Using the reaction quotient

WEBVTT
Kind: captions
Language: en

00:00:00.250 --> 00:00:01.240
- [Instructor] The reaction quotient

00:00:01.240 --> 00:00:03.800
is symbolized by the capital letter Q.

00:00:03.800 --> 00:00:07.840
And it tells us whether a
reaction is at equilibrium or not.

00:00:07.840 --> 00:00:09.900
If the reaction is not at equilibrium,

00:00:09.900 --> 00:00:11.730
it also allows us to predict

00:00:11.730 --> 00:00:14.160
which direction the net reaction will go

00:00:14.160 --> 00:00:15.900
to reach equilibrium.

00:00:15.900 --> 00:00:17.770
For example, let's look
at the hypothetical

00:00:17.770 --> 00:00:21.700
reaction where A gas turns into gas B.

00:00:21.700 --> 00:00:25.200
Gas A will be represented by
red circles or red spheres,

00:00:25.200 --> 00:00:28.580
and gas B will be
represented by blue spheres.

00:00:28.580 --> 00:00:31.320
The equilibrium constant for
this hypothetical reaction

00:00:31.320 --> 00:00:35.280
is equal to 3 at 25 degrees Celsius.

00:00:35.280 --> 00:00:36.880
Let's start by writing out the expression

00:00:36.880 --> 00:00:38.400
for the reaction quotient.

00:00:38.400 --> 00:00:41.450
So we would write out here QC is equal to,

00:00:41.450 --> 00:00:42.840
and this has the same form

00:00:42.840 --> 00:00:45.600
as the equilibrium constant expression.

00:00:45.600 --> 00:00:49.460
So we would put concentration
of B to the first power,

00:00:49.460 --> 00:00:54.273
divided by the concentration
of A, also to the first power.

00:00:55.640 --> 00:00:58.080
Let's look at our first
particulate diagram here,

00:00:58.080 --> 00:00:59.600
and let's think about each particle

00:00:59.600 --> 00:01:02.720
representing 0.1 moles of a substance.

00:01:02.720 --> 00:01:06.960
And the volume of the
container is one liter.

00:01:06.960 --> 00:01:10.220
So let's first find the concentration of B

00:01:10.220 --> 00:01:12.980
so we can plug it into
our expression for Q.

00:01:12.980 --> 00:01:14.550
B are the blue spheres.

00:01:14.550 --> 00:01:18.370
So we count up one,
two, three blue spheres.

00:01:18.370 --> 00:01:21.260
Each sphere or each particle
represents 0.1 moles.

00:01:21.260 --> 00:01:24.340
So three times 0.1 is 0.3.

00:01:24.340 --> 00:01:26.390
And then we divide that by the volume

00:01:26.390 --> 00:01:30.380
of one liter to get a
concentration of 0.3 moles.

00:01:30.380 --> 00:01:33.370
So we can go ahead and plug in 0.3 mole

00:01:33.370 --> 00:01:35.800
for the concentration of B.

00:01:35.800 --> 00:01:38.660
Next, we divide that by
the concentration of A,

00:01:38.660 --> 00:01:41.950
and since there are five red particles,

00:01:41.950 --> 00:01:44.310
and each particle represents 0.1 moles,

00:01:44.310 --> 00:01:47.650
five times 0.1 is 0.5 moles of A

00:01:47.650 --> 00:01:49.490
divided by a volume of one liter,

00:01:49.490 --> 00:01:54.080
it gives a concentration
of 0.5 mole for A.

00:01:54.080 --> 00:01:55.610
Notice that we could have just counted

00:01:55.610 --> 00:01:58.820
the number of particles,
three blues and five reds,

00:01:58.820 --> 00:02:00.930
and just done three divided by five

00:02:00.930 --> 00:02:05.010
and get the same value for
the reaction quotient Q.

00:02:05.010 --> 00:02:08.500
So QC is equal to three fists or 0.6.

00:02:08.500 --> 00:02:11.860
And KC remember was equal to three.

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So Q is not equal to K.

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In this case, QC is less than KC.

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Since Q is not equal to K,
at this moment and time,

00:02:21.480 --> 00:02:24.420
the reaction is not at equilibrium.

00:02:24.420 --> 00:02:26.860
In order for this reaction
to reach equilibrium,

00:02:26.860 --> 00:02:29.360
Q needs to be equal to K.

00:02:29.360 --> 00:02:32.560
And since Q is a lot smaller than K,

00:02:32.560 --> 00:02:33.440
if you think about it,

00:02:33.440 --> 00:02:36.130
we need to increase the numerator

00:02:36.130 --> 00:02:38.620
and decrease the denominator.

00:02:38.620 --> 00:02:41.650
So that means, we have too many reactants

00:02:41.650 --> 00:02:43.220
and not enough products.

00:02:43.220 --> 00:02:46.970
And so the net reaction,
if go back up here

00:02:46.970 --> 00:02:48.480
to the equation here,

00:02:48.480 --> 00:02:51.490
the net reaction is going
to move to the right

00:02:51.490 --> 00:02:53.460
to make more products.

00:02:53.460 --> 00:02:55.810
So the net reaction moves to the right

00:02:55.810 --> 00:02:57.920
to make more blue particles

00:02:57.920 --> 00:03:00.650
and therefore the number of
red particles would decrease.

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We can see that comparing these

00:03:02.080 --> 00:03:04.020
first two particulate diagrams here.

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So let's compare the
first particular diagram

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to the second particular diagram.

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And the first particulate diagram,

00:03:10.190 --> 00:03:12.920
we had three blues and five reds.

00:03:12.920 --> 00:03:15.220
And the second particulate diagram,

00:03:15.220 --> 00:03:19.210
we have five blues and only three reds.

00:03:19.210 --> 00:03:21.460
So that shows the reaction
has moved to the right

00:03:21.460 --> 00:03:23.010
to increase the amount of products

00:03:23.010 --> 00:03:25.590
and to decrease the amount of reactants.

00:03:25.590 --> 00:03:28.490
Let's calculate QC at this moment and time

00:03:28.490 --> 00:03:30.310
for our second particular diagram

00:03:30.310 --> 00:03:32.860
to see if we've reached equilibrium yet.

00:03:32.860 --> 00:03:36.230
Well, we have five blue
particles and only three reds.

00:03:36.230 --> 00:03:40.890
So QC would be equal to five
over three or five thirds.

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Remember that the equilibrium constant KC,

00:03:43.980 --> 00:03:45.690
is equal to three.

00:03:45.690 --> 00:03:49.750
Therefore, QC is still not equal to KC.

00:03:49.750 --> 00:03:53.050
And therefore the reaction
is not at equilibrium.

00:03:53.050 --> 00:03:56.300
And Q is actually still less than K.

00:03:56.300 --> 00:03:57.850
Therefore the net reaction

00:03:57.850 --> 00:04:00.290
is going to move to the right again

00:04:00.290 --> 00:04:01.930
to increase the amount of products

00:04:01.930 --> 00:04:04.700
and to decrease the amount of reactants.

00:04:04.700 --> 00:04:06.990
Let's compare our second
particulate diagram

00:04:06.990 --> 00:04:09.110
to our third particulate diagram.

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And the second particulate diagram,

00:04:10.820 --> 00:04:12.750
we had five blues and three reds.

00:04:12.750 --> 00:04:14.150
And then the third one here,

00:04:14.150 --> 00:04:18.070
we have one, two, three,
four, five, six blues,

00:04:18.070 --> 00:04:19.460
and two reds.

00:04:19.460 --> 00:04:21.890
So we've increased in the amount of blue

00:04:21.890 --> 00:04:24.480
and we've decreased in the amount of red.

00:04:24.480 --> 00:04:27.510
Let's calculate QC for the moment of time

00:04:27.510 --> 00:04:30.450
represented by our third
particulate diagram.

00:04:30.450 --> 00:04:33.610
Well, there are six
blues and only two reds.

00:04:33.610 --> 00:04:36.080
So QC is equal to six divided by two,

00:04:36.080 --> 00:04:38.283
which is equal to three.

00:04:39.490 --> 00:04:41.840
So QC is equal to three

00:04:41.840 --> 00:04:44.570
and remember K is also equal to three.

00:04:44.570 --> 00:04:49.330
So QC is equal to KC and
therefore this reaction

00:04:49.330 --> 00:04:51.520
is now at equilibrium.

00:04:51.520 --> 00:04:52.750
And at equilibrium,

00:04:52.750 --> 00:04:54.520
the reactions turn into the products

00:04:54.520 --> 00:04:57.890
at the same rate the products
turn back into the reactants.

00:04:57.890 --> 00:05:00.420
And therefore the
concentration of both reactants

00:05:00.420 --> 00:05:03.580
and products remained
constant at equilibrium.

00:05:03.580 --> 00:05:05.300
So when Q is less than K,

00:05:05.300 --> 00:05:07.710
the reaction is not at equilibrium.

00:05:07.710 --> 00:05:10.620
There are too many reactants
and not enough products.

00:05:10.620 --> 00:05:13.230
Therefore the net
reaction goes to the right

00:05:13.230 --> 00:05:15.870
to increase the amount of products.

00:05:15.870 --> 00:05:18.120
The net reaction continues
to go to the right

00:05:18.120 --> 00:05:20.790
until Q is equal to K and the reaction

00:05:20.790 --> 00:05:22.600
has reached equilibrium.

00:05:22.600 --> 00:05:25.180
At that point, the concentrations
of reactants and products

00:05:25.180 --> 00:05:27.720
stop changing and they remain constant.

00:05:27.720 --> 00:05:31.140
It's also possible for
Q to be greater than K.

00:05:31.140 --> 00:05:34.400
And if that's true, the
reaction is not at equilibrium,

00:05:34.400 --> 00:05:36.590
but in this case you
have too many products

00:05:36.590 --> 00:05:38.450
and not enough reactants.

00:05:38.450 --> 00:05:41.610
Therefore the net
reaction goes to the left

00:05:41.610 --> 00:05:43.970
to increase the amount of reactants

00:05:43.970 --> 00:05:46.140
and to decrease the amount of products.

00:05:46.140 --> 00:05:48.540
The net reaction will
continue to go to the left

00:05:48.540 --> 00:05:53.170
until Q is equal to K and the
reaction reaches equilibrium.

00:05:53.170 --> 00:05:54.420
Let's look at another reaction,

00:05:54.420 --> 00:05:56.800
which is the decomposition of phosgene

00:05:56.800 --> 00:06:00.250
to form carbon monoxide and chlorine gas.

00:06:00.250 --> 00:06:03.300
KC for this reaction is equal to 2.2 times

00:06:03.300 --> 00:06:06.850
10 to the negative 10th
at 100 degrees Celsius.

00:06:06.850 --> 00:06:08.820
And let's say we're given concentrations

00:06:08.820 --> 00:06:11.730
of phosgene carbon monoxide and chlorine

00:06:11.730 --> 00:06:12.760
at a moment in time,

00:06:12.760 --> 00:06:16.050
and asked if the reaction
is at equilibrium or not.

00:06:16.050 --> 00:06:18.170
And if the reaction's not at equilibrium,

00:06:18.170 --> 00:06:21.440
we need to predict which
direction the net reaction will go

00:06:21.440 --> 00:06:23.470
to reach equilibrium.

00:06:23.470 --> 00:06:26.070
Our approach is gonna be to calculate QC

00:06:26.070 --> 00:06:30.410
at that moment in time,
and then compare QC to KC.

00:06:30.410 --> 00:06:34.180
So first we need to
write our QC expression,

00:06:34.180 --> 00:06:38.720
and this is equal to the
concentration of carbon monoxide

00:06:38.720 --> 00:06:40.140
raised to the first power

00:06:40.140 --> 00:06:43.030
times the concentration of chlorine

00:06:43.030 --> 00:06:45.070
raised to the first power

00:06:45.070 --> 00:06:49.070
and that's divided by the
concentration of phosgene.

00:06:49.070 --> 00:06:52.283
So the concentration of COCL2.

00:06:53.170 --> 00:06:54.310
So at this moment of time,

00:06:54.310 --> 00:06:57.210
the concentration of carbon
monoxide is 3.4 times

00:06:57.210 --> 00:06:58.950
10 to the negative six mole.

00:06:58.950 --> 00:07:01.470
The concentration of
chlorine is 6.0 times,

00:07:01.470 --> 00:07:03.150
times 10 to the negative six mole,

00:07:03.150 --> 00:07:04.590
and the concentration of phosgene

00:07:04.590 --> 00:07:08.160
is equal to 2.0 times 10
to the negative third mole.

00:07:08.160 --> 00:07:11.150
When we plug all those into
our Q expression and solve,

00:07:11.150 --> 00:07:13.570
we get that QC is equal to 1.0

00:07:13.570 --> 00:07:15.483
times 10 to the negative eighth.

00:07:16.530 --> 00:07:21.460
So in this case, QC is greater than KC

00:07:21.460 --> 00:07:24.790
because QC is equal to 1.0
times 10 to the negative eight

00:07:24.790 --> 00:07:28.610
and KC is equal to 2.2 times
10 to the negative 10th.

00:07:28.610 --> 00:07:30.950
And when QC is greater than KC,

00:07:30.950 --> 00:07:33.870
we have too many products
and not enough reactants.

00:07:33.870 --> 00:07:37.270
Therefore the net reaction
is going to go to the left

00:07:37.270 --> 00:07:39.150
and there's going to be an increase

00:07:39.150 --> 00:07:41.200
in the amount of phosgene.

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The reaction will
continue to go to the left

00:07:43.850 --> 00:07:48.850
until Q is equal to K, and the
reaction reaches equilibrium.


Worked example: Calculating equilibrium concentrations from initial concentrations

WEBVTT
Kind: captions
Language: en

00:00:00.700 --> 00:00:02.400
- [Instructor] For the
reaction bromine gas

00:00:02.400 --> 00:00:05.670
plus chlorine gas goes to BrCl,

00:00:05.670 --> 00:00:10.290
Kc is equal to 7.0 at 400 Kelvin.

00:00:10.290 --> 00:00:13.840
If the initial concentration
of bromine is 0.60 molar

00:00:13.840 --> 00:00:15.740
and the initial concentration of chlorine

00:00:15.740 --> 00:00:19.140
is also 0.60 molar, our
goal is to calculate

00:00:19.140 --> 00:00:24.140
the equilibrium concentrations
of Br2, Cl2 and BrCl.

00:00:25.970 --> 00:00:28.220
To help us find the
equilibrium concentrations,

00:00:28.220 --> 00:00:30.030
we're gonna use an ICE table,

00:00:30.030 --> 00:00:33.280
where I stands for the
initial concentration,

00:00:33.280 --> 00:00:35.740
C stands for the change in concentration

00:00:35.740 --> 00:00:39.750
and E stands for
equilibrium concentration.

00:00:39.750 --> 00:00:41.380
For the initial concentrations,

00:00:41.380 --> 00:00:43.423
we have 0.60 molar for bromine,

00:00:43.423 --> 00:00:46.700
0.60 molar for chlorine, and if we assume

00:00:46.700 --> 00:00:48.440
the reaction hasn't started yet,

00:00:48.440 --> 00:00:52.600
then we're gonna put a zero
in here for our product, BrCl.

00:00:52.600 --> 00:00:57.040
Next, we think about Br2
reacting with Cl2 to form BrCl.

00:00:58.200 --> 00:01:00.300
Some of the bromine is going to react,

00:01:00.300 --> 00:01:01.930
but we don't know how much,

00:01:01.930 --> 00:01:04.650
so we're gonna call that amount x,

00:01:04.650 --> 00:01:06.760
and we're gonna lose some of that bromine

00:01:06.760 --> 00:01:10.340
when we form our product,
so we're gonna write minus x

00:01:10.340 --> 00:01:13.090
under bromine in our ICE table.

00:01:13.090 --> 00:01:14.980
Next, we think about mole ratios.

00:01:14.980 --> 00:01:18.490
In the balanced equation,
it's a one to one mole ratio

00:01:18.490 --> 00:01:20.360
of bromine to chlorine.

00:01:20.360 --> 00:01:22.790
Therefore, if we're losing x for bromine,

00:01:22.790 --> 00:01:25.290
we're also going to lose x for chlorine.

00:01:25.290 --> 00:01:29.730
So I can write here minus x
under chlorine in the ICE table.

00:01:29.730 --> 00:01:33.640
When Br2 and Cl2 react
together, we lose our reactants,

00:01:33.640 --> 00:01:36.630
and that means we're gonna
gain some of our products.

00:01:36.630 --> 00:01:39.800
To figure out how much, we
need to look at mole ratios.

00:01:39.800 --> 00:01:44.730
So the mole ratio of bromine
to BrCl is one to two,

00:01:44.730 --> 00:01:48.270
therefore if we're losing x for Br2,

00:01:48.270 --> 00:01:52.050
we must be gaining two x for BrCl.

00:01:52.050 --> 00:01:55.800
So I can go ahead and write
plus two x under BrCl.

00:01:55.800 --> 00:01:58.870
Next, let's think about
equilibrium concentrations.

00:01:58.870 --> 00:02:01.450
If the initial concentration
of bromine is 0.6

00:02:01.450 --> 00:02:04.780
and we're losing x, the
equilibrium concentration

00:02:04.780 --> 00:02:08.870
must be 0.60 minus x.

00:02:08.870 --> 00:02:10.340
And the same thing for chlorine.

00:02:10.340 --> 00:02:14.640
It would be 0.60 minus x.

00:02:14.640 --> 00:02:19.430
For BrCl, we start off with
zero, and we gained two x.

00:02:19.430 --> 00:02:22.840
Therefore at equilibrium,
the equilibrium concentration

00:02:22.840 --> 00:02:25.730
would be equal to just two x.

00:02:25.730 --> 00:02:27.730
The next step is to use
the balanced equation

00:02:27.730 --> 00:02:31.150
to write an equilibrium
constant expression.

00:02:31.150 --> 00:02:33.920
So we would write Kc is equal to,

00:02:33.920 --> 00:02:35.690
and then we look at our balanced equation,

00:02:35.690 --> 00:02:38.410
and for our product we have BrCl

00:02:38.410 --> 00:02:39.900
with a two as a coefficient,

00:02:39.900 --> 00:02:44.807
so Kc would be equal to the
concentration of BrCl squared,

00:02:46.530 --> 00:02:49.200
and we're gonna divide
that by the concentration

00:02:49.200 --> 00:02:51.570
of our reactants, which would be Br2,

00:02:51.570 --> 00:02:55.980
so the concentration of Br2
raised to the first power,

00:02:55.980 --> 00:02:57.550
because the coefficient of one,

00:02:57.550 --> 00:03:00.370
times the concentration of Cl2

00:03:00.370 --> 00:03:03.970
also raised to the first power.

00:03:03.970 --> 00:03:06.960
The concentrations in an
equilibrium constant expression

00:03:06.960 --> 00:03:10.560
are equilibrium concentrations,
therefore we can plug in

00:03:10.560 --> 00:03:13.720
the equilibrium concentrations
from our ICE table.

00:03:13.720 --> 00:03:18.560
So the equilibrium concentration
for BrCl was two x,

00:03:18.560 --> 00:03:23.560
the equilibrium concentration
for Br2 was 0.60 minus x,

00:03:23.600 --> 00:03:27.830
and the same for chlorine, so
we can plug that in as well.

00:03:27.830 --> 00:03:30.130
Here we have our
equilibrium concentrations

00:03:30.130 --> 00:03:33.080
plugged into our equilibrium
constant expression,

00:03:33.080 --> 00:03:37.830
and also Kc was equal to 7.0
for this reaction at 400 Kelvin

00:03:37.830 --> 00:03:41.220
so 7.0 is plugged in for Kc.

00:03:41.220 --> 00:03:45.460
Our goal is to solve for x, and
I've re-written it down here

00:03:45.460 --> 00:03:49.730
because 0.60 minus x times 0.60 minus x

00:03:49.730 --> 00:03:52.970
is equal to 0.60 minus x squared.

00:03:52.970 --> 00:03:56.048
And if you write it this
way, it's a little bit easier

00:03:56.048 --> 00:03:57.430
to see that we can solve for x

00:03:57.430 --> 00:04:00.930
by taking the square root of both sides.

00:04:00.930 --> 00:04:04.260
So let's go ahead and take
the square root of both sides

00:04:04.260 --> 00:04:06.390
and solve for x.

00:04:06.390 --> 00:04:07.850
Taking the square root of both sides

00:04:07.850 --> 00:04:12.850
gives us 2.65 is equal to
two x over 0.60 minus x.

00:04:12.940 --> 00:04:15.190
To solve for x, we would
then multiply both sides

00:04:15.190 --> 00:04:18.760
by 0.60 minus x to give us this,

00:04:18.760 --> 00:04:20.520
and then after a little more algebra,

00:04:20.520 --> 00:04:24.317
we get 1.59 is equal to 4.65x.

00:04:24.317 --> 00:04:27.853
So x is equal to 1.59 divided by 4.65,

00:04:28.870 --> 00:04:30.880
which is equal to 0.34.

00:04:33.920 --> 00:04:36.260
Now that we know that x is equal to 0.34,

00:04:36.260 --> 00:04:38.830
we can plug that into our ICE table

00:04:38.830 --> 00:04:42.070
and solve for our
equilibrium concentrations.

00:04:42.070 --> 00:04:44.890
So for the equilibrium
concentration of Br2,

00:04:44.890 --> 00:04:49.513
it's 0.60 minus x, so
that's 0.60 minus 0.34,

00:04:52.400 --> 00:04:56.620
which is equal to 0.26 molar.

00:04:56.620 --> 00:05:00.980
So 0.26 molar is the equilibrium
concentration for bromine.

00:05:00.980 --> 00:05:02.970
For chlorine, it would
be the same calculation,

00:05:02.970 --> 00:05:07.350
0.60 minus x would be 0.60 minus 0.34,

00:05:07.350 --> 00:05:09.520
so the equilibrium
concentration of chlorine

00:05:09.520 --> 00:05:12.330
is also 0.26 molar.

00:05:12.330 --> 00:05:16.990
For BrCl, it's two times x
so that's two times 0.34,

00:05:19.030 --> 00:05:23.020
which is equal to 0.68 molar.

00:05:23.020 --> 00:05:27.733
So 0.68 molar is the equilibrium
concentration for BrCl.


Worked example: Calculating an equilibrium constant from initial and equilibrium pressures

WEBVTT
Kind: captions
Language: en

00:00:00.760 --> 00:00:02.370
- [Instructor] Let's say
we have a pure sample

00:00:02.370 --> 00:00:04.540
of phosphorus pentachloride,

00:00:04.540 --> 00:00:08.284
and we add the PCl5 to a
previously evacuated flask

00:00:08.284 --> 00:00:11.410
at 500 Kelvin.

00:00:11.410 --> 00:00:13.620
And let's say the initial
pressure of the PCl5

00:00:13.620 --> 00:00:18.200
is 1.66 atmospheres.

00:00:18.200 --> 00:00:23.159
Some of the PCl5 is going
to turn into PCl3 and Cl2.

00:00:23.159 --> 00:00:26.370
Once equilibrium is
reached, the total pressure,

00:00:26.370 --> 00:00:30.332
let's say, is measured
to be 2.35 atmospheres.

00:00:30.332 --> 00:00:33.930
Our goal is to calculate the
equilibrium partial pressures

00:00:33.930 --> 00:00:38.930
of these three substances,
so PCl5, PCl3 and Cl2.

00:00:39.250 --> 00:00:41.500
And from those equilibrium
partial pressures,

00:00:41.500 --> 00:00:44.140
we can also calculate the Kp value

00:00:44.140 --> 00:00:47.120
for this reaction at 500 Kelvin.

00:00:47.120 --> 00:00:49.430
To help us find the
equilibrium partial pressures,

00:00:49.430 --> 00:00:51.650
we're gonna use an ICE
table where I stands

00:00:51.650 --> 00:00:54.298
for the initial partial
pressure in atmospheres,

00:00:54.298 --> 00:00:56.770
C is the change in partial pressure,

00:00:56.770 --> 00:01:00.620
and E stands for the
equilibrium partial pressure.

00:01:00.620 --> 00:01:01.470
We already know we're starting

00:01:01.470 --> 00:01:06.340
with a partial pressure of
1.66 atmospheres for PCl5.

00:01:06.340 --> 00:01:08.960
And if we assume that the
reaction hasn't started yet,

00:01:08.960 --> 00:01:11.820
we're starting with zero
for our partial pressures

00:01:11.820 --> 00:01:14.473
of PCl3 and Cl2.

00:01:15.570 --> 00:01:18.690
Some of the PCl5 is going to decompose.

00:01:18.690 --> 00:01:20.380
And since we don't know how much,

00:01:20.380 --> 00:01:22.370
we're gonna call that amount x.

00:01:22.370 --> 00:01:24.630
So, we're gonna write minus x here,

00:01:24.630 --> 00:01:27.320
since we're gonna lose some PCl5.

00:01:27.320 --> 00:01:29.350
Next, we need to look at mole ratios.

00:01:29.350 --> 00:01:34.125
So, the mole ratio of PCl5
to PCl3 is one to one.

00:01:34.125 --> 00:01:37.070
So, for losing x for PCl5,

00:01:37.070 --> 00:01:40.755
we must be gaining x for PCl3.

00:01:40.755 --> 00:01:43.040
The same idea with Cl2,

00:01:43.040 --> 00:01:45.460
the coefficient in the
balanced equation is a one.

00:01:45.460 --> 00:01:48.280
So, if we're losing x for PCl5,

00:01:48.280 --> 00:01:52.220
that means we're gaining x for Cl2.

00:01:52.220 --> 00:01:55.161
Therefore, the equilibrium
partial pressure for PCl5

00:01:55.161 --> 00:01:58.900
would be 1.66 minus x.

00:01:58.900 --> 00:02:01.700
The equilibrium partial pressure for PCl3

00:02:01.700 --> 00:02:04.210
would be zero plus x, which is just x.

00:02:04.210 --> 00:02:06.446
And the equilibrium
partial pressure for Cl2

00:02:06.446 --> 00:02:09.780
would be zero plus x, which is also x.

00:02:09.780 --> 00:02:13.050
To figure out what x is, we're
going to use Dalton's law.

00:02:13.050 --> 00:02:15.700
And Dalton's law says
that the total pressure

00:02:15.700 --> 00:02:18.430
of a mixture of gases is equal to the sum

00:02:18.430 --> 00:02:20.560
of the individual partial pressures

00:02:20.560 --> 00:02:22.850
of the gases in the mixture.

00:02:22.850 --> 00:02:24.360
So, we said that the total pressure

00:02:24.360 --> 00:02:26.856
of all the gases at equilibrium

00:02:26.856 --> 00:02:30.870
is equal to 2.35 atmospheres.

00:02:30.870 --> 00:02:33.600
So, we can plug that into Dalton's law.

00:02:33.600 --> 00:02:36.290
And then, we can take the
equilibrium partial pressures

00:02:36.290 --> 00:02:40.800
from our ICE table and plug
those into Dalton's law as well.

00:02:40.800 --> 00:02:43.220
So, we're gonna plug in 1.66 minus x

00:02:43.220 --> 00:02:46.154
for the equilibrium
partial pressure of PCl5,

00:02:46.154 --> 00:02:49.962
x for the equilibrium
partial pressure of PCl3,

00:02:49.962 --> 00:02:53.370
and x for the equilibrium
partial pressure of PCl2.

00:02:54.560 --> 00:02:59.110
So, let's plug in 1.66 minus x.

00:02:59.110 --> 00:03:03.340
And then, we have plus x and plus x.

00:03:03.340 --> 00:03:04.450
And let's solve for x.

00:03:04.450 --> 00:03:07.110
Notice how we have a minus x and a plus x.

00:03:07.110 --> 00:03:08.630
So, that cancels out.

00:03:08.630 --> 00:03:13.490
So, we simply subtract 1.66 from 2.35

00:03:13.490 --> 00:03:16.880
and we find that x is equal to .69.

00:03:16.880 --> 00:03:20.220
So, if x is equal to .69,

00:03:20.220 --> 00:03:25.220
the equilibrium partial pressure
of Cl2 is .69 atmospheres.

00:03:25.733 --> 00:03:29.061
And the equilibrium partial pressure PCl3

00:03:29.061 --> 00:03:31.061
is also .69 atmospheres.

00:03:32.854 --> 00:03:35.104
And 1.66 minus .69 gives us

00:03:36.162 --> 00:03:39.335
the equilibrium partial pressure of PCl5

00:03:39.335 --> 00:03:42.335
and that's equal to .97 atmospheres.

00:03:44.860 --> 00:03:47.180
Now that we have our
equilibrium partial pressures

00:03:47.180 --> 00:03:48.680
for all three gases,

00:03:48.680 --> 00:03:51.930
we can calculate the value
for the equilibrium constant

00:03:51.930 --> 00:03:54.470
for this reaction at 500 Kelvin.

00:03:54.470 --> 00:03:57.170
First, we need to write an
equilibrium constant expression.

00:03:57.170 --> 00:04:01.630
So, we would write Kp is equal
to, and for our products,

00:04:01.630 --> 00:04:05.847
we have PCl3, so this would be
the partial pressure of PCl3

00:04:07.446 --> 00:04:10.930
times the partial pressure
of our other product,

00:04:10.930 --> 00:04:12.720
which is Cl2, so let's put in there

00:04:12.720 --> 00:04:15.130
the partial pressure of Cl2.

00:04:15.130 --> 00:04:20.130
And all of that is divided by
the partial pressure of PCl5.

00:04:21.210 --> 00:04:24.423
So, this would be the
partial pressure of PCl5.

00:04:25.890 --> 00:04:28.730
Next, we plug in our
equilibrium partial pressures.

00:04:28.730 --> 00:04:33.730
So, the equilibrium partial
pressure of PCl3 is .69.

00:04:33.790 --> 00:04:38.220
The equilibrium partial
pressure of Cl2 is also .69.

00:04:38.220 --> 00:04:42.664
And the equilibrium partial
pressure of PCl5 is .97.

00:04:42.664 --> 00:04:45.090
Once we plug our numbers in and we solve,

00:04:45.090 --> 00:04:50.090
we get that Kp is equal
to .49 at 500 Kelvin.


Properties of the equilibrium constant

WEBVTT
Kind: captions
Language: en

00:00:00.400 --> 00:00:02.900
- [Instructor] An equilibrium
constant has one value

00:00:02.900 --> 00:00:05.990
for a particular reaction
at a certain temperature.

00:00:05.990 --> 00:00:07.640
For example, for this reaction,

00:00:07.640 --> 00:00:11.940
we have oxalic acid turning
into two H plus ions

00:00:11.940 --> 00:00:14.320
and the oxalate anion.

00:00:14.320 --> 00:00:17.740
The equilibrium constant,
K C, for this reaction

00:00:17.740 --> 00:00:20.490
is equal to 3.8 times
10 to the negative six

00:00:20.490 --> 00:00:22.910
at 25 degrees Celsius.

00:00:22.910 --> 00:00:25.830
However, the value of the
equilibrium constant changes

00:00:25.830 --> 00:00:28.300
if we change how we write
the balanced equation.

00:00:28.300 --> 00:00:30.250
For example, let's reverse everything.

00:00:30.250 --> 00:00:34.100
So, instead of having oxalic
acid on the reactant side,

00:00:34.100 --> 00:00:35.810
let's have it on the product side.

00:00:35.810 --> 00:00:38.970
And instead of having H plus
ions in the product side

00:00:38.970 --> 00:00:40.760
and oxalate on the product side,

00:00:40.760 --> 00:00:43.800
let's put them on the reactant side.

00:00:43.800 --> 00:00:46.050
Since we've reversed the reaction,

00:00:46.050 --> 00:00:48.270
there's a new equilibrium constant.

00:00:48.270 --> 00:00:51.570
So we're going to write K
C for the reverse reaction,

00:00:51.570 --> 00:00:52.403
is equal to,

00:00:52.403 --> 00:00:54.100
and to figure out the value

00:00:54.100 --> 00:00:56.180
for the new equilibrium constant,

00:00:56.180 --> 00:00:57.600
since we reversed everything,

00:00:57.600 --> 00:01:01.630
we take the inverse of the
original equilibrium constant.

00:01:01.630 --> 00:01:03.250
So to find the new K C value,

00:01:03.250 --> 00:01:06.790
we take one over the
original equilibrium constant

00:01:06.790 --> 00:01:11.790
of 3.8 times 10 to the negative six.

00:01:13.400 --> 00:01:18.400
And this is equal to 2.6
times 10 to the fifth,

00:01:20.360 --> 00:01:23.490
also at 25 degrees Celsius.

00:01:23.490 --> 00:01:24.980
Let's look at another example of how

00:01:24.980 --> 00:01:26.700
changing how we write the equation

00:01:26.700 --> 00:01:29.600
changes the value for
the equilibrium constant.

00:01:29.600 --> 00:01:31.330
And we start by looking at

00:01:31.330 --> 00:01:33.560
the ionization of hydrofluoric acid,

00:01:33.560 --> 00:01:37.350
turning into an H plus
ion and a fluoride anion.

00:01:37.350 --> 00:01:39.450
The K C value for this
reaction is equal to

00:01:39.450 --> 00:01:41.700
6.8 times 10 to the negative fourth

00:01:41.700 --> 00:01:44.570
at 25 degrees Celsius.

00:01:44.570 --> 00:01:46.410
This time we're going to
multiply everything through

00:01:46.410 --> 00:01:47.800
by a factor of two.

00:01:47.800 --> 00:01:50.120
So it would be two H F

00:01:50.120 --> 00:01:54.790
turning into two H plus, plus two F minus.

00:01:54.790 --> 00:01:56.210
Our goal is to find the value

00:01:56.210 --> 00:01:58.920
for the new equilibrium constant K C.

00:01:58.920 --> 00:02:01.609
And since we multiplied everything through

00:02:01.609 --> 00:02:02.830
by a factor of two,

00:02:02.830 --> 00:02:05.070
we're going to take the
old equilibrium constant

00:02:05.070 --> 00:02:08.220
and we're going to raise
it to the second power.

00:02:08.220 --> 00:02:10.120
So this is K C is equal to

00:02:10.120 --> 00:02:15.050
6.8 times 10 to the negative fourth.

00:02:15.050 --> 00:02:16.893
And we're going to square that.

00:02:18.010 --> 00:02:23.010
So this is equal to 4.6 times
10 to the negative seventh

00:02:24.140 --> 00:02:27.010
at 25 degrees Celsius.

00:02:27.010 --> 00:02:30.050
So since we multiplied
through by a factor of two,

00:02:30.050 --> 00:02:32.940
we raised the equilibrium
constant to the second power.

00:02:32.940 --> 00:02:35.490
If we had multiplied through
by a factor of three,

00:02:35.490 --> 00:02:37.820
we would have raised the
old equilibrium constant

00:02:37.820 --> 00:02:39.602
to the third power.

00:02:39.602 --> 00:02:42.800
Let's use the two reactions
we've just looked at here.

00:02:42.800 --> 00:02:46.640
So we have oxalic acid turning
into H plus and oxalate.

00:02:46.640 --> 00:02:50.010
And the other one we looked
at was hydrofluoric acid

00:02:50.010 --> 00:02:52.830
ionizing to turn into H plus and F minus.

00:02:52.830 --> 00:02:54.270
Let's use those two reactions

00:02:54.270 --> 00:02:56.610
to calculate the equilibrium constant

00:02:56.610 --> 00:02:58.470
for a third reaction.

00:02:58.470 --> 00:03:00.450
And this third reaction shows oxalate

00:03:00.450 --> 00:03:02.550
reacting with hydrofluoric acid

00:03:02.550 --> 00:03:06.800
to form oxalic acid
and the fluoride anion.

00:03:06.800 --> 00:03:09.260
Our first step is to get
these first two reactions

00:03:09.260 --> 00:03:11.360
to look more like the third reaction.

00:03:11.360 --> 00:03:15.051
For example, we have oxalic
acid on the reactant side,

00:03:15.051 --> 00:03:17.120
in the first reaction here,

00:03:17.120 --> 00:03:19.500
and oxalic acid is on the product side

00:03:19.500 --> 00:03:21.230
for the third reaction.

00:03:21.230 --> 00:03:23.240
Also notice, oxalate
is on the product side

00:03:23.240 --> 00:03:25.170
for the first and over here for the third,

00:03:25.170 --> 00:03:26.860
it's on the reactant side.

00:03:26.860 --> 00:03:29.610
So we need to reverse the first reaction

00:03:29.610 --> 00:03:32.060
to make it look more like the third one.

00:03:32.060 --> 00:03:34.060
Here we have that reaction reversed.

00:03:34.060 --> 00:03:37.136
And since we reversed the
reaction, we would have to take

00:03:37.136 --> 00:03:40.210
the inverse of the equilibrium constant,

00:03:40.210 --> 00:03:42.610
which we did in the earlier example.

00:03:42.610 --> 00:03:45.530
So let's go ahead and cross
out this first one here,

00:03:45.530 --> 00:03:48.215
because we're not going
to need it anymore.

00:03:48.215 --> 00:03:50.700
And let's move on to our next reaction

00:03:50.700 --> 00:03:53.510
and try to make it look
more like the third one.

00:03:53.510 --> 00:03:55.100
So here we have hydrofluoric acid

00:03:55.100 --> 00:03:58.210
turning into H plus and F minus.

00:03:58.210 --> 00:04:00.140
And for our third reaction,

00:04:00.140 --> 00:04:02.650
we have hydrofluoric acid
on the reactant side,

00:04:02.650 --> 00:04:05.040
but notice there is a coefficient of two

00:04:05.040 --> 00:04:06.400
in the balanced equation,

00:04:06.400 --> 00:04:09.970
and we have only a one in
how we have it written.

00:04:09.970 --> 00:04:12.720
So we need to multiply everything through

00:04:12.720 --> 00:04:14.609
by a factor of two

00:04:14.609 --> 00:04:19.609
to give us two H F and two
H plus and two F minus.

00:04:21.630 --> 00:04:23.270
And since we multiplied everything through

00:04:23.270 --> 00:04:25.040
by a factor of two,

00:04:25.040 --> 00:04:27.760
we would need to square
the equilibrium constant

00:04:27.760 --> 00:04:29.310
for the original reaction.

00:04:29.310 --> 00:04:32.670
And once again, this is
from our previous example.

00:04:32.670 --> 00:04:35.583
So we already know that
equilibrium constant value.

00:04:36.430 --> 00:04:38.970
Next, let's add these
two reactions together

00:04:38.970 --> 00:04:43.140
and make sure they give
us the third reaction.

00:04:43.140 --> 00:04:46.726
So we have all of our
reactants on one side

00:04:46.726 --> 00:04:51.726
and we have all of our products
on the other side here.

00:04:52.000 --> 00:04:53.830
And we look for things that are the same

00:04:53.830 --> 00:04:56.050
on both the reactants
and the products side.

00:04:56.050 --> 00:04:58.170
Well, there's two H plus on the left

00:04:58.170 --> 00:05:00.050
and there's two H plus on the right,

00:05:00.050 --> 00:05:01.700
so that would cancel out.

00:05:01.700 --> 00:05:04.640
And now, notice that we would have

00:05:04.640 --> 00:05:09.640
oxalate plus two H F going to
oxalic acid plus two F minus.

00:05:13.620 --> 00:05:15.610
So adding our two reactions together

00:05:15.610 --> 00:05:18.800
does give us our third reaction.

00:05:18.800 --> 00:05:21.980
Once we've confirmed that adding
our two reactions together

00:05:21.980 --> 00:05:24.350
gives us our third reaction,

00:05:24.350 --> 00:05:26.240
we can use the equilibrium constants

00:05:26.240 --> 00:05:27.670
for the first two reactions

00:05:27.670 --> 00:05:30.950
to figure out the equilibrium
constant for the third.

00:05:30.950 --> 00:05:32.370
So the equilibrium constant

00:05:32.370 --> 00:05:34.960
for this first reaction
we're gonna call K one,

00:05:34.960 --> 00:05:37.720
for the second reaction
we're gonna call K two,

00:05:37.720 --> 00:05:39.090
and the equilibrium constant

00:05:39.090 --> 00:05:41.260
that we're trying to find
for the third reaction

00:05:41.260 --> 00:05:42.653
we'll call K C.

00:05:43.530 --> 00:05:47.150
To find the equilibrium constant
for the third reaction K C,

00:05:47.150 --> 00:05:49.530
we need to multiply the
equilibrium constants

00:05:49.530 --> 00:05:51.720
for the first two reactions together.

00:05:51.720 --> 00:05:56.720
So K C is equal to K one times K two.

00:05:58.040 --> 00:06:00.240
We've already calculated
the value for our K one

00:06:00.240 --> 00:06:02.298
from an earlier example,

00:06:02.298 --> 00:06:03.650
it was 2.6 times 10 to the fifth,

00:06:03.650 --> 00:06:05.880
and we also calculated K two,

00:06:05.880 --> 00:06:09.830
and it was 4.6 times 10
to the negative seven.

00:06:09.830 --> 00:06:10.870
So to find K C,

00:06:10.870 --> 00:06:12.980
the equilibrium constant
for the third reaction,

00:06:12.980 --> 00:06:15.230
we simply multiply those two together

00:06:15.230 --> 00:06:20.130
and we get that K C is
equal to point one two

00:06:20.130 --> 00:06:23.410
at 25 degrees Celsius.

00:06:23.410 --> 00:06:25.350
So whenever you add reactions together

00:06:25.350 --> 00:06:26.930
to get a new reaction,

00:06:26.930 --> 00:06:29.280
to find the equilibrium
constant for the new reaction,

00:06:29.280 --> 00:06:32.206
you simply need to multiply
the equilibrium constants

00:06:32.206 --> 00:06:34.853
of the reactions that you added together.


Magnitude of the equilibrium constant

WEBVTT
Kind: captions
Language: en

00:00:00.080 --> 00:00:02.070
- [Instructor] The magnitude
of the equilibrium constant

00:00:02.070 --> 00:00:04.480
tells us the relative amounts of products

00:00:04.480 --> 00:00:06.840
and reactants at equilibrium.

00:00:06.840 --> 00:00:09.270
For example, let's look
at a hypothetical reaction

00:00:09.270 --> 00:00:13.110
where gas A turns into gas B.

00:00:13.110 --> 00:00:14.350
And for the first example,

00:00:14.350 --> 00:00:17.640
let's say that gas A is
represented by a red sphere

00:00:17.640 --> 00:00:20.740
and gas B is represented by a blue sphere.

00:00:20.740 --> 00:00:22.900
And down here, we have
a particulate diagram

00:00:22.900 --> 00:00:26.910
showing an equilibrium mixture
of our hypothetical reaction.

00:00:26.910 --> 00:00:29.160
Let's write the equilibrium
constant expression

00:00:29.160 --> 00:00:31.440
for this hypothetical reaction.

00:00:31.440 --> 00:00:34.310
So we're gonna write Kc is equal to,

00:00:34.310 --> 00:00:36.820
and we think products over reactants.

00:00:36.820 --> 00:00:38.600
So our product is B.

00:00:38.600 --> 00:00:41.520
So this can be the concentration of B.

00:00:41.520 --> 00:00:44.200
And since the coefficient is a
one in the balanced equation,

00:00:44.200 --> 00:00:47.000
it's the concentration of
B raised to the first power

00:00:47.000 --> 00:00:50.530
divided by the concentration
of our reactant, which is A.

00:00:50.530 --> 00:00:52.540
And A in the balanced equation

00:00:52.540 --> 00:00:54.070
also has a coefficient of one.

00:00:54.070 --> 00:00:57.160
So this is the concentration
of A raised to the first power.

00:00:57.160 --> 00:01:00.910
If we assume that each particle
in our particulate diagram

00:01:00.910 --> 00:01:04.510
represents 0.1 moles of a substance,

00:01:04.510 --> 00:01:08.610
and the volume is one liter,

00:01:08.610 --> 00:01:13.160
we can calculate the
concentration of both A and B.

00:01:13.160 --> 00:01:16.950
For example, for B, there
are five blue spheres.

00:01:16.950 --> 00:01:21.360
So that'll be five times
0.1 moles or 0.5 moles.

00:01:21.360 --> 00:01:23.510
So for the concentration of B,

00:01:23.510 --> 00:01:28.510
we have 0.5 moles divided
by a volume of one liter.

00:01:28.650 --> 00:01:32.670
So 0.5 divided by one is 0.5 molar.

00:01:32.670 --> 00:01:34.060
So we can go ahead and plug that in

00:01:34.060 --> 00:01:36.120
for our concentration of B.

00:01:36.120 --> 00:01:38.730
It's 0.5 molar.

00:01:38.730 --> 00:01:41.440
Next, we can do the same thing for A.

00:01:41.440 --> 00:01:44.430
There are also five red spheres.

00:01:44.430 --> 00:01:48.270
And so therefore the concentration
of A is also 0.5 molar.

00:01:48.270 --> 00:01:49.180
So we can plug that

00:01:49.180 --> 00:01:52.240
into our equilibrium constant expression.

00:01:52.240 --> 00:01:54.410
0.5 divided by 0.5 is equal to one.

00:01:54.410 --> 00:01:57.080
So therefore, Kc, the equilibrium constant

00:01:57.080 --> 00:02:00.060
is equal to one at whatever
temperature we have

00:02:00.060 --> 00:02:02.130
for our hypothetical reaction.

00:02:02.130 --> 00:02:05.130
So our equilibrium constant
Kc is equal to one.

00:02:05.130 --> 00:02:08.800
And we saw in our particulate
diagram at equilibrium,

00:02:08.800 --> 00:02:12.170
we have equal amounts of
reactants and products.

00:02:12.170 --> 00:02:13.720
Therefore just by knowing the value

00:02:13.720 --> 00:02:15.530
for the equilibrium constant,

00:02:15.530 --> 00:02:17.757
we know about the relative
amounts of reactants

00:02:17.757 --> 00:02:20.620
and products at equilibrium.

00:02:20.620 --> 00:02:22.680
Let's look at another
hypothetical reaction,

00:02:22.680 --> 00:02:25.990
which also has gas A turning into gas B.

00:02:25.990 --> 00:02:30.310
However, this time gas A
is green and gas B is red.

00:02:30.310 --> 00:02:33.250
And let's calculate the
equilibrium constant Kc

00:02:33.250 --> 00:02:34.570
for this reaction.

00:02:34.570 --> 00:02:36.620
And once again, our particulate diagram

00:02:36.620 --> 00:02:38.970
shows an equilibrium mixture.

00:02:38.970 --> 00:02:40.960
So Kc is equal to the concentration of B

00:02:40.960 --> 00:02:42.917
over the concentration of A.

00:02:42.917 --> 00:02:46.540
And it's a lot faster to
simply count our particles.

00:02:46.540 --> 00:02:50.420
So for B, B is red, we
have one red particle here,

00:02:50.420 --> 00:02:53.400
so we can go ahead and put in one.

00:02:53.400 --> 00:02:57.740
And then for gas A, we
have one, two, three, four,

00:02:57.740 --> 00:03:01.260
five, six, seven, eight,
nine, 10 particles.

00:03:01.260 --> 00:03:04.700
So one divided by 10 is equal to 0.1.

00:03:04.700 --> 00:03:08.530
So Kc is equal to 0.1 for
this hypothetical reaction

00:03:08.530 --> 00:03:10.100
at a certain temperature.

00:03:10.100 --> 00:03:13.120
So the magnitude of the
equilibrium constant tells us

00:03:13.120 --> 00:03:16.840
about the reaction mixture at equilibrium.

00:03:16.840 --> 00:03:19.950
For this reaction, Kc is equal to 0.1.

00:03:19.950 --> 00:03:22.660
So K is less than one.

00:03:22.660 --> 00:03:24.930
And if we think about what that means,

00:03:24.930 --> 00:03:27.320
K is equal to products over reactants.

00:03:27.320 --> 00:03:28.910
So if K is less than one,

00:03:28.910 --> 00:03:32.120
that means we have a smaller
number in the numerator

00:03:32.120 --> 00:03:34.080
and a larger number in the denominator,

00:03:34.080 --> 00:03:37.610
which means there are more
reactants than products

00:03:37.610 --> 00:03:39.530
at equilibrium.

00:03:39.530 --> 00:03:41.520
Let's look at another
hypothetical reaction

00:03:41.520 --> 00:03:43.750
where gas A turns into gas B.

00:03:43.750 --> 00:03:47.370
This time gas A is
yellow and gas B is blue.

00:03:47.370 --> 00:03:49.360
If we look at our particulate diagram,

00:03:49.360 --> 00:03:52.100
showing our reaction
mixture at equilibrium,

00:03:52.100 --> 00:03:57.100
there are 10 blue particles
and only one yellow particle.

00:03:58.240 --> 00:04:02.070
So when plug into our
equilibrium constant expression,

00:04:02.070 --> 00:04:04.980
this time it's going to be 10 over one.

00:04:04.980 --> 00:04:09.460
Therefore the equilibrium
constant Kc is equal to 10

00:04:09.460 --> 00:04:12.920
for this particular reaction
at a certain temperature.

00:04:12.920 --> 00:04:15.410
Once again, the magnitude
of the equilibrium constant

00:04:15.410 --> 00:04:18.270
tells us something about
the reaction mixture

00:04:18.270 --> 00:04:19.730
at equilibrium.

00:04:19.730 --> 00:04:23.260
For this hypothetical
reaction, Kc is equal to 10.

00:04:23.260 --> 00:04:25.830
So K is greater than one.

00:04:25.830 --> 00:04:27.480
And when K is greater than one,

00:04:27.480 --> 00:04:29.720
once again, we have
products over reactants.

00:04:29.720 --> 00:04:33.520
So the numerator must be
larger than the denominator,

00:04:33.520 --> 00:04:35.410
which means we have a lot more products

00:04:35.410 --> 00:04:38.190
than reactants at equilibrium.

00:04:38.190 --> 00:04:40.610
Let's look at the reaction
of carbon monoxide

00:04:40.610 --> 00:04:44.030
and chlorine gas to form phosgene.

00:04:44.030 --> 00:04:46.530
At 100 degrees Celsius,
the equilibrium constant

00:04:46.530 --> 00:04:50.653
for this reaction is 4.56
times 10 to the ninth.

00:04:51.830 --> 00:04:55.460
Since the equilibrium constant
K is greater than one,

00:04:55.460 --> 00:04:57.870
we know there are more
products than reactants

00:04:57.870 --> 00:04:59.310
at equilibrium.

00:04:59.310 --> 00:05:01.410
And with the extremely large value for K,

00:05:01.410 --> 00:05:02.800
like 10 to the ninth,

00:05:02.800 --> 00:05:04.300
we could even assume this reaction

00:05:04.300 --> 00:05:06.610
essentially goes to completion.

00:05:06.610 --> 00:05:09.240
For the reaction of
hydrogen gas and iodine gas

00:05:09.240 --> 00:05:13.010
to form hydrogen iodine,
the equilibrium constant Kc

00:05:13.010 --> 00:05:17.120
is equal to 51 at 448 degrees Celsius.

00:05:17.120 --> 00:05:20.540
Since the equilibrium constant
is relatively close to one.

00:05:20.540 --> 00:05:21.850
This means at equilibrium,

00:05:21.850 --> 00:05:24.920
we have appreciable amounts
of both our reactants

00:05:24.920 --> 00:05:27.010
and our products.

00:05:27.010 --> 00:05:30.010
Let's look at the reaction of
nitrogen gas plus oxygen gas

00:05:30.010 --> 00:05:32.700
plus bromine gas to form NOBr.

00:05:32.700 --> 00:05:36.370
At 298 Kelvin, the equilibrium
constant for this reaction

00:05:36.370 --> 00:05:40.490
is 9.5 times 10 to the negative 31st.

00:05:40.490 --> 00:05:43.960
Since the equilibrium
constant K is less than one,

00:05:43.960 --> 00:05:45.260
we know at equilibrium,

00:05:45.260 --> 00:05:48.200
there are more reactants
than there are products.

00:05:48.200 --> 00:05:50.360
And with an extremely small K value,

00:05:50.360 --> 00:05:53.518
like 9.5 times 10 to the negative 31st,

00:05:53.518 --> 00:05:56.720
this reaction barely proceeds at all.

00:05:56.720 --> 00:05:58.080
So at equilibrium, you're gonna have

00:05:58.080 --> 00:06:00.770
almost all nitrogen, oxygen and bromine

00:06:00.770 --> 00:06:02.833
and very little NOBr.


Worked examples: Calculating equilibrium constants

WEBVTT
Kind: captions
Language: en

00:00:00.220 --> 00:00:03.200
- [Instructor] An equilibrium
constant can be calculated

00:00:03.200 --> 00:00:06.130
from experimentally
measured concentrations

00:00:06.130 --> 00:00:08.930
or partial pressures of
reactants and products

00:00:08.930 --> 00:00:09.963
at equilibrium.

00:00:10.800 --> 00:00:12.700
As an example, let's look at the reaction

00:00:12.700 --> 00:00:16.040
where N2O4 in the gaseous
state turns into 2NO2

00:00:16.930 --> 00:00:18.820
also in the gaseous state.

00:00:18.820 --> 00:00:20.490
And let's say we do an experiment

00:00:20.490 --> 00:00:23.920
and we allow this reaction
to come to equilibrium

00:00:23.920 --> 00:00:26.888
and the temperature is
100 degrees Celsius.

00:00:26.888 --> 00:00:28.650
And at equilibrium,

00:00:28.650 --> 00:00:33.650
the concentration of NO2 0.017 molar

00:00:33.710 --> 00:00:38.710
and the concentration of
N2O4 is 0.00140 molar.

00:00:40.350 --> 00:00:43.590
To calculate the equilibrium
constant for this reaction

00:00:43.590 --> 00:00:46.210
at 100 degrees Celsius,
we first need to write

00:00:46.210 --> 00:00:48.950
the equilibrium constant expression.

00:00:48.950 --> 00:00:51.170
We can write the equilibrium
constant expression

00:00:51.170 --> 00:00:53.170
by using the balanced equation.

00:00:53.170 --> 00:00:55.410
We start by writing the
equilibrium constant,

00:00:55.410 --> 00:00:57.110
which is symbolized by K.

00:00:57.110 --> 00:00:58.950
And since we're dealing
with concentrations,

00:00:58.950 --> 00:01:00.760
we're calculating Kc.

00:01:00.760 --> 00:01:04.690
And Kc is equal to, we do
products over reactants.

00:01:04.690 --> 00:01:08.740
So this would be the concentration of NO2.

00:01:08.740 --> 00:01:12.490
And since there is a coefficient
of two in front of NO2,

00:01:12.490 --> 00:01:15.994
this is the concentration of
NO2 raised to the second power

00:01:15.994 --> 00:01:19.823
divided by the concentration
of our reactant, N2O4.

00:01:21.650 --> 00:01:24.630
And since there's an implied
one in front of N2O4,

00:01:24.630 --> 00:01:28.530
this is the concentration of
N2O4 raised to the first power.

00:01:28.530 --> 00:01:31.470
Next, we plug in our
equilibrium concentrations.

00:01:31.470 --> 00:01:36.470
So the equilibrium
concentration of NO2 is 0.0172.

00:01:36.770 --> 00:01:38.130
So let's plug that in.

00:01:38.130 --> 00:01:42.463
So this is equal to 0.0172 squared

00:01:45.084 --> 00:01:48.643
divided by the equilibrium
concentration of N2O4,

00:01:49.780 --> 00:01:51.553
which was 0.00140.

00:01:52.750 --> 00:01:54.920
So we plug that in as well.

00:01:54.920 --> 00:01:59.480
So 0.00140.

00:01:59.480 --> 00:02:03.033
When we solve this, we get
that Kc is equal to 0.211,

00:02:05.460 --> 00:02:08.660
and this is at 100 degrees Celsius.

00:02:08.660 --> 00:02:10.960
It's important to always
give the temperature

00:02:10.960 --> 00:02:13.960
when you're giving a value
for an equilibrium constant,

00:02:13.960 --> 00:02:17.110
because an equilibrium
constant is only constant

00:02:17.110 --> 00:02:20.930
for a particular reaction
at a particular temperature.

00:02:20.930 --> 00:02:22.470
And it's also important to note

00:02:22.470 --> 00:02:26.090
that the equilibrium constant
doesn't have any units.

00:02:26.090 --> 00:02:29.850
So we would just say
that Kc is equal to 0.211

00:02:29.850 --> 00:02:34.100
at 100 degrees Celsius for
this particular reaction.

00:02:34.100 --> 00:02:36.080
Let's calculate the equilibrium constant

00:02:36.080 --> 00:02:37.720
for another reaction.

00:02:37.720 --> 00:02:42.000
In this reaction, carbon
dioxide reacts with hydrogen gas

00:02:42.000 --> 00:02:45.410
to produce carbon monoxide and H2O.

00:02:45.410 --> 00:02:47.920
And since everything is
in the gaseous state,

00:02:47.920 --> 00:02:50.960
experimentally, it's easier
to work with partial pressures

00:02:50.960 --> 00:02:53.150
than it is to work with concentrations.

00:02:53.150 --> 00:02:55.490
So instead of calculating Kc,

00:02:55.490 --> 00:02:59.110
we're gonna calculate Kp or
the p stands for pressure.

00:02:59.110 --> 00:03:03.840
So we're trying to find Kp at
500 Kelvin for this reaction.

00:03:03.840 --> 00:03:07.080
To help us find Kp, we're
going to use an ICE table

00:03:07.080 --> 00:03:10.150
where I stands for the
initial partial pressure

00:03:10.150 --> 00:03:12.070
in atmospheres, C stands for the

00:03:12.070 --> 00:03:15.020
change in the partial
pressure, also in atmospheres

00:03:15.020 --> 00:03:18.070
and E is the equilibrium partial pressure.

00:03:18.070 --> 00:03:20.640
Let's say that a mixture
of carbon dioxide,

00:03:20.640 --> 00:03:22.940
hydrogen gas and H2O are placed

00:03:22.940 --> 00:03:25.150
in a previously evacuated flask

00:03:25.150 --> 00:03:29.320
and allowed to come to
equilibrium at 500 Kelvin.

00:03:29.320 --> 00:03:32.510
And let's say the initial
measured partial pressures

00:03:32.510 --> 00:03:36.540
are 4.10 atmospheres for carbon dioxide,

00:03:36.540 --> 00:03:40.160
1.80 atmospheres for hydrogen gas

00:03:40.160 --> 00:03:44.090
and 3.20 atmospheres for H2O.

00:03:44.090 --> 00:03:46.420
And since we didn't
add any carbon monoxide

00:03:46.420 --> 00:03:48.320
in the beginning, the
initial partial pressure

00:03:48.320 --> 00:03:49.888
of that would be zero.

00:03:49.888 --> 00:03:52.850
And after the reaction
comes to equilibrium,

00:03:52.850 --> 00:03:55.440
we measure the partial pressure of H2O

00:03:55.440 --> 00:03:58.440
to be 3.40 atmospheres.

00:03:58.440 --> 00:04:01.885
So that's why we have 3.40
in the equilibrium parts

00:04:01.885 --> 00:04:04.780
on the ICE table under H2O.

00:04:04.780 --> 00:04:09.150
So the initial partial pressure
of H2O is 3.20 atmospheres

00:04:09.150 --> 00:04:13.370
and the equilibrium
partial pressure is 3.40.

00:04:13.370 --> 00:04:16.790
So H2O has increased in partial pressure.

00:04:16.790 --> 00:04:19.080
We can go ahead in here and write plus X

00:04:19.080 --> 00:04:21.750
for an increase in the
partial pressure of H2O

00:04:21.750 --> 00:04:26.750
and 3.20 plus X must be equal to 3.40.

00:04:28.920 --> 00:04:33.080
So X is equal to 0.20.

00:04:33.080 --> 00:04:37.050
So the partial pressure of
water increased by 0.20.

00:04:37.050 --> 00:04:40.115
And we could either write plus
X in here on our ICE table,

00:04:40.115 --> 00:04:43.193
or we could just write plus 0.20.

00:04:45.290 --> 00:04:47.710
Now that we know that change
in the partial pressure

00:04:47.710 --> 00:04:50.900
for H2O, we can use this information

00:04:50.900 --> 00:04:53.350
to fill out the rest of our ICE table.

00:04:53.350 --> 00:04:56.270
For example, the mole
ratio of carbon monoxide

00:04:56.270 --> 00:04:58.970
to H2O is 1:1.

00:04:58.970 --> 00:05:02.810
So if we gained plus 0.20 for H2O,

00:05:02.810 --> 00:05:07.810
we're also gonna gain plus
0.20 for carbon monoxide.

00:05:08.010 --> 00:05:10.120
And if we're gaining for
our two products here,

00:05:10.120 --> 00:05:12.410
the net reaction is moving to the right

00:05:12.410 --> 00:05:14.270
to increase the amount of products,

00:05:14.270 --> 00:05:16.990
which means we're losing reactants.

00:05:16.990 --> 00:05:19.460
And we can figure out
by how much by looking

00:05:19.460 --> 00:05:21.340
at the mole ratios again.

00:05:21.340 --> 00:05:22.900
So for both of our reactants,

00:05:22.900 --> 00:05:26.460
we have ones as coefficients
in the balanced equation.

00:05:26.460 --> 00:05:29.310
So if it's plus X for
both of our products,

00:05:29.310 --> 00:05:32.400
it must be minus X for
both of our reactants.

00:05:32.400 --> 00:05:36.877
And since X is 0.20, it'd be minus 0.20

00:05:38.000 --> 00:05:39.960
for the change in the partial pressure

00:05:39.960 --> 00:05:42.143
for both of our reactants.

00:05:43.250 --> 00:05:46.830
Therefore the equilibrium partial
pressure of carbon dioxide

00:05:46.830 --> 00:05:51.670
would be 4.10 minus 0.20, which is 3.90

00:05:53.720 --> 00:05:58.720
and for H2, it'd be 1.80
minus 0.20, which is 1.60.

00:06:00.750 --> 00:06:03.650
And for carbon monoxide,
we started off with zero

00:06:03.650 --> 00:06:06.260
and we gained positive 0.20.

00:06:06.260 --> 00:06:11.020
Therefore the equilibrium
partial pressure is 0.20.

00:06:11.020 --> 00:06:13.900
So as the net reaction moved to the right,

00:06:13.900 --> 00:06:16.140
we lost some of our reactants

00:06:16.140 --> 00:06:18.300
and we gained some of our products

00:06:18.300 --> 00:06:20.870
until the reaction reached equilibrium

00:06:20.870 --> 00:06:23.920
and we got our equilibrium
partial pressures.

00:06:23.920 --> 00:06:27.310
In our equilibrium, the
rate of the forward reaction

00:06:28.410 --> 00:06:30.300
is equal to the rate
of the reverse reaction

00:06:30.300 --> 00:06:33.040
and therefore these
equilibrium partial pressures

00:06:33.040 --> 00:06:35.130
remain constant.

00:06:35.130 --> 00:06:37.470
Now that we know our
equilibrium partial pressures,

00:06:37.470 --> 00:06:41.040
we're ready to calculate
the equilibrium constant Kp.

00:06:41.040 --> 00:06:43.900
So we need to write an
equilibrium constant expression.

00:06:43.900 --> 00:06:48.460
So Kp is equal to, we think
about products over reactants.

00:06:48.460 --> 00:06:49.820
And for our products,

00:06:49.820 --> 00:06:53.270
we would have the partial
pressure of carbon monoxide.

00:06:53.270 --> 00:06:56.100
And since the coefficient is a one

00:06:56.100 --> 00:06:59.010
in front of carbon monoxide
in the balanced equation,

00:06:59.010 --> 00:07:01.410
it would be the partial
pressure of carbon monoxide

00:07:01.410 --> 00:07:04.530
raised to the first power
times the partial pressure

00:07:04.530 --> 00:07:07.440
of our other product, which is H2O.

00:07:07.440 --> 00:07:09.830
And once again, the coefficient is a one.

00:07:09.830 --> 00:07:13.300
So that's the partial pressure
raised to the first power.

00:07:13.300 --> 00:07:15.470
All of this is divided by,

00:07:15.470 --> 00:07:17.300
we think about our reactants next,

00:07:17.300 --> 00:07:19.270
and they both have coefficients of one

00:07:19.270 --> 00:07:20.770
in the balanced equation.

00:07:20.770 --> 00:07:24.940
So it would be the partial
pressure of carbon dioxide

00:07:24.940 --> 00:07:29.940
times the partial
pressure of hydrogen gas.

00:07:30.780 --> 00:07:34.090
The partial pressures in our
equilibrium constant expression

00:07:34.090 --> 00:07:36.370
are the equilibrium partial pressures,

00:07:36.370 --> 00:07:38.266
which we can get from the ICE table.

00:07:38.266 --> 00:07:41.590
So the equilibrium partial
pressure of carbon monoxide

00:07:41.590 --> 00:07:46.590
is 0.20, the equilibrium
partial pressure of H2O is 3.40.

00:07:48.070 --> 00:07:50.270
We can plug in the
equilibrium partial pressures

00:07:50.270 --> 00:07:53.369
for carbon dioxide and the
equilibrium partial pressure

00:07:53.369 --> 00:07:56.030
for hydrogen gas as well.

00:07:56.030 --> 00:07:58.240
And here we have the
equilibrium partial pressures

00:07:58.240 --> 00:08:01.230
plugged into our equilibrium
constant expression.

00:08:01.230 --> 00:08:02.600
And when we solve this,

00:08:02.600 --> 00:08:06.767
we get that Kp is equal
to 0.11 at 500 Kelvin.


Writing equilibrium constant and reaction quotient expressions

WEBVTT
Kind: captions
Language: en

00:00:00.480 --> 00:00:02.820
- [Instructor] The equilibrium
constant is symbolized

00:00:02.820 --> 00:00:07.410
by the letter K, and
equilibrium constant tells us

00:00:07.410 --> 00:00:10.110
about the relative
concentrations of reactants

00:00:10.110 --> 00:00:12.590
and products at equilibrium.

00:00:12.590 --> 00:00:14.822
Let's say we have a hypothetical reaction

00:00:14.822 --> 00:00:19.822
where reactants A and B
turn into products C and D.

00:00:20.070 --> 00:00:21.510
And in the balanced equation,

00:00:21.510 --> 00:00:23.840
the lowercase letters
are the coefficients.

00:00:23.840 --> 00:00:26.380
So we have a lowercase a, a lowercase b,

00:00:26.380 --> 00:00:29.230
lowercase c and lowercase
d as coefficients

00:00:29.230 --> 00:00:31.020
in our balanced equation.

00:00:31.020 --> 00:00:33.530
If we were to write an
equilibrium constant expression

00:00:33.530 --> 00:00:35.420
for this hypothetical reaction,

00:00:35.420 --> 00:00:37.700
we'd start by writing the
equilibrium constant K

00:00:37.700 --> 00:00:39.760
and then we have a subscript c here

00:00:39.760 --> 00:00:41.670
because we're dealing with concentrations

00:00:41.670 --> 00:00:43.930
in our equilibrium constant expression.

00:00:43.930 --> 00:00:47.330
And the equilibrium
constant Kc is equal to,

00:00:47.330 --> 00:00:51.180
and in the numerator, we
have the concentrations

00:00:51.180 --> 00:00:53.420
of our two products multiplied together.

00:00:53.420 --> 00:00:55.200
And the concentration of each product

00:00:55.200 --> 00:00:58.810
is raised to the power of the coefficient.

00:00:58.810 --> 00:00:59.940
In the denominator,

00:00:59.940 --> 00:01:03.260
we have the concentrations
of the two reactants

00:01:03.260 --> 00:01:06.720
multiplied by each other
and raised to the power,

00:01:06.720 --> 00:01:08.400
each concentration is raised to the power

00:01:08.400 --> 00:01:11.220
of the coefficient in
the balanced equation.

00:01:11.220 --> 00:01:13.351
It's important to emphasize
that the concentrations

00:01:13.351 --> 00:01:14.184
that we're plugging

00:01:14.184 --> 00:01:16.730
into our equilibrium constant expression

00:01:16.730 --> 00:01:19.550
are equilibrium concentrations.

00:01:19.550 --> 00:01:22.020
And when we plug in our
equilibrium concentrations

00:01:22.020 --> 00:01:24.430
into our equilibrium constant expression,

00:01:24.430 --> 00:01:27.800
we get a value for the
equilibrium constant K.

00:01:27.800 --> 00:01:30.960
And K is constant for
a particular reaction

00:01:30.960 --> 00:01:32.940
at a certain temperature.

00:01:32.940 --> 00:01:35.360
Let's write an equilibrium
constant expression

00:01:35.360 --> 00:01:36.890
for the following reaction,

00:01:36.890 --> 00:01:39.450
which shows the synthesis of ammonia

00:01:39.450 --> 00:01:41.910
from nitrogen and hydrogen,

00:01:41.910 --> 00:01:43.940
and everything is in the gaseous state.

00:01:43.940 --> 00:01:46.910
We start by writing the
equilibrium constant Kc,

00:01:46.910 --> 00:01:49.180
c because we're dealing
with concentrations,

00:01:49.180 --> 00:01:51.700
and we start with our
product, which is ammonia.

00:01:51.700 --> 00:01:55.340
So we write the concentration of ammonia

00:01:55.340 --> 00:01:58.490
and we raise the concentration of ammonia

00:01:58.490 --> 00:02:00.370
to the power of the coefficient

00:02:00.370 --> 00:02:02.400
in the balanced equation, which is a two.

00:02:02.400 --> 00:02:06.600
So this is the concentration
of ammonia to the second power.

00:02:06.600 --> 00:02:10.150
Then, in the denominator, we
think about our reactants.

00:02:10.150 --> 00:02:11.730
So we have nitrogen.

00:02:11.730 --> 00:02:14.620
So we write the concentration of nitrogen.

00:02:14.620 --> 00:02:17.620
And since the coefficient is a
one in the balanced equation,

00:02:17.620 --> 00:02:20.700
that'd be the concentration
of nitrogen to the first power

00:02:20.700 --> 00:02:23.540
multiplied by the concentration
of our other reactants,

00:02:23.540 --> 00:02:24.880
which is hydrogens.

00:02:24.880 --> 00:02:26.650
We write in here H2.

00:02:26.650 --> 00:02:28.500
And because there's a coefficient of three

00:02:28.500 --> 00:02:29.910
in the balanced equation,

00:02:29.910 --> 00:02:34.140
we raise the concentration of
hydrogen to the third power.

00:02:34.140 --> 00:02:35.940
For gases, it's often more convenient

00:02:35.940 --> 00:02:37.820
to measure partial pressures

00:02:37.820 --> 00:02:40.340
instead of measuring concentrations.

00:02:40.340 --> 00:02:44.750
So let's say that A, B,
C and D are all gases.

00:02:44.750 --> 00:02:47.780
We could write an equilibrium
constant expression

00:02:47.780 --> 00:02:52.260
using partial pressures
instead of concentrations.

00:02:52.260 --> 00:02:54.720
And if we did that, instead of writing Kc,

00:02:54.720 --> 00:02:58.290
we would write Kp where
p stands for pressure.

00:02:58.290 --> 00:03:03.200
And Kc and Kp usually have
different values from each other.

00:03:03.200 --> 00:03:05.560
So if we go back to our previous reaction

00:03:05.560 --> 00:03:07.580
where everything was in the gaseous state,

00:03:07.580 --> 00:03:10.020
we could write a Kp expression.

00:03:10.020 --> 00:03:13.130
So we would write Kp is equal to,

00:03:13.130 --> 00:03:15.540
we think about products over reactants.

00:03:15.540 --> 00:03:19.020
So this would be the partial
pressure of our product,

00:03:19.020 --> 00:03:23.770
ammonia, raised to the second power

00:03:23.770 --> 00:03:27.600
divided by the partial
pressure of nitrogen

00:03:27.600 --> 00:03:32.000
raised to the first power
times the partial pressure

00:03:32.000 --> 00:03:37.000
of hydrogen raised to the third power.

00:03:37.620 --> 00:03:39.360
For the synthesis of ammonia,

00:03:39.360 --> 00:03:41.650
everything was in the gaseous state.

00:03:41.650 --> 00:03:44.660
And when all substances,
reactants and products

00:03:44.660 --> 00:03:46.390
are in the same phase,

00:03:46.390 --> 00:03:50.460
we call this a homogeneous equilibrium.

00:03:50.460 --> 00:03:52.850
When the substances are
in different phases,

00:03:52.850 --> 00:03:56.630
we call it a heterogeneous equilibrium.

00:03:56.630 --> 00:03:59.860
For example, in the decomposition
of calcium carbonate

00:03:59.860 --> 00:04:03.060
to turn into calcium
oxide and carbon dioxide,

00:04:03.060 --> 00:04:07.450
calcium carbonate is a solid
and calcium oxide is a solid,

00:04:07.450 --> 00:04:09.210
but carbon dioxide is a gas.

00:04:09.210 --> 00:04:12.910
So we have substances in different phases.

00:04:12.910 --> 00:04:15.310
When we write an equilibrium
constant expression

00:04:15.310 --> 00:04:18.090
for a heterogeneous equilibrium,

00:04:18.090 --> 00:04:20.980
we leave pure solids and pure liquids

00:04:20.980 --> 00:04:23.740
out of the equilibrium
constant expression.

00:04:23.740 --> 00:04:26.160
So if we write an equilibrium
constant expression

00:04:26.160 --> 00:04:28.850
for the decomposition
of calcium carbonate,

00:04:28.850 --> 00:04:31.350
let's write a Kc expression first here.

00:04:31.350 --> 00:04:33.620
So we write Kc is equal to,

00:04:33.620 --> 00:04:36.850
and we think about
products over reactants.

00:04:36.850 --> 00:04:40.030
For products, we have carbon
dioxide in the gaseous state.

00:04:40.030 --> 00:04:41.670
So it's okay to include that

00:04:41.670 --> 00:04:43.710
in our equilibrium constant expression.

00:04:43.710 --> 00:04:46.880
So we write the concentration of CO2.

00:04:46.880 --> 00:04:49.560
And since the coefficient is a
one in the balanced equation,

00:04:49.560 --> 00:04:51.400
this would be the concentration of CO2

00:04:51.400 --> 00:04:53.280
raised to the first power.

00:04:53.280 --> 00:04:55.090
Our other products is a solid.

00:04:55.090 --> 00:04:56.150
So we're gonna leave that

00:04:56.150 --> 00:04:59.450
out of our equilibrium
constant expression.

00:04:59.450 --> 00:05:03.510
And for our reactant calcium
carbonate, that's also a solid,

00:05:03.510 --> 00:05:06.600
so that's also left out of our expression.

00:05:06.600 --> 00:05:09.670
If we were to write a Kp expression here,

00:05:09.670 --> 00:05:12.640
we would include the
partial pressure of our gas,

00:05:12.640 --> 00:05:14.180
which is carbon dioxide.

00:05:14.180 --> 00:05:16.470
So this would be the partial
pressure of carbon dioxide

00:05:16.470 --> 00:05:17.720
to the first power.

00:05:17.720 --> 00:05:19.920
And once again, we would
leave the two solids

00:05:19.920 --> 00:05:23.600
out of our equilibrium
constant expression.

00:05:23.600 --> 00:05:25.200
The reason why we leave pure solids

00:05:25.200 --> 00:05:28.150
and pure liquids out of
equilibrium constant expressions

00:05:28.150 --> 00:05:31.690
for heterogeneous equilibria
is because the concentration

00:05:31.690 --> 00:05:35.930
of a pure solid or a pure liquid
remains constant over time.

00:05:35.930 --> 00:05:38.120
So it doesn't help us to include it

00:05:38.120 --> 00:05:40.530
in our equilibrium expression.

00:05:40.530 --> 00:05:43.420
Finally, let's talk about
the reaction quotient,

00:05:43.420 --> 00:05:46.680
which is symbolized by the letter Q.

00:05:46.680 --> 00:05:48.560
A Q expression has the same form

00:05:48.560 --> 00:05:50.800
as an equilibrium constant expression.

00:05:50.800 --> 00:05:54.120
And Q tells us the relative
concentrations of reactants

00:05:54.120 --> 00:05:57.950
and products at any moment in time.

00:05:57.950 --> 00:06:02.290
And just like we could write
a Kc or a Kp expression,

00:06:02.290 --> 00:06:05.960
we could write a Qc or a Qp expression.

00:06:05.960 --> 00:06:09.940
Let's go back to our reaction
for the synthesis of ammonia

00:06:09.940 --> 00:06:12.880
from nitrogen gas and hydrogen gas.

00:06:12.880 --> 00:06:16.410
Notice how the Qc
expression has the same form

00:06:16.410 --> 00:06:18.660
as the Kc expression.

00:06:18.660 --> 00:06:21.560
The difference is, for the Kc expression,

00:06:21.560 --> 00:06:25.430
all of our concentrations are
equilibrium concentrations.

00:06:25.430 --> 00:06:27.500
So I could put an eq here

00:06:27.500 --> 00:06:32.233
for the concentrations of
ammonia, nitrogen, and hydrogen.

00:06:33.170 --> 00:06:34.780
So for the Kc expression,

00:06:34.780 --> 00:06:37.240
it's only equilibrium concentrations,

00:06:37.240 --> 00:06:39.430
but for the Qc expression,

00:06:39.430 --> 00:06:42.850
it's the concentrations
at any moment in time.

00:06:42.850 --> 00:06:46.150
So that moment in time
might be at equilibrium

00:06:46.150 --> 00:06:49.630
or it might not be at equilibrium.

00:06:49.630 --> 00:06:54.450
If Qc is equal to KC, the
reaction is at equilibrium,

00:06:54.450 --> 00:06:59.450
but if Qc is greater than
Kc, or if Qc is less than Kc,

00:06:59.600 --> 00:07:02.233
the reaction is not at equilibrium.


Direction of reversible reactions

WEBVTT
Kind: captions
Language: en

00:00:00.190 --> 00:00:02.660
- [Instructor] As an example
of a reversible reaction,

00:00:02.660 --> 00:00:04.900
let's look at the hypothetical reaction

00:00:04.900 --> 00:00:09.900
where diatomic gas X2 turns
into its individual atoms X,

00:00:10.050 --> 00:00:11.340
and it would turn into two of them.

00:00:11.340 --> 00:00:13.650
So X2 goes to 2X.

00:00:13.650 --> 00:00:18.160
The forward reaction
is X2 turning into 2X.

00:00:18.160 --> 00:00:23.160
And the reverse reaction
is 2X combining to form X2.

00:00:23.210 --> 00:00:26.930
And let's say the X2
is a reddish brown gas.

00:00:26.930 --> 00:00:30.850
If we assume that both the
forward and the reverse reactions

00:00:30.850 --> 00:00:32.690
are elementary reactions,

00:00:32.690 --> 00:00:34.570
we can actually write the rate law

00:00:34.570 --> 00:00:36.350
from the balanced equation.

00:00:36.350 --> 00:00:38.330
So for the forward reaction,

00:00:38.330 --> 00:00:41.680
let's go ahead and write the
rate of the forward reaction

00:00:41.680 --> 00:00:45.320
is equal to the rate constant
for the forward reaction,

00:00:45.320 --> 00:00:49.340
which we will symbolize
as K with the subscript F,

00:00:49.340 --> 00:00:51.560
times the concentration of,

00:00:51.560 --> 00:00:53.880
if we're going in the forward direction,

00:00:53.880 --> 00:00:56.190
the reactants would be X2.

00:00:56.190 --> 00:00:59.490
So times the concentration of X2.

00:00:59.490 --> 00:01:02.930
And since we have a coefficient
of 1 in front of X2,

00:01:02.930 --> 00:01:04.313
for this elementary reaction,

00:01:04.313 --> 00:01:07.083
this would be raised to the first power.

00:01:07.940 --> 00:01:11.090
Next we can write the rate
law for the reverse reaction.

00:01:11.090 --> 00:01:13.320
So the rate of the reverse reaction

00:01:13.320 --> 00:01:15.040
is equal to the rate constant,

00:01:15.040 --> 00:01:17.300
and we'll put in a subscript R here.

00:01:17.300 --> 00:01:19.840
So that's the rate constant
for the reverse reaction.

00:01:19.840 --> 00:01:24.710
And in the reverse reaction,
2X combines to form X2.

00:01:24.710 --> 00:01:28.190
So this would be times
the concentration of X.

00:01:28.190 --> 00:01:30.870
And since we have a
two as our coefficient,

00:01:30.870 --> 00:01:34.743
we need to raise the concentration
of X to the second power.

00:01:35.980 --> 00:01:38.130
Next, let's look at these
particulate diagrams

00:01:38.130 --> 00:01:41.520
and think about what happens
for the forward reaction.

00:01:41.520 --> 00:01:43.210
So we start at time is equal to zero,

00:01:43.210 --> 00:01:45.060
and we start with only X2.

00:01:45.060 --> 00:01:47.680
So here are five particles of X2.

00:01:47.680 --> 00:01:49.130
If we wait 10 seconds,

00:01:49.130 --> 00:01:51.260
now we've gone from five particles of X2

00:01:51.260 --> 00:01:54.120
to only three particles of X2.

00:01:54.120 --> 00:01:57.010
So overall, two of those particles of X2

00:01:57.010 --> 00:01:58.720
have turned into X.

00:01:58.720 --> 00:02:01.180
And so there are four particles of X

00:02:01.180 --> 00:02:03.480
in this second particular diagram.

00:02:03.480 --> 00:02:04.990
We wait another 10 seconds

00:02:04.990 --> 00:02:07.120
for a total of time is
equal to 20 seconds.

00:02:07.120 --> 00:02:09.980
And we've gone from three particles of X2

00:02:09.980 --> 00:02:13.240
to only two particles of X2,

00:02:13.240 --> 00:02:16.050
and we've increased in the particles of X.

00:02:16.050 --> 00:02:19.650
So now we're up to six particles of X.

00:02:19.650 --> 00:02:22.650
So the concentration of X2 has decreased.

00:02:22.650 --> 00:02:27.650
We went from five particles
of X2 to three particles of X2

00:02:28.220 --> 00:02:30.530
to only two particles of X2.

00:02:30.530 --> 00:02:34.240
And if we look at the rate
law for the forward reaction,

00:02:34.240 --> 00:02:37.500
the rate of the forward
reaction is proportional

00:02:37.500 --> 00:02:40.470
to the concentration of X2.

00:02:40.470 --> 00:02:43.760
So if the concentration of X2 decreases,

00:02:43.760 --> 00:02:48.450
the rate of the forward
reaction also decreases.

00:02:48.450 --> 00:02:49.810
We can see the same concept

00:02:49.810 --> 00:02:53.560
if we look at a graph of rate versus time.

00:02:53.560 --> 00:02:56.100
So if we look at this line right here,

00:02:56.100 --> 00:02:59.250
we're starting on a certain
rate for the forward reaction.

00:02:59.250 --> 00:03:01.730
And as the concentration of X2 decreases,

00:03:01.730 --> 00:03:04.760
we can see the rate of
the reaction decrease.

00:03:04.760 --> 00:03:07.130
So the rate of the
reaction stops decreasing

00:03:07.130 --> 00:03:10.430
when we get to time is
equal to 20 seconds.

00:03:10.430 --> 00:03:13.960
Next, let's think about the
rate of the reverse reaction.

00:03:13.960 --> 00:03:15.850
Well, when time is equal to zero,

00:03:15.850 --> 00:03:18.450
the rate of the reverse reaction is zero.

00:03:18.450 --> 00:03:21.410
And that's because when we
start out, we have only X2,

00:03:21.410 --> 00:03:23.330
we don't have any X present.

00:03:23.330 --> 00:03:25.760
So the reverse reaction doesn't happen.

00:03:25.760 --> 00:03:28.430
But as soon as some of
that X2 turns into X,

00:03:28.430 --> 00:03:31.400
it's possible for the
reverse reaction to happen.

00:03:31.400 --> 00:03:34.310
And as we increase in the amount of X,

00:03:34.310 --> 00:03:37.444
and we look at our rate law
here for the reverse reaction,

00:03:37.444 --> 00:03:40.590
as we increase in the concentration of X,

00:03:40.590 --> 00:03:44.530
the rate of the reverse reaction
should increase as well.

00:03:44.530 --> 00:03:46.120
And so that's why we see,

00:03:46.120 --> 00:03:49.550
that's why we see the rate of
the reverse reaction increase

00:03:49.550 --> 00:03:52.220
as time increases.

00:03:52.220 --> 00:03:54.760
So as the forward reaction is happening,

00:03:54.760 --> 00:03:59.490
the reverse reaction is also
occurring at the same time.

00:03:59.490 --> 00:04:00.820
However, we don't really see that

00:04:00.820 --> 00:04:03.120
when we look at our particular diagram.

00:04:03.120 --> 00:04:04.410
In our particular diagrams,

00:04:04.410 --> 00:04:07.520
we see a net conversion of X2 into 2X,

00:04:07.520 --> 00:04:10.030
for example, looking from
the first particular diagram

00:04:10.030 --> 00:04:11.410
to the second,

00:04:11.410 --> 00:04:13.950
we see that two particles of X2

00:04:13.950 --> 00:04:17.130
have turned into four particles of X.

00:04:17.130 --> 00:04:20.600
And going from the second
diagram to the third diagram,

00:04:20.600 --> 00:04:23.170
we see that another particle of X2

00:04:23.170 --> 00:04:24.700
has turned into 2X.

00:04:24.700 --> 00:04:27.860
And therefore we have six particles of X

00:04:27.860 --> 00:04:30.780
at time is equal to 20 seconds.

00:04:30.780 --> 00:04:34.050
So since we see a net conversion
of reactants to products

00:04:34.050 --> 00:04:35.770
in our particular diagram,

00:04:35.770 --> 00:04:38.840
the rate of the forward
reaction must be greater

00:04:38.840 --> 00:04:41.250
than the rate of the reverse reaction.

00:04:41.250 --> 00:04:42.490
And we can see that.

00:04:42.490 --> 00:04:46.460
So before time is equal
to 20 seconds here,

00:04:46.460 --> 00:04:48.720
if we look at our rates, let's just pick,

00:04:48.720 --> 00:04:51.391
for example, time is equal to 10 seconds,

00:04:51.391 --> 00:04:53.140
for the forward reaction,

00:04:53.140 --> 00:04:57.473
there's a higher rate than
for the reverse reaction.

00:04:58.683 --> 00:05:00.428
So at times equal to 20 seconds,

00:05:00.428 --> 00:05:03.030
the rate of the forward reaction

00:05:03.030 --> 00:05:06.480
becomes equal to the rate
of the reverse reaction.

00:05:06.480 --> 00:05:09.070
So here's the line on our graph,

00:05:09.070 --> 00:05:11.010
where the rates become equal,

00:05:11.010 --> 00:05:14.480
and also notice the rates
become constant at this point.

00:05:14.480 --> 00:05:16.520
And when the rate of the forward reaction

00:05:16.520 --> 00:05:19.070
is equal to the rate of
the reverse reaction,

00:05:19.070 --> 00:05:21.220
the reaction has reached equilibrium.

00:05:21.220 --> 00:05:23.330
So to the right of the dotted line,

00:05:23.330 --> 00:05:25.830
the reaction is at equilibrium.

00:05:25.830 --> 00:05:27.780
And to the left of the dotted line,

00:05:27.780 --> 00:05:31.573
the reaction is not at equilibrium.

00:05:34.210 --> 00:05:36.550
And since the rate of the forward reaction

00:05:36.550 --> 00:05:39.983
is equal to the rate of the
reverse reaction at equilibrium,

00:05:39.983 --> 00:05:41.900
X2 is turning into 2X

00:05:41.900 --> 00:05:45.180
at the same rate that 2X
is turning back into X2.

00:05:45.180 --> 00:05:47.690
Therefore, the concentrations of X2

00:05:47.690 --> 00:05:50.480
and X at equilibrium remain constant.

00:05:50.480 --> 00:05:53.260
And we can see that when we
look at the particular diagrams

00:05:53.260 --> 00:05:54.710
where time is equal to 20 seconds,

00:05:54.710 --> 00:05:56.780
and time is equal to 30 seconds.

00:05:56.780 --> 00:06:01.780
So both of these particular
diagrams have two X2 particles

00:06:01.960 --> 00:06:05.733
and six X particles.

00:06:06.660 --> 00:06:09.530
Let's look at a summary of
what the rates of the forward

00:06:09.530 --> 00:06:11.110
and reverse reaction mean

00:06:11.110 --> 00:06:13.590
in terms of reactants and products.

00:06:13.590 --> 00:06:16.280
If the rate of the forward reaction

00:06:16.280 --> 00:06:19.450
is greater than the rate
of the reverse reaction,

00:06:19.450 --> 00:06:21.290
that means there's a net conversion

00:06:21.290 --> 00:06:23.170
of reactants to products.

00:06:23.170 --> 00:06:25.070
So therefore over time,

00:06:25.070 --> 00:06:26.970
the amount of reactants would decrease

00:06:26.970 --> 00:06:29.193
and the amount of products would increase.

00:06:30.040 --> 00:06:32.880
Eventually, the rate
of the forward reaction

00:06:32.880 --> 00:06:36.200
becomes equal to the rate
of the reverse reaction.

00:06:36.200 --> 00:06:38.860
And that means the
reaction is at equilibrium

00:06:38.860 --> 00:06:40.430
and there's no net change

00:06:40.430 --> 00:06:42.820
in the amounts of reactants or products.

00:06:42.820 --> 00:06:45.640
So reactants are turning into
products at the same rate

00:06:45.640 --> 00:06:48.850
the products are turning
back into reactants.

00:06:48.850 --> 00:06:51.610
And then finally, if the
rate of the reverse reaction

00:06:51.610 --> 00:06:54.710
is greater than the rate
of the forward reaction,

00:06:54.710 --> 00:06:57.430
there's a net conversion
of products to reactants.

00:06:57.430 --> 00:07:00.890
So products are turning into reactants

00:07:00.890 --> 00:07:04.150
faster than reactants are
turning into products.

00:07:04.150 --> 00:07:06.240
And the example that we looked at,

00:07:06.240 --> 00:07:08.270
the rate of the forward reaction

00:07:08.270 --> 00:07:10.910
was greater than the rate
of the reverse reaction.

00:07:10.910 --> 00:07:12.770
And eventually the rates became equal

00:07:12.770 --> 00:07:15.100
and the reaction reached equilibrium.

00:07:15.100 --> 00:07:17.750
If we had looked at an example
of this third case here

00:07:17.750 --> 00:07:19.920
where the rate of the
reverse reaction is greater

00:07:19.920 --> 00:07:22.300
than the rate of the forward reaction,

00:07:22.300 --> 00:07:24.530
eventually the two
rates would become equal

00:07:24.530 --> 00:07:27.393
and this reaction would
reach equilibrium too.


Dynamic equilibrium

WEBVTT
Kind: captions
Language: en

00:00:00.650 --> 00:00:02.940
- [Instructor] To illustrate
the concept of equilibrium,

00:00:02.940 --> 00:00:04.970
let's say that we have a beaker

00:00:04.970 --> 00:00:08.840
and we put some water into our beaker.

00:00:08.840 --> 00:00:13.840
And also we make sure that
our beaker has a lid on it.

00:00:14.220 --> 00:00:17.210
Some of those water molecules
are going to evaporate

00:00:17.210 --> 00:00:19.160
and turn into a gas.

00:00:19.160 --> 00:00:22.050
And eventually once we
have enough gaseous water,

00:00:22.050 --> 00:00:24.650
some of the gaseous water
is going to condense

00:00:24.650 --> 00:00:27.610
and turn back into liquid water.

00:00:27.610 --> 00:00:29.250
To represent these two processes,

00:00:29.250 --> 00:00:31.120
we can show liquid water on the left

00:00:31.120 --> 00:00:33.370
and gaseous water on the right.

00:00:33.370 --> 00:00:35.220
So in the forward process,

00:00:35.220 --> 00:00:37.630
liquid water turns into gaseous water.

00:00:37.630 --> 00:00:39.460
And this forward arrow here

00:00:39.460 --> 00:00:42.490
represents the process of vaporization.

00:00:42.490 --> 00:00:45.490
And when gaseous water turns
back into liquid water,

00:00:45.490 --> 00:00:47.930
that's represented by this
arrow here on the bottom,

00:00:47.930 --> 00:00:51.240
and so that's the process of condensation.

00:00:51.240 --> 00:00:53.430
Since we start with liquid water,

00:00:53.430 --> 00:00:55.960
at first, the rate of vaporization

00:00:55.960 --> 00:00:59.100
is greater than the rate of condensation.

00:00:59.100 --> 00:01:01.860
But eventually we reach
a point where the rate of

00:01:01.860 --> 00:01:05.800
vaporization is equal to
the rate of condensation.

00:01:05.800 --> 00:01:09.340
And when that happens, if
you're turning liquid water

00:01:09.340 --> 00:01:11.470
into gaseous water at the same rate,

00:01:11.470 --> 00:01:14.890
you're turning gaseous water
back into liquid water,

00:01:14.890 --> 00:01:18.330
the number of water molecules
in a liquid and gaseous state

00:01:18.330 --> 00:01:20.420
would remain constant.

00:01:20.420 --> 00:01:23.570
So when the rate of vaporization
is equal to the rate of

00:01:23.570 --> 00:01:27.700
condensation, we've reached
a state of equilibrium.

00:01:27.700 --> 00:01:31.720
And this is a dynamic equilibrium
because if we zoom in and

00:01:31.720 --> 00:01:35.170
we look at this, water
molecules are being converted

00:01:35.170 --> 00:01:38.220
from the liquid state to the
gaseous state all the time

00:01:38.220 --> 00:01:40.970
and molecules are going from
the gaseous state back to the

00:01:40.970 --> 00:01:42.190
liquid state all the time.

00:01:42.190 --> 00:01:44.890
However, since the rates are equal,

00:01:44.890 --> 00:01:46.670
the number of molecules in the liquid

00:01:46.670 --> 00:01:48.660
and gaseous state remained constant.

00:01:48.660 --> 00:01:51.340
And if we look at it from a
macroscopic point of view,

00:01:51.340 --> 00:01:54.640
the level of water wouldn't change at all.

00:01:54.640 --> 00:01:57.660
Now let's apply this concept
of dynamic equilibrium

00:01:57.660 --> 00:02:00.440
to a hypothetical chemical reaction.

00:02:00.440 --> 00:02:04.140
In our hypothetical reaction
X2 which is a reddish brown

00:02:04.140 --> 00:02:09.140
gas, decomposes into its
individual atoms to form 2X

00:02:09.740 --> 00:02:12.610
and individual atoms are colorless.

00:02:12.610 --> 00:02:17.610
So in the forward reaction,
we're going from X2 to 2X.

00:02:18.414 --> 00:02:20.960
So X2 is decomposing to 2X.

00:02:20.960 --> 00:02:25.140
And in the reverse
reaction, the two atoms of X

00:02:25.140 --> 00:02:28.870
are combining together to form X2.

00:02:28.870 --> 00:02:32.040
When we have a forward
reaction and a reverse reaction

00:02:32.040 --> 00:02:35.190
by convention, we say,
what's on the left side,

00:02:35.190 --> 00:02:39.760
are the reactants, and
what's on the right side,

00:02:39.760 --> 00:02:41.150
are the products.

00:02:41.150 --> 00:02:45.023
And by using these terms,
we can avoid confusion.

00:02:46.000 --> 00:02:48.990
Let's say that we start our
reaction with only reactants.

00:02:48.990 --> 00:02:53.040
So only X2 is present in
this first container here.

00:02:53.040 --> 00:02:57.250
And there are five particles of X2.

00:02:57.250 --> 00:03:01.060
If every particle represents 0.1 moles,

00:03:01.060 --> 00:03:03.580
since we have five particles of X2,

00:03:03.580 --> 00:03:07.230
we have 0.5 moles of X2.

00:03:07.230 --> 00:03:10.325
And let's say, this is
a one liter container.

00:03:10.325 --> 00:03:14.290
0.5 divided by one would be 0.5 molar.

00:03:14.290 --> 00:03:19.290
So the initial concentration
of gas, X2 is 0.5 molar.

00:03:21.290 --> 00:03:24.250
And since we don't have any of the X,

00:03:24.250 --> 00:03:27.220
there are no white dots
in this box, right?

00:03:27.220 --> 00:03:30.723
The initial concentration
of X would be zero molar.

00:03:32.170 --> 00:03:34.748
So let me just write that in here, 0M.

00:03:34.748 --> 00:03:36.330
Next we wait 10 seconds.

00:03:36.330 --> 00:03:38.077
So we start off at time's
equal to zero seconds,

00:03:38.077 --> 00:03:40.650
and now we're at time's
equal to 10 seconds.

00:03:40.650 --> 00:03:41.860
And now we can see, there are

00:03:41.860 --> 00:03:46.400
three particles of X2 in our box.

00:03:46.400 --> 00:03:49.260
And so that would be 0.3M,

00:03:49.260 --> 00:03:51.940
so let's go ahead and write 0.3M in

00:03:51.940 --> 00:03:53.490
for our concentration.

00:03:53.490 --> 00:03:56.370
And now we have some particles of X.

00:03:56.370 --> 00:04:00.230
There are one, two, three, four particles.

00:04:00.230 --> 00:04:04.060
And once again, if each
particle represents 0.1 moles

00:04:04.060 --> 00:04:08.133
that's 0.4 moles of X
divided by one or 0.4M.

00:04:11.930 --> 00:04:13.310
We wait another 10 seconds,

00:04:13.310 --> 00:04:15.940
so when time is equal to 20 seconds,

00:04:15.940 --> 00:04:19.130
now there are two particles of X2

00:04:19.130 --> 00:04:23.820
and one, two, three, four,
five, six particles of X.

00:04:23.820 --> 00:04:28.820
So now the concentrations are
0.2M for X2 and 0.6M for X.

00:04:33.640 --> 00:04:36.900
We wait another 10 seconds
for a total of 30 seconds.

00:04:36.900 --> 00:04:40.330
And there are still two particles of X2

00:04:40.330 --> 00:04:43.170
and six particles of X.

00:04:43.170 --> 00:04:46.000
And so the concentrations
after 30 seconds,

00:04:46.000 --> 00:04:49.160
the concentration of X2
is 0.2M and of X is 0.6M.

00:04:54.350 --> 00:04:57.728
Notice how the concentration
of X2 went from 0.5M

00:04:57.728 --> 00:05:02.728
to 0.3M to 0.2, and then it
was also 0.2 after 30 seconds.

00:05:03.430 --> 00:05:07.860
So it became constant when
time is equal to 20 seconds.

00:05:07.860 --> 00:05:10.410
The concentration of X went
from zero to 0.4 to 0.6

00:05:11.760 --> 00:05:15.560
and then it was also 0.6 after 30 seconds.

00:05:15.560 --> 00:05:18.650
So the concentrations became
constant when time is equal to

00:05:18.650 --> 00:05:22.440
20 seconds, which means the
reaction reached equilibrium

00:05:22.440 --> 00:05:24.160
after 20 seconds.

00:05:24.160 --> 00:05:26.770
So at time is equal to zero
was not at equilibrium.

00:05:26.770 --> 00:05:29.900
When time is equal to 10 seconds,
it was not at equilibrium.

00:05:29.900 --> 00:05:32.220
Only when time was equal to 20 seconds,

00:05:32.220 --> 00:05:33.700
did it reach equilibrium.

00:05:33.700 --> 00:05:37.470
And that equilibrium, the
rate of the forward reaction

00:05:37.470 --> 00:05:40.530
is equal to the rate of
the reverse reaction.

00:05:40.530 --> 00:05:44.440
And if that's true, then
X2 is being turned into 2X

00:05:44.440 --> 00:05:49.080
at the same rate that 2X is
being turned back into X2.

00:05:49.080 --> 00:05:53.490
And if those rates are equal,
the concentrations of X2

00:05:53.490 --> 00:05:56.893
and X at equilibrium
would remain constant.

00:05:57.800 --> 00:05:59.280
We can see the same concept,

00:05:59.280 --> 00:06:03.440
if you look at a graph of
concentration versus time.

00:06:03.440 --> 00:06:06.840
Concentration of X2 starts
at 0.5M when time is

00:06:06.840 --> 00:06:10.270
equal to zero seconds,
and then drops to 0.3M

00:06:10.270 --> 00:06:11.400
after 10 seconds.

00:06:11.400 --> 00:06:14.220
And after 20 seconds, it's at 0.2M

00:06:14.220 --> 00:06:16.480
and then stays constant after that.

00:06:16.480 --> 00:06:19.650
For the concentration of
X, we start out at 0M,

00:06:19.650 --> 00:06:22.560
we increase 2.4 after 10 seconds,

00:06:22.560 --> 00:06:25.640
then we're at 0.6 and
then we are constant.

00:06:25.640 --> 00:06:27.660
So if you think about a line,

00:06:27.660 --> 00:06:31.280
if we just draw a dash
line here at 20 seconds.

00:06:31.280 --> 00:06:34.014
That's the dividing line
between on the left,

00:06:34.014 --> 00:06:36.750
where we're not at equilibrium.

00:06:36.750 --> 00:06:40.000
And so the concentrations
are always changing.

00:06:40.000 --> 00:06:42.290
And then to the right of that dotted line,

00:06:42.290 --> 00:06:44.260
we are at equilibrium where

00:06:44.260 --> 00:06:47.000
the concentrations remain constant.

00:06:47.000 --> 00:06:51.873
So the equilibrium concentration
of X2 gas is equal to

00:06:51.873 --> 00:06:56.220
0.2M and the equilibrium concentration

00:06:56.220 --> 00:07:00.470
of X gas is equal to 0.6M.

00:07:00.470 --> 00:07:02.850
Finally, let's use these
particular diagrams

00:07:02.850 --> 00:07:04.450
to think about what we would see

00:07:04.450 --> 00:07:06.630
at a macroscopic level as the

00:07:06.630 --> 00:07:08.980
reaction proceeds to equilibrium.

00:07:08.980 --> 00:07:10.520
And the first particulate diagram

00:07:10.520 --> 00:07:12.752
we see only red particles.

00:07:12.752 --> 00:07:17.752
So only our reactants X2 are
present in the beginning.

00:07:18.010 --> 00:07:20.097
However, as time goes on,
the number of red particles

00:07:20.097 --> 00:07:23.620
decreases from five in the
first particular diagram

00:07:23.620 --> 00:07:26.650
to three in the second to two

00:07:26.650 --> 00:07:30.140
and then the number stays at
two because remember we reach

00:07:30.140 --> 00:07:33.500
equilibrium after 20 seconds.

00:07:33.500 --> 00:07:36.360
So what we would see from a
macroscopic point of view,

00:07:36.360 --> 00:07:40.223
is we'd start out with a
reaction vessel that is a darker

00:07:40.223 --> 00:07:44.340
brown red, and then it would
be a lighter brown red.

00:07:44.340 --> 00:07:46.670
And then finally, when
we reach equilibrium

00:07:46.670 --> 00:07:48.180
and even lighter brown red,

00:07:48.180 --> 00:07:51.680
and it would stay that
same light brown red,

00:07:51.680 --> 00:07:54.207
because we've reached equilibrium
and the concentrations

00:07:54.207 --> 00:07:58.400
of reactants and products
remain constant at equilibrium.

00:07:58.400 --> 00:08:02.520
Even though our reactants are
turning into our products.

00:08:02.520 --> 00:08:05.360
Our products are turning back
into our reactants at the same

00:08:05.360 --> 00:08:07.440
rate and therefore the concentrations

00:08:07.440 --> 00:08:10.143
of both reactants and
products are constant.