[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:00.75,0:00:03.01,Default,,0000,0000,0000,,- [Instructor] Let's say we\Nhave a hypothetical reaction
Dialogue: 0,0:00:03.01,0:00:06.21,Default,,0000,0000,0000,,where reactant A turns into products
Dialogue: 0,0:00:06.21,0:00:10.42,Default,,0000,0000,0000,,and that the reaction is\Nfirst-order with respect to A.
Dialogue: 0,0:00:11.50,0:00:15.30,Default,,0000,0000,0000,,If the reaction is first-order\Nwith respect to reactant A,
Dialogue: 0,0:00:15.30,0:00:17.85,Default,,0000,0000,0000,,for the rate law we can write\Nthe rate of the reaction
Dialogue: 0,0:00:17.85,0:00:19.65,Default,,0000,0000,0000,,is equal to the rate constant K
Dialogue: 0,0:00:19.65,0:00:23.41,Default,,0000,0000,0000,,times the concentration\Nof A to the first power.
Dialogue: 0,0:00:23.41,0:00:25.70,Default,,0000,0000,0000,,We can also write that\Nthe rate of the reaction
Dialogue: 0,0:00:25.70,0:00:29.24,Default,,0000,0000,0000,,is equal to the negative of\Nthe change in the concentration
Dialogue: 0,0:00:29.24,0:00:31.46,Default,,0000,0000,0000,,of A over the change in time.
Dialogue: 0,0:00:32.53,0:00:35.20,Default,,0000,0000,0000,,By setting both of these\Nequal to each other,
Dialogue: 0,0:00:35.20,0:00:36.90,Default,,0000,0000,0000,,and by doing some calculus,
Dialogue: 0,0:00:36.90,0:00:39.23,Default,,0000,0000,0000,,including the concept of integration,
Dialogue: 0,0:00:39.23,0:00:41.20,Default,,0000,0000,0000,,we arrive at the integrated rate law
Dialogue: 0,0:00:41.20,0:00:43.20,Default,,0000,0000,0000,,for a first-order reaction,
Dialogue: 0,0:00:43.20,0:00:46.02,Default,,0000,0000,0000,,which says that the natural\Nlog of the concentration of A
Dialogue: 0,0:00:46.02,0:00:49.78,Default,,0000,0000,0000,,at some time T, is equal to negative KT,
Dialogue: 0,0:00:49.78,0:00:51.53,Default,,0000,0000,0000,,where K is the rate constant
Dialogue: 0,0:00:51.53,0:00:55.13,Default,,0000,0000,0000,,plus the natural log of the\Ninitial concentration of A.
Dialogue: 0,0:00:56.89,0:00:58.45,Default,,0000,0000,0000,,Notice how the integrated rate law
Dialogue: 0,0:00:58.45,0:01:03.45,Default,,0000,0000,0000,,has the form of Y is equal to mx plus b,
Dialogue: 0,0:01:03.63,0:01:07.18,Default,,0000,0000,0000,,which is the equation for a straight line.
Dialogue: 0,0:01:07.18,0:01:10.98,Default,,0000,0000,0000,,So if we were to graph the\Nnatural log of the concentration
Dialogue: 0,0:01:10.98,0:01:15.04,Default,,0000,0000,0000,,of A on the Y axis, so let's\Ngo ahead and put that in here,
Dialogue: 0,0:01:15.04,0:01:19.13,Default,,0000,0000,0000,,the natural log of the concentration of A,
Dialogue: 0,0:01:19.13,0:01:23.66,Default,,0000,0000,0000,,and on the X axis we put the time,
Dialogue: 0,0:01:23.66,0:01:26.33,Default,,0000,0000,0000,,we would get a straight line
Dialogue: 0,0:01:26.33,0:01:28.90,Default,,0000,0000,0000,,and the slope of that straight line
Dialogue: 0,0:01:28.90,0:01:31.68,Default,,0000,0000,0000,,would be equal to negative K.
Dialogue: 0,0:01:31.68,0:01:33.59,Default,,0000,0000,0000,,So the slope of this line,
Dialogue: 0,0:01:34.52,0:01:36.67,Default,,0000,0000,0000,,the slope would be equal to the negative
Dialogue: 0,0:01:36.67,0:01:39.17,Default,,0000,0000,0000,,of the rate constant K,
Dialogue: 0,0:01:39.17,0:01:43.20,Default,,0000,0000,0000,,and the Y intercept would\Nbe equal to the natural log
Dialogue: 0,0:01:43.20,0:01:45.11,Default,,0000,0000,0000,,of the initial concentration of A.
Dialogue: 0,0:01:45.11,0:01:49.15,Default,,0000,0000,0000,,So right where this line meets the Y axis,
Dialogue: 0,0:01:49.15,0:01:51.56,Default,,0000,0000,0000,,that point is equal to the natural log
Dialogue: 0,0:01:51.56,0:01:54.74,Default,,0000,0000,0000,,of the initial concentration of A.
Dialogue: 0,0:01:56.25,0:01:59.97,Default,,0000,0000,0000,,The conversion of methyl\Nisonitrile to acetonitrile
Dialogue: 0,0:01:59.97,0:02:02.15,Default,,0000,0000,0000,,is a first-order reaction.
Dialogue: 0,0:02:02.15,0:02:05.34,Default,,0000,0000,0000,,And these two molecules\Nare isomers of each other.
Dialogue: 0,0:02:05.34,0:02:08.22,Default,,0000,0000,0000,,Let's use the data that's\Nprovided to us in this data table
Dialogue: 0,0:02:08.22,0:02:12.68,Default,,0000,0000,0000,,to show that this conversion\Nis a first-order reaction.
Dialogue: 0,0:02:12.68,0:02:15.72,Default,,0000,0000,0000,,Since the coefficient in\Nfront of methyl isonitrile
Dialogue: 0,0:02:15.72,0:02:19.87,Default,,0000,0000,0000,,is a one, we can use this form\Nof the integrated rate law
Dialogue: 0,0:02:19.87,0:02:21.87,Default,,0000,0000,0000,,where the slope is equal to the negative
Dialogue: 0,0:02:21.87,0:02:23.56,Default,,0000,0000,0000,,of the rate constant K.
Dialogue: 0,0:02:24.82,0:02:28.10,Default,,0000,0000,0000,,If our balanced equation\Nhad a two as a coefficient
Dialogue: 0,0:02:28.10,0:02:31.13,Default,,0000,0000,0000,,in front of our reactant, we\Nwould have had to include 1/2
Dialogue: 0,0:02:32.07,0:02:35.57,Default,,0000,0000,0000,,as a stoichiometric coefficient.
Dialogue: 0,0:02:35.57,0:02:38.54,Default,,0000,0000,0000,,And when we set our two\Nrates equal to each other now
Dialogue: 0,0:02:38.54,0:02:42.61,Default,,0000,0000,0000,,and go through the calculus,\Ninstead of getting negative KT,
Dialogue: 0,0:02:42.61,0:02:45.51,Default,,0000,0000,0000,,we have gotten negative two KT.
Dialogue: 0,0:02:46.40,0:02:51.10,Default,,0000,0000,0000,,However for our reaction we\Ndon't have a coefficient of two.
Dialogue: 0,0:02:51.10,0:02:54.34,Default,,0000,0000,0000,,We have a coefficient of one and therefore
Dialogue: 0,0:02:54.34,0:02:58.11,Default,,0000,0000,0000,,we can use this form of\Nthe integrated rate law.
Dialogue: 0,0:02:59.66,0:03:02.07,Default,,0000,0000,0000,,Also notice that this form\Nof the integrated rate law
Dialogue: 0,0:03:02.07,0:03:05.76,Default,,0000,0000,0000,,is in terms of the concentration of A
Dialogue: 0,0:03:05.76,0:03:08.93,Default,,0000,0000,0000,,but we don't have the\Nconcentration of methyl isonitrile
Dialogue: 0,0:03:08.93,0:03:10.09,Default,,0000,0000,0000,,in our data table,
Dialogue: 0,0:03:10.09,0:03:13.35,Default,,0000,0000,0000,,we have the pressure of methyl isonitrile.
Dialogue: 0,0:03:13.35,0:03:16.08,Default,,0000,0000,0000,,But pressure is related to concentration
Dialogue: 0,0:03:16.08,0:03:21.08,Default,,0000,0000,0000,,from the ideal gas law,\Nso PV is equal to nRT.
Dialogue: 0,0:03:21.36,0:03:25.47,Default,,0000,0000,0000,,If we divide both sides by V,
Dialogue: 0,0:03:25.47,0:03:28.63,Default,,0000,0000,0000,,then we can see that pressure is equal to,
Dialogue: 0,0:03:28.63,0:03:30.73,Default,,0000,0000,0000,,n is moles and V is volumes,
Dialogue: 0,0:03:30.73,0:03:34.02,Default,,0000,0000,0000,,so moles divided by\Nvolume would be molarity,
Dialogue: 0,0:03:34.02,0:03:38.21,Default,,0000,0000,0000,,so molarity times R times T.
Dialogue: 0,0:03:38.21,0:03:41.34,Default,,0000,0000,0000,,And therefore pressure\Nis directly proportional
Dialogue: 0,0:03:41.34,0:03:45.52,Default,,0000,0000,0000,,to concentration, and for a\Ngas it's easier to measure
Dialogue: 0,0:03:45.52,0:03:48.39,Default,,0000,0000,0000,,the pressure than to\Nget the concentration.
Dialogue: 0,0:03:48.39,0:03:50.31,Default,,0000,0000,0000,,And so you'll often see data
Dialogue: 0,0:03:50.31,0:03:53.67,Default,,0000,0000,0000,,for gases in terms of the pressure.
Dialogue: 0,0:03:53.67,0:03:55.78,Default,,0000,0000,0000,,Therefore, we can imagine this form of the
Dialogue: 0,0:03:55.78,0:04:00.29,Default,,0000,0000,0000,,integrated rate law as the\Nnatural log of the pressure
Dialogue: 0,0:04:00.29,0:04:03.53,Default,,0000,0000,0000,,of our gas at time T\Nis equal to negative KT
Dialogue: 0,0:04:03.53,0:04:08.16,Default,,0000,0000,0000,,plus the natural log of the\Ninitial pressure of the gas.
Dialogue: 0,0:04:08.16,0:04:10.35,Default,,0000,0000,0000,,Therefore, to show that this reaction
Dialogue: 0,0:04:10.35,0:04:14.06,Default,,0000,0000,0000,,is a first-order reaction we\Nneed to graph the natural log
Dialogue: 0,0:04:14.06,0:04:18.38,Default,,0000,0000,0000,,of the pressure of methyl\Nisonitrile on the Y axis
Dialogue: 0,0:04:18.38,0:04:20.96,Default,,0000,0000,0000,,and time on the X axis.
Dialogue: 0,0:04:20.96,0:04:24.28,Default,,0000,0000,0000,,So we need a new column in our data table.
Dialogue: 0,0:04:25.18,0:04:27.95,Default,,0000,0000,0000,,We need to put in the natural log
Dialogue: 0,0:04:27.95,0:04:32.95,Default,,0000,0000,0000,,of the pressure of methyl isonitrile.
Dialogue: 0,0:04:33.69,0:04:36.46,Default,,0000,0000,0000,,So for example, when time is equal to zero
Dialogue: 0,0:04:36.46,0:04:41.46,Default,,0000,0000,0000,,the pressure of methyl\Nisonitrile is 502 torrs.
Dialogue: 0,0:04:41.67,0:04:45.35,Default,,0000,0000,0000,,So we need to take the natural log of 502.
Dialogue: 0,0:04:45.35,0:04:49.31,Default,,0000,0000,0000,,And the natural log of\N502 is equal to 6.219.
Dialogue: 0,0:04:55.48,0:04:56.73,Default,,0000,0000,0000,,To save time, I've gone ahead
Dialogue: 0,0:04:56.73,0:04:58.60,Default,,0000,0000,0000,,and filled in this last column here,
Dialogue: 0,0:04:58.60,0:05:01.84,Default,,0000,0000,0000,,the natural log of the\Npressure methyl isonitrile.
Dialogue: 0,0:05:01.84,0:05:04.79,Default,,0000,0000,0000,,Notice what happens as\Ntime increases, right,
Dialogue: 0,0:05:04.79,0:05:09.23,Default,,0000,0000,0000,,as time increases the\Npressure of methyl isonitrile
Dialogue: 0,0:05:09.23,0:05:13.90,Default,,0000,0000,0000,,decreases since it's being\Nturned into acetonitrile.
Dialogue: 0,0:05:13.90,0:05:16.69,Default,,0000,0000,0000,,So for our graph, we're\Ngonna have the natural log
Dialogue: 0,0:05:16.69,0:05:20.37,Default,,0000,0000,0000,,of the pressure of methyl\Nisonitrile on the y-axis.
Dialogue: 0,0:05:20.37,0:05:23.93,Default,,0000,0000,0000,,And we're gonna have time on the X axis.
Dialogue: 0,0:05:23.93,0:05:25.54,Default,,0000,0000,0000,,So notice our first point here
Dialogue: 0,0:05:25.54,0:05:28.46,Default,,0000,0000,0000,,when time is equal to zero seconds,
Dialogue: 0,0:05:28.46,0:05:33.46,Default,,0000,0000,0000,,the natural log of the\Npressure as equal to 6.219.
Dialogue: 0,0:05:33.80,0:05:37.05,Default,,0000,0000,0000,,So let's go down and\Nlet's look at the graph.
Dialogue: 0,0:05:37.05,0:05:39.56,Default,,0000,0000,0000,,All right, so I've\Nalready graphed it here.
Dialogue: 0,0:05:39.56,0:05:43.49,Default,,0000,0000,0000,,And we just saw when time\Nis equal to zero seconds,
Dialogue: 0,0:05:43.49,0:05:47.56,Default,,0000,0000,0000,,the first point is equal to 6.219.
Dialogue: 0,0:05:47.56,0:05:52.56,Default,,0000,0000,0000,,And here I have the other data\Npoints already on the graph.
Dialogue: 0,0:05:52.81,0:05:55.82,Default,,0000,0000,0000,,Here's the integrated rate\Nlaw for a first-order reaction
Dialogue: 0,0:05:55.82,0:06:00.72,Default,,0000,0000,0000,,and I put pressures in there\Ninstead of concentrations.
Dialogue: 0,0:06:00.72,0:06:03.89,Default,,0000,0000,0000,,And so we have the natural\Nlog of the pressure
Dialogue: 0,0:06:03.89,0:06:06.85,Default,,0000,0000,0000,,of methyl isonitrile on the y-axis
Dialogue: 0,0:06:06.85,0:06:10.30,Default,,0000,0000,0000,,and we have time on the X axis,
Dialogue: 0,0:06:10.30,0:06:12.70,Default,,0000,0000,0000,,and the slope of this line should be equal
Dialogue: 0,0:06:12.70,0:06:15.95,Default,,0000,0000,0000,,to the negative of the rate constant K.
Dialogue: 0,0:06:15.95,0:06:18.87,Default,,0000,0000,0000,,So there are many ways to\Nfind the slope of this line,
Dialogue: 0,0:06:18.87,0:06:22.90,Default,,0000,0000,0000,,one way would be to use\Na graphing calculator.
Dialogue: 0,0:06:22.90,0:06:25.21,Default,,0000,0000,0000,,So I used a graphing calculator
Dialogue: 0,0:06:25.21,0:06:28.03,Default,,0000,0000,0000,,and I put in the data from the data table
Dialogue: 0,0:06:28.03,0:06:30.70,Default,,0000,0000,0000,,and I found that the slope of this line
Dialogue: 0,0:06:30.70,0:06:33.85,Default,,0000,0000,0000,,is equal to negative 2.08
Dialogue: 0,0:06:37.23,0:06:42.13,Default,,0000,0000,0000,,times 10 to the negative fourth.
Dialogue: 0,0:06:42.13,0:06:46.36,Default,,0000,0000,0000,,And since if I go ahead and\Nwrite y is equal to mx plus b,
Dialogue: 0,0:06:46.36,0:06:49.90,Default,,0000,0000,0000,,I need to remember to take\Nthe negative of that slope
Dialogue: 0,0:06:49.90,0:06:52.43,Default,,0000,0000,0000,,to find the rate constant K.
Dialogue: 0,0:06:52.43,0:06:56.35,Default,,0000,0000,0000,,Therefore K is equal to positive 2.08
Dialogue: 0,0:06:58.23,0:07:02.15,Default,,0000,0000,0000,,times 10 to the negative fourth.
Dialogue: 0,0:07:02.15,0:07:03.80,Default,,0000,0000,0000,,To get the units for the rate constant,
Dialogue: 0,0:07:03.80,0:07:06.04,Default,,0000,0000,0000,,we can remember that slope is equal to
Dialogue: 0,0:07:06.04,0:07:10.46,Default,,0000,0000,0000,,change in Y over change in X.
Dialogue: 0,0:07:10.46,0:07:14.10,Default,,0000,0000,0000,,So change in Y would be the\Nnatural log of the pressure,
Dialogue: 0,0:07:14.10,0:07:19.07,Default,,0000,0000,0000,,which has no unit, and X\Nthe unit is in seconds.
Dialogue: 0,0:07:19.07,0:07:24.07,Default,,0000,0000,0000,,So we would have one over\Nseconds for the units for K.
Dialogue: 0,0:07:24.95,0:07:27.64,Default,,0000,0000,0000,,And finally, since we got a straight line
Dialogue: 0,0:07:27.64,0:07:32.64,Default,,0000,0000,0000,,when we graphed the natural log\Nof the pressure versus time,
Dialogue: 0,0:07:33.13,0:07:36.77,Default,,0000,0000,0000,,we know that this data is\Nfor a first-order reaction.
Dialogue: 0,0:07:36.77,0:07:39.07,Default,,0000,0000,0000,,And therefore we've proved\Nthat the transformation
Dialogue: 0,0:07:39.07,0:07:41.40,Default,,0000,0000,0000,,of methyl isonitrile to acetonitrile
Dialogue: 0,0:07:41.40,0:07:43.28,Default,,0000,0000,0000,,is a first-order reaction.