Using matrices to represent data: Networks

WEBVTT
Kind: captions
Language: en

00:00:00.620 --> 00:00:02.330
- [Instructor] We're
told this network diagram

00:00:02.330 --> 00:00:06.040
represents the different train
routes between three cities.

00:00:06.040 --> 00:00:09.690
Each node is a city
and each directed arrow

00:00:09.690 --> 00:00:14.160
represents a direct bus
route from city to city.

00:00:14.160 --> 00:00:19.160
So for example, this
arrow right over here,

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I guess would represent a direct bus route

00:00:21.840 --> 00:00:25.990
that starts in city three
and ends in city one,

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while this arrow that has
an arrow on both sides

00:00:30.360 --> 00:00:33.120
shows a route that both starts

00:00:33.120 --> 00:00:35.320
in city three and ends in city one,

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and a route that starts in city
one and ends in city three.

00:00:40.430 --> 00:00:43.080
So it says, complete the
matrix so it represents

00:00:43.080 --> 00:00:46.120
the number of direct
routes between the cities,

00:00:46.120 --> 00:00:50.270
where rows are starting points
and columns are end points.

00:00:50.270 --> 00:00:52.250
So this is the matrix right over here.

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I encourage you if you feel so inspired,

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and I encourage you to feel so inspired,

00:00:55.940 --> 00:00:57.480
pause this video and see if you can

00:00:57.480 --> 00:00:59.580
fill out this matrix right over here.

00:00:59.580 --> 00:01:01.960
You have nine entries in this matrix

00:01:01.960 --> 00:01:03.330
for each of these combinations

00:01:03.330 --> 00:01:05.430
between the starting city and ending city.

00:01:06.280 --> 00:01:07.640
All right, now let's do it together.

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So, what would go here?

00:01:09.360 --> 00:01:11.350
This would be the number of direct routes

00:01:11.350 --> 00:01:15.550
that start at city one
and end at city one.

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So if we start at city one,

00:01:18.690 --> 00:01:22.230
are there any things that
then end at city one?

00:01:22.230 --> 00:01:23.063
Well, no.

00:01:23.063 --> 00:01:24.300
It doesn't look like there's anything

00:01:24.300 --> 00:01:26.860
that starts at city one
and ends at city one.

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So I'll put a zero there.

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What about this one right over here?

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Well it needs to start at
city one and end at city two.

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So let's see.

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This starts at city one and
ends at city two, so that's one.

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We get two.

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And then we get three.

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And then we get four, 'cause
you can start at city one here

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and then end at city two.

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So we get four.

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Now how many start at city
one and end at city three?

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Pause this video and think
about that, actually.

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All right, we're gonna start at city one

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and end at city three.

00:02:02.100 --> 00:02:03.800
I'm going to get another color out here.

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So I could start here and go on this route

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and 'cause this arrow ends
at city three, so that's one.

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This middle one does not start at city one

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and end at city three.

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It goes the other way around,
so I'm not gonna count that.

00:02:16.100 --> 00:02:19.030
This one right over here,
I can go either way,

00:02:19.030 --> 00:02:21.510
so I could start at city
one and end at city three,

00:02:21.510 --> 00:02:23.304
'cause we have that arrow there.

00:02:23.304 --> 00:02:25.330
And those look like the only two

00:02:25.330 --> 00:02:29.760
that start at city one
and end at city three.

00:02:29.760 --> 00:02:32.980
So that looks like, go
back to the original color,

00:02:32.980 --> 00:02:35.080
two routes right over there.

00:02:35.080 --> 00:02:38.623
Now what about starting at city
two and ending at city one?

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Well, if we start at city
two and end at city one,

00:02:43.260 --> 00:02:44.500
these three over here,

00:02:44.500 --> 00:02:46.870
all of these start at one and end at two.

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They don't go the other way.

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But this one on top with the double arrows

00:02:50.580 --> 00:02:51.413
you can go either way.

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So you could start at city
two and end at city one.

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So there's one route here.

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Let's see.

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Start at city two, end at city two.

00:03:02.420 --> 00:03:04.450
Well, I don't see anything that looks

00:03:04.450 --> 00:03:06.460
like that for city two,

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so this is going to be a zero.

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And then starts at city
two, ends at city three.

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So starts at city two, ends at city three.

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This arrow doesn't count, 'cause this

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starts at three, ends at two,
not the other way around.

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So we get a zero there as well.

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And then let's go to city three.

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How many start at three and end at one?

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So start at three and end at one?

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So this two-way arrow, you could do that.

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You could start at three and
end at one, so that's one.

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Then this one right over
here starts at three

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and ends at one because
we can see the arrow

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points to one right over there.

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And then it looks like,
and actually this one,

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I have so much that I've written here

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that I actually can't see
too well the original.

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Let me erase it actually,
so I can make sure

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I'm seeing things properly.

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Yeah, that one, too, looks like,

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so this one I can do, and then this one

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I can start at city three
and end at city one, as well.

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So it looks like we
have three paths there.

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Now start at city three, end at city two.

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That one's a little bit
more straight forward.

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That's that path there, so that is one.

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And then starts at city
three, ends at city three.

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Well, we have this one right over here.

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That's the only one, so I would put one.

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So there you have it.

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We have filled in this matrix.

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So which city has the
most incoming routes?

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Pause the video and think about that.

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So the city with the most incoming routes,

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we can look at the cities
that are the end points,

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and so city one has a total
of zero plus one plus three,

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has four incoming routes.

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City two has a total of four
plus one, five incoming routes.

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And city three has a total
of two plus zero plus one,

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has three incoming routes.

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So it looks like this would be city two

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with five incoming routes.

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Which city has the most outgoing routes?

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Well, then we would
just look the other way.

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Actually, pause the video
and think about that.

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Well, it looks like city one

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has six outgoing routes.

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City two only has one outgoing route.

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I'm just adding up along the row.

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And city three has, looks
like five outgoing routes.

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So city one was zero plus four plus two.

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There's a total of six
routes that start at city one

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and go out of the city, so that
is city one with six routes.


Human population dynamics

WEBVTT
Kind: captions
Language: en

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- [Instructor] What we have here

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is a really interesting visual

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that shows world population growth

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from 1750 all the way to 2100.

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And obviously this isn't 2100 yet.

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So it's doing some projecting
for roughly the next 80 years.

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It also shows the
absolute world population

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over that time period.

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So just to make sure we're
understanding this graphic,

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our horizontal axis here, we
can see the years going by.

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You can view that as our time
axis, starting with the 1750

00:00:30.280 --> 00:00:33.100
right over here, going
all the way to 2100.

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Now we are here in, when
I'm making this video

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is 2021 but this graphic was made in 2015.

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So everything beyond 2015 is
considered to be a projection.

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Now on this left-hand vertical axis,

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this is really the rate of
world population growth,

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the growth rate that it has here.

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And I could write that down
just to make it more clear.

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This is the growth rate axis

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and we have a right vertical axis here

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because we're really showing
two different things.

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We're showing the annual growth rate

00:01:06.240 --> 00:01:07.800
of world population in red.

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And then we're showing the
absolute world population

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in this blue curve right over here.

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And so we could view this
axis as world population.

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Now there's a bunch of really
interesting things here.

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One, you might just wonder, well

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what was the world population around 1750?

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And if you look at this and
these are obviously estimates

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because you did not have a
global census back in 1750,

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it was around seven or 800 million people.

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Now what's interesting is
that's not a lot larger

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than the number of people
that we had at say the time

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of Jesus, which is also the time

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of the Roman empire and Han
China, where it's estimated

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that at around that time
period around 2000 years ago

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the world population was
around 300 million people.

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But then we see something
interesting happening

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over the next several centuries from 1750

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it seems to start growing
pretty dramatically.

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And we could see why that is.

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We can see the growth rate
of population for a while.

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If we look at this red
curve, the growth rate

00:02:11.850 --> 00:02:16.660
of world population was
around 0.6%, but it looks

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like in the early 20th
century, right around here,

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the rate of growth of
world population starts

00:02:24.380 --> 00:02:28.320
to really increase where
at least based on this

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it looks like around the 1970s,

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world population the growth
rate peaked out at 2.1%.

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Now the question might be,
why was this happening?

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Why did the rate of
growth increased so much?

00:02:43.480 --> 00:02:46.390
Well, the main argument
is as we became a more

00:02:46.390 --> 00:02:49.560
industrial society, healthcare
would have improved.

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So child mortality would have gone down.

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People would have died of fewer diseases.

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At the same time

00:02:55.630 --> 00:02:58.480
we would have gotten more
efficient with agriculture.

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We would have had farming
methods, so more people

00:03:02.310 --> 00:03:03.510
so food would get cheaper.

00:03:03.510 --> 00:03:04.410
It would be more abundant.

00:03:04.410 --> 00:03:06.530
More people would have access to nutrition

00:03:06.530 --> 00:03:10.630
and all of those would
drive the growth rate up.

00:03:10.630 --> 00:03:12.650
Now, an interesting question
that folks have been thinking

00:03:12.650 --> 00:03:15.990
about for many hundreds
of years, is there a limit

00:03:15.990 --> 00:03:19.720
to how much human population
the earth can sustain?

00:03:19.720 --> 00:03:21.700
And so they've introduced this idea

00:03:21.700 --> 00:03:24.130
known as carrying capacity.

00:03:24.130 --> 00:03:27.680
Why do we use the letter K for
capacity when capacity starts

00:03:27.680 --> 00:03:31.660
with a C, because a version
of that word in German starts

00:03:31.660 --> 00:03:33.970
with a K and there's
always been this notion

00:03:33.970 --> 00:03:35.240
that for a given species

00:03:35.240 --> 00:03:38.430
there's gotta be some maximum capacity

00:03:38.430 --> 00:03:41.120
that an ecosystem can sustain.

00:03:41.120 --> 00:03:42.940
And there's folks like Thomas Malthus,

00:03:42.940 --> 00:03:45.300
who's theorized that
there must be some limit

00:03:45.300 --> 00:03:47.440
to how many people there
could be on the planet.

00:03:47.440 --> 00:03:48.940
Just based on how much land there is,

00:03:48.940 --> 00:03:52.450
how much nutrition or how
much resources there are.

00:03:52.450 --> 00:03:53.760
But the carrying capacity

00:03:53.760 --> 00:03:56.750
for humanity is really an open question

00:03:56.750 --> 00:03:59.610
because our technology is
constantly on the move.

00:03:59.610 --> 00:04:01.360
We constantly are getting better

00:04:01.360 --> 00:04:04.190
at using our resources more efficiently

00:04:04.190 --> 00:04:05.940
going into new ecosystems.

00:04:05.940 --> 00:04:07.010
Now it's a huge debate.

00:04:07.010 --> 00:04:08.550
How sustainable is that?

00:04:08.550 --> 00:04:10.110
And as we become more efficient

00:04:10.110 --> 00:04:13.150
are we also trashing certain
parts of our ecosystem

00:04:13.150 --> 00:04:14.840
but the carrying capacity of humans,

00:04:14.840 --> 00:04:16.640
we really don't know how many people

00:04:16.640 --> 00:04:19.080
the planet earth could sustain.

00:04:19.080 --> 00:04:21.780
But we could see that the
rate of population growth

00:04:21.780 --> 00:04:24.810
starts to decline after 1970.

00:04:24.810 --> 00:04:26.880
And the main argument for this,

00:04:26.880 --> 00:04:28.970
and we've seen this and
recover this in other videos

00:04:28.970 --> 00:04:31.830
when we study countries
that are still developing

00:04:31.830 --> 00:04:33.810
and countries that would
be considered industrial

00:04:33.810 --> 00:04:35.290
or even post-industrial,

00:04:35.290 --> 00:04:38.710
is that as countries
develop and become wealthier

00:04:38.710 --> 00:04:41.680
you might have more women
entering into the workforce.

00:04:41.680 --> 00:04:43.820
You might have more family planning.

00:04:43.820 --> 00:04:45.780
Women have more control
over their destiny,

00:04:45.780 --> 00:04:49.890
get education longer, and
people just have fewer children.

00:04:49.890 --> 00:04:52.660
And so as people have fewer children

00:04:52.660 --> 00:04:54.330
you could see that the world that,

00:04:54.330 --> 00:04:56.490
especially we see this in industrial

00:04:56.490 --> 00:04:58.030
and post-industrial countries,

00:04:58.030 --> 00:05:00.280
the growth rate starts to come down.

00:05:00.280 --> 00:05:02.980
And so the aggregate growth
rate is coming down, arguably

00:05:02.980 --> 00:05:05.410
because more and more countries
are becoming wealthier,

00:05:05.410 --> 00:05:08.000
have better healthcare,
better rights for women.

00:05:08.000 --> 00:05:11.160
And we even see today that
when you look at countries

00:05:11.160 --> 00:05:16.160
that are developed being their
growth rate is around 1.2%

00:05:16.410 --> 00:05:18.680
while more economically developed nations

00:05:18.680 --> 00:05:20.767
aren't around 0.2%.

00:05:20.767 --> 00:05:23.120
So there's definitely that
correlation between the two.

00:05:23.120 --> 00:05:25.810
And of course, this growth
rate that you see in red

00:05:25.810 --> 00:05:28.290
is going to affect the
absolute population.

00:05:28.290 --> 00:05:31.070
And so it's no coincidence
that at the same time

00:05:31.070 --> 00:05:34.060
that the growth rate
started to go up like that.

00:05:34.060 --> 00:05:36.120
We see that the population

00:05:36.120 --> 00:05:38.630
in absolute terms starts to accelerate.

00:05:38.630 --> 00:05:42.320
And just over, let's say
the last hundred years

00:05:42.320 --> 00:05:45.300
we've gone from 1.6, 5 billion.

00:05:45.300 --> 00:05:47.260
I guess if we go back
a hundred years to 1920

00:05:47.260 --> 00:05:49.710
it looks like we're
approaching 2 billion folks.

00:05:49.710 --> 00:05:50.700
And now we're sitting

00:05:50.700 --> 00:05:53.140
at roughly seven and a half billion folks.

00:05:53.140 --> 00:05:55.540
So the world population roughly doubled

00:05:55.540 --> 00:05:59.810
over the, last let's 2000
years prior to this chart.

00:05:59.810 --> 00:06:04.560
But then over this chart
over just the last 270 years

00:06:04.560 --> 00:06:07.750
our world population has grown pretty much

00:06:07.750 --> 00:06:09.750
by a factor of 10.

00:06:09.750 --> 00:06:11.660
Now we don't know how
sustainable that would be

00:06:11.660 --> 00:06:13.540
if our growth rate were to continue,

00:06:13.540 --> 00:06:16.950
but it does look like that
growth rate is moderating

00:06:16.950 --> 00:06:19.140
and at least the projections in this chart

00:06:19.140 --> 00:06:22.790
have a starting to approach
11 billion, maybe over time

00:06:22.790 --> 00:06:26.090
12 billion and maybe stabilize
someplace around there.

00:06:26.090 --> 00:06:28.583
But it's an interesting
thing to think about.


Khan Kickoff Pep Talk: Brandi Chastain

WEBVTT
Kind: captions
Language: en

00:00:00.521 --> 00:00:02.860
- Hello, Khan Academy students.

00:00:02.860 --> 00:00:04.038
This is Brandi Chastain,

00:00:04.038 --> 00:00:07.090
former U.S. Women's National Team member,

00:00:07.090 --> 00:00:10.068
Olympic gold medalist
and World Cup champion.

00:00:10.068 --> 00:00:13.991
And I'm here to say to you that

00:00:13.991 --> 00:00:17.090
what you're doing is not easy.

00:00:17.090 --> 00:00:19.970
Trying to learn in an adverse environment,

00:00:19.970 --> 00:00:23.700
to be a champion takes
a lot of determination,

00:00:23.700 --> 00:00:28.016
desire, resiliency, it
takes a lot of falling down

00:00:28.016 --> 00:00:29.496
and getting back up

00:00:29.496 --> 00:00:32.533
and I want you to know that anything

00:00:32.533 --> 00:00:35.988
that is worthwhile is worth fighting for,

00:00:35.988 --> 00:00:40.988
is worth getting dirty, it's
worth getting a couple scrapes.

00:00:41.070 --> 00:00:43.974
And if it were easy,
everybody would be doing it

00:00:43.974 --> 00:00:45.950
and it wouldn't be a big deal,

00:00:45.950 --> 00:00:47.420
but what you're doing is a big deal.

00:00:47.420 --> 00:00:50.555
You're challenging yourselves
to learn like champions,

00:00:50.555 --> 00:00:54.447
and to be a champion takes
time, it's a process,

00:00:54.447 --> 00:00:56.874
it's not always going to be smooth

00:00:56.874 --> 00:01:01.060
and most of the time success
isn't this upward ascension,

00:01:01.060 --> 00:01:04.310
straight line to gold medals.

00:01:04.310 --> 00:01:07.030
So remember that what you're doing

00:01:08.869 --> 00:01:12.540
is really a life's work
and a life's journey.

00:01:12.540 --> 00:01:15.560
And when you get to be my age, (chuckles)

00:01:15.560 --> 00:01:17.465
you'll understand that
maybe a little bit more,

00:01:17.465 --> 00:01:20.500
but you can look back and
say, "I stuck with it,

00:01:20.500 --> 00:01:24.479
it wasn't easy, sometimes
I wanted to give up".

00:01:24.479 --> 00:01:28.820
I know that along my journey
to playing in the Olympics

00:01:28.820 --> 00:01:31.501
and being on the World Cup team

00:01:31.501 --> 00:01:34.120
and scoring that final penalty kick goal,

00:01:34.120 --> 00:01:37.263
where I got to do a little celebrating,

00:01:39.690 --> 00:01:40.580
it wasn't easy.

00:01:40.580 --> 00:01:44.452
I was on the team, cut from the team.

00:01:44.452 --> 00:01:46.493
I made the team in a different position

00:01:46.493 --> 00:01:48.450
which I'd never played before.

00:01:48.450 --> 00:01:52.437
And when that was presented
to me, I had a choice.

00:01:52.437 --> 00:01:54.503
Did I wanna be on the team

00:01:54.503 --> 00:01:59.489
and try to make a contribution
to a positive outcome,

00:01:59.489 --> 00:02:02.610
or was I more prideful and say,
"No, that's not my position,

00:02:02.610 --> 00:02:03.930
I wanna play my position".

00:02:03.930 --> 00:02:06.550
And I realized that being on the team

00:02:06.550 --> 00:02:08.050
and having a chance to get

00:02:08.050 --> 00:02:10.052
into the game was really what I wanted.

00:02:10.052 --> 00:02:11.925
So I want you to remember

00:02:11.925 --> 00:02:16.925
that when things just
get tough, you got this.

00:02:17.770 --> 00:02:21.963
You got to remember, you
are talented, you're smart.

00:02:23.040 --> 00:02:26.062
We all have a competitor inside of us

00:02:26.062 --> 00:02:28.750
and you have to be willing to allow it

00:02:28.750 --> 00:02:32.555
to come out and don't
settle, never settle.

00:02:32.555 --> 00:02:35.448
Becoming a gold medalist is
definitely not about settling,

00:02:35.448 --> 00:02:39.038
and you in 2021, we'll be champions.

00:02:39.038 --> 00:02:42.730
So I'm going to hopefully give you

00:02:42.730 --> 00:02:44.555
your first gold medal ever.

00:02:44.555 --> 00:02:48.490
So if you bow your head,
here comes the gold medal.

00:02:48.490 --> 00:02:50.320
There it is.

00:02:50.320 --> 00:02:54.480
You are now on team
Brandi and we are strong

00:02:54.480 --> 00:02:57.960
and we are resilient
and we get the job done.

00:02:57.960 --> 00:02:59.551
So all the best to you in 2021.

00:02:59.551 --> 00:03:02.486
Hopefully at some point
we'll meet on a soccer field

00:03:02.486 --> 00:03:04.670
or somewhere else near you.

00:03:04.670 --> 00:03:05.503
Take care.


Types of catalysts

WEBVTT
Kind: captions
Language: en

00:00:00.520 --> 00:00:02.240
- [Instructor] A catalyst
speeds up a reaction

00:00:02.240 --> 00:00:04.550
by lowering the activation energy.

00:00:04.550 --> 00:00:05.740
And there are many types of catalysts.

00:00:05.740 --> 00:00:07.630
And first we're going to look at enzymes

00:00:07.630 --> 00:00:10.984
which are biological catalysts.

00:00:10.984 --> 00:00:13.884
Let's say that this represents our enzyme,

00:00:13.884 --> 00:00:17.294
and the place where the reaction occurs

00:00:17.294 --> 00:00:19.677
is called the active site of the enzyme.

00:00:19.677 --> 00:00:23.160
So right in here, let's say
this is our active site.

00:00:23.160 --> 00:00:26.165
And then the substance that
reacts at the active site

00:00:26.165 --> 00:00:28.908
is called the substrate.

00:00:28.908 --> 00:00:33.298
So this little picture here
with two triangles together,

00:00:33.298 --> 00:00:36.382
that's the substrate for our reaction.

00:00:36.382 --> 00:00:37.700
In the next step,

00:00:37.700 --> 00:00:42.640
the substrate binds to the
enzyme at the active site.

00:00:42.640 --> 00:00:44.560
And when the substrate binds,

00:00:44.560 --> 00:00:46.940
the substrate can produce
changes in the shape

00:00:46.940 --> 00:00:49.732
of the active site, that
allow for better binding.

00:00:49.732 --> 00:00:52.366
So on the left here, we
can see how the shape

00:00:52.366 --> 00:00:55.367
of the active site changes slightly

00:00:55.367 --> 00:00:58.688
when the substrate binds to it.

00:00:58.688 --> 00:01:02.686
This formation of the
enzyme substrate complex

00:01:02.686 --> 00:01:06.562
is called the induced fit model.

00:01:06.562 --> 00:01:10.322
The substrate interacts with the enzyme

00:01:10.322 --> 00:01:13.926
through non-covalent
interactions in the active site.

00:01:13.926 --> 00:01:15.827
So things like hydrogen bonding

00:01:15.827 --> 00:01:18.299
or dipole-dipole interactions.

00:01:18.299 --> 00:01:20.618
Perhaps some of these
non-covalent interactions

00:01:20.618 --> 00:01:22.990
cause a shift in electron density

00:01:22.990 --> 00:01:24.001
which make it easier

00:01:24.001 --> 00:01:27.770
to reach the transition
state for the reaction.

00:01:27.770 --> 00:01:30.346
Therefore lowering the activation energy

00:01:30.346 --> 00:01:34.030
and speeding up the overall reaction.

00:01:34.030 --> 00:01:37.050
Next let's say the bond between
the two triangles breaks

00:01:37.050 --> 00:01:39.410
and we get our two products here.

00:01:39.410 --> 00:01:41.520
So two individual triangles.

00:01:41.520 --> 00:01:44.725
And the enzyme active site
goes back to the original shape

00:01:44.725 --> 00:01:48.750
and it's ready to
catalyze another reaction.

00:01:48.750 --> 00:01:51.145
Next, let's talk about
a homogeneous catalyst,

00:01:51.145 --> 00:01:52.813
which is a catalyst that's present

00:01:52.813 --> 00:01:57.256
in the same phase as the
reactants in a reaction mixture.

00:01:57.256 --> 00:02:01.014
So let's look at the hydrolysis of sucrose

00:02:01.014 --> 00:02:04.803
to turn into glucose and fructose.

00:02:04.803 --> 00:02:06.850
This reaction can be catalyzed

00:02:06.850 --> 00:02:09.620
by the hydronium ion H30+.

00:02:09.620 --> 00:02:13.080
And since sucrose, our reactant
is an aqueous solution,

00:02:13.080 --> 00:02:14.922
and so as the hydronium ion,

00:02:14.922 --> 00:02:19.435
we can say that the hydronium
ion is a homogeneous catalyst.

00:02:19.435 --> 00:02:21.220
And it's a source of protons

00:02:21.220 --> 00:02:23.840
to catalyze this hydrolysis reaction.

00:02:23.840 --> 00:02:26.660
This is a drawing of the sucrose molecule,

00:02:26.660 --> 00:02:30.740
which is a disaccharide
composed of two monosaccharides.

00:02:30.740 --> 00:02:32.300
Glucose is over here on the left

00:02:32.300 --> 00:02:34.978
and fructose is over here on the right.

00:02:34.978 --> 00:02:37.804
And these two monosaccharides are joined

00:02:37.804 --> 00:02:39.418
by an ether linkage.

00:02:39.418 --> 00:02:41.586
So we can see, we can see this,

00:02:41.586 --> 00:02:43.290
this connection here, alright?

00:02:43.290 --> 00:02:46.227
This oxygen in between
our two monosaccharides

00:02:46.227 --> 00:02:47.740
isn't ether linkage.

00:02:47.740 --> 00:02:49.823
And ethers are fairly nonreactive.

00:02:50.965 --> 00:02:54.116
Since ethers are generally unreactive,

00:02:54.116 --> 00:02:57.700
the hydrolysis of sucrose
is a pretty slow reaction.

00:02:57.700 --> 00:03:00.870
And to speed it up, we need
to add an acid catalyst.

00:03:00.870 --> 00:03:02.360
So if we add an acid catalyst

00:03:02.360 --> 00:03:05.238
and we have hydronium
ions in aqueous solution,

00:03:05.238 --> 00:03:08.874
a lone pair of electrons
on the oxygen on the ether,

00:03:08.874 --> 00:03:10.258
will take this proton

00:03:10.258 --> 00:03:13.653
and these electrons move in to form water.

00:03:14.500 --> 00:03:15.805
Protonation of the oxygen,

00:03:15.805 --> 00:03:18.760
gives the oxygen a plus one formal charge.

00:03:18.760 --> 00:03:22.526
And allows an acid catalyzed
mechanism to proceed.

00:03:22.526 --> 00:03:24.530
And there are more steps to the mechanism

00:03:24.530 --> 00:03:28.860
but ultimately sucrose is
broken down to form glucose

00:03:28.860 --> 00:03:33.860
and fructose in this acid
catalyzed hydrolysis of sucrose.

00:03:34.690 --> 00:03:38.374
Honey bees actually have the
enzyme to convert sucrose

00:03:38.374 --> 00:03:43.374
which is table sugar,
into glucose and fructose.

00:03:43.450 --> 00:03:47.524
And since fructose is
sweeter than sucrose,

00:03:47.524 --> 00:03:51.220
honey is sweeter than table sugar.

00:03:51.220 --> 00:03:54.440
A heterogeneous catalyst is
a catalyst that's present

00:03:54.440 --> 00:03:55.590
in a different phase

00:03:55.590 --> 00:03:58.397
from the reactants in a reaction mixture.

00:03:58.397 --> 00:04:02.370
As an example, let's look
at a hydrogenation reaction.

00:04:02.370 --> 00:04:05.715
And this reaction ethene
reacts with hydrogen

00:04:05.715 --> 00:04:09.903
on a surface of platinum to form ethane.

00:04:09.903 --> 00:04:13.080
Now, since the platinum
is in the solid form

00:04:13.080 --> 00:04:15.843
and our reactants are
in the gaseous state,

00:04:15.843 --> 00:04:20.220
the platinum is an example
of a heterogeneous catalyst.

00:04:20.220 --> 00:04:23.610
So here in our picture, we have
our piece of platinum metal

00:04:23.610 --> 00:04:28.610
and both the ethene molecule
and hydrogen are adsorbed

00:04:29.165 --> 00:04:32.319
to the surface of the platinum metal.

00:04:32.319 --> 00:04:36.580
Next the bond between the
two hydrogen atoms breaks

00:04:36.580 --> 00:04:39.470
and we get the two individual
hydrogen atoms bonded

00:04:39.470 --> 00:04:41.930
to the surface of the platinum metal.

00:04:41.930 --> 00:04:43.360
Eventually these two hydrogens

00:04:43.360 --> 00:04:46.040
add across the double bond of ethene

00:04:46.040 --> 00:04:50.490
and form the ethane molecules C2H6.

00:04:50.490 --> 00:04:53.922
So the hydrogenation of
ethene to form ethane

00:04:53.922 --> 00:04:58.153
is catalyzed by the presence
of the platinum metal.


Multistep reaction energy profiles

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:01.440
- [Instructor] Let's consider a reaction

00:00:01.440 --> 00:00:04.010
with the following multi-step mechanism.

00:00:04.010 --> 00:00:09.010
In step one, A reacts
with BC to form AC plus B.

00:00:09.670 --> 00:00:14.670
And in step two, AC reacts
with D to form A plus CD.

00:00:16.870 --> 00:00:19.050
If we add the two steps
of our mechanism together

00:00:19.050 --> 00:00:20.640
we can find the balanced equation

00:00:20.640 --> 00:00:23.130
for this hypothetical reaction.

00:00:23.130 --> 00:00:25.120
So we're gonna put all of our reactants

00:00:25.120 --> 00:00:27.560
on the left side here,
and we're gonna have

00:00:27.560 --> 00:00:30.800
all of our products on the right side.

00:00:30.800 --> 00:00:33.130
And we can see that AC is on the left

00:00:33.130 --> 00:00:36.450
and it's on the right side,
so we can cancel that out.

00:00:36.450 --> 00:00:38.600
A is also in the left and the right side,

00:00:38.600 --> 00:00:40.530
so we can cancel that out.

00:00:40.530 --> 00:00:45.360
So the overall equation would be BC plus D

00:00:46.940 --> 00:00:51.673
goes to B plus CD.

00:00:56.002 --> 00:00:58.850
We've just seen that BC
and D are our reactants

00:01:01.220 --> 00:01:05.480
and B and CD are the products

00:01:05.480 --> 00:01:07.980
for this hypothetical reaction.

00:01:07.980 --> 00:01:11.550
If we look at the mechanism,
A is there in the beginning

00:01:11.550 --> 00:01:13.330
and A is there in the end.

00:01:13.330 --> 00:01:16.040
But A is not a reactant or a product,

00:01:16.040 --> 00:01:19.363
therefore A must be a catalyst.

00:01:20.600 --> 00:01:24.260
Something else that's not a
reactant or a product is AC.

00:01:24.260 --> 00:01:26.150
You notice how AC was generated

00:01:26.150 --> 00:01:27.740
in the first step of our mechanism,

00:01:27.740 --> 00:01:31.630
and then AC is used up in the
second step of the mechanism.

00:01:31.630 --> 00:01:36.630
Therefore AC must be the
intermediate for this reaction.

00:01:41.860 --> 00:01:43.810
Next, let's look at the energy profile

00:01:43.810 --> 00:01:45.720
for this multi-step reaction.

00:01:45.720 --> 00:01:48.820
Energy profiles usually have
potential energy in the y-axis

00:01:48.820 --> 00:01:51.640
and then reaction progress on the x-axis.

00:01:51.640 --> 00:01:55.270
So as we move to the right on the x-axis,

00:01:55.270 --> 00:01:56.963
the reaction is occurring.

00:01:58.310 --> 00:02:00.750
This first line on our energy profile

00:02:00.750 --> 00:02:03.750
represents the energy
level of our reactants,

00:02:03.750 --> 00:02:06.330
which are BC and D.

00:02:06.330 --> 00:02:10.890
So let's go ahead and show
the bond between B and C.

00:02:11.750 --> 00:02:14.980
And then we also have D present.

00:02:14.980 --> 00:02:17.140
Our catalyst is also present

00:02:17.140 --> 00:02:19.210
at the very beginning of our reactions.

00:02:19.210 --> 00:02:23.883
So I'll go ahead and draw in
A above our two reactants.

00:02:26.408 --> 00:02:29.870
We can see in our energy
profile that we have two hills.

00:02:29.870 --> 00:02:32.440
The first Hill corresponds
to the first step

00:02:32.440 --> 00:02:34.190
of the mechanism and the second hill

00:02:34.190 --> 00:02:37.090
corresponds to the second step.

00:02:37.090 --> 00:02:40.950
So the peak of the first
hill is the transition state

00:02:40.950 --> 00:02:44.250
for the first step of the mechanism.

00:02:44.250 --> 00:02:47.330
And we can see in the first
step that the catalyst A,

00:02:47.330 --> 00:02:50.510
is colliding with BC or reacting with BC

00:02:50.510 --> 00:02:53.490
to form our intermediate AC.

00:02:53.490 --> 00:02:58.490
So A must collide with BC
and at the transition state,

00:02:59.030 --> 00:03:02.370
the bond between B and C is breaking,

00:03:02.370 --> 00:03:07.370
and at the same time, the bond
between A and C is forming.

00:03:10.380 --> 00:03:12.760
We would still have reactant D present

00:03:12.760 --> 00:03:14.330
at the top of this hill too.

00:03:14.330 --> 00:03:17.410
So I'll go ahead and draw in D here.

00:03:17.410 --> 00:03:22.410
When a collides with BC,
the collision has to have

00:03:22.660 --> 00:03:27.660
enough kinetic energy to
overcome the activation energy

00:03:28.280 --> 00:03:31.270
necessary for this reaction to occur.

00:03:31.270 --> 00:03:33.530
And on this energy profile,

00:03:33.530 --> 00:03:36.520
the activation energy is
the difference in energy

00:03:36.520 --> 00:03:41.520
between the reactants
and the transition state,

00:03:42.440 --> 00:03:44.000
so the very peak of the hill.

00:03:44.000 --> 00:03:46.900
So this difference in energy,

00:03:46.900 --> 00:03:49.310
corresponds to the activation energy

00:03:49.310 --> 00:03:53.403
for the first step of the
mechanism which we will call Ea1.

00:03:54.890 --> 00:03:57.760
If we assume that the collision
has enough kinetic energy

00:03:57.760 --> 00:04:00.260
to overcome the activation energy,

00:04:00.260 --> 00:04:04.610
we'll form our intermediate
AC, and we'd also form B.

00:04:04.610 --> 00:04:06.650
So let's go ahead and show the bond

00:04:06.650 --> 00:04:09.960
between A and C has now been formed.

00:04:09.960 --> 00:04:12.380
So this valley here between our two hills

00:04:12.380 --> 00:04:16.330
represents the energy
level of the intermediate.

00:04:16.330 --> 00:04:18.120
We would also have be present,

00:04:18.120 --> 00:04:20.990
so I can go ahead and
I'll just write in B here.

00:04:20.990 --> 00:04:23.130
And then we still have some D present,

00:04:23.130 --> 00:04:25.290
D still hasn't reacted yet.

00:04:25.290 --> 00:04:27.803
So I'll go ahead and draw in D as well.

00:04:30.440 --> 00:04:31.970
Next we're ready for the second hill

00:04:31.970 --> 00:04:34.360
or the second step of our mechanism.

00:04:34.360 --> 00:04:39.060
In the second step, AC the
intermediate AC reacts with D

00:04:39.060 --> 00:04:41.840
to form A and CD.

00:04:41.840 --> 00:04:45.160
So the top of this second hill
would be the transition state

00:04:45.160 --> 00:04:47.240
for this second step.

00:04:47.240 --> 00:04:51.970
So we can show the bond
between A and C braking,

00:04:51.970 --> 00:04:56.970
and at the same time the bond
between C and D is forming.

00:04:58.980 --> 00:05:02.120
The difference in energy
between the energy

00:05:02.120 --> 00:05:05.540
of the intermediate and the
energy of the transition state

00:05:05.540 --> 00:05:07.660
represents the activation energy

00:05:07.660 --> 00:05:09.900
for the second step of the mechanism,

00:05:09.900 --> 00:05:12.007
which we will call Ea2.

00:05:14.650 --> 00:05:19.390
So AC and D must collide
with enough kinetic energy

00:05:19.390 --> 00:05:24.390
to overcome the activation
energy for this second step.

00:05:24.450 --> 00:05:28.100
If AC and D collide with
enough kinetic energy,

00:05:28.100 --> 00:05:31.760
we would produce A and CD.

00:05:31.760 --> 00:05:35.130
So this line at the end here
represents the energy level

00:05:35.130 --> 00:05:37.910
of our products.

00:05:37.910 --> 00:05:42.910
So CD is one of our products,
so we'll write that in here.

00:05:43.530 --> 00:05:46.600
And remember B is our other products,

00:05:46.600 --> 00:05:51.330
which we formed from the
first step of the mechanisms.

00:05:51.330 --> 00:05:55.640
So let's go ahead and
write in here B plus CD.

00:05:55.640 --> 00:05:58.180
And we also reformed our catalyst,

00:05:58.180 --> 00:06:01.540
so A would be present here as well.

00:06:01.540 --> 00:06:04.910
Next let's compare the
first activation energy Ea1

00:06:04.910 --> 00:06:08.040
with the second activation energy Ea2.

00:06:08.040 --> 00:06:10.490
Looking at the energy
profile we can see that Ea1

00:06:11.741 --> 00:06:14.587
has a much greater
activation energy than Ea2.

00:06:15.550 --> 00:06:20.550
So let's go ahead and write
Ea1 is greater than Ea2.

00:06:20.890 --> 00:06:24.490
The smaller the activation
energy, the faster the reaction,

00:06:24.490 --> 00:06:26.910
and since there's a
smaller activation energy

00:06:26.910 --> 00:06:28.440
for the second step,

00:06:28.440 --> 00:06:32.320
the second step must be
the faster of the two.

00:06:32.320 --> 00:06:35.440
Since the first step has the
higher activation energy,

00:06:35.440 --> 00:06:40.440
the first step must be slow
compared to the second step.

00:06:42.150 --> 00:06:45.070
Since the first step of the
mechanism is the slow step,

00:06:45.070 --> 00:06:48.083
the first step is the
rate determining step.

00:06:48.960 --> 00:06:51.620
Finally, let's Find the
overall change in energy

00:06:51.620 --> 00:06:54.020
for our reaction.

00:06:54.020 --> 00:06:57.320
So to find the overall change
in energy, that's Delta E,

00:06:57.320 --> 00:06:59.640
which is final minus initial.

00:06:59.640 --> 00:07:02.220
So that would be the
energy of the products

00:07:02.220 --> 00:07:05.870
minus the energy of the reactants.

00:07:05.870 --> 00:07:09.320
So the energy level of
the products is right here

00:07:09.320 --> 00:07:11.100
and then the energy level of the reactants

00:07:11.100 --> 00:07:11.933
is at the beginning.

00:07:11.933 --> 00:07:14.210
So let me just extend
this dashed line here

00:07:14.210 --> 00:07:16.570
so we can better compare the two.

00:07:16.570 --> 00:07:18.540
Representing Delta E on a graph,

00:07:18.540 --> 00:07:20.300
it would be the difference in energy

00:07:20.300 --> 00:07:22.760
between these two lines.

00:07:22.760 --> 00:07:24.750
And since the energy of the products

00:07:24.750 --> 00:07:27.650
is greater than the
energy of the reactants,

00:07:27.650 --> 00:07:29.930
we would be subtracting a smaller number

00:07:29.930 --> 00:07:31.210
from a larger number

00:07:31.210 --> 00:07:34.360
and therefore Delta E would be positive

00:07:34.360 --> 00:07:37.010
for this hypothetical reaction.

00:07:37.010 --> 00:07:39.290
And since Delta E is positive,

00:07:39.290 --> 00:07:43.563
we know that this reaction
is an endothermic reaction.


Collision theory and the Maxwell–Boltzmann distribution

WEBVTT
Kind: captions
Language: en

00:00:01.400 --> 00:00:02.233
- [Instructor] Collision theory

00:00:02.233 --> 00:00:05.130
can be related to
Maxwell-Boltzmann distributions.

00:00:05.130 --> 00:00:07.590
And first we'll start
with collision theory.

00:00:07.590 --> 00:00:09.970
Collision theory says that
particles must collide

00:00:09.970 --> 00:00:13.860
in the proper orientation and
with enough kinetic energy

00:00:13.860 --> 00:00:17.370
to overcome the activation energy barrier.

00:00:17.370 --> 00:00:21.130
So let's look at the reaction
where A reacts with B and C

00:00:21.130 --> 00:00:24.420
to form AB plus C.

00:00:24.420 --> 00:00:28.240
On an energy profile, we
have the reactants over here

00:00:28.240 --> 00:00:29.580
in the left.

00:00:29.580 --> 00:00:33.660
So A, atom A is colored red,

00:00:33.660 --> 00:00:36.590
and we have molecule BC over here,

00:00:36.590 --> 00:00:39.790
So these two particles must collide

00:00:39.790 --> 00:00:43.150
for the reaction to occur,

00:00:43.150 --> 00:00:45.400
and they must collide with enough energy

00:00:45.400 --> 00:00:48.070
to overcome the activation energy barrier.

00:00:48.070 --> 00:00:51.210
So the activation energy
on an energy profile

00:00:51.210 --> 00:00:52.250
is the difference in energy

00:00:52.250 --> 00:00:56.010
between the peak here, which
is the transition state

00:00:56.010 --> 00:00:57.610
and the energy of the reactants.

00:00:57.610 --> 00:01:01.810
So this energy here is
our activation energy.

00:01:01.810 --> 00:01:04.030
The minimum amount of energy necessary

00:01:04.030 --> 00:01:06.770
for the reaction to occur.

00:01:06.770 --> 00:01:09.830
So if these particles
collide with enough energy,

00:01:09.830 --> 00:01:13.920
we can just get over this
activation energy barrier

00:01:13.920 --> 00:01:17.653
and the reactions can turn
into our two products.

00:01:20.020 --> 00:01:22.090
If our reactant particles
don't hit each other

00:01:22.090 --> 00:01:25.340
with enough energy, they
simply bounce off of each other

00:01:25.340 --> 00:01:27.070
and our reaction never occurs.

00:01:27.070 --> 00:01:30.820
We never overcome this
activation energy barrier.

00:01:30.820 --> 00:01:33.420
As an analogy, let's think
about hitting a golf ball.

00:01:33.420 --> 00:01:35.310
So let's imagine we have a hill,

00:01:35.310 --> 00:01:37.150
and on the right side of the hill,

00:01:37.150 --> 00:01:39.050
somewhere is the hole down here,

00:01:39.050 --> 00:01:42.700
and the left side of the
hill is our golf ball.

00:01:42.700 --> 00:01:45.720
So we know we have to hit this
golf ball with enough force

00:01:45.720 --> 00:01:47.920
to give it enough kinetic energy

00:01:47.920 --> 00:01:49.970
for it to reach the top of the hill

00:01:49.970 --> 00:01:53.050
and to roll over the hill
and go into the hole.

00:01:53.050 --> 00:01:55.320
So we can imagine this hill

00:01:55.320 --> 00:01:58.580
as being a hill of potential energy.

00:01:58.580 --> 00:02:02.080
And this golf ball needs to
have enough kinetic energy

00:02:02.080 --> 00:02:06.353
to turn into enough potential
energy to go over the hill.

00:02:09.390 --> 00:02:11.230
If we don't hit our golf ball hard enough,

00:02:11.230 --> 00:02:13.730
it might not have enough
energy to go over the hill.

00:02:13.730 --> 00:02:16.160
So if we hit it softly,
it might just roll halfway

00:02:16.160 --> 00:02:18.470
up the hill and roll back down again.

00:02:18.470 --> 00:02:23.470
Kinetic energy is equal to 1/2 MV squared.

00:02:24.770 --> 00:02:26.630
And so M would be the
mass of the golf ball

00:02:26.630 --> 00:02:28.790
and V would be the velocity.

00:02:28.790 --> 00:02:30.350
So we have to hit it with enough force

00:02:30.350 --> 00:02:33.770
so it has enough as a high enough velocity

00:02:33.770 --> 00:02:35.190
to have a high enough kinetic energy

00:02:35.190 --> 00:02:37.083
to get over the hill.

00:02:38.740 --> 00:02:40.160
Let's apply collision theory

00:02:40.160 --> 00:02:42.810
to a Maxwell-Boltzmann distribution.

00:02:42.810 --> 00:02:45.370
Usually a Maxwell-Boltzmann distribution

00:02:45.370 --> 00:02:48.400
has fractional particles or
relative numbers of particles

00:02:48.400 --> 00:02:52.790
on the y-axis and particle
speed on the x-axis.

00:02:52.790 --> 00:02:55.160
And a Maxwell-Boltzmann distribution

00:02:55.160 --> 00:03:00.160
shows us the range of speeds
available to the particles

00:03:01.380 --> 00:03:02.840
in a sample of gas.

00:03:02.840 --> 00:03:04.160
So let's say we have,

00:03:04.160 --> 00:03:06.420
here's a particulate diagram over here.

00:03:06.420 --> 00:03:07.820
Let's say we have a sample of gas

00:03:07.820 --> 00:03:10.290
at a particular temperature T.

00:03:10.290 --> 00:03:12.980
These particles aren't
traveling at the same speed,

00:03:12.980 --> 00:03:15.790
there's a range of
speeds available to them.

00:03:15.790 --> 00:03:19.820
So one particle might be
traveling really slowly

00:03:19.820 --> 00:03:22.090
so we'll draw a very short arrow here.

00:03:22.090 --> 00:03:24.450
A few more might be
traveling a little faster,

00:03:24.450 --> 00:03:28.190
so we'll draw the arrow longer
to indicate a faster speed.

00:03:28.190 --> 00:03:31.310
And maybe one particle
is traveling the fastest.

00:03:31.310 --> 00:03:34.293
So we'll give this
particle the longest arrow.

00:03:36.130 --> 00:03:38.400
We can think about the
area under the curve

00:03:38.400 --> 00:03:40.310
for a Maxwell-Boltzmann distribution

00:03:40.310 --> 00:03:43.850
as representing all of the
particles in our sample.

00:03:43.850 --> 00:03:47.920
So we had this one particle
here moving very slowly,

00:03:47.920 --> 00:03:50.450
and so if we look at
our curve and we think

00:03:50.450 --> 00:03:52.120
about the area under the curve

00:03:52.120 --> 00:03:54.073
that's at a low particle speed,

00:03:54.920 --> 00:03:57.410
this area is smaller than
other parts of the curve.

00:03:57.410 --> 00:03:59.640
So that's represented here
by only this one particle

00:03:59.640 --> 00:04:01.490
moving very slowly.

00:04:01.490 --> 00:04:03.620
We think about this
next part of the curve,

00:04:03.620 --> 00:04:06.610
most, this is a large
amount of area in here

00:04:06.610 --> 00:04:09.640
and these particles are
traveling at a higher speed.

00:04:09.640 --> 00:04:14.640
So maybe these three
particles here would represent

00:04:15.550 --> 00:04:17.870
the particles moving at a higher speed.

00:04:17.870 --> 00:04:21.670
And then finally, we had
this one particle here,

00:04:21.670 --> 00:04:23.610
We drew this arrow longer than the others.

00:04:23.610 --> 00:04:26.440
So this particle's traveling
faster than the other one.

00:04:26.440 --> 00:04:29.750
So maybe this area under the curve up here

00:04:29.750 --> 00:04:31.853
is represented by that one particle.

00:04:33.520 --> 00:04:34.890
We know from collision theory,

00:04:34.890 --> 00:04:37.760
that particles have to
have enough kinetic energy

00:04:37.760 --> 00:04:42.760
to overcome the activation
energy for a reaction to occur.

00:04:42.850 --> 00:04:46.610
So we can draw a line
representing the activation energy

00:04:46.610 --> 00:04:48.840
on a Maxwell-Boltzmann distribution.

00:04:48.840 --> 00:04:52.722
So if I draw this line,
this dotted line right here,

00:04:52.722 --> 00:04:57.060
this represents my activation energy.

00:04:57.060 --> 00:04:59.600
And instead of particle
speed, you could think

00:04:59.600 --> 00:05:02.400
about the x-axis as being
kinetic energy if you want.

00:05:02.400 --> 00:05:05.480
So the faster a particle is traveling,

00:05:05.480 --> 00:05:07.690
the higher its kinetic energy.

00:05:07.690 --> 00:05:11.390
And so the area under the curve

00:05:11.390 --> 00:05:13.010
to the right of this dash line,

00:05:13.010 --> 00:05:14.640
this represents all of the particles

00:05:14.640 --> 00:05:19.640
that have enough kinetic energy
for this reaction to occur.

00:05:21.820 --> 00:05:24.300
Next, let's think about what
happens to the particles

00:05:24.300 --> 00:05:27.770
in our sample when we
increase the temperature.

00:05:27.770 --> 00:05:29.240
So when we increase the temperature,

00:05:29.240 --> 00:05:32.350
the Maxwell-Boltzmann
distribution changes.

00:05:32.350 --> 00:05:36.030
So what happens is the peak height drops

00:05:36.030 --> 00:05:40.360
and our Maxwell-Boltzmann
distribution curve gets broader.

00:05:40.360 --> 00:05:43.373
So it looks something like
this at a higher temperature.

00:05:45.380 --> 00:05:47.210
So we still have some particles traveling

00:05:47.210 --> 00:05:48.820
at relatively low speeds, right?

00:05:48.820 --> 00:05:50.440
Remember it's the area under the curve.

00:05:50.440 --> 00:05:53.583
So maybe that's represented
by this one particle here,

00:05:54.440 --> 00:05:56.520
and next, let's think about the area

00:05:56.520 --> 00:06:00.180
to the left of this dash line for Ea.

00:06:00.180 --> 00:06:02.750
So we want to make these
particles green here

00:06:02.750 --> 00:06:05.180
as we have some particles traveling

00:06:05.180 --> 00:06:06.510
a little bit of faster speeds.

00:06:06.510 --> 00:06:09.250
So let me go ahead and draw
these arrows a little bit longer

00:06:09.250 --> 00:06:12.330
but notice what happens to
the right of this dash line.

00:06:12.330 --> 00:06:14.880
We think about the area under the curve

00:06:14.880 --> 00:06:17.960
for the magenta curve.

00:06:17.960 --> 00:06:21.350
Notice how the area is bigger
than in the previous example.

00:06:21.350 --> 00:06:24.450
So maybe this time we have
these two particles here

00:06:24.450 --> 00:06:25.900
traveling at a faster speed.

00:06:25.900 --> 00:06:28.150
So I'm gonna draw these arrows longer

00:06:28.150 --> 00:06:30.270
to indicate they're
traveling at a faster speed.

00:06:30.270 --> 00:06:33.780
And since they're to the
right of this dash line here,

00:06:33.780 --> 00:06:36.690
both of these particles
have enough kinetic energy

00:06:36.690 --> 00:06:40.840
to overcome the activation
energy for our reaction.

00:06:40.840 --> 00:06:44.530
So we can see when you
increase the temperature,

00:06:44.530 --> 00:06:46.730
you increase the number of particles

00:06:46.730 --> 00:06:48.580
that have enough kinetic energy

00:06:48.580 --> 00:06:51.323
to overcome the activation energy.

00:06:52.840 --> 00:06:54.090
It's important to point out

00:06:54.090 --> 00:06:56.770
that since the number of
particles hasn't changed,

00:06:56.770 --> 00:06:59.480
all we've done is increase
the temperature here,

00:06:59.480 --> 00:07:02.360
the area under the curve remains the same.

00:07:02.360 --> 00:07:06.420
So the area under the curve
for the curve in yellow,

00:07:06.420 --> 00:07:08.880
is the same as the area under the curve

00:07:08.880 --> 00:07:12.010
for the one drawn in magenta.

00:07:12.010 --> 00:07:14.640
The difference of course
is the one in magenta

00:07:14.640 --> 00:07:15.950
is at a higher temperature,

00:07:15.950 --> 00:07:18.370
and therefore there are more
particles with enough energy

00:07:18.370 --> 00:07:20.790
to overcome the activation energy.

00:07:20.790 --> 00:07:23.030
So increasing the temperature

00:07:23.030 --> 00:07:25.783
increases the rate of reaction.


The pre-equilibrium approximation

WEBVTT
Kind: captions
Language: en

00:00:00.320 --> 00:00:02.370
- [Instructor] The
Pre-equilibrium Approximation

00:00:02.370 --> 00:00:03.770
is used to find the rate law

00:00:03.770 --> 00:00:06.740
for a mechanism with a fast initial step.

00:00:06.740 --> 00:00:08.940
As an example, let's look
at the reaction between

00:00:08.940 --> 00:00:11.383
nitric oxide and bromine.

00:00:12.770 --> 00:00:14.310
In the first step of the mechanism,

00:00:14.310 --> 00:00:17.323
nitric oxide combines with
bromine to form NOBr2.

00:00:18.663 --> 00:00:20.560
And in the second step of the mechanism,

00:00:20.560 --> 00:00:25.323
NOBr2 reacts with NO to
form our product 2NOBr.

00:00:28.570 --> 00:00:30.530
NOBr2 is generated

00:00:30.530 --> 00:00:33.400
from the first elementary
step of the mechanism.

00:00:33.400 --> 00:00:36.960
And then NOBr2 is used
up in the second step.

00:00:36.960 --> 00:00:39.790
Since NOBr2 wasn't there in the beginning

00:00:39.790 --> 00:00:41.270
and it's not there in the end,

00:00:41.270 --> 00:00:44.973
we call NOBr2 an intermediate.

00:00:50.310 --> 00:00:53.100
The first step of the mechanism is fast.

00:00:53.100 --> 00:00:55.360
And the second step of
the mechanism is slow.

00:00:55.360 --> 00:00:57.830
Since the second step of
the mechanism is slow,

00:00:57.830 --> 00:01:00.640
this is the rate determining step.

00:01:00.640 --> 00:01:04.420
And we can write the rate
law for the overall reaction

00:01:04.420 --> 00:01:07.540
by writing the rate law for
this elementary reaction

00:01:07.540 --> 00:01:10.750
that makes up step two of our mechanism.

00:01:10.750 --> 00:01:14.760
So we can write the rate
of reaction is equal to

00:01:14.760 --> 00:01:19.220
for step two, our rate constant is K2.

00:01:19.220 --> 00:01:21.940
And we multiply K2, the rate constant,

00:01:21.940 --> 00:01:24.460
by the concentration of our two reactants,

00:01:24.460 --> 00:01:27.253
which would be the concentration of NOBr2,

00:01:29.320 --> 00:01:32.323
and the concentration of NO.

00:01:33.370 --> 00:01:36.300
Since the coefficients
in our balanced equation

00:01:36.300 --> 00:01:40.400
are ones for NOBr2 and
one and a one for NO,

00:01:40.400 --> 00:01:41.790
we can take the coefficients

00:01:41.790 --> 00:01:44.860
and turn them into
exponents in our rate law.

00:01:44.860 --> 00:01:48.763
So we can do this because this
is an elementary reaction.

00:01:50.490 --> 00:01:52.320
However, we can't leave the rate law

00:01:52.320 --> 00:01:53.730
for the overall reaction,

00:01:53.730 --> 00:01:57.427
in terms of the concentration
of our intermediate, NOBr2.

00:01:58.360 --> 00:02:00.430
It's preferable to have rate laws

00:02:00.430 --> 00:02:03.230
written in terms of the
concentration of our reactants,

00:02:03.230 --> 00:02:04.447
which were NO and Br2.

00:02:07.430 --> 00:02:09.550
So we need some way of substituting N

00:02:09.550 --> 00:02:11.923
for the concentration of NOBr2.

00:02:13.832 --> 00:02:15.860
And we can do that by assuming that

00:02:15.860 --> 00:02:19.110
the first elementary step in our mechanism

00:02:19.110 --> 00:02:22.430
comes to a fast equilibrium.

00:02:22.430 --> 00:02:27.430
So if we assume the first step
comes to a fast equilibrium,

00:02:27.470 --> 00:02:31.283
we can use the
Pre-equilibrium Approximation.

00:02:32.370 --> 00:02:34.070
If we assume that the first step

00:02:34.070 --> 00:02:39.070
comes to a fast equilibrium,
or a pre equilibrium,

00:02:39.080 --> 00:02:40.090
we know at equilibrium,

00:02:40.090 --> 00:02:42.340
the rate of the forward reaction

00:02:42.340 --> 00:02:45.630
is equal to the rate of
the reverse reaction.

00:02:45.630 --> 00:02:48.390
So in the forward reaction for step one,

00:02:48.390 --> 00:02:50.953
NO combines with Br2 to form NOBr2.

00:02:52.350 --> 00:02:54.216
And in the reverse reaction,

00:02:54.216 --> 00:02:59.216
NOBr2 breaks apart to form NO and Br2.

00:02:59.310 --> 00:03:03.750
So if the rate of the forward reaction

00:03:03.750 --> 00:03:08.750
is equal to the rate of the
reverse reaction at equilibrium,

00:03:11.180 --> 00:03:13.770
let's go ahead and write the rate laws

00:03:13.770 --> 00:03:16.453
for the forward and the reverse reaction.

00:03:17.610 --> 00:03:21.250
The rate constant for the
forward reaction is K1.

00:03:21.250 --> 00:03:22.340
So we can go ahead and write

00:03:22.340 --> 00:03:25.810
the rate of the forward
reaction is equal to K1.

00:03:25.810 --> 00:03:29.730
And our two reactants are NO and Br2.

00:03:29.730 --> 00:03:33.110
So we have K1 times the
concentration of NO,

00:03:33.110 --> 00:03:36.260
times the concentration of Br2.

00:03:36.260 --> 00:03:38.640
Since the coefficients
in our balanced equation

00:03:38.640 --> 00:03:40.670
are both ones for these two reactants,

00:03:40.670 --> 00:03:43.040
we can raise the power of
these two concentrations

00:03:43.040 --> 00:03:44.070
to the first power.

00:03:44.070 --> 00:03:47.620
Since this is an elementary
reaction, we can do this.

00:03:47.620 --> 00:03:50.210
And we set this rate
of the forward reaction

00:03:50.210 --> 00:03:52.710
equal to the rate of the reverse reaction.

00:03:52.710 --> 00:03:56.600
The reverse reaction has a
rate constant of K minus one,

00:03:56.600 --> 00:04:00.750
and we have only NOBr2 with
a coefficient of one in it.

00:04:00.750 --> 00:04:02.360
So we multiply K minus one

00:04:02.360 --> 00:04:07.360
times the concentration of
NOBr2, to the first power.

00:04:09.150 --> 00:04:11.440
Next, our goal is to substitute N

00:04:11.440 --> 00:04:14.780
for the concentration of our intermediate.

00:04:14.780 --> 00:04:17.380
And so we can divide both sides

00:04:17.380 --> 00:04:20.480
of the equation by K minus one.

00:04:20.480 --> 00:04:22.330
So if we decide if we divide

00:04:22.330 --> 00:04:24.980
both sides of the equation by K minus one,

00:04:24.980 --> 00:04:27.460
on the right side, K
minus one cancels out.

00:04:27.460 --> 00:04:32.300
And we get that the concentration
of our intermediate NOBr2,

00:04:32.300 --> 00:04:33.780
is equal to K one

00:04:33.780 --> 00:04:36.870
times the concentration
of NO to the first power,

00:04:36.870 --> 00:04:39.520
times the concentration
of Br2 to the first power,

00:04:39.520 --> 00:04:42.410
divided by K minus one.

00:04:42.410 --> 00:04:44.730
Next, we can substitute all of this in

00:04:44.730 --> 00:04:48.083
for the concentration of our intermediate.

00:04:49.240 --> 00:04:52.930
That gives us the rate
of reaction is equal to,

00:04:52.930 --> 00:04:54.670
we still have this K2 in here,

00:04:54.670 --> 00:04:57.180
so we need to make sure to include it.

00:04:57.180 --> 00:04:59.890
And we're gonna substitute everything in

00:04:59.890 --> 00:05:02.410
all of this in for the
concentration of our intermediate.

00:05:02.410 --> 00:05:05.773
So that would be times K one,

00:05:07.000 --> 00:05:10.440
times the concentration
of NO to the first power,

00:05:10.440 --> 00:05:14.680
times the concentration
of Br2 to the first power,

00:05:14.680 --> 00:05:17.350
divided by K minus one.

00:05:17.350 --> 00:05:18.720
And then we still have this

00:05:18.720 --> 00:05:20.930
concentration of NO to the first power.

00:05:20.930 --> 00:05:25.310
So we have to make sure to
include that in our rate law.

00:05:25.310 --> 00:05:27.810
Let's think about what we would get

00:05:27.810 --> 00:05:30.650
if we multiply two constants together

00:05:30.650 --> 00:05:32.410
and then divide by a third constant.

00:05:32.410 --> 00:05:35.150
So multiplying K2 times K1,

00:05:35.150 --> 00:05:37.120
and then we divide by K minus one,

00:05:37.120 --> 00:05:39.310
that would just give us another constant,

00:05:39.310 --> 00:05:42.040
which we could just call K.

00:05:42.040 --> 00:05:46.350
So K is now the rate constant
for the overall reaction.

00:05:46.350 --> 00:05:48.990
So we have the rate law
for the overall reaction

00:05:48.990 --> 00:05:50.210
is equal to K,

00:05:50.210 --> 00:05:52.160
times the concentration of

00:05:52.160 --> 00:05:56.030
we have NO to the first power
times NO to the first power,

00:05:56.030 --> 00:06:00.470
which is just the concentration
of NO to the second power.

00:06:00.470 --> 00:06:02.310
And we still have to include

00:06:02.310 --> 00:06:06.130
the concentration of
bromine to the first power.

00:06:06.130 --> 00:06:10.600
So now we have a rate law
for our overall reaction

00:06:10.600 --> 00:06:13.640
in terms of the concentration
of our two reactants.

00:06:13.640 --> 00:06:16.500
The rate of reaction is
equal to the rate constant K,

00:06:16.500 --> 00:06:18.730
times the concentration of NO squared,

00:06:18.730 --> 00:06:21.553
times the concentration of
bromine to the first power.

00:06:22.980 --> 00:06:25.190
The experimentally determined rate law

00:06:25.190 --> 00:06:27.350
matches the rate law that we found

00:06:27.350 --> 00:06:31.157
using the Pre-equilibrium Approximation.

00:06:31.157 --> 00:06:33.190
And if you look at the coefficients

00:06:33.190 --> 00:06:34.880
for the overall equation,

00:06:34.880 --> 00:06:37.630
there's a two in front of NO
and a one in front of Br2,

00:06:38.480 --> 00:06:39.867
it might be tempting just to say,

00:06:39.867 --> 00:06:41.520
"Can't we just take those coefficients

00:06:41.520 --> 00:06:44.450
and turn them into exponents,
because in this case,

00:06:44.450 --> 00:06:47.640
they happen to match the
exponent in our rate law?"

00:06:47.640 --> 00:06:50.170
That's just a coincidence
for this reaction.

00:06:50.170 --> 00:06:53.060
We can't just take the coefficients
for an overall equation

00:06:53.060 --> 00:06:55.140
and turn them into
exponents in the rate law.

00:06:55.140 --> 00:06:57.750
We can only do that for
elementary reactions.

00:06:57.750 --> 00:06:59.350
Like the elementary reactions

00:06:59.350 --> 00:07:02.043
in the two steps of our mechanism.

00:07:02.960 --> 00:07:03.980
It's important to point out

00:07:03.980 --> 00:07:06.160
that if the rate of the forward reaction

00:07:06.160 --> 00:07:08.320
is equal to the rate of
the reverse reaction,

00:07:08.320 --> 00:07:10.461
the concentration of our intermediate,

00:07:10.461 --> 00:07:12.920
NOBr2 remains constant.

00:07:12.920 --> 00:07:17.920
And therefore we can use this
Pre-equilibrium Approximation

00:07:18.290 --> 00:07:20.580
to find the rate law for a reaction

00:07:20.580 --> 00:07:22.713
with a fast initial step.


Kinetics of radioactive decay

WEBVTT
Kind: captions
Language: en

00:00:01.240 --> 00:00:03.920
- [Instructor] Strontium-90
is a radioactive isotope

00:00:03.920 --> 00:00:05.743
that undergoes beta decay.

00:00:06.960 --> 00:00:10.230
Because radioactive decay
is a first-order process,

00:00:10.230 --> 00:00:13.773
radioactive isotopes
have constant half-lives.

00:00:15.450 --> 00:00:18.940
Half-life is symbolized by t1/2,

00:00:18.940 --> 00:00:22.580
and it's the time required
for 1/2 of a sample

00:00:22.580 --> 00:00:26.640
of a particular radioactive
isotope to decay.

00:00:26.640 --> 00:00:29.570
For example, the half-life of Strontium-90

00:00:29.570 --> 00:00:34.513
is equal to 28.8 years.

00:00:38.090 --> 00:00:40.250
Let's say we start with 10 grams

00:00:40.250 --> 00:00:45.250
of our Strontium-90 radioactive isotope.

00:00:45.270 --> 00:00:46.830
And on the y-axis,

00:00:46.830 --> 00:00:50.220
we're gonna put the mass of
our Strontium isotope in grams.

00:00:50.220 --> 00:00:53.260
And on the x-axis, we're
going to have time.

00:00:53.260 --> 00:00:56.230
So when time is equal to zero,

00:00:56.230 --> 00:00:59.483
we have 10 grams of our isotope.

00:01:00.810 --> 00:01:05.300
Since the half-life of
Strontium-90 is 28.8 years,

00:01:05.300 --> 00:01:09.130
if we wait 28.8 years,

00:01:09.130 --> 00:01:12.820
we'll go from 10 grams to five grams.

00:01:12.820 --> 00:01:17.820
So the next point on our
graph will be at five grams,

00:01:18.750 --> 00:01:23.260
and this time should be 28.8 years.

00:01:27.530 --> 00:01:30.720
If we wait another 28.8 years,

00:01:30.720 --> 00:01:33.300
we're gonna go from five
grams to half of that,

00:01:33.300 --> 00:01:35.350
which would be 2.5 grams.

00:01:35.350 --> 00:01:39.020
So the next point would be here at 2.5.

00:01:44.410 --> 00:01:47.230
If we wait another 28.8 years,

00:01:47.230 --> 00:01:50.393
we'd go from 2.5 grams down to 1.25,

00:01:52.350 --> 00:01:56.113
so approximately here on our graph.

00:01:57.180 --> 00:02:01.713
So this graph shows exponential decay.

00:02:03.130 --> 00:02:05.380
Let's say we were asked to find out

00:02:05.380 --> 00:02:08.300
how much of our radioactive
isotope Strontium is left

00:02:08.300 --> 00:02:11.940
after 115.2 years.

00:02:11.940 --> 00:02:15.790
So what we would do is take 115.2 years

00:02:15.790 --> 00:02:20.790
and divide that by the
half-life of 28.8 years.

00:02:21.960 --> 00:02:22.920
And by doing that,

00:02:22.920 --> 00:02:27.920
we realize that 115.2 is
really just four half-lives.

00:02:28.740 --> 00:02:30.270
So one approach with this problem

00:02:30.270 --> 00:02:32.360
would be starting with our 10 grams.

00:02:32.360 --> 00:02:36.030
And we think about one half-life
taking us to five grams,

00:02:36.030 --> 00:02:38.970
another half-life taking us to 2.5,

00:02:38.970 --> 00:02:41.250
another half-life taking us to 1.25,

00:02:42.520 --> 00:02:44.480
and then finally, one more half-life

00:02:44.480 --> 00:02:49.260
that takes us to 0.625 grams.

00:02:49.260 --> 00:02:51.330
So just doing the problem this way,

00:02:51.330 --> 00:02:55.760
we can see that's one, two,
three, four half-lives.

00:02:55.760 --> 00:02:56.920
So our final answer

00:02:56.920 --> 00:03:01.920
is 0.625 grams remain after 115.2 years.

00:03:03.770 --> 00:03:06.180
Another approach to do the same problem

00:03:06.180 --> 00:03:08.120
would be to start with our 10 grams,

00:03:08.120 --> 00:03:10.180
and we multiply that by 1/2

00:03:10.180 --> 00:03:13.230
to get the amount remaining
after one half-life.

00:03:13.230 --> 00:03:15.750
And we could do that three more times

00:03:15.750 --> 00:03:20.200
to get the amount that
remains after four half-lives.

00:03:20.200 --> 00:03:21.480
We could have also have written this

00:03:21.480 --> 00:03:26.480
as 10 times 1/2 to the fourth power

00:03:26.480 --> 00:03:29.500
since we needed to wait four half-lives.

00:03:29.500 --> 00:03:33.810
All these approaches will get
you the answer of 0.625 grams

00:03:33.810 --> 00:03:38.810
of our radioactive isotope
remaining after 115.2 years.

00:03:40.910 --> 00:03:44.300
For a chemical reaction with
reactant A that's first-order,

00:03:44.300 --> 00:03:46.710
the rate law says that
the rate of reaction

00:03:46.710 --> 00:03:48.540
is equal to the rate constant k

00:03:48.540 --> 00:03:52.623
times the concentration
of A to the first power.

00:03:53.580 --> 00:03:56.130
Since radioactive decay
is a first-order process,

00:03:56.130 --> 00:03:58.090
we can write that the rate of decay

00:03:58.090 --> 00:04:01.760
is equal to the rate constant
k times N to the first power,

00:04:01.760 --> 00:04:06.760
where N is the number of
radioactive nuclei in a sample.

00:04:07.420 --> 00:04:10.390
Since radioactive decay
is a first-order process,

00:04:10.390 --> 00:04:14.380
we can also use this equation
for the rate constant,

00:04:14.380 --> 00:04:16.610
which comes from first-order kinetics,

00:04:16.610 --> 00:04:20.320
which says that the rate
constant k is equal to 0.693

00:04:20.320 --> 00:04:22.283
divided by the half-life.

00:04:23.390 --> 00:04:26.250
For example, if we wanted
to find the rate constant

00:04:26.250 --> 00:04:29.840
for the radioactive decay of Strontium-90,

00:04:29.840 --> 00:04:34.840
the rate constant would be equal to 0.693

00:04:34.840 --> 00:04:38.060
divided by the half-life of Strontium-90,

00:04:38.060 --> 00:04:42.353
which we saw was 28.8 years.

00:04:43.510 --> 00:04:45.370
So when we do that math,

00:04:45.370 --> 00:04:49.597
we find that k is equal to 0.0241

00:04:51.200 --> 00:04:52.963
one over years.

00:04:54.020 --> 00:04:56.420
Another equation from first-order kinetics

00:04:56.420 --> 00:05:00.550
is the integrated rate law
for a first-order reaction.

00:05:00.550 --> 00:05:03.200
And the integrated rate law
says that the natural log

00:05:03.200 --> 00:05:06.400
of the concentration of
reactant A at some time t

00:05:06.400 --> 00:05:09.770
is equal to negative
kt plus the natural log

00:05:09.770 --> 00:05:13.860
of the initial
concentration of reactant A.

00:05:13.860 --> 00:05:15.020
Since we're using N,

00:05:15.020 --> 00:05:18.200
which is the number of
radioactive nuclei in our sample,

00:05:18.200 --> 00:05:20.700
instead of the concentration of A,

00:05:20.700 --> 00:05:22.570
we can write the integrated rate law

00:05:22.570 --> 00:05:26.060
for our first-order
radioactive decay process,

00:05:26.060 --> 00:05:27.800
which says the natural log

00:05:27.800 --> 00:05:31.180
of the number of radioactive
nuclei at some time t

00:05:31.180 --> 00:05:34.300
is equal to negative
kt plus the natural log

00:05:34.300 --> 00:05:37.043
of the initial number
of radioactive nuclei.

00:05:38.080 --> 00:05:41.523
So let's say we start with 1.000 grams

00:05:44.170 --> 00:05:47.590
of our Strontium-90 radioactive isotope,

00:05:47.590 --> 00:05:49.610
and our goal is to find out

00:05:49.610 --> 00:05:53.380
how much remains after two years.

00:05:53.380 --> 00:05:55.220
So we're gonna wait two years

00:05:55.220 --> 00:05:59.150
and find out how much of our
radioactive isotope remains.

00:05:59.150 --> 00:06:02.060
So we're going to use our equation

00:06:02.060 --> 00:06:04.850
for the integrated rate law.

00:06:04.850 --> 00:06:08.073
Let's go ahead and plug in what we know.

00:06:09.200 --> 00:06:10.510
We already know what k is,

00:06:10.510 --> 00:06:12.500
we found that in our earlier problem,

00:06:12.500 --> 00:06:16.980
so we can write this is
equal to negative times k,

00:06:16.980 --> 00:06:19.080
which is 0.0241.

00:06:22.340 --> 00:06:25.700
Next, we also know the time
period we're interested in.

00:06:25.700 --> 00:06:28.580
So we know what t is, t is two years,

00:06:28.580 --> 00:06:33.580
so let's go ahead and write
2.00 in here for the time.

00:06:34.520 --> 00:06:37.580
Next, we're gonna add the natural log

00:06:37.580 --> 00:06:42.380
of the initial number of
radioactive nuclei in our sample.

00:06:42.380 --> 00:06:44.550
And while we don't have
the initial number,

00:06:44.550 --> 00:06:45.990
we do have the mass.

00:06:45.990 --> 00:06:48.020
And since the mass is proportional

00:06:48.020 --> 00:06:50.180
to the number of radioactive nuclei,

00:06:50.180 --> 00:06:53.020
it's okay to go ahead and
plug that into our equation.

00:06:53.020 --> 00:06:56.763
So we're gonna plug in
the natural log of one.

00:06:58.750 --> 00:07:01.730
And all of this is
equal to the natural log

00:07:01.730 --> 00:07:05.300
of the initial number of
radioactive nuclei at some time t.

00:07:05.300 --> 00:07:08.253
So we have the natural log of that.

00:07:10.370 --> 00:07:13.720
The natural log of one is zero,

00:07:13.720 --> 00:07:17.740
so now we have the natural log of N

00:07:17.740 --> 00:07:19.450
is equal to,

00:07:19.450 --> 00:07:24.313
and when we do this math, we're
gonna get negative 0.0482.

00:07:26.290 --> 00:07:29.170
Next, we have to get
rid of the natural log.

00:07:29.170 --> 00:07:31.920
And we can do that by
taking e to both sides.

00:07:31.920 --> 00:07:36.050
So if we take e to both sides,
the natural log cancels out,

00:07:36.050 --> 00:07:39.650
and we get that N is equal

00:07:39.650 --> 00:07:44.650
to 0.953 grams.

00:07:45.650 --> 00:07:50.240
So that's how much of our
radioactive isotope remains

00:07:50.240 --> 00:07:51.973
after two years.


Zero-order reactions

WEBVTT
Kind: captions
Language: en

00:00:00.125 --> 00:00:02.970
- [Instructor] Let's say we
have a hypothetical reaction

00:00:02.970 --> 00:00:06.440
where reactant A turns into products.

00:00:06.440 --> 00:00:10.970
And let's say the reaction is
zero order with respect to A.

00:00:10.970 --> 00:00:13.660
If it's zero order with
respect to A, we can write

00:00:13.660 --> 00:00:15.970
that the rate of the reaction is equal to

00:00:15.970 --> 00:00:19.440
the rate constant k, times
the concentration of A

00:00:19.440 --> 00:00:22.010
to the zero power.

00:00:22.010 --> 00:00:26.420
And since any number to the
zero power is equal to one,

00:00:26.420 --> 00:00:27.810
then the rate of the reaction

00:00:27.810 --> 00:00:31.413
would just be equal to
the rate constant k.

00:00:32.390 --> 00:00:34.580
We can also write that
the rate of the reaction

00:00:34.580 --> 00:00:36.760
is equal to the negative and the change

00:00:36.760 --> 00:00:40.143
in the concentration of A,
over the change in time.

00:00:41.290 --> 00:00:44.200
If we set these two ways of
writing the rate of reaction

00:00:44.200 --> 00:00:47.100
equal to each other, and
we use some calculus,

00:00:47.100 --> 00:00:49.990
including the concept of
integration, we will arrive

00:00:49.990 --> 00:00:53.920
at the integrated rate law
for a zero-order reaction,

00:00:53.920 --> 00:00:56.980
which says that the concentration
of A at time t is equal

00:00:56.980 --> 00:01:00.640
to the negative of the rate
constant k times the time,

00:01:00.640 --> 00:01:02.783
plus the initial concentration of A.

00:01:04.380 --> 00:01:05.980
Notice that the integrated rate law

00:01:05.980 --> 00:01:10.840
is in the form of Y is equal to mx plus b,

00:01:10.840 --> 00:01:13.680
which is the equation for a straight line.

00:01:13.680 --> 00:01:18.630
So if we graph the concentration
of A on the Y axis,

00:01:18.630 --> 00:01:21.830
and the time on the X axis,

00:01:21.830 --> 00:01:26.330
we will get a straight line
if the reaction is zero order.

00:01:26.330 --> 00:01:29.830
So if we write the concentration
of A on the Y axis,

00:01:29.830 --> 00:01:33.070
and time on the X axis,

00:01:33.070 --> 00:01:35.850
the graph will be a straight line,

00:01:35.850 --> 00:01:40.160
and the slope of that line
is equal to the negative

00:01:40.160 --> 00:01:42.970
of the rate constant k,

00:01:42.970 --> 00:01:46.703
so the slope is equal to -k,

00:01:48.130 --> 00:01:51.070
and the Y intercept of that line,

00:01:51.070 --> 00:01:56.070
so right where the line
intersects with the y-axis,

00:01:57.690 --> 00:02:01.900
this point is the initial
concentration of A.

00:02:01.900 --> 00:02:04.630
So everything we've talked
about assumes that there's a

00:02:04.630 --> 00:02:08.803
coefficient of 1 in front
of the concentration of A.

00:02:10.160 --> 00:02:12.890
However, let's say we
have a coefficient of 2

00:02:12.890 --> 00:02:15.620
in front of A in our balanced equation.

00:02:15.620 --> 00:02:20.620
That means we need a
stoichiometric coefficient of 1/2,

00:02:20.902 --> 00:02:23.152
which changes the math.

00:02:23.152 --> 00:02:26.985
Now, instead of getting
-kt, we would get -2kt

00:02:28.730 --> 00:02:33.400
after we integrate, which means
that the slope of the line,

00:02:33.400 --> 00:02:37.560
when we graph the
concentration of A versus time,

00:02:37.560 --> 00:02:39.760
the slope of the line
would be equal to -2k.

00:02:42.630 --> 00:02:45.020
It's important to note
that textbooks often just

00:02:45.020 --> 00:02:48.040
assume the coefficient
in front of A as a 1,

00:02:48.040 --> 00:02:51.700
which would give the slope as equal to -k.

00:02:51.700 --> 00:02:54.740
However, if the coefficient
in front of A is a 2,

00:02:54.740 --> 00:02:57.100
then technically the slope of the line

00:02:57.100 --> 00:02:59.670
should be equal to -2k

00:03:01.073 --> 00:03:03.510
As an example of a zero-order
reaction, let's look at

00:03:03.510 --> 00:03:08.380
the decomposition of ammonia
on a hot platinum surface

00:03:08.380 --> 00:03:11.970
to form nitrogen gas and hydrogen gas.

00:03:11.970 --> 00:03:16.210
In our diagram, we have
four ammonia molecules

00:03:16.210 --> 00:03:19.020
on the surface of our platinum catalyst,

00:03:19.020 --> 00:03:21.320
and then we have another four

00:03:21.320 --> 00:03:24.850
that are above the
surface of the catalyst.

00:03:24.850 --> 00:03:27.500
Only the ammonia molecules on
the surface of the catalyst

00:03:27.500 --> 00:03:30.770
can react and turn into
nitrogen and hydrogen,

00:03:30.770 --> 00:03:34.180
the ammonia molecules above
the surface can't react.

00:03:34.180 --> 00:03:38.010
And even if we were to add in
some more ammonia molecules,

00:03:38.010 --> 00:03:40.430
so let's just add in some more here,

00:03:40.430 --> 00:03:43.360
those molecules still can't
react, and therefore the rate

00:03:43.360 --> 00:03:46.730
of the reaction doesn't
change as we increase

00:03:46.730 --> 00:03:49.670
the concentration of ammonia.

00:03:49.670 --> 00:03:52.900
So we can write that the
rate of the reaction is equal

00:03:52.900 --> 00:03:57.900
to the rate constant k times
the concentration of ammonia,

00:03:58.280 --> 00:04:01.510
but since increasing the
concentration of ammonia

00:04:01.510 --> 00:04:03.760
has no effect on rate,

00:04:03.760 --> 00:04:07.520
that's why this is
raised to the zero power.

00:04:07.520 --> 00:04:09.480
And therefore we get
the rate of the reaction

00:04:09.480 --> 00:04:12.003
is just equal to the rate constant k.

00:04:12.960 --> 00:04:16.010
Normally, increasing the
concentration of a reactant

00:04:16.010 --> 00:04:18.770
increases the rate of the reaction.

00:04:18.770 --> 00:04:21.280
However, for this reaction,
since we're limited

00:04:21.280 --> 00:04:24.000
by the surface area of the catalyst,

00:04:24.000 --> 00:04:26.550
if the catalyst is covered
with ammonia molecules,

00:04:26.550 --> 00:04:29.070
increasing the concentration
of ammonia molecules

00:04:29.070 --> 00:04:32.850
will have no effect on
the rate of the reaction.

00:04:32.850 --> 00:04:34.800
And therefore this reaction,

00:04:34.800 --> 00:04:38.120
the decomposition of ammonia
on a hot platinum surface,

00:04:38.120 --> 00:04:41.263
is an example of a zero-order reaction.


Second-order reactions

WEBVTT
Kind: captions
Language: en

00:00:00.710 --> 00:00:01.543
- [Instructor] Let's say we have

00:00:01.543 --> 00:00:02.880
a hypothetical reaction

00:00:02.880 --> 00:00:06.120
where reactant A turns into products.

00:00:06.120 --> 00:00:08.080
And let's say the reaction is second order

00:00:08.080 --> 00:00:09.970
with respect to A.

00:00:09.970 --> 00:00:12.140
If the reaction is second
order with respect to A,

00:00:12.140 --> 00:00:14.840
then we can write the rate of the reaction

00:00:14.840 --> 00:00:18.230
is equal to the rate constant
k times the concentration

00:00:18.230 --> 00:00:20.460
of A to the second power

00:00:20.460 --> 00:00:23.293
since this is a second order reaction.

00:00:24.550 --> 00:00:26.960
We can also write that
the rate of the reaction

00:00:26.960 --> 00:00:29.160
is equal to the negative of the change

00:00:29.160 --> 00:00:32.543
in the concentration of A
over the change in time.

00:00:33.900 --> 00:00:36.220
If we set these two
ways of writing the rate

00:00:36.220 --> 00:00:38.210
of reaction equal to each other,

00:00:38.210 --> 00:00:40.030
and use of calculus,

00:00:40.030 --> 00:00:42.580
including the concept of integration,

00:00:42.580 --> 00:00:45.290
we will arrive at the integrated rate law

00:00:45.290 --> 00:00:47.553
for a second order reaction.

00:00:48.750 --> 00:00:51.730
The integrated rate law
for a second order reaction

00:00:51.730 --> 00:00:54.060
says that one over the concentration

00:00:54.060 --> 00:00:56.590
of reactant A at some time t,

00:00:56.590 --> 00:00:59.240
is equal to the rate
constant k times the time

00:00:59.240 --> 00:01:04.123
plus one over the initial
concentration of A.

00:01:05.270 --> 00:01:06.790
Notice how the integrated rate law

00:01:06.790 --> 00:01:10.907
has the form of y is equal to mx plus b,

00:01:12.360 --> 00:01:15.360
which is the equation for a straight line.

00:01:15.360 --> 00:01:18.990
So if we graph one over
the concentration of A

00:01:18.990 --> 00:01:23.530
on the y axis, and time on the x axis,

00:01:23.530 --> 00:01:25.160
so let's go ahead and put that in here,

00:01:25.160 --> 00:01:29.040
one over the concentration
of A on the y axis,

00:01:29.040 --> 00:01:33.670
and time on the x axis, we
will get a straight line

00:01:33.670 --> 00:01:35.910
and the slope of that line

00:01:35.910 --> 00:01:39.760
is equal to the rate constant k.

00:01:39.760 --> 00:01:44.760
So slope is equal to K and the y intercept

00:01:46.000 --> 00:01:50.210
is equal to one over the
initial concentration of A.

00:01:50.210 --> 00:01:54.520
So the point where our
line intersects the y axis

00:01:54.520 --> 00:01:59.500
is equal to one over the
initial concentration of A.

00:02:00.460 --> 00:02:04.000
Let's look at an example
of a second order reaction.

00:02:04.000 --> 00:02:09.000
C5H6 is cyclopentadiene, and two molecules

00:02:09.570 --> 00:02:12.960
of cyclopentadiene will
react with each other

00:02:12.960 --> 00:02:14.913
to form dicyclopentadiene.

00:02:16.120 --> 00:02:19.200
Our goal is to use the
data from this data table

00:02:19.200 --> 00:02:22.083
to prove that this
reaction is second order.

00:02:23.150 --> 00:02:24.530
However, we have to be careful

00:02:24.530 --> 00:02:26.140
because in our balanced equation,

00:02:26.140 --> 00:02:28.870
we have a two as a coefficient in front

00:02:28.870 --> 00:02:29.893
of cyclopentadiene.

00:02:31.660 --> 00:02:34.100
Going back to our hypothetical reaction

00:02:34.100 --> 00:02:36.730
where reactant A turned into products,

00:02:36.730 --> 00:02:40.020
there's also one as a
coefficient in front of the A.

00:02:40.020 --> 00:02:42.450
And if there's a one as a
coefficient in front of the A,

00:02:42.450 --> 00:02:45.250
we can use this form of
the integrated rate law

00:02:45.250 --> 00:02:47.370
for a second order reaction.

00:02:47.370 --> 00:02:50.110
However, for our problem, we have a two

00:02:50.110 --> 00:02:54.180
as a coefficient in
front of cyclopentadiene.

00:02:54.180 --> 00:02:55.740
And that means we need to have

00:02:55.740 --> 00:03:00.000
a stoichiometric
coefficient of 1/2 in here,

00:03:00.000 --> 00:03:01.590
and that changes the math.

00:03:01.590 --> 00:03:05.140
Now when we set these two
rates of reactions equal

00:03:05.140 --> 00:03:08.990
to each other, and we use
calculus and we integrate

00:03:08.990 --> 00:03:10.770
to get our integrated rate law,

00:03:10.770 --> 00:03:13.100
because of this one half,

00:03:13.100 --> 00:03:17.500
we end up with a two in front of the K.

00:03:17.500 --> 00:03:22.500
So thinking about y is equal to mx plus b,

00:03:23.320 --> 00:03:28.103
now the slope of the
line is equal to two K.

00:03:29.490 --> 00:03:30.940
So now for our reaction,

00:03:30.940 --> 00:03:32.660
we can write our integrated rate law

00:03:32.660 --> 00:03:36.240
as one over the concentration
of cyclopentadiene

00:03:36.240 --> 00:03:40.470
at some time t, is
equal to two kt plus one

00:03:40.470 --> 00:03:43.913
over the initial concentration
of cyclopentadiene.

00:03:45.340 --> 00:03:46.720
So if we look at our data table,

00:03:46.720 --> 00:03:50.750
we have time in seconds and
we have the concentration

00:03:50.750 --> 00:03:54.450
of cyclopentadiene,
but we need to have one

00:03:54.450 --> 00:03:57.430
over the concentration of cyclopentadiene

00:03:57.430 --> 00:03:59.900
so we need a new column.

00:03:59.900 --> 00:04:04.420
So we're going to calculate
one over the concentration

00:04:04.420 --> 00:04:06.513
of cyclopentadiene.

00:04:08.900 --> 00:04:11.790
So if the concentration of cyclopentadiene

00:04:11.790 --> 00:04:16.790
when time is equal to zero
seconds is 0.0400 molar,

00:04:16.930 --> 00:04:19.980
if we take one divided by 0.0400,

00:04:23.260 --> 00:04:24.830
we would get 25.0.

00:04:29.810 --> 00:04:31.450
To save some time I've gone ahead

00:04:31.450 --> 00:04:34.200
and filled in the rest of this column.

00:04:34.200 --> 00:04:37.480
Notice as time increases so
as we go from zero seconds

00:04:37.480 --> 00:04:40.610
to 50 to 100 to 150 to 200,

00:04:40.610 --> 00:04:43.640
the concentration of
cyclopentadiene decreases

00:04:43.640 --> 00:04:47.983
because cyclopentadiene is
turning into dicyclopentadiene.

00:04:49.290 --> 00:04:50.760
Next, we need to graph our data.

00:04:50.760 --> 00:04:52.840
So we're gonna have one
over the concentration

00:04:52.840 --> 00:04:55.110
of cyclopentadiene on the y axis

00:04:55.110 --> 00:04:57.890
and time on the x axis.

00:04:57.890 --> 00:05:00.840
Our first point, so when
time is equal to zero,

00:05:00.840 --> 00:05:04.770
one over the concentration
of cyclopentadiene is 25.0.

00:05:04.770 --> 00:05:08.113
So if we go down to our graph,

00:05:09.880 --> 00:05:12.210
we can see that when
time is equal to zero,

00:05:12.210 --> 00:05:15.570
our first point here is 25.0.

00:05:15.570 --> 00:05:20.403
And plotting the other points
gives us a straight line.

00:05:21.860 --> 00:05:26.860
Next, we need to find the
slope of this straight line.

00:05:26.860 --> 00:05:28.730
And there are many ways to do that.

00:05:28.730 --> 00:05:32.440
One way to do it is to
use a graphing calculator.

00:05:32.440 --> 00:05:34.890
And when I used a graphing calculator

00:05:34.890 --> 00:05:36.670
to find the slope of this line,

00:05:36.670 --> 00:05:41.207
I found that the slope is equal to 0.1634.

00:05:46.350 --> 00:05:51.090
Thinking about y is equal to mx plus b,

00:05:52.090 --> 00:05:56.163
our slope should be equal to two k.

00:05:57.280 --> 00:05:59.140
So to find the rate constant k,

00:05:59.140 --> 00:06:03.230
we need to divide the slope by two,

00:06:03.230 --> 00:06:06.890
which gives us 0.0817.

00:06:12.480 --> 00:06:15.320
To find the units for
K, remember that slope

00:06:15.320 --> 00:06:20.320
is equal to change in y over change in x,

00:06:20.740 --> 00:06:24.000
and on our Y axis, our
units are one over molar,

00:06:24.000 --> 00:06:27.180
and the x axis the units are seconds.

00:06:27.180 --> 00:06:29.740
So therefore, we can
write the rate constant k

00:06:29.740 --> 00:06:32.443
is equal to 0.0817.

00:06:34.786 --> 00:06:37.270
It's to be one over
molar divided by seconds

00:06:37.270 --> 00:06:42.270
which is the same thing as
one over molar times seconds.

00:06:43.370 --> 00:06:45.630
It's important to point
out that most textbooks

00:06:45.630 --> 00:06:50.360
don't cover how the two
as a coefficient changes

00:06:50.360 --> 00:06:53.050
the integrated rate law.

00:06:53.050 --> 00:06:55.800
And so a lot of textbooks will simply say

00:06:55.800 --> 00:06:57.970
that the slope of the line

00:06:57.970 --> 00:07:00.970
for the second order integrated rate law,

00:07:00.970 --> 00:07:02.100
is equal to K.

00:07:02.100 --> 00:07:04.590
So a lot of books would
just say the final answer

00:07:04.590 --> 00:07:07.570
for the rate constant is 0.163.

00:07:07.570 --> 00:07:10.000
So you'll see you'll see
a lot of textbooks say

00:07:10.000 --> 00:07:13.290
that the rate constant would be 0.163,

00:07:13.290 --> 00:07:17.743
one over molar times seconds.

00:07:18.820 --> 00:07:20.800
However, since the coefficient

00:07:20.800 --> 00:07:23.570
in front of cyclopentadiene is a two,

00:07:23.570 --> 00:07:28.063
technically this rate
constant is the correct one.

00:07:29.390 --> 00:07:32.700
Finally, since we got a straight line,

00:07:32.700 --> 00:07:36.140
when we graphed one over the concentration

00:07:36.140 --> 00:07:38.830
of cyclopentadiene versus time,

00:07:38.830 --> 00:07:42.403
we know that this reaction
has second order kinetics.


First-order reactions

WEBVTT
Kind: captions
Language: en

00:00:00.750 --> 00:00:03.010
- [Instructor] Let's say we
have a hypothetical reaction

00:00:03.010 --> 00:00:06.210
where reactant A turns into products

00:00:06.210 --> 00:00:10.423
and that the reaction is
first-order with respect to A.

00:00:11.500 --> 00:00:15.300
If the reaction is first-order
with respect to reactant A,

00:00:15.300 --> 00:00:17.850
for the rate law we can write
the rate of the reaction

00:00:17.850 --> 00:00:19.650
is equal to the rate constant K

00:00:19.650 --> 00:00:23.410
times the concentration
of A to the first power.

00:00:23.410 --> 00:00:25.700
We can also write that
the rate of the reaction

00:00:25.700 --> 00:00:29.240
is equal to the negative of
the change in the concentration

00:00:29.240 --> 00:00:31.463
of A over the change in time.

00:00:32.530 --> 00:00:35.200
By setting both of these
equal to each other,

00:00:35.200 --> 00:00:36.900
and by doing some calculus,

00:00:36.900 --> 00:00:39.230
including the concept of integration,

00:00:39.230 --> 00:00:41.200
we arrive at the integrated rate law

00:00:41.200 --> 00:00:43.200
for a first-order reaction,

00:00:43.200 --> 00:00:46.020
which says that the natural
log of the concentration of A

00:00:46.020 --> 00:00:49.780
at some time T, is equal to negative KT,

00:00:49.780 --> 00:00:51.530
where K is the rate constant

00:00:51.530 --> 00:00:55.130
plus the natural log of the
initial concentration of A.

00:00:56.890 --> 00:00:58.450
Notice how the integrated rate law

00:00:58.450 --> 00:01:03.450
has the form of Y is equal to mx plus b,

00:01:03.630 --> 00:01:07.180
which is the equation for a straight line.

00:01:07.180 --> 00:01:10.980
So if we were to graph the
natural log of the concentration

00:01:10.980 --> 00:01:15.040
of A on the Y axis, so let's
go ahead and put that in here,

00:01:15.040 --> 00:01:19.130
the natural log of the concentration of A,

00:01:19.130 --> 00:01:23.660
and on the X axis we put the time,

00:01:23.660 --> 00:01:26.330
we would get a straight line

00:01:26.330 --> 00:01:28.900
and the slope of that straight line

00:01:28.900 --> 00:01:31.680
would be equal to negative K.

00:01:31.680 --> 00:01:33.593
So the slope of this line,

00:01:34.520 --> 00:01:36.670
the slope would be equal to the negative

00:01:36.670 --> 00:01:39.170
of the rate constant K,

00:01:39.170 --> 00:01:43.200
and the Y intercept would
be equal to the natural log

00:01:43.200 --> 00:01:45.110
of the initial concentration of A.

00:01:45.110 --> 00:01:49.150
So right where this line meets the Y axis,

00:01:49.150 --> 00:01:51.560
that point is equal to the natural log

00:01:51.560 --> 00:01:54.740
of the initial concentration of A.

00:01:56.250 --> 00:01:59.970
The conversion of methyl
isonitrile to acetonitrile

00:01:59.970 --> 00:02:02.150
is a first-order reaction.

00:02:02.150 --> 00:02:05.340
And these two molecules
are isomers of each other.

00:02:05.340 --> 00:02:08.220
Let's use the data that's
provided to us in this data table

00:02:08.220 --> 00:02:12.680
to show that this conversion
is a first-order reaction.

00:02:12.680 --> 00:02:15.720
Since the coefficient in
front of methyl isonitrile

00:02:15.720 --> 00:02:19.870
is a one, we can use this form
of the integrated rate law

00:02:19.870 --> 00:02:21.870
where the slope is equal to the negative

00:02:21.870 --> 00:02:23.563
of the rate constant K.

00:02:24.820 --> 00:02:28.100
If our balanced equation
had a two as a coefficient

00:02:28.100 --> 00:02:31.130
in front of our reactant, we
would have had to include 1/2

00:02:32.070 --> 00:02:35.570
as a stoichiometric coefficient.

00:02:35.570 --> 00:02:38.540
And when we set our two
rates equal to each other now

00:02:38.540 --> 00:02:42.610
and go through the calculus,
instead of getting negative KT,

00:02:42.610 --> 00:02:45.513
we have gotten negative two KT.

00:02:46.400 --> 00:02:51.100
However for our reaction we
don't have a coefficient of two.

00:02:51.100 --> 00:02:54.340
We have a coefficient of one and therefore

00:02:54.340 --> 00:02:58.113
we can use this form of
the integrated rate law.

00:02:59.660 --> 00:03:02.070
Also notice that this form
of the integrated rate law

00:03:02.070 --> 00:03:05.760
is in terms of the concentration of A

00:03:05.760 --> 00:03:08.930
but we don't have the
concentration of methyl isonitrile

00:03:08.930 --> 00:03:10.090
in our data table,

00:03:10.090 --> 00:03:13.350
we have the pressure of methyl isonitrile.

00:03:13.350 --> 00:03:16.080
But pressure is related to concentration

00:03:16.080 --> 00:03:21.080
from the ideal gas law,
so PV is equal to nRT.

00:03:21.360 --> 00:03:25.470
If we divide both sides by V,

00:03:25.470 --> 00:03:28.630
then we can see that pressure is equal to,

00:03:28.630 --> 00:03:30.730
n is moles and V is volumes,

00:03:30.730 --> 00:03:34.020
so moles divided by
volume would be molarity,

00:03:34.020 --> 00:03:38.210
so molarity times R times T.

00:03:38.210 --> 00:03:41.340
And therefore pressure
is directly proportional

00:03:41.340 --> 00:03:45.520
to concentration, and for a
gas it's easier to measure

00:03:45.520 --> 00:03:48.390
the pressure than to
get the concentration.

00:03:48.390 --> 00:03:50.310
And so you'll often see data

00:03:50.310 --> 00:03:53.670
for gases in terms of the pressure.

00:03:53.670 --> 00:03:55.780
Therefore, we can imagine this form of the

00:03:55.780 --> 00:04:00.290
integrated rate law as the
natural log of the pressure

00:04:00.290 --> 00:04:03.530
of our gas at time T
is equal to negative KT

00:04:03.530 --> 00:04:08.160
plus the natural log of the
initial pressure of the gas.

00:04:08.160 --> 00:04:10.350
Therefore, to show that this reaction

00:04:10.350 --> 00:04:14.060
is a first-order reaction we
need to graph the natural log

00:04:14.060 --> 00:04:18.380
of the pressure of methyl
isonitrile on the Y axis

00:04:18.380 --> 00:04:20.960
and time on the X axis.

00:04:20.960 --> 00:04:24.283
So we need a new column in our data table.

00:04:25.184 --> 00:04:27.950
We need to put in the natural log

00:04:27.950 --> 00:04:32.950
of the pressure of methyl isonitrile.

00:04:33.690 --> 00:04:36.460
So for example, when time is equal to zero

00:04:36.460 --> 00:04:41.460
the pressure of methyl
isonitrile is 502 torrs.

00:04:41.670 --> 00:04:45.350
So we need to take the natural log of 502.

00:04:45.350 --> 00:04:49.313
And the natural log of
502 is equal to 6.219.

00:04:55.480 --> 00:04:56.730
To save time, I've gone ahead

00:04:56.730 --> 00:04:58.600
and filled in this last column here,

00:04:58.600 --> 00:05:01.840
the natural log of the
pressure methyl isonitrile.

00:05:01.840 --> 00:05:04.790
Notice what happens as
time increases, right,

00:05:04.790 --> 00:05:09.230
as time increases the
pressure of methyl isonitrile

00:05:09.230 --> 00:05:13.900
decreases since it's being
turned into acetonitrile.

00:05:13.900 --> 00:05:16.690
So for our graph, we're
gonna have the natural log

00:05:16.690 --> 00:05:20.370
of the pressure of methyl
isonitrile on the y-axis.

00:05:20.370 --> 00:05:23.930
And we're gonna have time on the X axis.

00:05:23.930 --> 00:05:25.540
So notice our first point here

00:05:25.540 --> 00:05:28.460
when time is equal to zero seconds,

00:05:28.460 --> 00:05:33.460
the natural log of the
pressure as equal to 6.219.

00:05:33.800 --> 00:05:37.050
So let's go down and
let's look at the graph.

00:05:37.050 --> 00:05:39.560
All right, so I've
already graphed it here.

00:05:39.560 --> 00:05:43.490
And we just saw when time
is equal to zero seconds,

00:05:43.490 --> 00:05:47.560
the first point is equal to 6.219.

00:05:47.560 --> 00:05:52.560
And here I have the other data
points already on the graph.

00:05:52.810 --> 00:05:55.820
Here's the integrated rate
law for a first-order reaction

00:05:55.820 --> 00:06:00.720
and I put pressures in there
instead of concentrations.

00:06:00.720 --> 00:06:03.890
And so we have the natural
log of the pressure

00:06:03.890 --> 00:06:06.850
of methyl isonitrile on the y-axis

00:06:06.850 --> 00:06:10.300
and we have time on the X axis,

00:06:10.300 --> 00:06:12.700
and the slope of this line should be equal

00:06:12.700 --> 00:06:15.950
to the negative of the rate constant K.

00:06:15.950 --> 00:06:18.870
So there are many ways to
find the slope of this line,

00:06:18.870 --> 00:06:22.900
one way would be to use
a graphing calculator.

00:06:22.900 --> 00:06:25.210
So I used a graphing calculator

00:06:25.210 --> 00:06:28.030
and I put in the data from the data table

00:06:28.030 --> 00:06:30.700
and I found that the slope of this line

00:06:30.700 --> 00:06:33.850
is equal to negative 2.08

00:06:37.230 --> 00:06:42.130
times 10 to the negative fourth.

00:06:42.130 --> 00:06:46.360
And since if I go ahead and
write y is equal to mx plus b,

00:06:46.360 --> 00:06:49.900
I need to remember to take
the negative of that slope

00:06:49.900 --> 00:06:52.430
to find the rate constant K.

00:06:52.430 --> 00:06:56.350
Therefore K is equal to positive 2.08

00:06:58.230 --> 00:07:02.150
times 10 to the negative fourth.

00:07:02.150 --> 00:07:03.800
To get the units for the rate constant,

00:07:03.800 --> 00:07:06.040
we can remember that slope is equal to

00:07:06.040 --> 00:07:10.460
change in Y over change in X.

00:07:10.460 --> 00:07:14.100
So change in Y would be the
natural log of the pressure,

00:07:14.100 --> 00:07:19.070
which has no unit, and X
the unit is in seconds.

00:07:19.070 --> 00:07:24.070
So we would have one over
seconds for the units for K.

00:07:24.950 --> 00:07:27.640
And finally, since we got a straight line

00:07:27.640 --> 00:07:32.640
when we graphed the natural log
of the pressure versus time,

00:07:33.130 --> 00:07:36.770
we know that this data is
for a first-order reaction.

00:07:36.770 --> 00:07:39.070
And therefore we've proved
that the transformation

00:07:39.070 --> 00:07:41.400
of methyl isonitrile to acetonitrile

00:07:41.400 --> 00:07:43.283
is a first-order reaction.