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PROFESSOR: OK, welcome back.

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We're going to be looking at
problem number 11 in the fall

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2009 final exam.

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This is one of those problems on
the exam that kind of tests

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a whole bunch of other
stuff we couldn't get

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into one big question.

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So it's looking at a bunch
of different areas.

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But the main themes are going
to be bonding and reaction

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00:00:40,790 --> 00:00:42,380
rates and half lives.

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00:00:42,380 --> 00:00:45,360
So the things I think you need
to know, the W I N list as I

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00:00:45,360 --> 00:00:47,740
usually call it, you
need to understand

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00:00:47,740 --> 00:00:49,710
reaction rate orders.

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00:00:49,710 --> 00:00:52,930
What they mean, how they
apply to reactions.

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00:00:52,930 --> 00:00:55,960
What a half life is and
how you use that.

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00:00:55,960 --> 00:01:00,780
And the third thing is melting
points and how they are

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00:01:00,780 --> 00:01:03,040
related to intermolecular forces
and how intermolecular

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00:01:03,040 --> 00:01:05,070
forces affect melting points.

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00:01:05,070 --> 00:01:08,060
So review these three things and
once you've got them under

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00:01:08,060 --> 00:01:11,620
your belt then let's attempt
the problem.

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00:01:11,620 --> 00:01:13,150
So let's start the problem.

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00:01:13,150 --> 00:01:16,130
The first part is asking us to
determine the half life for a

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00:01:16,130 --> 00:01:17,660
particular reaction.

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00:01:17,660 --> 00:01:19,060
So there's two places
you could start.

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00:01:19,060 --> 00:01:20,330
We saw both answers.

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00:01:20,330 --> 00:01:22,190
You could start with the
definition of half life.

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00:01:22,190 --> 00:01:24,190
If you've got it down on your
equation sheet, you could just

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00:01:24,190 --> 00:01:26,840
say, well I know this is the
equation which is the natural

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00:01:26,840 --> 00:01:28,640
log of the starting
concentration--

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00:01:28,640 --> 00:01:29,630
c naught--

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00:01:29,630 --> 00:01:31,430
divided by the current
concentration that you're

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00:01:31,430 --> 00:01:33,450
looking at-- c--

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00:01:33,450 --> 00:01:40,490
equals kt, where t is time and
k is your constant for k.

40
00:01:40,490 --> 00:01:42,600
Or your reaction
rate constant.

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00:01:42,600 --> 00:01:44,470
Or-- and this is the place
that I would recommend

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00:01:44,470 --> 00:01:47,680
starting because it's
universally useful to use for

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00:01:47,680 --> 00:01:49,940
doing kinetics problems
and also for

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00:01:49,940 --> 00:01:51,340
doing half life decays--

45
00:01:51,340 --> 00:01:53,310
you start with your
rate equation.

46
00:01:53,310 --> 00:01:55,380
This is a very general rate
equation, which I'm going to

47
00:01:55,380 --> 00:01:58,360
show you how it can be
transformed to solve this very

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00:01:58,360 --> 00:02:00,240
simple problem.

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00:02:00,240 --> 00:02:07,160
We're told here that we're
looking at the half life of a

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00:02:07,160 --> 00:02:09,810
particular reagent,
it's not listed.

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00:02:09,810 --> 00:02:11,310
And we're also told
that it falls by

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00:02:11,310 --> 00:02:14,710
factor of 11 in 11 minutes.

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00:02:14,710 --> 00:02:17,690
Because we're looking at one
material, we don't need two

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00:02:17,690 --> 00:02:20,110
different items. So we're
just going to look at a.

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00:02:20,110 --> 00:02:20,760
Let's forget about b.

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00:02:20,760 --> 00:02:21,650
There's no b.

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00:02:21,650 --> 00:02:25,430
OK, we're just looking at
how a changes over time.

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00:02:25,430 --> 00:02:26,805
We're also told that
it's first order.

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00:02:26,805 --> 00:02:28,930
So we're given this information
in the problem.

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00:02:28,930 --> 00:02:34,030
First order means that you
have an n equal to 1.

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00:02:34,030 --> 00:02:45,150
So n equals 1, which turns this
equation into k times the

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00:02:45,150 --> 00:02:47,980
concentration of species a.

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00:02:47,980 --> 00:02:51,930
We also know that rate is
nothing more than the changing

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00:02:51,930 --> 00:02:53,960
concentration with respect
to the change in time.

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00:02:53,960 --> 00:02:59,510
So in calculus that
would be da / dt.

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00:03:02,390 --> 00:03:04,340
So looking at this now, we've
already got a differential

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00:03:04,340 --> 00:03:06,950
equation, which we can solve
and given certain boundary

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00:03:06,950 --> 00:03:10,100
conditions we can apply as
a naught as our starting

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00:03:10,100 --> 00:03:13,260
concentration, we can create
a new equation.

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So the way we're going to do
this is we're going to do the

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00:03:15,940 --> 00:03:18,050
calculus-- which I'm not going
to do for you here--

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00:03:18,050 --> 00:03:23,440
it's very simply you're going
to find this equation.

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00:03:23,440 --> 00:03:27,740
This is all coming from the
very general rate equation

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00:03:27,740 --> 00:03:28,990
that we started with.

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00:03:35,210 --> 00:03:36,360
So there we are.

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00:03:36,360 --> 00:03:38,680
That's where we're at and you'll
notice if I bring this

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00:03:38,680 --> 00:03:41,200
a to the other side, I take
the natural log, I put a

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00:03:41,200 --> 00:03:44,330
negative sign in there, I return
back to my original

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equation that I sort
of luckily--

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00:03:46,440 --> 00:03:48,560
or didn't luckily-- have
on my equation sheet.

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00:03:48,560 --> 00:03:50,030
So now we've got
this equation.

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And now we can actually go ahead
and solve the first part

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00:03:51,760 --> 00:03:54,850
of the problem very easily.

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00:03:54,850 --> 00:03:57,090
So I'm going to move
over here.

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00:03:57,090 --> 00:03:57,740
We're going to start off.

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We're going to say we're told
that in 11 minutes we have now

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00:04:02,420 --> 00:04:05,440
1/11 of the concentration
we originally had.

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00:04:05,440 --> 00:04:08,030
So let me write that
mathematically for you.

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00:04:08,030 --> 00:04:09,930
I'm going to write a naught
as my starting

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00:04:09,930 --> 00:04:11,920
concentration of a.

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00:04:11,920 --> 00:04:21,510
So in 11 minutes we're going to
have 1/11 a naught a naught

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00:04:21,510 --> 00:04:23,510
e to the negative k.

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00:04:23,510 --> 00:04:26,490
And that occurs in 11 minutes.

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00:04:26,490 --> 00:04:29,200
Just express the statement
mathematically.

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00:04:29,200 --> 00:04:31,930
We've can cancel out these a
naughts so they don't matter.

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00:04:31,930 --> 00:04:33,180
Which is good because we
don't know what it is

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00:04:33,180 --> 00:04:34,250
in the first place.

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00:04:34,250 --> 00:04:37,770
And now we can rearrange this
equation pretty simply.

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00:04:37,770 --> 00:04:40,400
We can just do it as,
solve for k, our

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00:04:40,400 --> 00:04:42,160
reaction rate constant.

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00:04:42,160 --> 00:04:48,530
That's going to be negative
1/11, natural log of 1/11.

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00:04:48,530 --> 00:04:49,550
That's k.

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00:04:49,550 --> 00:04:51,740
And now we have a value for k.

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00:04:51,740 --> 00:04:54,360
Now that we know the value
for k we can go back and

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00:04:54,360 --> 00:04:55,810
find the half life.

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00:04:55,810 --> 00:04:58,610
And half life is simply defined
very, very similar to

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00:04:58,610 --> 00:04:59,900
how we approached the problem
to begin with.

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00:04:59,900 --> 00:05:04,460
Half life is how long it takes
for the concentration to drop

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00:05:04,460 --> 00:05:07,270
from its original value to half
of that original value.

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00:05:07,270 --> 00:05:22,230
So let me write that now
in mathematical terms.

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00:05:22,230 --> 00:05:24,950
In this problem we know
k now but we don't

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00:05:24,950 --> 00:05:25,610
know the half life.

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00:05:25,610 --> 00:05:26,790
We don't know t 1/2.

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00:05:26,790 --> 00:05:29,450
I'm going to call it t 1/2 to
designate it and to mark it

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00:05:29,450 --> 00:05:31,220
differently than before.

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Again we can cancel these off.

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00:05:33,660 --> 00:05:34,500
And now we can solve.

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00:05:34,500 --> 00:05:35,540
Because we know k.

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00:05:35,540 --> 00:05:38,040
So let's plug in everything.

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00:05:38,040 --> 00:05:39,150
We can reverse everything.

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00:05:39,150 --> 00:05:40,590
So let me do that for you.

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00:05:43,470 --> 00:05:44,720
We're going to have this.

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00:05:47,160 --> 00:05:57,810
Natural log of 1/2 being equal
to negative k, t to the 1/2.

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00:05:57,810 --> 00:06:01,980
And now we can solve for our t
1/2 because we know that k--

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00:06:01,980 --> 00:06:08,250
let's bring k to the
other side, and the

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00:06:08,250 --> 00:06:11,430
negative sign as well--

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00:06:11,430 --> 00:06:20,630
k is going to be negative
1/11 natural log 1/11.

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00:06:20,630 --> 00:06:23,560
We can then change natural log
1/2 into natural log 2.

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00:06:23,560 --> 00:06:26,110
That actually looks like our
original half life equation if

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00:06:26,110 --> 00:06:27,970
you had that on your
equation sheet.

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So when you solve this out,
you're solving for t 1/2, t

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1/2 the long and the short
of it is 3.18 minutes.

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The main takeaway from this
problem is that you don't need

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to have memorized any number
of equations to do it.

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You can start from a very
general equation, which is the

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00:06:46,170 --> 00:06:48,980
rate equation, which we showed
in the beginning, and go from

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00:06:48,980 --> 00:06:51,930
there using logic and then just
mathematically represent

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00:06:51,930 --> 00:06:53,750
what the problem is telling you
to do in the first place.

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00:06:53,750 --> 00:06:55,780
So that's part A and that's the
majority of the points on

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00:06:55,780 --> 00:06:57,940
this problem.

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00:06:57,940 --> 00:07:01,100
Moving over, part B and C are
very quick, very short.

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00:07:01,100 --> 00:07:02,900
Just to check if you
know the facts.

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00:07:02,900 --> 00:07:04,280
Let's do them right now.

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00:07:04,280 --> 00:07:05,530
We're asking you--

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00:07:08,360 --> 00:07:12,140
we're going to do B 1
and B 2 together--

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00:07:12,140 --> 00:07:15,200
to tell us which one of
the compounds has a

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00:07:15,200 --> 00:07:17,840
higher boiling point.

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00:07:17,840 --> 00:07:19,970
And we give you two
different cases.

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In each case there's
two compounds.

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So number 1, you're given
silene and phosphene.

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00:07:26,070 --> 00:07:27,840
So the first thing you should
do is actually draw those

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00:07:27,840 --> 00:07:30,000
molecules out.

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So that's what I did.

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Silene looks like this.

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00:07:39,330 --> 00:07:40,590
And then phosphene--

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00:07:40,590 --> 00:07:42,320
put the bonds in there--

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00:07:42,320 --> 00:07:43,899
phosphene looks something
like this.

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So now we have to sort of
identify which one of these

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00:07:52,870 --> 00:07:53,950
has a higher melting point.

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00:07:53,950 --> 00:07:56,650
Now in order to do that you need
to have reviewed and know

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00:07:56,650 --> 00:07:59,640
that melting point is related
to the intermolecular forces

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00:07:59,640 --> 00:08:03,250
between the between
the molecules.

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00:08:03,250 --> 00:08:05,320
The forces between the molecules
are dependent on

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00:08:05,320 --> 00:08:08,150
both their geometry and
their chemistry.

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00:08:08,150 --> 00:08:11,670
So knowing that, if something
has stronger intermolecular

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00:08:11,670 --> 00:08:14,320
bonds, it'll have a higher
melting point.

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00:08:14,320 --> 00:08:15,710
So let's take a look at
these two molecules.

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00:08:15,710 --> 00:08:18,610
Which one has stronger
intermolecular bonds?

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00:08:18,610 --> 00:08:21,090
We have in silene case, we have
van der Waals forces,

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00:08:21,090 --> 00:08:22,380
they're generally very weak.

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00:08:22,380 --> 00:08:24,890
We've got those tpo
for the phosphene.

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00:08:24,890 --> 00:08:28,100
But in phosphene we additionally
have a dipole.

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00:08:28,100 --> 00:08:30,290
We have partial--

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00:08:30,290 --> 00:08:37,360
we put maybe delta minus here
and delta plus here and here

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00:08:37,360 --> 00:08:38,150
on this side.

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00:08:38,150 --> 00:08:39,515
So our molecule is actually
polarized.

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00:08:39,515 --> 00:08:42,020
There's a plus region
and a minus region.

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00:08:42,020 --> 00:08:44,610
And because of that you have
much stronger intermolecular

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00:08:44,610 --> 00:08:47,640
forces between the phosphenes
then you do the silenes.

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00:08:47,640 --> 00:08:49,350
And because these have stronger
intermolecular

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00:08:49,350 --> 00:08:54,750
forces, this one has a
higher melting point.

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00:08:54,750 --> 00:08:58,910
The last part of the
problem, part 2.

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00:08:58,910 --> 00:09:01,330
We try to throw a little trick
in there with you.

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00:09:01,330 --> 00:09:03,600
We look at ammonia
and phosphene.

185
00:09:03,600 --> 00:09:05,930
So now we're doing
NH3 and PH3.

186
00:09:05,930 --> 00:09:07,850
And they actually both
look identical in

187
00:09:07,850 --> 00:09:09,310
terms of their geometry.

188
00:09:09,310 --> 00:09:10,560
So let me do that now.

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00:09:13,300 --> 00:09:15,710
It actually looks tetragonal
if you were to draw the

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00:09:15,710 --> 00:09:17,910
structure out.

191
00:09:17,910 --> 00:09:22,450
There's ammonia and here's
our phosphene again.

192
00:09:22,450 --> 00:09:23,700
You can tell they
look the same.

193
00:09:26,300 --> 00:09:28,930
So I think you have to sort of
reason out which one has the

194
00:09:28,930 --> 00:09:30,970
higher melting point, which is
the same thing as saying which

195
00:09:30,970 --> 00:09:34,570
one has stronger intermolecular
forces.

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00:09:34,570 --> 00:09:39,650
A very common answer we got
was that the nitrogen is a

197
00:09:39,650 --> 00:09:41,470
smaller atom, therefore
everything is held more

198
00:09:41,470 --> 00:09:43,830
closely, therefore the
forces are stronger.

199
00:09:43,830 --> 00:09:47,260
That's not really the answer
that is most dominant and it's

200
00:09:47,260 --> 00:09:49,670
also not the answer we
were looking for.

201
00:09:49,670 --> 00:09:53,000
The correct answer here is that
even though both of these

202
00:09:53,000 --> 00:09:54,080
have the same--

203
00:09:54,080 --> 00:09:56,620
probably very similar-- van der
Waals forces, they're both

204
00:09:56,620 --> 00:10:01,560
polarized molecules, nitrogen
has hydrogen bonding.

205
00:10:01,560 --> 00:10:03,450
So remember the hydrogen bonding
elements you should

206
00:10:03,450 --> 00:10:09,560
probably know are fluorine,
oxygen, chlorine, nitrogen,

207
00:10:09,560 --> 00:10:11,220
four of them.

208
00:10:11,220 --> 00:10:14,080
So nitrogen will hydrogen bond,
which means that this

209
00:10:14,080 --> 00:10:17,100
nitrogen will have a hydrogen
bond with some other ammonia

210
00:10:17,100 --> 00:10:18,630
molecule over here.

211
00:10:18,630 --> 00:10:21,980
This H might be bonded to
an N somewhere else.

212
00:10:21,980 --> 00:10:24,570
And because there are additional
hydrogen bonding

213
00:10:24,570 --> 00:10:27,070
force in the system, the
intermolecular forces are

214
00:10:27,070 --> 00:10:29,720
stronger and more prevalent in
ammonia, which means that

215
00:10:29,720 --> 00:10:33,065
ammonia will have a higher
boiling point, which is true.

216
00:10:35,820 --> 00:10:37,500
The answer here is ammonia.

217
00:10:37,500 --> 00:10:40,540
So all you have to take way
from this is that the

218
00:10:40,540 --> 00:10:44,140
mechanical properties as well
as the melting and boiling

219
00:10:44,140 --> 00:10:47,030
properties of a material are
very much governed by the

220
00:10:47,030 --> 00:10:50,610
intermolecular forces that
exist in that particular

221
00:10:50,610 --> 00:10:53,040
material, that the liquid
or the solid state.

222
00:10:53,040 --> 00:10:56,890
And so in this case we're
looking at these compounds and

223
00:10:56,890 --> 00:11:00,040
we've identified this one as
being a higher melting point

224
00:11:00,040 --> 00:11:02,740
because it has a polarity
whereas this one doesn't.

225
00:11:02,740 --> 00:11:06,330
And in this case where we try to
trick you up, we identified

226
00:11:06,330 --> 00:11:07,790
that nitrogen has
this additional

227
00:11:07,790 --> 00:11:09,210
hydrogen bonding force.

228
00:11:09,210 --> 00:11:11,430
And that's all there is to it.