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JOCELYN: Hi.

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00:00:23,450 --> 00:00:27,130
Jocelyn here and today we're
going to go over fall 2009

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00:00:27,130 --> 00:00:30,440
exam three problem
number four.

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00:00:30,440 --> 00:00:31,720
Starting with part A--

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00:00:34,520 --> 00:00:36,580
let's read the problem.

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00:00:36,580 --> 00:00:40,940
In the 1920s, Jack Breitbart of
Revlon laboratories found

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that acne could be treated by
the use of benzoyl peroxide.

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00:00:44,830 --> 00:00:49,240
The oxygen-oxygen bond in
peroxide is weak and under the

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00:00:49,240 --> 00:00:52,200
influence of modest heating,
benzoyl peroxide readily

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00:00:52,200 --> 00:00:55,870
decompose to form free radicals
according to the

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00:00:55,870 --> 00:00:57,650
reaction given.

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00:00:57,650 --> 00:01:02,150
The rate of decomposition is
measured at 92 degrees C at

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00:01:02,150 --> 00:01:05,660
various concentrations
and found to be--

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00:01:05,660 --> 00:01:08,220
and you're given two different
concentrations and two

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00:01:08,220 --> 00:01:09,870
different rates.

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00:01:09,870 --> 00:01:13,270
So part A asks, determine the
order of reaction for the

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00:01:13,270 --> 00:01:14,980
decomposition of benzoyl
peroxide.

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00:01:18,560 --> 00:01:24,370
The first thing to do is, when
asked for the order of the

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00:01:24,370 --> 00:01:29,420
decomposition reaction, we
want to think about the

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00:01:29,420 --> 00:01:31,220
general rate loss, right?

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00:01:31,220 --> 00:01:32,155
We're given concentrations.

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00:01:32,155 --> 00:01:34,390
We're given rates.

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00:01:34,390 --> 00:01:36,030
We're asked for the order.

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00:01:36,030 --> 00:01:39,350
Something you might want to
have written down on your

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00:01:39,350 --> 00:01:46,290
equation sheet or look up is the
general rate law equation.

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00:01:46,290 --> 00:01:48,630
Remember that students taking
these exams did have an

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00:01:48,630 --> 00:01:52,715
equation sheet so they could
have this ready.

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00:01:52,715 --> 00:01:55,220
The important part is that
you know when to use this

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00:01:55,220 --> 00:01:58,130
equation, what each
of the terms mean.

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00:01:58,130 --> 00:02:01,060
So r is the rate.

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00:02:04,220 --> 00:02:05,790
k is a rate constant.

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00:02:12,030 --> 00:02:14,620
It is a function of temperature,
but if you're at

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00:02:14,620 --> 00:02:17,520
the same temperature, that rate
constant will not change.

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00:02:17,520 --> 00:02:19,560
It's constant.

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00:02:19,560 --> 00:02:20,810
c is your concentration.

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00:02:26,430 --> 00:02:31,660
And n is your reaction order.

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00:02:37,060 --> 00:02:41,670
So this would be the place to
start for this question.

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00:02:41,670 --> 00:02:44,690
And now we know that we're
solving for n.

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00:02:44,690 --> 00:02:49,050
So what is n in this case?

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00:02:51,700 --> 00:02:54,190
Looking at what we're
given, we can write

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00:02:54,190 --> 00:02:55,440
two different equations.

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00:02:58,540 --> 00:03:04,770
The first line gives us a
specific rate for a specific

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00:03:04,770 --> 00:03:07,140
concentration.

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00:03:07,140 --> 00:03:11,830
The second line gives us another
rate for a second

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00:03:11,830 --> 00:03:14,100
concentration.

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00:03:14,100 --> 00:03:18,490
So now we have two equations
and two unknowns.

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00:03:18,490 --> 00:03:24,050
That means we can solve for one
of the unknowns and then

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00:03:24,050 --> 00:03:26,170
could solve for the other one
if we wanted to, but here we

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00:03:26,170 --> 00:03:27,290
only care about n.

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00:03:27,290 --> 00:03:30,000
So how I'm going to choose to
solve this and you may choose

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00:03:30,000 --> 00:03:31,250
to do it differently--

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00:03:31,250 --> 00:03:37,500
I'm going to divide the first
line by the second line and I

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00:03:37,500 --> 00:03:49,390
get rate one over rate two
equals c1 divided by c2 all to

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00:03:49,390 --> 00:03:52,090
the n power.

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00:03:52,090 --> 00:03:55,010
So that made the k drop out,
which is one of the unknowns

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00:03:55,010 --> 00:03:58,040
that I don't care about because
the problem isn't

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00:03:58,040 --> 00:04:00,350
asking for that.

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00:04:00,350 --> 00:04:02,720
Now all we need to do
is solve for n.

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00:04:02,720 --> 00:04:06,320
So if we remember algebra, we're
going to take the log of

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00:04:06,320 --> 00:04:07,570
both sides.

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00:04:14,930 --> 00:04:18,250
Because of the properties
of the log, we can

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00:04:18,250 --> 00:04:19,500
bring the n out front.

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00:04:25,990 --> 00:04:27,952
So n equals--

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00:04:40,110 --> 00:04:43,390
and plugging in the numbers
given in the problem--

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00:04:43,390 --> 00:04:45,370
they're given in the same
units so everything will

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00:04:45,370 --> 00:04:47,650
cancel out and that's good--

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00:04:47,650 --> 00:04:51,300
we get that the n equals one.

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00:04:51,300 --> 00:04:54,970
So this is a first
order reaction.

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00:04:54,970 --> 00:04:57,480
Now you could've probably done
that by inspection if you're

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00:04:57,480 --> 00:05:02,250
familiar with the rate law and
how to determine that by just

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00:05:02,250 --> 00:05:04,855
seeing that as the concentration
went down by a

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00:05:04,855 --> 00:05:07,680
fourth, the rate also went
down by a fourth.

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00:05:07,680 --> 00:05:11,730
However, the question asks for
you to determine the order of

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00:05:11,730 --> 00:05:15,910
reaction and so we were
looking for more of a

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00:05:15,910 --> 00:05:19,770
derivation of the problem
instead of just an inspection

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00:05:19,770 --> 00:05:23,760
because that shows that you have
a familiarity with the

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00:05:23,760 --> 00:05:27,780
general rate law equation.

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00:05:27,780 --> 00:05:29,380
So now moving to part B--

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00:05:34,210 --> 00:05:37,850
we are asked, on the plot below,
sketch the variation in

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00:05:37,850 --> 00:05:41,180
energy with extent of reaction
for the decomposition of

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00:05:41,180 --> 00:05:42,770
benzoyl peroxide.

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00:05:42,770 --> 00:05:48,580
Assume that the ratio of
activation energy to--

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00:05:48,580 --> 00:05:51,290
delta e, which is the energy
of the reaction--

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00:05:51,290 --> 00:05:54,170
equals -2.5.

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00:05:54,170 --> 00:06:00,200
Label the energy states and
label the delta e of reaction,

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00:06:00,200 --> 00:06:03,710
the activation energy for the
forward reaction and the

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00:06:03,710 --> 00:06:07,200
activation energy for the
backward reaction.

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00:06:07,200 --> 00:06:09,260
So that's asking
us to do a lot.

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00:06:09,260 --> 00:06:12,230
So first we're going to write
down, what are we actually

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00:06:12,230 --> 00:06:13,480
asked to do?

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00:06:19,430 --> 00:06:26,490
So the main thing is that we're
asked for the variation

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00:06:26,490 --> 00:06:34,240
in energy with the extent
of reaction.

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00:06:34,240 --> 00:06:37,280
So as the reaction progresses,
what's the

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00:06:37,280 --> 00:06:39,090
energy of the molecule?

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00:06:39,090 --> 00:06:42,560
The energy state that
it's going through?

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00:06:42,560 --> 00:06:54,920
In that plot, we're asked to
label the energy states of the

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00:06:54,920 --> 00:07:09,330
reactant and product and
we're asked to label

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00:07:09,330 --> 00:07:12,210
the delta e of reaction.

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00:07:12,210 --> 00:07:17,450
So what's the net change in
energy as well as the

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00:07:17,450 --> 00:07:22,820
activation, both forward
and backward?

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00:07:27,430 --> 00:07:30,040
I would always write this to
the side or scrap piece of

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00:07:30,040 --> 00:07:32,400
paper or something because this
is a lot of things to put

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00:07:32,400 --> 00:07:35,700
on one plot and this will be a
nice reference to help us to

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00:07:35,700 --> 00:07:40,630
make sure we answered all the
things the question is asking.

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00:07:40,630 --> 00:07:45,730
So on the plot given, which I'll
reproduce here, it has

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00:07:45,730 --> 00:07:53,992
axes of extent of reaction
and energy.

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00:07:58,660 --> 00:08:02,900
The first thing to do would be
to label the beginning and

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00:08:02,900 --> 00:08:05,740
start energy states.

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00:08:05,740 --> 00:08:12,310
So because we know at
92 degrees C, this--

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00:08:12,310 --> 00:08:14,610
the benzoyl peroxide--

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00:08:14,610 --> 00:08:18,010
decomposes readily, we can
assume that there's

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00:08:18,010 --> 00:08:20,640
a decrease in energy.

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00:08:20,640 --> 00:08:22,620
So we'll have our products
start at a

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00:08:22,620 --> 00:08:25,760
higher energy than our--

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00:08:25,760 --> 00:08:27,440
sorry-- our reactant
start at a higher

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00:08:27,440 --> 00:08:29,430
energy than our products.

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00:08:29,430 --> 00:08:32,525
And because the question asks us
to label that, we're going

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00:08:32,525 --> 00:08:34,593
to put in the actual labels.

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00:09:00,750 --> 00:09:03,350
And so that's your reactant
and then we

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00:09:03,350 --> 00:09:05,250
have the radical product.

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00:09:21,330 --> 00:09:26,480
Now that we have our beginning
and end energy states, we need

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00:09:26,480 --> 00:09:29,230
to think about what's happening
in the middle.

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00:09:29,230 --> 00:09:33,160
A clue is that he talks about
the activation energy, right?

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00:09:33,160 --> 00:09:35,370
And we know that most processes
that we talk about

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00:09:35,370 --> 00:09:37,850
have a certain activation
energy.

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00:09:37,850 --> 00:09:41,650
You can't just go straight
from the

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00:09:41,650 --> 00:09:43,930
molecule to the radical.

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00:09:43,930 --> 00:09:47,900
You have to have a little
bit of energy cuffed.

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00:09:47,900 --> 00:09:52,410
And so we know we're going to
have some type of hill that we

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00:09:52,410 --> 00:09:54,650
have to go over.

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00:09:54,650 --> 00:09:59,160
Furthermore, in the problem he
states that the activation

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00:09:59,160 --> 00:10:05,700
energy divided by the
delta e is -2.5.

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00:10:05,700 --> 00:10:11,520
So we know that because this
is a negative delta e--

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00:10:11,520 --> 00:10:17,580
so this is our delta e here
and it's negative--

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00:10:17,580 --> 00:10:21,060
our activation energy is two and
a half times as large as

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00:10:21,060 --> 00:10:24,110
that net change in energy.

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00:10:24,110 --> 00:10:26,435
So we want to kind of just
eyeball that in here.

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00:10:31,130 --> 00:10:33,820
You didn't have to be exact,
but relatively--

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00:10:33,820 --> 00:10:36,320
it's going to be larger
than twice the

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00:10:36,320 --> 00:10:38,820
distance between here.

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00:10:38,820 --> 00:10:43,450
And remember that activation
energy is the hill, the energy

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00:10:43,450 --> 00:10:47,390
costs that you have to go
through to get to your lower

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00:10:47,390 --> 00:10:49,796
energy state in the
product size.

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00:10:49,796 --> 00:10:53,670
So now that we have all of our
energy states, we want to draw

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00:10:53,670 --> 00:10:55,630
a smooth curve to show
the variation of

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00:10:55,630 --> 00:10:57,380
energy with the reaction.

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00:10:57,380 --> 00:11:05,300
So connecting all these points,
we get two valleys and

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00:11:05,300 --> 00:11:05,940
a hill, right?

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00:11:05,940 --> 00:11:08,450
We have two stable energy
states and we have an

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00:11:08,450 --> 00:11:11,430
activated complex corresponding
to the

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00:11:11,430 --> 00:11:14,940
activation energy there.

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00:11:14,940 --> 00:11:19,120
So we have our energy diagram,
but we need to go back to what

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00:11:19,120 --> 00:11:24,490
the question is asking and see
that we've labeled the energy

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00:11:24,490 --> 00:11:27,910
states, we've labeled the delta
e of reaction, but we

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00:11:27,910 --> 00:11:29,950
need to label both
the forward and

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00:11:29,950 --> 00:11:32,040
backwards activation energy.

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00:11:32,040 --> 00:11:34,650
And this is one of the things
people had issues with--

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00:11:34,650 --> 00:11:38,120
the backward activation.

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00:11:38,120 --> 00:11:39,900
So forward activation
is pretty self

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00:11:39,900 --> 00:11:41,360
explanatory, right?

168
00:11:41,360 --> 00:11:47,170
It's the energy we need to
overcome the activation of

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00:11:47,170 --> 00:11:53,910
this process and so this is
the Ea of the forward.

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00:11:53,910 --> 00:11:58,500
Now the activation energy of the
backward reaction is just

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00:11:58,500 --> 00:12:01,100
the same concept, but
going backwards.

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00:12:01,100 --> 00:12:05,820
So if we started down here, what
is the energy we need to

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00:12:05,820 --> 00:12:09,360
get over this activation
barrier?

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00:12:09,360 --> 00:12:18,090
And so that's this full energy,
this full amount of

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00:12:18,090 --> 00:12:20,330
energy here.

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00:12:20,330 --> 00:12:23,650
And so we can see that it's
bigger than the activation

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00:12:23,650 --> 00:12:26,910
energy for the forward
reaction by the

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00:12:26,910 --> 00:12:30,110
delta e of the reaction.

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00:12:30,110 --> 00:12:34,510
So going back to our checklist
here, we've answered all over

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00:12:34,510 --> 00:12:39,500
the questions and so we
can move to part C.

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00:12:39,500 --> 00:12:42,580
So part C asks us, on the
same plot above--

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00:12:42,580 --> 00:12:45,980
which we have over here
to the right--

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00:12:45,980 --> 00:12:48,910
sketch the variation in energy
with extent of reaction for

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00:12:48,910 --> 00:12:52,370
the decomposition of benzoyl
peroxide under the influence

185
00:12:52,370 --> 00:12:54,810
of a catalyst.

186
00:12:54,810 --> 00:12:57,420
So with a catalyst--

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00:12:57,420 --> 00:12:59,610
first to answer this question,
we need to know what a

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00:12:59,610 --> 00:13:02,200
catalyst does.

189
00:13:02,200 --> 00:13:04,970
And in class, we learned that
a catalyst will lower the

190
00:13:04,970 --> 00:13:09,470
activation barrier, thus
allowing for the reaction to

191
00:13:09,470 --> 00:13:10,570
proceed faster.

192
00:13:10,570 --> 00:13:15,750
You don't necessarily get more
product because that's

193
00:13:15,750 --> 00:13:19,890
governed by a different set of
laws, but you do get the

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00:13:19,890 --> 00:13:21,510
product faster.

195
00:13:21,510 --> 00:13:31,210
And so the way we can change
this is that instead of having

196
00:13:31,210 --> 00:13:34,005
our activation barrier up there,
we can say it's going

197
00:13:34,005 --> 00:13:37,956
to be somewhere here.

198
00:13:37,956 --> 00:13:44,390
It's going to be lowered by the
presence of the catalyst.

199
00:13:44,390 --> 00:13:48,870
So we want to make a smooth
curve again and connect this

200
00:13:48,870 --> 00:13:52,770
lower activation barrier.

201
00:13:52,770 --> 00:13:55,920
And you'd want to label this--

202
00:13:55,920 --> 00:13:59,540
I'll label it in white--

203
00:13:59,540 --> 00:14:08,920
this is the catalyzed
curve there.

204
00:14:08,920 --> 00:14:11,380
As long as you've showed that
the activation barrier was

205
00:14:11,380 --> 00:14:15,480
lowered by the presence of
a catalyst, that's fine.

206
00:14:15,480 --> 00:14:17,610
That was what the question
was asking for.

207
00:14:17,610 --> 00:14:21,720
So we've labeled everything,
answered all the questions and

208
00:14:21,720 --> 00:14:23,550
we're done with this problem.