Start":"03:46.375 ","End":"03:48.280","Text":"which are the products."},{"Start":"03:48.280 ","End":"03:51.700","Text":"Here\u0027s another example of transition state."},{"Start":"03:51.700 ","End":"03:58.020","Text":"Here we have 2NOCl reacting to give 2NO plus Cl_2."},{"Start":"03:58.020 ","End":"04:00.405","Text":"The green is the Cl,"},{"Start":"04:00.405 ","End":"04:07.245","Text":"the blue is N and the red is O."},{"Start":"04:07.245 ","End":"04:10.050","Text":"Here\u0027s our transition state."},{"Start":"04:10.050 ","End":"04:12.810","Text":"N, Cl, Cl,"},{"Start":"04:12.810 ","End":"04:16.830","Text":"N are all very weak bonds."},{"Start":"04:16.830 ","End":"04:18.480","Text":"A weak bond between N and Cl,"},{"Start":"04:18.480 ","End":"04:19.710","Text":"between Cl and Cl,"},{"Start":"04:19.710 ","End":"04:25.800","Text":"between Cl and N and as we proceed towards the products,"},{"Start":"04:25.800 ","End":"04:30.255","Text":"then this will break, this will break,"},{"Start":"04:30.255 ","End":"04:33.120","Text":"and Cl will come closer to Cl,"},{"Start":"04:33.120 ","End":"04:35.985","Text":"to give us Cl_2, here\u0027s Cl_2."},{"Start":"04:35.985 ","End":"04:38.590","Text":"Eventually, we get 2NO,"},{"Start":"04:39.230 ","End":"04:43.620","Text":"another NO and Cl."},{"Start":"04:43.620 ","End":"04:50.330","Text":"Eventually we get 2NO and Cl_2."},{"Start":"04:50.330 ","End":"04:51.620","Text":"In this video,"},{"Start":"04:51.620 ","End":"04:55.500","Text":"we learned about transition state theory."}],"ID":30323},{"Watched":false,"Name":"Exercise 1","Duration":"1m 54s","ChapterTopicVideoID":31676,"CourseChapterTopicPlaylistID":292233,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33912},{"Watched":false,"Name":"Exercise 2","Duration":"5m 16s","ChapterTopicVideoID":31677,"CourseChapterTopicPlaylistID":292233,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33913},{"Watched":false,"Name":"Exercise 3","Duration":"2m 45s","ChapterTopicVideoID":31675,"CourseChapterTopicPlaylistID":292233,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33914}],"Thumbnail":null,"ID":292233},{"Name":"Catalysis","TopicPlaylistFirstVideoID":0,"Duration":null,"Videos":[{"Watched":false,"Name":"Homogeneous Catalysis","Duration":"4m 5s","ChapterTopicVideoID":28834,"CourseChapterTopicPlaylistID":292234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:02.385","Text":"In the next 3 videos,"},{"Start":"00:02.385 ","End":"00:04.560","Text":"we\u0027ll learn about catalysis."},{"Start":"00:04.560 ","End":"00:06.585","Text":"What\u0027s a catalyst?"},{"Start":"00:06.585 ","End":"00:09.735","Text":"A catalyst increases the rate of a chemical reaction"},{"Start":"00:09.735 ","End":"00:13.005","Text":"without itself being consumed in the reaction."},{"Start":"00:13.005 ","End":"00:16.010","Text":"It increases the rate of both the forward and reverse"},{"Start":"00:16.010 ","End":"00:19.879","Text":"reactions and so as no effect on the equilibrium constant,"},{"Start":"00:19.879 ","End":"00:23.120","Text":"remember that the equilibrium constant is equal to"},{"Start":"00:23.120 ","End":"00:28.325","Text":"the ratio of the rate constant for the forward and backward reactions."},{"Start":"00:28.325 ","End":"00:32.870","Text":"Forward divided by backward or reverse."},{"Start":"00:32.870 ","End":"00:37.780","Text":"It increases the rate by providing an alternative pathway,"},{"Start":"00:37.780 ","End":"00:43.145","Text":"an alternative reaction mechanism with a lower activation energy."},{"Start":"00:43.145 ","End":"00:48.785","Text":"Here\u0027s an illustration here without a catalyst is the activation barrier,"},{"Start":"00:48.785 ","End":"00:55.070","Text":"the activation energy E_a and it\u0027s lower in the presence of the catalyst."},{"Start":"00:55.070 ","End":"00:57.480","Text":"Let\u0027s call it E_a prime."},{"Start":"00:58.220 ","End":"01:05.420","Text":"Reaction with a lower activation energy is faster so it\u0027s a particular temperature"},{"Start":"01:05.420 ","End":"01:06.950","Text":"more reactants can cross"},{"Start":"01:06.950 ","End":"01:12.575","Text":"a lower activation barrier than the one in the absence of a catalyst."},{"Start":"01:12.575 ","End":"01:16.010","Text":"We can see that from the equation of the rate constant is"},{"Start":"01:16.010 ","End":"01:21.010","Text":"proportional to e^ of minus E_a divided by RT."},{"Start":"01:21.010 ","End":"01:26.660","Text":"K increases as E_a decreases."},{"Start":"01:26.660 ","End":"01:32.495","Text":"There are various types of catalysts and one of them is called a homogeneous catalyst."},{"Start":"01:32.495 ","End":"01:37.415","Text":"Now, homogeneous catalyst is in the same phase as the reactants."},{"Start":"01:37.415 ","End":"01:39.590","Text":"If the reactants are in liquid solution,"},{"Start":"01:39.590 ","End":"01:45.050","Text":"the catalyst will be dissolved in the solution and if the reactants are in the gas phase,"},{"Start":"01:45.050 ","End":"01:47.960","Text":"the catalysts will also be in the gas phase."},{"Start":"01:47.960 ","End":"01:49.640","Text":"Let\u0027s take an example,"},{"Start":"01:49.640 ","End":"01:52.715","Text":"the decomposition of formic acid."},{"Start":"01:52.715 ","End":"01:55.115","Text":"If there\u0027s no catalyst,"},{"Start":"01:55.115 ","End":"02:01.370","Text":"we see that in order to form the water which results from the decomposition of"},{"Start":"02:01.370 ","End":"02:09.294","Text":"the formic acid this water then this H has to get to the other side of the molecule,"},{"Start":"02:09.294 ","End":"02:14.365","Text":"H has to get to here in order to form water."},{"Start":"02:14.365 ","End":"02:19.700","Text":"The activation energy is high and the reaction is slow."},{"Start":"02:19.700 ","End":"02:23.240","Text":"If H plus is the catalyst,"},{"Start":"02:23.240 ","End":"02:25.835","Text":"we\u0027ll see that there are 3 steps in the reaction,"},{"Start":"02:25.835 ","End":"02:28.580","Text":"each with its own activation energy."},{"Start":"02:28.580 ","End":"02:31.430","Text":"The highest activation energy is lower than"},{"Start":"02:31.430 ","End":"02:36.544","Text":"the activation energy without the catalyst and so the reaction is faster."},{"Start":"02:36.544 ","End":"02:38.824","Text":"Here\u0027s our picture."},{"Start":"02:38.824 ","End":"02:41.405","Text":"We see that there are 3 stages,"},{"Start":"02:41.405 ","End":"02:45.780","Text":"each with its own activation energy."},{"Start":"02:45.780 ","End":"02:53.520","Text":"The highest one, E_a prime is still lower than E_a when there\u0027s no catalysts."},{"Start":"02:53.520 ","End":"02:59.625","Text":"E_a prime with a catalyst is lower than E_a without a catalyst."},{"Start":"02:59.625 ","End":"03:02.820","Text":"Here are the 3 stages, 1,"},{"Start":"03:02.820 ","End":"03:07.155","Text":"2, and here\u0027s our catalyst, H plus."},{"Start":"03:07.155 ","End":"03:13.215","Text":"H plus goes in here and that\u0027s our first intermediate."},{"Start":"03:13.215 ","End":"03:19.940","Text":"Then it breaks into 2 and these are our second intermediate,"},{"Start":"03:19.940 ","End":"03:24.380","Text":"HCO plus and water."},{"Start":"03:24.380 ","End":"03:26.960","Text":"Then in the third step,"},{"Start":"03:26.960 ","End":"03:30.110","Text":"the H plus comes off here,"},{"Start":"03:30.110 ","End":"03:33.850","Text":"and we\u0027re left with the CO carbon monoxide."},{"Start":"03:33.850 ","End":"03:36.975","Text":"Here\u0027s our H plus is formed again."},{"Start":"03:36.975 ","End":"03:40.625","Text":"We started with H plus and the end,"},{"Start":"03:40.625 ","End":"03:42.515","Text":"we\u0027ve got the H plus back,"},{"Start":"03:42.515 ","End":"03:46.795","Text":"which means H plus was not consumed in the reaction."},{"Start":"03:46.795 ","End":"03:51.769","Text":"There\u0027s 3 stages here so there are 3 activation energy."},{"Start":"03:51.769 ","End":"03:54.905","Text":"Here they are 1, 2, 3."},{"Start":"03:54.905 ","End":"03:59.680","Text":"We can see that the catalyst is not consumed in the reaction."},{"Start":"03:59.680 ","End":"04:04.830","Text":"In this video, we learned about homogeneous catalysis."}],"ID":30429},{"Watched":false,"Name":"Heterogeneous Catalysis","Duration":"3m 4s","ChapterTopicVideoID":28833,"CourseChapterTopicPlaylistID":292234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.000 ","End":"00:01.900","Text":"In the previous video,"},{"Start":"00:01.900 ","End":"00:04.720","Text":"we talked about homogeneous catalysis."},{"Start":"00:04.720 ","End":"00:08.365","Text":"This video we\u0027ll talk about heterogeneous catalysis."},{"Start":"00:08.365 ","End":"00:11.305","Text":"We talked about homogeneous catalysis,"},{"Start":"00:11.305 ","End":"00:16.910","Text":"and we said that a homogeneous catalyst is in the same phase as the reactants."},{"Start":"00:16.910 ","End":"00:21.340","Text":"How is this different from a heterogeneous catalyst?"},{"Start":"00:21.340 ","End":"00:25.810","Text":"A heterogeneous catalyst is in a different phase from the reactants."},{"Start":"00:25.810 ","End":"00:27.355","Text":"Let\u0027s take an example."},{"Start":"00:27.355 ","End":"00:31.434","Text":"The conversion of carbon monoxide to carbon dioxide."},{"Start":"00:31.434 ","End":"00:37.525","Text":"Carbon monoxide is 1 of the pollutants emitted by an internal combustion engine."},{"Start":"00:37.525 ","End":"00:46.300","Text":"Since 1975, cars must have a catalytic converter that converts CO into CO_2."},{"Start":"00:46.300 ","End":"00:48.330","Text":"It also does other conversions,"},{"Start":"00:48.330 ","End":"00:53.330","Text":"but here we\u0027ll talk about the conversion of CO into CO_2."},{"Start":"00:53.330 ","End":"00:59.065","Text":"Here\u0027s the equation, CO plus 1/2 O_2 to give us CO_2."},{"Start":"00:59.065 ","End":"01:03.950","Text":"Possible catalysts are small quantities of precious metals,"},{"Start":"01:03.950 ","End":"01:09.110","Text":"such as platinum, rhodium and palladium on a surface."},{"Start":"01:09.110 ","End":"01:11.515","Text":"Here\u0027s our surface."},{"Start":"01:11.515 ","End":"01:13.879","Text":"In the first stage,"},{"Start":"01:13.879 ","End":"01:20.080","Text":"the oxygen molecule is adsorbed onto the surface."},{"Start":"01:20.080 ","End":"01:25.205","Text":"The carbon monoxide is also adsorbed onto the surface."},{"Start":"01:25.205 ","End":"01:29.315","Text":"Then as a result of the adsorption,"},{"Start":"01:29.315 ","End":"01:34.700","Text":"the O_2 molecule breaks up into 2 oxygen atoms."},{"Start":"01:34.700 ","End":"01:37.355","Text":"Then 1 of the oxygen atoms migrates,"},{"Start":"01:37.355 ","End":"01:44.195","Text":"migration to the CO molecule and then it reacts with it."},{"Start":"01:44.195 ","End":"01:46.750","Text":"That\u0027s reaction, the third stage,"},{"Start":"01:46.750 ","End":"01:51.395","Text":"and we have O plus CO to give us CO_2."},{"Start":"01:51.395 ","End":"01:54.740","Text":"The final stage is desorption,"},{"Start":"01:54.740 ","End":"01:58.135","Text":"where the CO_2 molecule is released,"},{"Start":"01:58.135 ","End":"02:01.475","Text":"desorbed from the surface."},{"Start":"02:01.475 ","End":"02:07.645","Text":"Now the surface enables the reaction to occur with a lower activation energy."},{"Start":"02:07.645 ","End":"02:12.410","Text":"Now there\u0027s enormous amounts of research to produce converters with"},{"Start":"02:12.410 ","End":"02:18.450","Text":"lower quantities of precious metals or alternatives to precious metals."},{"Start":"02:18.450 ","End":"02:23.060","Text":"One of the challenges of modern catalytic converters is that"},{"Start":"02:23.060 ","End":"02:27.470","Text":"they must work at lower temperatures because as the engines are improved,"},{"Start":"02:27.470 ","End":"02:29.945","Text":"the engines work at lower temperatures."},{"Start":"02:29.945 ","End":"02:32.705","Text":"Here\u0027s an example of an improvement."},{"Start":"02:32.705 ","End":"02:38.580","Text":"Single platinum atoms on a copper oxide support near room temperature."},{"Start":"02:38.580 ","End":"02:44.450","Text":"The single platinum atoms hold the CO molecules in place,"},{"Start":"02:44.450 ","End":"02:51.245","Text":"while the CuO, the copper oxide supplies the oxygen to convert it to CO_2."},{"Start":"02:51.245 ","End":"02:58.895","Text":"That\u0027s just 1 example of modern research on improving catalytic converters."},{"Start":"02:58.895 ","End":"03:04.260","Text":"In this video, we learned about heterogeneous catalysis."}],"ID":30430},{"Watched":false,"Name":"Enzymes as Catalysts","Duration":"6m 46s","ChapterTopicVideoID":28832,"CourseChapterTopicPlaylistID":292234,"HasSubtitles":true,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[{"Start":"00:00.440 ","End":"00:05.925","Text":"In previous videos, we learned about catalysis of chemical reactions."},{"Start":"00:05.925 ","End":"00:14.200","Text":"In this video, we\u0027ll learn about enzymes that act as catalysts in biochemical reactions."},{"Start":"00:14.570 ","End":"00:20.550","Text":"You\u0027re all probably familiar with a lock and key model of enzymes."},{"Start":"00:20.550 ","End":"00:22.245","Text":"Here\u0027s an example."},{"Start":"00:22.245 ","End":"00:26.610","Text":"The enzyme in this case is sucrase and here\u0027s"},{"Start":"00:26.610 ","End":"00:32.085","Text":"its active site into which the substrate which is sucrose fits."},{"Start":"00:32.085 ","End":"00:35.920","Text":"Then an enzyme-substrate complex is formed."},{"Start":"00:35.920 ","End":"00:37.750","Text":"We call that ES."},{"Start":"00:37.750 ","End":"00:45.890","Text":"The substrate then reacts to form an enzyme product complex and"},{"Start":"00:45.890 ","End":"00:49.610","Text":"then the products are released from the enzyme and"},{"Start":"00:49.610 ","End":"00:54.470","Text":"then we get glucose and fructose separately."},{"Start":"00:54.470 ","End":"01:00.500","Text":"Now in 1923, Michaelis-Menten proposed"},{"Start":"01:00.500 ","End":"01:04.370","Text":"a reaction mechanism for the lock and key model and"},{"Start":"01:04.370 ","End":"01:09.865","Text":"they assumed a steady-state of enzyme-substrate complex."},{"Start":"01:09.865 ","End":"01:16.885","Text":"This model is rather similar to the steady-state model we spoke about a few videos ago."},{"Start":"01:16.885 ","End":"01:19.070","Text":"Here is our mechanism."},{"Start":"01:19.070 ","End":"01:25.280","Text":"The enzyme reacts with a substrate to give an enzyme-substrate complex ES and"},{"Start":"01:25.280 ","End":"01:28.350","Text":"the rate constants are k_1 for"},{"Start":"01:28.350 ","End":"01:32.970","Text":"the forward reaction and k_ minus 1 for the reverse reaction."},{"Start":"01:32.970 ","End":"01:42.490","Text":"The complex then gives us the products and the rate constant for that is k_2."},{"Start":"01:43.900 ","End":"01:47.405","Text":"Now, let\u0027s work out the rate equation."},{"Start":"01:47.405 ","End":"01:52.940","Text":"If we look at the last stage, ES to products,"},{"Start":"01:52.940 ","End":"01:56.435","Text":"we can work out the rate of the production of the product,"},{"Start":"01:56.435 ","End":"02:03.255","Text":"which we\u0027ll call v is k_2 times the concentration of ES from here."},{"Start":"02:03.255 ","End":"02:07.250","Text":"If we make the steady-state approximation for ES,"},{"Start":"02:07.250 ","End":"02:09.950","Text":"there we get that the rate of appearance of ES will"},{"Start":"02:09.950 ","End":"02:12.860","Text":"be equal to the rate of disappearance of ES."},{"Start":"02:12.860 ","End":"02:17.090","Text":"Now the rate of appearance is k_1 times the concentration of E times"},{"Start":"02:17.090 ","End":"02:22.280","Text":"the concentration of S. That comes from the forward reaction here."},{"Start":"02:22.280 ","End":"02:30.865","Text":"The rate of disappearance is k_ minus 1 plus k_2 times ES the concentration of ES."},{"Start":"02:30.865 ","End":"02:36.680","Text":"The k_ minus 1 concentration of ES comes from the reverse reaction"},{"Start":"02:36.680 ","End":"02:43.610","Text":"here and the k_2 ES comes from the third reaction, ES to product."},{"Start":"02:43.610 ","End":"02:46.730","Text":"Now the total concentration of the enzyme,"},{"Start":"02:46.730 ","End":"02:51.290","Text":"which we\u0027ll call E_0 is equal to the concentration of the enzyme when it\u0027s"},{"Start":"02:51.290 ","End":"02:56.440","Text":"free plus the enzyme when it\u0027s in a complex with a substrate."},{"Start":"02:56.440 ","End":"03:00.435","Text":"E_0 is equal to E plus ES."},{"Start":"03:00.435 ","End":"03:09.550","Text":"E can be written as E_0 concentration of E_0 minus concentration of ES."},{"Start":"03:09.760 ","End":"03:13.460","Text":"Now we\u0027re going to substitute this expression for"},{"Start":"03:13.460 ","End":"03:17.060","Text":"the concentration of E in the steady-state expression."},{"Start":"03:17.060 ","End":"03:21.780","Text":"Here\u0027s the steady-state expression that we wrote before."},{"Start":"03:21.860 ","End":"03:28.625","Text":"We\u0027re going to substitute the expression for the concentration of E in this expression."},{"Start":"03:28.625 ","End":"03:36.500","Text":"We get k_1 times the concentration of E_0 minus the concentration of"},{"Start":"03:36.500 ","End":"03:40.370","Text":"ES times the concentration of S equal"},{"Start":"03:40.370 ","End":"03:45.240","Text":"to k_ minus 1 plus k_2 times the concentration of ES."},{"Start":"03:46.130 ","End":"03:50.960","Text":"Then we can isolate the concentration of ES and"},{"Start":"03:50.960 ","End":"03:55.490","Text":"that\u0027s equal to k_1 times the concentration of E. Initially,"},{"Start":"03:55.490 ","End":"04:00.010","Text":"total concentration times the concentration of S"},{"Start":"04:00.010 ","End":"04:05.670","Text":"divided by k_ minus 1 plus k_2 plus k_1 concentration"},{"Start":"04:05.670 ","End":"04:10.595","Text":"of S. Now if we substitute"},{"Start":"04:10.595 ","End":"04:16.555","Text":"the expression for ES in each equation,"},{"Start":"04:16.555 ","End":"04:18.330","Text":"we\u0027ll get the V,"},{"Start":"04:18.330 ","End":"04:23.030","Text":"which is equal to k_2 times concentration of ES is equal"},{"Start":"04:23.030 ","End":"04:29.670","Text":"to k_2 here times this expression here."},{"Start":"04:30.110 ","End":"04:33.130","Text":"We have k_1, k_2,"},{"Start":"04:33.130 ","End":"04:36.470","Text":"then the total concentration of the enzyme times"},{"Start":"04:36.470 ","End":"04:40.060","Text":"the concentration of S divided by k_ minus 1 plus"},{"Start":"04:40.060 ","End":"04:48.530","Text":"k_2 plus k_1 times the concentration of S. Now if we divide this equation by k_1,"},{"Start":"04:48.530 ","End":"04:55.880","Text":"we get k_2 times the total concentration of enzyme times the concentration of"},{"Start":"04:55.880 ","End":"05:04.905","Text":"the substrate here divided by k_M which is after Michaelis-Menten,"},{"Start":"05:04.905 ","End":"05:13.099","Text":"k_M is k_ minus 1 plus k_2 divided by k_1 plus S. This is our final expression."},{"Start":"05:13.099 ","End":"05:19.790","Text":"Now if we look at the plot of V versus the substrate concentration,"},{"Start":"05:19.790 ","End":"05:23.060","Text":"we will see that it agrees with this equation."},{"Start":"05:23.060 ","End":"05:29.090","Text":"If the concentration of S is much less than k_M so that it can be ignored,"},{"Start":"05:29.090 ","End":"05:38.100","Text":"we get k_2 is 0 times S divided by k_M, this expression."},{"Start":"05:38.100 ","End":"05:43.040","Text":"This is first-order in S and where the concentration of S is much"},{"Start":"05:43.040 ","End":"05:48.590","Text":"greater than k_M so that we can ignore the k_M this time."},{"Start":"05:48.590 ","End":"05:54.635","Text":"Then the S will cancel with this S and we get k_2 E_0."},{"Start":"05:54.635 ","End":"05:58.220","Text":"V will then be equal to k_2, E_0."},{"Start":"05:58.220 ","End":"06:00.785","Text":"That doesn\u0027t depend on S at all,"},{"Start":"06:00.785 ","End":"06:06.875","Text":"zero order in S. We expect that our plot will go from being"},{"Start":"06:06.875 ","End":"06:13.870","Text":"first-order in S at low concentrations of S to zero order in S at high concentrations."},{"Start":"06:13.870 ","End":"06:16.319","Text":"That\u0027s exactly what happens."},{"Start":"06:16.319 ","End":"06:19.335","Text":"Here\u0027s the experimental plot,"},{"Start":"06:19.335 ","End":"06:26.640","Text":"V versus the concentration of S. It goes from first-order,"},{"Start":"06:26.640 ","End":"06:32.885","Text":"the linear plot here at low concentrations of S to zero order"},{"Start":"06:32.885 ","End":"06:40.470","Text":"here at high concentrations of S. It goes from first to zero order."},{"Start":"06:40.470 ","End":"06:45.210","Text":"In this video, we learned about the Michaelis-Menten mechanism."}],"ID":30431},{"Watched":false,"Name":"Exercise 1","Duration":"1m 22s","ChapterTopicVideoID":31680,"CourseChapterTopicPlaylistID":292234,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33915},{"Watched":false,"Name":"Exercise 2","Duration":"3m 15s","ChapterTopicVideoID":31678,"CourseChapterTopicPlaylistID":292234,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33916},{"Watched":false,"Name":"Exercise 3","Duration":"3m 52s","ChapterTopicVideoID":31679,"CourseChapterTopicPlaylistID":292234,"HasSubtitles":false,"ThumbnailPath":null,"UploadDate":null,"DurationForVideoObject":null,"Description":null,"MetaTitle":null,"MetaDescription":null,"Canonical":null,"VideoComments":[],"Subtitles":[],"ID":33917}],"Thumbnail":null,"ID":292234}]

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