Worked example: Calculating solubility from Kₛₚ

WEBVTT
Kind: captions
Language: en

00:00:00.170 --> 00:00:01.003
- [Instructor] Let's calculate

00:00:01.003 --> 00:00:03.270
the molar solubility of calcium fluoride

00:00:03.270 --> 00:00:05.620
if the Ksp value for calcium fluoride is

00:00:05.620 --> 00:00:10.540
3.9 times 10 to the negative
11th at 25 degrees Celsius.

00:00:10.540 --> 00:00:11.400
The first step is

00:00:11.400 --> 00:00:15.710
to write the dissolution
equation for calcium fluoride.

00:00:15.710 --> 00:00:20.170
So, solid calcium fluoride
will dissolve in solution

00:00:20.170 --> 00:00:25.170
to form aqueous calcium two
plus ions and fluoride anions.

00:00:26.100 --> 00:00:28.960
And to balance that out,
we need to make sure

00:00:28.960 --> 00:00:32.840
and include a two in front
of the fluoride anions.

00:00:32.840 --> 00:00:34.900
The next step is to set up an ICE table,

00:00:34.900 --> 00:00:38.260
where I stands for initial concentration,

00:00:38.260 --> 00:00:40.930
C stands for the change in concentration,

00:00:40.930 --> 00:00:44.910
and E stands for
equilibrium concentration.

00:00:44.910 --> 00:00:47.420
Before any of the solid
calcium fluoride dissolves,

00:00:47.420 --> 00:00:50.190
the initial concentrations
of calcium two plus ions

00:00:50.190 --> 00:00:52.970
and fluoride anions in solution is zero.

00:00:52.970 --> 00:00:54.830
So we can go ahead and put a zero in here

00:00:54.830 --> 00:00:58.740
for the initial concentration
of the ions in solution.

00:00:58.740 --> 00:01:00.680
Some of the calcium
fluoride will dissolve,

00:01:00.680 --> 00:01:02.070
and we don't know how much.

00:01:02.070 --> 00:01:06.500
So I like to represent that by
writing -X on the ICE table,

00:01:06.500 --> 00:01:09.090
where X is the concentration
of calcium fluoride

00:01:09.090 --> 00:01:10.400
that dissolves.

00:01:10.400 --> 00:01:13.350
Looking at the mole ratios,
it's a one-to-one mole ratio

00:01:13.350 --> 00:01:16.720
between calcium fluoride
and calcium two plus ions.

00:01:16.720 --> 00:01:17.990
So if we're losing X

00:01:17.990 --> 00:01:20.100
for the concentration of calcium fluoride,

00:01:20.100 --> 00:01:21.570
we must be gaining X

00:01:21.570 --> 00:01:24.650
for the concentration of
calcium two plus ions.

00:01:24.650 --> 00:01:27.180
And since it's a one-to-two mole ratio

00:01:27.180 --> 00:01:30.300
for calcium two plus
ions to fluoride anions,

00:01:30.300 --> 00:01:33.250
if we're gaining +X for calcium two plus,

00:01:33.250 --> 00:01:35.350
we must gain plus +2X for fluoride anions.

00:01:36.890 --> 00:01:40.790
So the equilibrium concentration
of calcium two plus ions is

00:01:40.790 --> 00:01:42.930
zero plus X, or just X,

00:01:42.930 --> 00:01:45.870
and the equilibrium concentration
of fluoride anions will be

00:01:45.870 --> 00:01:49.070
zero plus 2X, or just 2X.

00:01:49.070 --> 00:01:51.730
The next step is to
write the Ksp expression

00:01:51.730 --> 00:01:53.460
from the balanced equation.

00:01:53.460 --> 00:01:55.360
So Ksp is equal to

00:01:55.360 --> 00:01:58.910
the concentration of
calcium two plus ions,

00:01:58.910 --> 00:02:00.800
and since there's a coefficient of one

00:02:00.800 --> 00:02:01.883
in the balanced equation,

00:02:01.883 --> 00:02:04.015
that's the concentration
of calcium two plus ions

00:02:04.015 --> 00:02:05.865
raised to the first power,

00:02:05.865 --> 00:02:08.906
times the concentration
of fluoride anions,

00:02:08.906 --> 00:02:10.615
and since there is a coefficient of two

00:02:10.615 --> 00:02:12.136
in the balanced equation,

00:02:12.136 --> 00:02:14.610
it's the concentration of
fluoride anions raised to

00:02:14.610 --> 00:02:16.040
the second power.

00:02:16.040 --> 00:02:17.790
Pure solids are not included

00:02:17.790 --> 00:02:19.760
in equilibrium constant expression.

00:02:19.760 --> 00:02:21.830
So we're going to leave calcium fluoride

00:02:21.830 --> 00:02:24.540
out of the Ksp expression.

00:02:24.540 --> 00:02:27.150
The concentration of ions
in our Ksp expression

00:02:27.150 --> 00:02:29.740
are equilibrium concentrations.

00:02:29.740 --> 00:02:31.220
Therefore we can plug in X

00:02:31.220 --> 00:02:34.360
for the equilibrium
concentration of calcium two plus

00:02:34.360 --> 00:02:39.000
and 2X for the equilibrium
concentration of fluoride anions.

00:02:39.000 --> 00:02:44.000
We can also plug in the Ksp
value for calcium fluoride.

00:02:44.490 --> 00:02:45.390
So that would give us

00:02:45.390 --> 00:02:50.390
3.9 times 10 to the
negative 11th is equal to

00:02:50.740 --> 00:02:55.330
X times 2X squared.

00:02:55.330 --> 00:02:57.190
Next we need to solve for X.

00:02:57.190 --> 00:03:01.410
So, 3.9 times 10 to the
negative 11th is equal to

00:03:01.410 --> 00:03:03.130
X times 2X squared.

00:03:03.130 --> 00:03:04.570
Well, 2X squared is equal to

00:03:04.570 --> 00:03:08.910
4X squared times X is equal to 4X cubed.

00:03:08.910 --> 00:03:13.410
So to solve for X, we need
to divide both sides by four

00:03:13.410 --> 00:03:17.980
and then take the cube root of both sides.

00:03:17.980 --> 00:03:21.100
So we'd take the cube
root of the left side

00:03:21.100 --> 00:03:25.683
and the cube root of X cubed.

00:03:26.620 --> 00:03:28.562
That gives us X is equal to

00:03:28.562 --> 00:03:33.562
2.1 times 10 to the negative fourth.

00:03:33.980 --> 00:03:36.100
And looking at our ICE table, X represents

00:03:36.100 --> 00:03:39.740
the equilibrium concentration
of calcium two plus ions.

00:03:39.740 --> 00:03:43.010
So 2.1 times 10 to the
negative fourth molar is

00:03:43.010 --> 00:03:46.700
the equilibrium concentration
of calcium two plus ions.

00:03:46.700 --> 00:03:48.110
For the fluoride anions,

00:03:48.110 --> 00:03:50.620
the equilibrium concentration is 2X.

00:03:50.620 --> 00:03:55.620
So two times 2.1 times 10 to
the negative fourth is 4.2,

00:03:55.690 --> 00:03:56.832
let me go ahead and write that down here,

00:03:56.832 --> 00:04:00.210
4.2 times 10 to the negative fourth molar

00:04:00.210 --> 00:04:03.860
for the equilibrium
concentration of fluoride anions.

00:04:03.860 --> 00:04:04.900
Our goal was to calculate

00:04:04.900 --> 00:04:07.410
the molar solubility of calcium fluoride.

00:04:07.410 --> 00:04:09.300
And molar solubility refers to

00:04:09.300 --> 00:04:12.470
the concentration of
our salt that dissolved

00:04:12.470 --> 00:04:16.030
to form a saturated
solution at equilibrium.

00:04:16.030 --> 00:04:17.460
So if X refers to

00:04:17.460 --> 00:04:21.430
the concentration of calcium
two plus ions at equilibrium,

00:04:21.430 --> 00:04:23.620
looking at our mole ratios, that's also

00:04:23.620 --> 00:04:26.790
the concentration of calcium
fluoride that dissolved.

00:04:26.790 --> 00:04:30.020
Therefore, 2.1 times 10 to
the negative fourth molar

00:04:30.020 --> 00:04:34.640
is also the molar solubility
of calcium fluoride.

00:04:34.640 --> 00:04:37.770
Technically at a constant
temperature of 25 degrees,

00:04:37.770 --> 00:04:40.690
the concentration of a
solid doesn't change.

00:04:40.690 --> 00:04:44.170
And so you'll see most
textbooks not to put in -X

00:04:44.170 --> 00:04:46.580
on the ICE table. I like
to just put it in though

00:04:46.580 --> 00:04:49.630
to remind me that X in
this case does refer to

00:04:49.630 --> 00:04:51.263
the molar solubility.


Introduction to solubility equilibria

WEBVTT
Kind: captions
Language: en

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- [Instructor] Let's say we have a beaker

00:00:01.870 --> 00:00:04.700
of distilled water at 25 degrees Celsius.

00:00:04.700 --> 00:00:07.750
And to the beaker, we
add some barium sulfate.

00:00:07.750 --> 00:00:10.810
Barium sulfate is a white solid.

00:00:10.810 --> 00:00:12.420
A small amount of the barium sulfate

00:00:12.420 --> 00:00:15.870
dissolves in the water and
forms Ba2+ ions in solution

00:00:15.870 --> 00:00:17.980
and sulfate ions in solution.

00:00:17.980 --> 00:00:20.350
So let me draw those in on our diagrams.

00:00:20.350 --> 00:00:22.520
So we're gonna form some Ba2+ ions

00:00:22.520 --> 00:00:25.530
and some sulfate anions.

00:00:25.530 --> 00:00:28.620
But most of the barium
sulfate remains undissolved

00:00:28.620 --> 00:00:31.180
and so we'll draw that
here sitting on the bottom

00:00:31.180 --> 00:00:33.040
of the beaker.

00:00:33.040 --> 00:00:36.650
So barium sulfate can
dissolve to form Ba2+ ions

00:00:36.650 --> 00:00:38.820
and sulfate anions in solution.

00:00:38.820 --> 00:00:41.440
And it's possible for the Ba2+ ion

00:00:41.440 --> 00:00:44.850
to combine with the sulfate
anion to form a precipitate,

00:00:44.850 --> 00:00:46.460
of barium sulfate.

00:00:46.460 --> 00:00:49.470
When the rate of dissolution
is equal to the rate of

00:00:49.470 --> 00:00:53.690
precipitation, the
system is at equilibrium.

00:00:53.690 --> 00:00:55.740
These types of equilibria are referred

00:00:55.740 --> 00:00:58.880
to as solubility equilibria.

00:00:58.880 --> 00:01:00.950
And when the system is at equilibrium,

00:01:00.950 --> 00:01:04.060
the concentrations of Ba2+
ions and sulfate anions

00:01:04.060 --> 00:01:05.630
solution are constant.

00:01:05.630 --> 00:01:08.780
And the amount of solid is constant too.

00:01:08.780 --> 00:01:11.170
And this forms a saturated solution.

00:01:11.170 --> 00:01:13.370
The balanced equation
shows the dissolution

00:01:13.370 --> 00:01:15.730
of a salt barium sulfate.

00:01:15.730 --> 00:01:17.070
And from the balanced equation,

00:01:17.070 --> 00:01:20.170
we can write an equilibrium
constant expression.

00:01:20.170 --> 00:01:23.015
So we would write the
equilibrium constant K

00:01:23.015 --> 00:01:27.410
is equal to the concentration of Ba2+

00:01:27.410 --> 00:01:29.470
and since there's a coefficient of one

00:01:29.470 --> 00:01:30.620
in the balanced equation,

00:01:30.620 --> 00:01:33.200
it'd be the concentration
raised to the first power

00:01:33.200 --> 00:01:35.640
times the concentration of sulfate

00:01:35.640 --> 00:01:39.000
also raised to the first power.

00:01:39.000 --> 00:01:42.110
And since pure solids are left
out of equilibrium constant

00:01:42.110 --> 00:01:46.510
expressions, we would not
include the solid barium sulfate.

00:01:46.510 --> 00:01:48.510
For solubility equilibria,

00:01:48.510 --> 00:01:53.510
we would write Ksp where sp
stands for solubility product.

00:01:53.630 --> 00:01:58.372
The solubility product constant
Ksp has only one value for a

00:01:58.372 --> 00:02:01.180
given salt at a specific temperature.

00:02:01.180 --> 00:02:04.340
That temperature is
usually 25 degrees Celsius.

00:02:04.340 --> 00:02:08.940
And Ksp indicates how much
of that salt will dissolve.

00:02:08.940 --> 00:02:11.430
For example, at 25 degrees Celsius,

00:02:11.430 --> 00:02:13.960
the Ksp value for barium sulfate

00:02:13.960 --> 00:02:16.930
is 1.1 times 10 to the negative 10th.

00:02:16.930 --> 00:02:19.670
When the Ksp value is much less than one,

00:02:19.670 --> 00:02:22.120
that indicates the salt
is not very soluble.

00:02:22.120 --> 00:02:25.440
So barium sulfate is not a soluble salt.

00:02:25.440 --> 00:02:28.610
If the Ksp value is greater than one,

00:02:28.610 --> 00:02:31.640
like it is for something
like sodium chloride,

00:02:31.640 --> 00:02:33.610
that indicates a soluble salt

00:02:33.610 --> 00:02:36.070
that dissolves easily in water.

00:02:36.070 --> 00:02:39.880
The solubility of a substance
refers to the amount of solid

00:02:39.880 --> 00:02:42.930
that dissolves to form
a saturated solution.

00:02:42.930 --> 00:02:47.930
Usually the units for solubility
are in grams per liter.

00:02:48.440 --> 00:02:52.140
Molar solubility refers to the
number of moles of the solid

00:02:52.140 --> 00:02:55.460
that dissolve to form one litter
of the saturated solution.

00:02:55.460 --> 00:02:59.550
And therefore the units
would be moles per one liter

00:02:59.550 --> 00:03:01.133
or you could just write M.

00:03:03.680 --> 00:03:07.030
Ksp values can be used to
predict the relative solubilities

00:03:07.030 --> 00:03:11.030
of salts that produce the same
number of ions in solution.

00:03:11.030 --> 00:03:13.560
For example, silver
chloride, silver bromide,

00:03:13.560 --> 00:03:17.750
and silver iodide all
produce two ions in solution.

00:03:17.750 --> 00:03:20.880
Let's look at the dissolution
equation for silver chloride

00:03:20.880 --> 00:03:22.470
to see why this is true.

00:03:22.470 --> 00:03:24.720
Solid silver chloride
turns into Ag+ and Cl-.

00:03:26.500 --> 00:03:30.370
So that's one Ag+ ion and
one Cl- ion for a total

00:03:30.370 --> 00:03:32.590
of two ions in solution.

00:03:32.590 --> 00:03:35.140
And we could write out similar
equations for silver bromide

00:03:35.140 --> 00:03:38.920
and silver iodide, so they all
produce two ions in solution.

00:03:38.920 --> 00:03:41.500
However, a salt like Lead(II)chloride

00:03:41.500 --> 00:03:43.940
produces three ions in solution.

00:03:43.940 --> 00:03:47.210
So Lead(II)chloride
would give one Pb2+ ion

00:03:47.210 --> 00:03:49.680
and two chloride anions in solution.

00:03:49.680 --> 00:03:51.900
One plus two is three ions.

00:03:51.900 --> 00:03:55.890
Since Lead(II)chloride produces
three ions in solution,

00:03:55.890 --> 00:03:59.360
we can determine its solubility
relative to the other

00:03:59.360 --> 00:04:02.040
three by comparing Ksp values.

00:04:02.040 --> 00:04:04.390
Here are the Ksp values
for the three salts

00:04:04.390 --> 00:04:06.250
at 25 degrees Celsius.

00:04:06.250 --> 00:04:09.530
For silver chloride, it's 1.8
times 10 to the negative 10th,

00:04:09.530 --> 00:04:12.950
for silver bromide, it's 5.0
times 10 to the negative 13th.

00:04:12.950 --> 00:04:15.246
And for silver iodide, it's 8.3 times 10

00:04:15.246 --> 00:04:17.270
to the negative 17th.

00:04:17.270 --> 00:04:20.130
When comparing salts that
produce the same number of ions,

00:04:20.130 --> 00:04:22.030
the higher the value of Ksp,

00:04:22.030 --> 00:04:25.020
the higher the solubility of the salt.

00:04:25.020 --> 00:04:29.190
And since silver chloride has
the highest Ksp value of these

00:04:29.190 --> 00:04:33.520
three, silver chloride is
the most soluble salts.

00:04:33.520 --> 00:04:35.340
For some insight into why this is true,

00:04:35.340 --> 00:04:38.200
let's look at the Ksp
expression for silver chloride,

00:04:38.200 --> 00:04:40.990
which we can get from
the balanced equation.

00:04:40.990 --> 00:04:43.010
The higher the value for Ksp,

00:04:43.010 --> 00:04:46.290
the higher the concentration
of these ions at equilibrium,

00:04:46.290 --> 00:04:49.490
which means that more of the
solid must have dissolved.

00:04:49.490 --> 00:04:52.540
Therefore, silver chloride
has the highest solubility

00:04:52.540 --> 00:04:54.680
out of these three salts.

00:04:54.680 --> 00:04:58.030
Let's say we have some solid
calcium fluoride that we add to

00:04:58.030 --> 00:05:01.300
pure water at 25 degrees Celsius.

00:05:01.300 --> 00:05:04.830
Eventually equilibrium is
reached and the equilibrium

00:05:04.830 --> 00:05:07.523
concentration of Ca2+ ions is measured

00:05:07.523 --> 00:05:11.290
to be 2.1 times 10 to the negative 4th M.

00:05:11.290 --> 00:05:15.040
Our goal is to calculate
the Ksp for calcium fluoride

00:05:15.040 --> 00:05:17.490
at 25 degrees Celsius.

00:05:17.490 --> 00:05:19.240
The first step is to write out the

00:05:19.240 --> 00:05:22.730
dissolution equation for calcium fluoride.

00:05:22.730 --> 00:05:27.180
So we would write CaF2 solid,
and we know that calcium forms

00:05:27.180 --> 00:05:32.180
a 2+ cation, so we would
write Ca2+ in aqueous solution

00:05:34.040 --> 00:05:37.210
and to balance everything
we'd need two fluoride anions.

00:05:37.210 --> 00:05:41.770
So 2F- also in aqueous solution.

00:05:41.770 --> 00:05:43.790
The next step is to use
the balanced equation

00:05:43.790 --> 00:05:45.950
to write the Ksp expression.

00:05:45.950 --> 00:05:50.180
So Ksp is equal to, there's
a one as a coefficient

00:05:50.180 --> 00:05:53.000
in front of Ca2+ so it'd
be the concentration

00:05:53.000 --> 00:05:56.360
of Ca2+ raise to the first power,

00:05:56.360 --> 00:05:59.220
times the concentration of fluoride anion

00:05:59.220 --> 00:06:01.160
and since there's a two as a coefficient,

00:06:01.160 --> 00:06:04.490
this is the concentration
of fluoride anion squared.

00:06:04.490 --> 00:06:08.800
For a Ksp expression, these
are equilibrium concentrations,

00:06:08.800 --> 00:06:12.260
and we already know the
concentration of Ca2+

00:06:12.260 --> 00:06:16.260
at equilibrium is 2.1 times
10 to the negative 4th.

00:06:16.260 --> 00:06:17.750
So that can be plugged in for the

00:06:17.750 --> 00:06:21.160
equilibrium concentration of Ca2+.

00:06:21.160 --> 00:06:23.760
So here's our expression with 2.1 times 10

00:06:23.760 --> 00:06:25.450
to the negative 4th plugged in,

00:06:25.450 --> 00:06:27.540
and next we need to
plug in the equilibrium

00:06:27.540 --> 00:06:30.090
concentration of fluoride anion.

00:06:30.090 --> 00:06:32.370
We're looking at the disillusion equation,

00:06:32.370 --> 00:06:37.370
the mole ratio of Ca2+
to fluoride anion is 1:2.

00:06:37.650 --> 00:06:40.300
Therefore, at equilibrium,
there's twice as many

00:06:40.300 --> 00:06:44.440
fluoride ions in solution
as there are Ca2+ ions.

00:06:44.440 --> 00:06:47.190
Therefore the equilibrium
concentration of the fluoride

00:06:47.190 --> 00:06:51.960
anion would just be twice
this concentration for Ca2+.

00:06:51.960 --> 00:06:55.490
So the equilibrium
concentration of fluoride anion

00:06:55.490 --> 00:06:59.533
must be 4.2 times 10
to the negative 4th M.

00:07:00.986 --> 00:07:03.360
And when you do the math, you get that Ksp

00:07:03.360 --> 00:07:06.750
for calcium fluoride is
equal to 3.7 times 10

00:07:06.750 --> 00:07:10.690
to the negative 11th
at 25 degrees Celsius.

00:07:10.690 --> 00:07:13.320
Ksp values can be difficult to measure

00:07:13.320 --> 00:07:16.060
and therefore different sources
often give different values

00:07:16.060 --> 00:07:18.990
for Ksp at the same temperature.

00:07:18.990 --> 00:07:22.690
For example, for calcium
fluoride at 25 degrees Celsius,

00:07:22.690 --> 00:07:26.110
one source had Ksp equal to 3.5 times 10

00:07:26.110 --> 00:07:27.520
to the negative 11th.

00:07:27.520 --> 00:07:32.090
Another one had 3.9 times
10 to the negative 11th.

00:07:32.090 --> 00:07:35.332
Since we got 3.7 times
10 to the negative 11th,

00:07:35.332 --> 00:07:38.140
this sounds like a pretty good
calculation for the numbers

00:07:38.140 --> 00:07:40.390
that we used for our problem.

00:07:40.390 --> 00:07:42.950
Finally, let's think
about the molar solubility

00:07:42.950 --> 00:07:44.260
of calcium fluoride.

00:07:44.260 --> 00:07:47.480
So how many moles of our
salt dissolve to form

00:07:47.480 --> 00:07:50.460
one litter of our saturated solution?

00:07:50.460 --> 00:07:53.267
Well, the mole ratio of Ca2+ ions

00:07:53.267 --> 00:07:56.349
to calcium fluoride is 1:1.

00:07:56.349 --> 00:08:01.349
Therefore, the concentration
of Ca2+ ions in solution 2.1

00:08:01.420 --> 00:08:03.770
times 10 to the negative 4th M,

00:08:03.770 --> 00:08:06.278
that number must also
be the molar solubility

00:08:06.278 --> 00:08:08.340
of calcium fluoride.

00:08:08.340 --> 00:08:10.420
Therefore, for this problem,
we could say that we use the

00:08:10.420 --> 00:08:13.100
molar solubility of calcium
fluoride to calculate

00:08:13.100 --> 00:08:15.673
the Ksp value for calcium fluoride.


Worked example: Using Le Chȃtelier’s principle to predict shifts in equilibrium

WEBVTT
Kind: captions
Language: en

00:00:00.490 --> 00:00:03.210
- [Instructor] Carbon monoxide
will react with hydrogen gas

00:00:03.210 --> 00:00:05.650
to produce methanol.

00:00:05.650 --> 00:00:07.770
Let's say that the
reaction is at equilibrium

00:00:07.770 --> 00:00:09.310
and our job is to figure out

00:00:09.310 --> 00:00:11.100
which direction the
equilibrium will shift,

00:00:11.100 --> 00:00:13.620
to the left, to the right, or not at all,

00:00:13.620 --> 00:00:17.390
as we try to make changes to
the reaction at equilibrium.

00:00:17.390 --> 00:00:19.640
For example, if we add some hydrogen gas

00:00:19.640 --> 00:00:21.650
to our reaction at equilibrium,

00:00:21.650 --> 00:00:25.690
we're increasing the concentration
of one of our reactants.

00:00:25.690 --> 00:00:27.360
According to the Le Chatelier's principle,

00:00:27.360 --> 00:00:29.130
the net reaction will
move in the direction

00:00:29.130 --> 00:00:32.420
that decreases the stress
placed on the system.

00:00:32.420 --> 00:00:34.170
So if the stress is increased amount

00:00:34.170 --> 00:00:35.860
of one of the reactants,

00:00:35.860 --> 00:00:37.900
the equilibrium will shift to the right

00:00:37.900 --> 00:00:40.610
to get rid of some of that reactant.

00:00:40.610 --> 00:00:43.370
In part B, some methanol is removed.

00:00:43.370 --> 00:00:47.790
So if we're decreasing the
concentration of our product,

00:00:47.790 --> 00:00:51.150
the equilibrium's gonna shift
to make more of our product,

00:00:51.150 --> 00:00:54.840
therefore, the equilibrium
will shift to the right.

00:00:54.840 --> 00:00:56.620
Next, the volume is increased

00:00:56.620 --> 00:00:58.780
on the reaction at equilibrium.

00:00:58.780 --> 00:01:03.660
And if we increase the volume,
we decrease the pressure,

00:01:03.660 --> 00:01:05.440
therefore, we could consider the stress

00:01:05.440 --> 00:01:07.280
to be decreased pressure.

00:01:07.280 --> 00:01:09.850
Le Chatelier's principle says
the net reaction is gonna go

00:01:09.850 --> 00:01:11.640
in the direction that relieves the stress.

00:01:11.640 --> 00:01:13.890
So if the stress is decreased pressure,

00:01:13.890 --> 00:01:18.140
the net reaction is going to
shift to increase the pressure.

00:01:18.140 --> 00:01:19.790
And we can figure out
which direction that is

00:01:19.790 --> 00:01:21.760
by looking at the balanced equation.

00:01:21.760 --> 00:01:23.000
On the reactant side,

00:01:23.000 --> 00:01:25.340
there's one mole of gas
and two moles of gas

00:01:25.340 --> 00:01:27.310
for a total of three moles of gas.

00:01:27.310 --> 00:01:30.040
On the product side, there's
only one mole of gas.

00:01:30.040 --> 00:01:32.280
So there's three moles of gas on the left

00:01:32.280 --> 00:01:34.760
and only one mole of gas on the right.

00:01:34.760 --> 00:01:36.940
Since the net reaction is going to try

00:01:36.940 --> 00:01:38.740
to increase the pressure,

00:01:38.740 --> 00:01:41.420
the equilibrium shifts to the left,

00:01:41.420 --> 00:01:44.520
toward the side that's gonna
form more moles of gas,

00:01:44.520 --> 00:01:47.300
therefore increasing the pressure.

00:01:47.300 --> 00:01:49.070
Next, we try adding some neon gas

00:01:49.070 --> 00:01:52.020
to our reaction mixture at equilibrium.

00:01:52.020 --> 00:01:53.850
Well, neon gas is an inert gas,

00:01:53.850 --> 00:01:55.010
which means it doesn't react

00:01:55.010 --> 00:01:57.200
with any of our reactants or products.

00:01:57.200 --> 00:01:59.010
And if we look at the expression

00:01:59.010 --> 00:02:01.160
for the reaction quotient Qp,

00:02:01.160 --> 00:02:03.160
neon gas is not included.

00:02:03.160 --> 00:02:05.300
Therefore, adding neon gas is not going

00:02:05.300 --> 00:02:07.590
to change the value for Qp,

00:02:07.590 --> 00:02:10.500
so the reaction remains at equilibrium.

00:02:10.500 --> 00:02:12.470
So the answer is there's no shift

00:02:12.470 --> 00:02:15.280
when an inert gas is added.

00:02:15.280 --> 00:02:17.190
And that might sound a
little strange at first

00:02:17.190 --> 00:02:18.980
because adding neon gas means

00:02:18.980 --> 00:02:21.540
that the total pressure would increase,

00:02:21.540 --> 00:02:24.080
the total pressure since
we're adding a gas.

00:02:24.080 --> 00:02:26.700
However, the partial
pressures stay the same.

00:02:26.700 --> 00:02:29.320
So the partial pressures for methanol

00:02:29.320 --> 00:02:32.290
and carbon monoxide and hydrogen
gas actually stay the same

00:02:32.290 --> 00:02:34.760
and therefore Q doesn't change.

00:02:34.760 --> 00:02:38.000
Next, we add a catalyst to
our reaction at equilibrium.

00:02:38.000 --> 00:02:39.300
Catalysts speed up reactions

00:02:39.300 --> 00:02:41.420
by lowering deactivation energy.

00:02:41.420 --> 00:02:43.970
However, the catalyst is gonna
speed up that the forward

00:02:43.970 --> 00:02:46.310
and the reverse reactions
by the same amount

00:02:46.310 --> 00:02:49.720
and therefore the reaction
remains at equilibrium.

00:02:49.720 --> 00:02:52.190
So there's no shift
when a catalyst is added

00:02:52.190 --> 00:02:54.700
to a reaction at equilibrium.

00:02:54.700 --> 00:02:55.690
And then in part F,

00:02:55.690 --> 00:02:57.580
let's try decreasing the temperature

00:02:57.580 --> 00:02:59.930
on the reaction at equilibrium.

00:02:59.930 --> 00:03:01.740
Well, this reaction is exothermic

00:03:01.740 --> 00:03:03.920
because Delta H is less than 0,

00:03:03.920 --> 00:03:06.680
so we can treat heat as a product.

00:03:06.680 --> 00:03:09.900
So we go ahead and write
heat on the product side.

00:03:09.900 --> 00:03:11.880
If we treat heat like a product,

00:03:11.880 --> 00:03:15.210
decreasing the temperature
is like decreasing the amount

00:03:15.210 --> 00:03:16.850
of our product, therefore,

00:03:16.850 --> 00:03:19.230
the net reaction will move to the right

00:03:19.230 --> 00:03:21.410
to make more of the product.

00:03:21.410 --> 00:03:23.110
Whether that reaction moves to the right,

00:03:23.110 --> 00:03:25.080
you can think about that being an increase

00:03:25.080 --> 00:03:26.550
in the amount of products

00:03:26.550 --> 00:03:29.740
and therefore a decrease
in the amount of reactants.

00:03:29.740 --> 00:03:31.200
And when you increase the products

00:03:31.200 --> 00:03:32.750
and decrease the reactants,

00:03:32.750 --> 00:03:36.130
you increase the value for
the equilibrium constant.

00:03:36.130 --> 00:03:40.020
Therefore, lowering the
temperature causes an increase

00:03:40.020 --> 00:03:44.390
in the equilibrium constant
for an exothermic reaction.

00:03:44.390 --> 00:03:45.740
Note that changing the temperature

00:03:45.740 --> 00:03:48.180
in part F is the only change

00:03:48.180 --> 00:03:50.710
that actually changed
the equilibrium constant.

00:03:50.710 --> 00:03:53.250
So in all the other ones, in A through E,

00:03:53.250 --> 00:03:57.233
the equilibrium constant
stayed the same value.


Worked example: Calculating the equilibrium total pressure after a change in volume

WEBVTT
Kind: captions
Language: en

00:00:00.220 --> 00:00:02.610
- [Instructor] Phosphorus
pentachloride will decompose

00:00:02.610 --> 00:00:06.220
into phosphorus trichloride
and chlorine gas.

00:00:06.220 --> 00:00:10.930
Kp for this reaction is
equal to .500 at 500 Kelvin.

00:00:10.930 --> 00:00:13.090
Let's say that this
reaction is at equilibrium

00:00:13.090 --> 00:00:16.370
and a reaction vessel that
has a volume of 2.0 liters

00:00:16.370 --> 00:00:18.450
and the equilibrium partial pressure,

00:00:18.450 --> 00:00:22.100
of PCl5 is equal to .980 atmospheres.

00:00:22.100 --> 00:00:24.670
The equilibrium partial pressure of PCl3

00:00:24.670 --> 00:00:27.310
is equal to .700 atmospheres.

00:00:27.310 --> 00:00:29.840
And the equilibrium partial
pressure of chlorine gas

00:00:29.840 --> 00:00:32.930
is equal to .700 atmospheres.

00:00:32.930 --> 00:00:35.090
If we add up those
three partial pressures,

00:00:35.090 --> 00:00:38.800
we get the total pressure of
the gas mixture at equilibrium,

00:00:38.800 --> 00:00:41.860
which is equal to 2.38 atmospheres.

00:00:41.860 --> 00:00:45.150
And we're gonna call
this total pressure, P1.

00:00:45.150 --> 00:00:47.730
If we decrease the volume from two liters

00:00:47.730 --> 00:00:48.950
down to one liter,

00:00:48.950 --> 00:00:52.840
and we keep the temperature
constant at 500 Kelvin,

00:00:52.840 --> 00:00:55.360
we decrease the volume by a factor of two,

00:00:55.360 --> 00:00:59.290
which means we increase the
pressure by a factor of two.

00:00:59.290 --> 00:01:02.930
So, all of the partial
pressures of our gases double,

00:01:02.930 --> 00:01:05.130
and there's a new total pressure,

00:01:05.130 --> 00:01:07.420
which is twice the
original total pressure.

00:01:07.420 --> 00:01:11.160
And this new total pressure
is equal to 4.76 atmospheres.

00:01:11.160 --> 00:01:14.580
And from now on, we'll call
this total pressure, P2.

00:01:14.580 --> 00:01:16.320
So, when the volume was two liters,

00:01:16.320 --> 00:01:18.780
the reaction started out at equilibrium

00:01:18.780 --> 00:01:21.430
and by decreasing the volume to one liter,

00:01:21.430 --> 00:01:22.850
and by doubling the pressure,

00:01:22.850 --> 00:01:25.870
we've introduced a stress to the system.

00:01:25.870 --> 00:01:27.300
And so, at this moment in time,

00:01:27.300 --> 00:01:28.910
when these are the partial pressures,

00:01:28.910 --> 00:01:32.173
the reaction is not at equilibrium.

00:01:33.210 --> 00:01:35.380
Le Chatelier's principle
says the net reaction

00:01:35.380 --> 00:01:38.290
will move in the direction
that decreases the stress.

00:01:38.290 --> 00:01:41.150
So, if the stress is an
increase in the pressure,

00:01:41.150 --> 00:01:43.210
the net reaction will
move in the direction

00:01:43.210 --> 00:01:45.500
that decreases the pressure.

00:01:45.500 --> 00:01:46.910
Looking at the balanced equation,

00:01:46.910 --> 00:01:49.320
there's one mole of gas
on the reactant side,

00:01:49.320 --> 00:01:52.960
and there are two moles of
gas on the product side.

00:01:52.960 --> 00:01:54.154
So, if the net reaction goes

00:01:54.154 --> 00:01:56.560
from the products to the reactants,

00:01:56.560 --> 00:01:58.581
if the net reaction goes to the left,

00:01:58.581 --> 00:02:00.960
the net reaction is going to the side

00:02:00.960 --> 00:02:03.890
with the smaller number of moles of gas,

00:02:03.890 --> 00:02:07.540
which would decrease the
pressure and relieve the stress.

00:02:07.540 --> 00:02:09.493
The net reaction keeps moving to the left

00:02:09.493 --> 00:02:13.110
until equilibrium is reestablished.

00:02:13.110 --> 00:02:15.160
And when equilibrium is reestablished,

00:02:15.160 --> 00:02:18.860
there'll be a new total
pressure, which we'll call P3.

00:02:18.860 --> 00:02:22.890
So, our goal is to calculate P3,

00:02:22.890 --> 00:02:26.670
so we can compare it to P1 and P2.

00:02:26.670 --> 00:02:29.210
And we're gonna do this
in a quantitative way

00:02:29.210 --> 00:02:31.730
and in a more qualitative way.

00:02:31.730 --> 00:02:32.920
Let's use an ICE table

00:02:32.920 --> 00:02:36.590
to help us figure out the
final total pressure, P3.

00:02:36.590 --> 00:02:37.423
In an ICE table,

00:02:37.423 --> 00:02:40.210
I stands for the initial
partial pressure in this case,

00:02:40.210 --> 00:02:43.700
C is the change and E is the
equilibrium partial pressure.

00:02:43.700 --> 00:02:46.360
The initial partial pressure of PCl5,

00:02:46.360 --> 00:02:48.667
after the volume was reduced to one liter,

00:02:48.667 --> 00:02:51.280
was 1.96 atmospheres.

00:02:51.280 --> 00:02:53.890
And the partial pressures of PCl3 and Cl2

00:02:53.890 --> 00:02:57.090
were both 1.40 atmospheres.

00:02:57.090 --> 00:02:58.938
We already used Le Chatelier's principle

00:02:58.938 --> 00:03:03.000
to realize the net reaction's
going to go to the left,

00:03:03.000 --> 00:03:04.880
which means we're going to decrease

00:03:04.880 --> 00:03:06.150
in the amount of our products,

00:03:06.150 --> 00:03:09.061
and we're going to increase
the amount of the reactants.

00:03:09.061 --> 00:03:12.310
So, if we're gonna increase
the amount of PCl5,

00:03:12.310 --> 00:03:13.880
we don't know how much
we're gonna increase,

00:03:13.880 --> 00:03:15.080
we're gonna call that x.

00:03:15.080 --> 00:03:16.570
But we know it's going to increase,

00:03:16.570 --> 00:03:20.110
so we write plus x under the
change part on the ICE table.

00:03:20.110 --> 00:03:24.450
And since our mole ratio of
PCl5 to PCl3 is one to one,

00:03:24.450 --> 00:03:26.930
if we're gaining x for PCl5,

00:03:26.930 --> 00:03:29.860
we must be losing x for PCl3.

00:03:29.860 --> 00:03:31.570
And the same goes for Cl2,

00:03:31.570 --> 00:03:33.170
since there's a coefficient of one.

00:03:33.170 --> 00:03:36.370
So, we write minus x in our ICE table.

00:03:36.370 --> 00:03:39.340
Therefore, the equilibrium
partial pressure of PCl5

00:03:39.340 --> 00:03:43.986
would be 1.96 plus x.

00:03:43.986 --> 00:03:46.820
The equilibrium partial pressure of PCl3

00:03:46.820 --> 00:03:50.140
would be 1.40 minus x.

00:03:50.140 --> 00:03:52.500
And the equilibrium
partial pressure of Cl2

00:03:52.500 --> 00:03:56.120
would also be 1.40 minus x.

00:03:56.120 --> 00:03:58.381
Next, we can plug in the
equilibrium partial pressures

00:03:58.381 --> 00:04:00.869
into our Kp expression.

00:04:00.869 --> 00:04:03.970
So, we can plug in 1.40 minus x

00:04:03.970 --> 00:04:06.793
for the equilibrium
partial pressure of PCl3,

00:04:06.793 --> 00:04:11.460
1.40 minus x for the equilibrium
partial pressure of Cl2,

00:04:11.460 --> 00:04:14.010
and 1.96 plus x

00:04:14.010 --> 00:04:17.514
for the equilibrium
partial pressure of PCl5.

00:04:17.514 --> 00:04:19.930
And we can also plug in the value

00:04:19.930 --> 00:04:22.630
for the equilibrium constant, Kp.

00:04:22.630 --> 00:04:25.360
Here's what our equilibrium
constant expression looks like

00:04:25.360 --> 00:04:27.130
with everything plugged in.

00:04:27.130 --> 00:04:29.780
And next, we would need to solve for x,

00:04:29.780 --> 00:04:32.610
which would involve the use
of a quadratic equation.

00:04:32.610 --> 00:04:34.100
And when you do all that math,

00:04:34.100 --> 00:04:36.890
you find that x is equal to .330.

00:04:40.140 --> 00:04:42.810
Now that we know x is equal to .33,

00:04:42.810 --> 00:04:45.440
we can solve for the
equilibrium partial pressures.

00:04:45.440 --> 00:04:49.760
1.96 plus .33 is equal to 2.29.

00:04:49.760 --> 00:04:52.660
Therefore, the equilibrium
partial pressure of PCl5

00:04:52.660 --> 00:04:54.970
is 2.29 atmospheres,

00:04:54.970 --> 00:04:58.003
for PCl3, it's 1.40 minus x,

00:04:58.003 --> 00:05:02.510
1.40 minus 0.33 is equal to 1.07.

00:05:02.510 --> 00:05:05.150
Therefore, the equilibrium
partial pressure of PCl3

00:05:05.150 --> 00:05:07.520
is equal to 1.07 atmospheres.

00:05:07.520 --> 00:05:09.990
And it's the same math for Cl2.

00:05:09.990 --> 00:05:13.000
So, the equilibrium
partial pressure of Cl2

00:05:13.000 --> 00:05:16.640
is also 1.07 atmospheres.

00:05:16.640 --> 00:05:18.840
So, to find the total pressure, P3,

00:05:18.840 --> 00:05:22.680
we simply need to add up the
individual partial pressures.

00:05:22.680 --> 00:05:27.680
So, 2.29 plus 1.07 plus 1.07
is equal to 4.43 atmospheres.

00:05:30.295 --> 00:05:33.878
Therefore, P3 is equal
to 4.43 atmospheres.

00:05:38.940 --> 00:05:40.830
Doing the math helps us realize that x

00:05:40.830 --> 00:05:42.295
is not a very large number.

00:05:42.295 --> 00:05:44.730
And the reason why x is
not a very large number

00:05:44.730 --> 00:05:49.210
is because the Kp value is
equal to .500 for this reaction.

00:05:49.210 --> 00:05:51.270
And when K is close to one,

00:05:51.270 --> 00:05:52.670
there's an appreciable amount

00:05:52.670 --> 00:05:55.410
of both reactants and
products at equilibrium.

00:05:55.410 --> 00:05:58.630
And we can see that with our
equilibrium partial pressures.

00:05:58.630 --> 00:06:01.200
There's an appreciable
amount of both of them.

00:06:01.200 --> 00:06:02.860
And since there has to be a decent amount

00:06:02.860 --> 00:06:05.133
of both reactants and
products at equilibrium,

00:06:05.133 --> 00:06:06.900
we're not gonna see huge change

00:06:06.900 --> 00:06:09.580
from these initial partial pressures.

00:06:09.580 --> 00:06:12.860
So, there will definitely
be a shift to the left

00:06:12.860 --> 00:06:14.950
to decrease the pressure.

00:06:14.950 --> 00:06:17.236
And when these were the partial pressures,

00:06:17.236 --> 00:06:21.150
if you remember P2 was equal to 4.76,

00:06:21.150 --> 00:06:23.260
so there's gonna be a
decrease of pressure.

00:06:23.260 --> 00:06:24.440
So, there's gonna be a decrease,

00:06:24.440 --> 00:06:27.550
so the pressure is going
to go down from 4.76,

00:06:27.550 --> 00:06:29.480
but since there's not a large change,

00:06:29.480 --> 00:06:31.660
it's not gonna be a huge change.

00:06:31.660 --> 00:06:36.480
And that's why we saw P3 only
dropped to 4.43 atmospheres.

00:06:36.480 --> 00:06:38.240
So, if you go back to
the original problem,

00:06:38.240 --> 00:06:40.800
and our goal is to figure out P3

00:06:40.800 --> 00:06:43.320
in relation to P1 and to P2,

00:06:43.320 --> 00:06:45.400
without doing all of that math,

00:06:45.400 --> 00:06:46.560
we could think to ourselves,

00:06:46.560 --> 00:06:49.120
okay, so, we decrease the
volume by a factor of two,

00:06:49.120 --> 00:06:51.470
which doubled the total pressure.

00:06:51.470 --> 00:06:53.340
But then, the net
reaction moves to the left

00:06:53.340 --> 00:06:54.873
to decrease the pressure.

00:06:54.873 --> 00:06:57.682
Since it's not gonna
move much to the left,

00:06:57.682 --> 00:07:00.270
it's not gonna decrease
the pressure by a lot.

00:07:00.270 --> 00:07:03.190
Therefore, the final pressure P3

00:07:03.190 --> 00:07:06.095
is going to be a little less than 4.76,

00:07:06.095 --> 00:07:09.953
but greater than 2.38 atmospheres.


Le Chȃtelier’s principle: Changing temperature

WEBVTT
Kind: captions
Language: en

00:00:00.260 --> 00:00:02.720
- [Instructor] Le Chateliers's
Principle says if a stress

00:00:02.720 --> 00:00:05.260
is applied to a reaction at equilibrium,

00:00:05.260 --> 00:00:07.710
the net reaction goes in
the direction that relieves

00:00:07.710 --> 00:00:09.060
the stress.

00:00:09.060 --> 00:00:11.580
One possible stress is
to change the temperature

00:00:11.580 --> 00:00:13.840
of the reaction at equilibrium.

00:00:13.840 --> 00:00:16.560
As an example, let's look
at the hypothetical reaction

00:00:16.560 --> 00:00:20.070
where gas A turns into gas B.

00:00:20.070 --> 00:00:22.980
Delta H for this reaction
is less than zero,

00:00:22.980 --> 00:00:27.170
which tells us this is
an exothermic reaction.

00:00:27.170 --> 00:00:30.070
And for an exothermic
reaction, heat is given off,

00:00:30.070 --> 00:00:31.110
heat is released.

00:00:31.110 --> 00:00:33.520
Therefore we can go
ahead and write plus heat

00:00:33.520 --> 00:00:35.620
on the product side.

00:00:35.620 --> 00:00:38.820
Let's say that our hypothetical
reaction is at equilibrium,

00:00:38.820 --> 00:00:40.560
and then we change the temperature,

00:00:40.560 --> 00:00:44.350
so we're going to
increase the temperature.

00:00:44.350 --> 00:00:45.980
According to Le Chatelier's Principal,

00:00:45.980 --> 00:00:47.750
the net reaction is
gonna go in the direction

00:00:47.750 --> 00:00:49.350
that decreases the stress.

00:00:49.350 --> 00:00:52.940
And if we treat heat if
we treat like a product

00:00:52.940 --> 00:00:55.140
and we've increased the temperature,

00:00:55.140 --> 00:00:58.260
it says if we've increased the
amount of one of our products

00:00:58.260 --> 00:01:01.100
and therefore the net reaction
is going to shift to the left

00:01:01.100 --> 00:01:03.670
to decrease a product.

00:01:03.670 --> 00:01:06.720
Let's use particulate diagrams
and the reaction quotient Q

00:01:06.720 --> 00:01:09.660
to explain what's going on when
we increase the temperature

00:01:09.660 --> 00:01:12.060
on our reaction at equilibrium.

00:01:12.060 --> 00:01:14.360
The first particular
diagram shows the reaction

00:01:14.360 --> 00:01:17.680
at equilibrium, and let's
prove that by calculating QC

00:01:17.680 --> 00:01:19.360
at this moment in time.

00:01:19.360 --> 00:01:21.820
We can get the expression
for the reaction quotient,

00:01:21.820 --> 00:01:24.680
QC by looking at the balanced equation.

00:01:24.680 --> 00:01:26.770
So we have coefficients
of one in front of A

00:01:26.770 --> 00:01:29.400
and in front of B, therefore QC is equal

00:01:29.400 --> 00:01:31.440
to the concentration
of B to the first power

00:01:31.440 --> 00:01:33.350
divided by the concentration of A,

00:01:33.350 --> 00:01:35.070
also to the first power.

00:01:35.070 --> 00:01:37.400
To find the concentration
of B, we know that B

00:01:37.400 --> 00:01:39.230
is represented by blue spheres,

00:01:39.230 --> 00:01:42.150
so there are 1, 2, 3 blue spheres,

00:01:42.150 --> 00:01:45.390
And if each sphere represents
0.1 moles of a substance

00:01:45.390 --> 00:01:49.430
three times 0.1 is equal to 0.3 moles of B

00:01:49.430 --> 00:01:52.644
and at the volumes equal to 1.0 liter,

00:01:52.644 --> 00:01:56.770
0.3 divided by 1.0 liter is 0.3 Molar.

00:01:56.770 --> 00:02:01.770
So the concentration of
B is equal to 0.3 Molar.

00:02:02.390 --> 00:02:05.210
There are also three particles of A

00:02:05.210 --> 00:02:07.030
therefore the concentration of A

00:02:07.030 --> 00:02:11.270
is also equal to 0.3 Molar.

00:02:11.270 --> 00:02:13.410
0.3 divided by 0.3 is equal to one.

00:02:13.410 --> 00:02:16.670
So QC at this moment of
time is equal to one.

00:02:16.670 --> 00:02:21.100
KC for this reaction is equal
to one at 25 degrees Celsius.

00:02:21.100 --> 00:02:24.070
So QC is equal to KC.

00:02:24.070 --> 00:02:28.970
And when QC is equal to KC,
the reaction is at equilibrium.

00:02:28.970 --> 00:02:31.570
So for this first particular diagram,

00:02:31.570 --> 00:02:33.353
the reaction's at equilibrium.

00:02:34.200 --> 00:02:36.300
Next, we introduced the
stress to the reaction

00:02:36.300 --> 00:02:39.700
at equilibrium and the
stress is an increase

00:02:39.700 --> 00:02:41.330
in the temperature.

00:02:41.330 --> 00:02:44.430
In general, for an exothermic reaction,

00:02:44.430 --> 00:02:46.840
increasing the temperature
lowers the value

00:02:46.840 --> 00:02:48.930
for the equilibrium constant.

00:02:48.930 --> 00:02:52.440
So for this hypothetical
reaction at 25 degrees Celsius,

00:02:52.440 --> 00:02:55.500
KC is equal to one, but
since we've increased

00:02:55.500 --> 00:02:59.120
the temperature, the value
for the equilibrium constant

00:02:59.120 --> 00:03:00.370
is going to decrease.

00:03:00.370 --> 00:03:04.990
So let's say it goes 2.5, if
we increase the temperature

00:03:04.990 --> 00:03:07.863
to 30 degrees Celsius.

00:03:08.770 --> 00:03:13.520
So if we calculate QC for our
second particular diagram,

00:03:13.520 --> 00:03:16.270
we still have three blues and three reds,

00:03:16.270 --> 00:03:17.730
and the volume is still the same,

00:03:17.730 --> 00:03:21.740
therefore QC is still equal to one,

00:03:21.740 --> 00:03:24.280
but the difference is KC has now changed,

00:03:24.280 --> 00:03:27.020
so QC is not equal to KC,

00:03:27.020 --> 00:03:30.170
so we are not at equilibrium.

00:03:30.170 --> 00:03:33.830
And in this case, QC is greater than KC

00:03:33.830 --> 00:03:36.583
'cause QC is equal to one
and KC is equal to 0.5.

00:03:37.660 --> 00:03:42.660
And when QC is greater than
KC, there are too many products

00:03:42.960 --> 00:03:44.890
and not enough reactants.

00:03:44.890 --> 00:03:49.890
And therefore the net
reaction goes to the left.

00:03:49.890 --> 00:03:52.030
When the net reaction goes to the left,

00:03:52.030 --> 00:03:55.430
we're going to have be turned into A.

00:03:55.430 --> 00:04:00.120
So we should see one blue
sphere turn into one red sphere.

00:04:00.120 --> 00:04:03.190
So if we have three blues and three reds,

00:04:03.190 --> 00:04:08.010
and one blue turns into a red,
that gives us only two blues

00:04:08.010 --> 00:04:10.550
and four reds now.

00:04:10.550 --> 00:04:13.810
So when we calculate QC for
our third particular diagram,

00:04:13.810 --> 00:04:17.440
the concentration of B would be 0.2 Molar,

00:04:17.440 --> 00:04:21.650
and the concentration
of A would be 0.4 Molar.

00:04:21.650 --> 00:04:26.650
So 0.2 divided by 0.4 is equal to 0.5.

00:04:27.010 --> 00:04:30.890
Well, KC is also equal to 0.5.

00:04:30.890 --> 00:04:34.580
Therefore QC is equal to KC,

00:04:34.580 --> 00:04:37.850
and the reaction is at equilibrium.

00:04:37.850 --> 00:04:39.740
And when a reaction is at equilibrium,

00:04:39.740 --> 00:04:43.830
the concentrations of reactants
and products are constant.

00:04:43.830 --> 00:04:46.980
Let's go back to our hypothetical
reaction at equilibrium,

00:04:46.980 --> 00:04:50.060
but this time we're going
to decrease the temperature.

00:04:50.060 --> 00:04:53.130
If we treat heat like a product
decrease in the temperature

00:04:53.130 --> 00:04:56.060
is like decreasing the amount
of one of the products.

00:04:56.060 --> 00:04:58.460
Therefore the net reaction
will go to the right

00:04:58.460 --> 00:05:00.490
to make more product.

00:05:00.490 --> 00:05:02.820
If we approach this problem
by thinking about the reaction

00:05:02.820 --> 00:05:06.100
quotient Q for an exothermic reaction,

00:05:06.100 --> 00:05:09.840
a decrease in temperature in
general causes an increase

00:05:09.840 --> 00:05:11.930
in the equilibrium constants.

00:05:11.930 --> 00:05:14.570
And if the equilibrium constant increases,

00:05:14.570 --> 00:05:17.130
then Q would be less than K.

00:05:17.130 --> 00:05:19.340
And when Q is less than K,

00:05:19.340 --> 00:05:22.850
the net reaction goes to the right.

00:05:22.850 --> 00:05:25.010
The net reaction would
continue to go to the right

00:05:25.010 --> 00:05:26.700
until Q is equal to K,

00:05:26.700 --> 00:05:29.990
and equilibrium has been re-established.

00:05:29.990 --> 00:05:32.240
Next let's look at an endothermic reaction

00:05:32.240 --> 00:05:34.800
where Delta H is greater than zero.

00:05:34.800 --> 00:05:39.140
When six water molecules
complex to a Cobalt two plus ion

00:05:39.140 --> 00:05:42.800
the resulting complex
ion is pink in color.

00:05:42.800 --> 00:05:46.440
And when for chloride anions
complex to a cobalt two plus

00:05:46.440 --> 00:05:50.910
ion, the resulting complex
ion is blue in color.

00:05:50.910 --> 00:05:53.830
When the pink ion reacts
with four chloride anions,

00:05:53.830 --> 00:05:55.850
the blue ion is formed.

00:05:55.850 --> 00:05:58.050
And since this reaction is endothermic,

00:05:58.050 --> 00:06:01.363
we can put heat on the reactant side.

00:06:02.500 --> 00:06:04.770
We're gonna use these
particular diagrams down here

00:06:04.770 --> 00:06:07.210
to help us understand what
happens to an endothermic

00:06:07.210 --> 00:06:10.340
reaction at equilibrium when
the temperature changes,

00:06:10.340 --> 00:06:12.640
however, these drawings aren't
designed to be completely

00:06:12.640 --> 00:06:14.560
accurate for this particular reaction.

00:06:14.560 --> 00:06:18.160
They're just to help us understand
what color we would see.

00:06:18.160 --> 00:06:21.410
For example, let's say that
this middle particulate diagram

00:06:21.410 --> 00:06:23.870
represents the reaction at equilibrium.

00:06:23.870 --> 00:06:26.710
And if there are decent
amounts of both the blue ion

00:06:26.710 --> 00:06:29.070
and the pink ion at equilibrium,

00:06:29.070 --> 00:06:31.280
the resulting equilibrium mixture,

00:06:31.280 --> 00:06:33.170
so this is an aqueous solution

00:06:33.170 --> 00:06:36.343
would appear to be purple or violet.

00:06:38.060 --> 00:06:40.950
If we were to increase the temperature

00:06:40.950 --> 00:06:42.650
for this endothermic reaction,

00:06:42.650 --> 00:06:45.270
we treat heat like a reactant.

00:06:45.270 --> 00:06:47.880
So increasing the temperature
is like increasing

00:06:47.880 --> 00:06:49.860
the amount of a reactant.

00:06:49.860 --> 00:06:53.730
And therefore the net reaction
will shift to the right

00:06:53.730 --> 00:06:56.620
to get rid of some of that reactant.

00:06:56.620 --> 00:06:58.210
Whether the net reaction
goes to the right,

00:06:58.210 --> 00:07:00.800
we're gonna increase in
the amount of blue ion,

00:07:00.800 --> 00:07:04.460
and we're going to decrease
in the amount of pink ion.

00:07:04.460 --> 00:07:08.240
Therefore looking at this
particular diagram on the right

00:07:08.240 --> 00:07:12.720
there are now more blue ions
than there are pink ions

00:07:12.720 --> 00:07:16.000
compared to our equilibrium
mixture in the middle.

00:07:16.000 --> 00:07:18.480
Therefore for this third
particular diagram,

00:07:18.480 --> 00:07:22.420
the resulting aqueous solution
is going to look blue.

00:07:22.420 --> 00:07:25.260
If we think about those
using Q in general,

00:07:25.260 --> 00:07:28.670
for an endothermic reaction,
an increase in temperature

00:07:28.670 --> 00:07:32.490
causes an increase in the
equilibrium constant K.

00:07:32.490 --> 00:07:37.490
And if K increases, then
the reaction quotient Q

00:07:37.500 --> 00:07:39.410
is less than K.

00:07:39.410 --> 00:07:40.970
And when Q is less than K,

00:07:40.970 --> 00:07:44.180
the net reaction goes to the right.

00:07:44.180 --> 00:07:47.720
Now let's go back to the
middle particular diagram.

00:07:47.720 --> 00:07:49.790
And so there are reactions at equilibrium

00:07:49.790 --> 00:07:53.000
and this time we're going
to decrease the temperature.

00:07:53.000 --> 00:07:55.090
If we treat heat as a reactant,

00:07:55.090 --> 00:07:56.660
and we decrease the temperature,

00:07:56.660 --> 00:07:59.000
it's like we're decreasing
one of our reactants.

00:07:59.000 --> 00:08:01.430
Therefore the net reaction
will shift to the left

00:08:01.430 --> 00:08:03.920
to make more of our reactant.

00:08:03.920 --> 00:08:05.840
And when that reaction goes to the left,

00:08:05.840 --> 00:08:08.320
we're gonna decrease in
the amount of the blue ion,

00:08:08.320 --> 00:08:12.110
and we're going to increase
in the amount of the pink ion.

00:08:12.110 --> 00:08:14.930
So when we compare the
middle particulate diagram

00:08:14.930 --> 00:08:17.110
to the one on the left
and the one that left,

00:08:17.110 --> 00:08:19.230
there's a lot more of the pink ion,

00:08:19.230 --> 00:08:20.970
then there is of the blue ion.

00:08:20.970 --> 00:08:23.860
Therefore the overall solution,

00:08:23.860 --> 00:08:27.380
the overall aqueous solution
is going to appear pink.

00:08:27.380 --> 00:08:29.990
If we think about what's happening using Q

00:08:29.990 --> 00:08:33.020
for an endothermic reaction in general,

00:08:33.020 --> 00:08:34.540
when you decrease the temperature,

00:08:34.540 --> 00:08:37.040
you decrease the equilibrium constant.

00:08:37.040 --> 00:08:40.030
And if the equilibrium constant decreases

00:08:40.030 --> 00:08:42.680
now, Q would be greater than K,

00:08:42.680 --> 00:08:45.410
which means too many products
and not enough reactance.

00:08:45.410 --> 00:08:48.670
Therefore the net reaction
would go to the left.

00:08:48.670 --> 00:08:51.400
Let's go back to our exothermic reaction.

00:08:51.400 --> 00:08:54.270
At this time let's pretend like
we're starting with only A,

00:08:54.270 --> 00:08:56.710
so we start with only A
and we have none of B.

00:08:56.710 --> 00:08:59.970
And our goal is to make
as much as we possibly can

00:08:59.970 --> 00:09:02.440
and to do it as fast as possible.

00:09:02.440 --> 00:09:05.160
One way to increase the rate of a reaction

00:09:05.160 --> 00:09:07.930
would be to increase the temperature.

00:09:07.930 --> 00:09:10.420
However, for an exothermic reaction,

00:09:10.420 --> 00:09:13.960
increasing the temperature
decreases the equilibrium

00:09:13.960 --> 00:09:15.170
constant K.

00:09:15.170 --> 00:09:17.870
And if you decrease the
equilibrium constant K,

00:09:17.870 --> 00:09:20.280
you would decrease the amount of B

00:09:20.280 --> 00:09:23.290
that you would have when
you reach equilibrium.

00:09:23.290 --> 00:09:25.490
So we can't run our hypothetical reaction

00:09:25.490 --> 00:09:27.770
at too high of a temperature
because that would decrease

00:09:27.770 --> 00:09:29.820
the equilibrium constant.

00:09:29.820 --> 00:09:32.250
So instead to speed up
the rate of the reaction,

00:09:32.250 --> 00:09:34.633
we could add a catalyst,

00:09:36.110 --> 00:09:39.760
let's look at a graph of
concentration of B versus time,

00:09:39.760 --> 00:09:41.970
and we're gonna start
with this blue curve here,

00:09:41.970 --> 00:09:44.890
which represents the hypothetical reaction

00:09:44.890 --> 00:09:48.530
without a catalyst, so
the uncatalyzed reaction.

00:09:48.530 --> 00:09:51.760
When time is equal to zero,
the concentration of B is zero

00:09:51.760 --> 00:09:53.620
because we start with only A.

00:09:53.620 --> 00:09:57.790
And as A turns into B the
concentration of B increases

00:09:57.790 --> 00:10:00.810
over time, and eventually
the concentration of B

00:10:00.810 --> 00:10:02.360
becomes constant.

00:10:02.360 --> 00:10:04.710
And when the concentration
of B becomes constant,

00:10:04.710 --> 00:10:06.710
the reaction reaches equilibrium.

00:10:06.710 --> 00:10:10.740
So this dotted line here
represents the concentration of B

00:10:10.740 --> 00:10:12.490
at equilibrium.

00:10:12.490 --> 00:10:14.950
The yellow line represents the reaction

00:10:14.950 --> 00:10:16.760
with a catalyst added.

00:10:16.760 --> 00:10:18.810
So once again, we're starting with only A,

00:10:18.810 --> 00:10:20.190
so when time is equal to zero,

00:10:20.190 --> 00:10:22.410
the concentration of B is equal to zero.

00:10:22.410 --> 00:10:24.850
And as time increases A turns into B,

00:10:24.850 --> 00:10:27.290
so the concentration of B increases

00:10:27.290 --> 00:10:30.860
and eventually the concentration
of B becomes constant.

00:10:30.860 --> 00:10:33.390
And the reaction reaches equilibrium.

00:10:33.390 --> 00:10:36.540
Notice that the reaction
reached equilibrium much faster

00:10:36.540 --> 00:10:38.410
with the addition of the catalyst

00:10:38.410 --> 00:10:40.900
than it did without the catalyst.

00:10:40.900 --> 00:10:43.730
So the addition of a
catalyst allows a reaction

00:10:43.730 --> 00:10:46.340
to reach equilibrium faster.

00:10:46.340 --> 00:10:49.810
However, notice that the final
equilibrium concentration

00:10:49.810 --> 00:10:53.220
of B is the same for both
the uncatalyzed reaction

00:10:53.220 --> 00:10:55.150
and the catalyzed reaction.

00:10:55.150 --> 00:10:57.730
Therefore, the addition of
a catalyst does not change

00:10:57.730 --> 00:11:01.570
the composition of the
equilibrium mixture.

00:11:01.570 --> 00:11:03.710
And that's because the catalyst speeds up

00:11:03.710 --> 00:11:07.920
both the forward reaction
and the reverse reaction,

00:11:07.920 --> 00:11:10.150
but the rates are still equal.

00:11:10.150 --> 00:11:11.490
And since the rates are equal,

00:11:11.490 --> 00:11:13.350
there's no change in the composition

00:11:13.350 --> 00:11:15.153
of the equilibrium mixture.


Le Chȃtelier’s principle: Changing volume

WEBVTT
Kind: captions
Language: en

00:00:00.955 --> 00:00:02.840
- [Tutor] Le Chatelier's
principle says that

00:00:02.840 --> 00:00:06.490
if a stress is applied to a
reaction mixture at equilibrium,

00:00:06.490 --> 00:00:08.500
the net reaction goes in the direction

00:00:08.500 --> 00:00:10.730
that relieves the stress.

00:00:10.730 --> 00:00:14.030
One possible stress that we
could do is to change the volume

00:00:14.030 --> 00:00:16.590
on our reaction at equilibrium.

00:00:16.590 --> 00:00:20.180
Let's say we have a hypothetical
reaction, where solid A,

00:00:20.180 --> 00:00:23.500
which is symbolized by red
in our particular diagram,

00:00:23.500 --> 00:00:27.370
turns into solid B, which
is symbolized by a blue

00:00:27.370 --> 00:00:29.620
and also C, which is a gas

00:00:29.620 --> 00:00:33.000
and C will be symbolized by white spheres.

00:00:33.000 --> 00:00:35.650
The equilibrium constant
Kc for this reaction,

00:00:35.650 --> 00:00:39.750
is equal to 0.4 at 25 degrees Celsius.

00:00:39.750 --> 00:00:40.930
The first particular diagram

00:00:40.930 --> 00:00:43.240
shows the reaction at equilibrium,

00:00:43.240 --> 00:00:46.380
and we can see there's some
solid A and some solid B

00:00:46.380 --> 00:00:48.070
at the bottom of the container,

00:00:48.070 --> 00:00:51.930
and there's also some
gaseous particles of C.

00:00:51.930 --> 00:00:53.330
Let's introduce a stress,

00:00:53.330 --> 00:00:55.970
to our reaction mixture at equilibrium.

00:00:55.970 --> 00:00:58.870
Let's decrease the
volume of the container.

00:00:58.870 --> 00:01:01.970
So looking from the first
particular diagram to the second,

00:01:01.970 --> 00:01:03.530
we can see there's been a decrease

00:01:03.530 --> 00:01:05.370
in the volume of the container.

00:01:05.370 --> 00:01:08.790
That's gonna cause an
increase in the pressure

00:01:08.790 --> 00:01:11.140
because pressure comes
from these gas particles

00:01:11.140 --> 00:01:13.830
slamming into the sides of the container.

00:01:13.830 --> 00:01:15.940
And if we decrease the volume,

00:01:15.940 --> 00:01:18.860
now there's less distance
for these particles to travel

00:01:18.860 --> 00:01:21.040
before they slam into the
sides of the container,

00:01:21.040 --> 00:01:23.970
which means we increase
the collision frequency.

00:01:23.970 --> 00:01:27.130
And therefore the pressure increases.

00:01:27.130 --> 00:01:29.060
According to Le Chatelier's principle,

00:01:29.060 --> 00:01:30.990
the net reaction is gonna
go in the direction,

00:01:30.990 --> 00:01:32.750
that relieves the stress.

00:01:32.750 --> 00:01:35.310
So if the stress is increased pressure,

00:01:35.310 --> 00:01:37.690
the net reaction is going to
try to move in the direction

00:01:37.690 --> 00:01:40.420
that decreases the pressure.

00:01:40.420 --> 00:01:43.070
Looking at the equation for
this hypothetical reaction,

00:01:43.070 --> 00:01:46.110
the two solids aren't really
contributing to the pressure.

00:01:46.110 --> 00:01:49.000
So it's only gas C that
we need to worry about.

00:01:49.000 --> 00:01:52.770
And there's one mole of
gas C on the product side,

00:01:52.770 --> 00:01:56.810
and there are zero moles of
gas on the reactant side.

00:01:56.810 --> 00:01:58.530
So if the net reaction went to the right,

00:01:58.530 --> 00:02:01.530
we'd be going from zero moles
of gas to one mole of gas.

00:02:01.530 --> 00:02:04.210
So going to the right would
increase the moles of gas,

00:02:04.210 --> 00:02:06.450
which would increase the pressure.

00:02:06.450 --> 00:02:08.720
However, that's not what
the reaction wants to do.

00:02:08.720 --> 00:02:11.350
The goal of the reaction
is to relieve the stress

00:02:11.350 --> 00:02:14.830
and therefore the reaction
wants to decrease the pressure.

00:02:14.830 --> 00:02:17.210
So the reactions gonna go to the left,

00:02:17.210 --> 00:02:19.250
to get rid of some of that gas

00:02:19.250 --> 00:02:23.250
and decreasing the amount of
gas will decrease the pressure,

00:02:23.250 --> 00:02:26.380
therefore relieving the stress.

00:02:26.380 --> 00:02:28.670
If the net reaction moves to the left,

00:02:28.670 --> 00:02:30.470
we're gonna lose some of our products.

00:02:30.470 --> 00:02:32.220
So we're gonna decrease the amount of C

00:02:32.220 --> 00:02:34.860
and we're gonna decrease
in the amount of B

00:02:34.860 --> 00:02:37.270
and we're going to gain
some of our reactants.

00:02:37.270 --> 00:02:39.630
So we're gonna increase
in the amount of A,

00:02:39.630 --> 00:02:42.810
and we can see all that and
the third particulate diagram.

00:02:42.810 --> 00:02:44.530
So going from the second
particular diagram

00:02:44.530 --> 00:02:46.150
to the third particular diagram,

00:02:46.150 --> 00:02:47.630
we've decreased the amount of C,

00:02:47.630 --> 00:02:49.370
we've gone from four particles of C

00:02:49.370 --> 00:02:51.110
to only two particles of C.

00:02:51.110 --> 00:02:52.900
We've also decreased the amount of B.

00:02:52.900 --> 00:02:54.790
You can see that this blue,

00:02:54.790 --> 00:02:56.870
this blue solid has gotten smaller,

00:02:56.870 --> 00:03:00.040
and we've increased the amount of A.

00:03:00.040 --> 00:03:02.870
So A, is this, A, is this red sphere here,

00:03:02.870 --> 00:03:04.390
and you can see how it's gotten bigger

00:03:04.390 --> 00:03:07.320
from the second particulate
diagram to the third.

00:03:07.320 --> 00:03:09.630
And by going from four particles of C,

00:03:09.630 --> 00:03:11.680
to only two particles of C,

00:03:11.680 --> 00:03:14.150
we've decreased the amount of the gas,

00:03:14.150 --> 00:03:17.000
and therefore we've
decreased the pressure.

00:03:17.000 --> 00:03:19.290
To better understand what
happens to a reaction mixture

00:03:19.290 --> 00:03:21.590
at equilibrium when a
stress is placed on it,

00:03:21.590 --> 00:03:24.140
let's calculate the reaction quotient Q

00:03:24.140 --> 00:03:26.160
for these three particular diagrams,

00:03:26.160 --> 00:03:28.820
for the same hypothetical reaction.

00:03:28.820 --> 00:03:31.550
The expression for the
reaction quotient Qc,

00:03:31.550 --> 00:03:35.010
has the same form as the
equilibrium constant expression

00:03:35.010 --> 00:03:36.330
for Kc.

00:03:36.330 --> 00:03:37.860
So since solids are left out

00:03:37.860 --> 00:03:40.320
of the equilibrium constant expression,

00:03:40.320 --> 00:03:43.420
we only need to include the
concentration of the gas.

00:03:43.420 --> 00:03:45.910
And since there's a coefficient
of one in front of C,

00:03:45.910 --> 00:03:48.070
Qc is equal to the concentration of C

00:03:48.070 --> 00:03:50.060
raised to the first power.

00:03:50.060 --> 00:03:51.610
Since there are four particles of C

00:03:51.610 --> 00:03:53.280
in the first particular diagram,

00:03:53.280 --> 00:03:56.430
and if each particle
represents 0.1 moles of C,

00:03:56.430 --> 00:03:58.046
four times, 0.1

00:03:58.046 --> 00:03:59.710
is equal to 0.4

00:03:59.710 --> 00:04:00.760
moles of C.

00:04:00.760 --> 00:04:02.650
So to find the concentration of C,

00:04:02.650 --> 00:04:04.100
we take the moles and divide that

00:04:04.100 --> 00:04:07.530
by the volume of the
container, which is 1.0 liter.

00:04:07.530 --> 00:04:09.730
So 0.4 divided by 1.0

00:04:09.730 --> 00:04:10.691
is equal to

00:04:10.691 --> 00:04:11.770
0.4

00:04:11.770 --> 00:04:12.603
molar.

00:04:12.603 --> 00:04:14.130
So that's the concentration of C

00:04:14.130 --> 00:04:16.200
in the first particular diagram.

00:04:16.200 --> 00:04:19.170
We plugged that into our expression for Q.

00:04:19.170 --> 00:04:20.240
So Qc

00:04:20.240 --> 00:04:21.073
is equal

00:04:21.073 --> 00:04:21.963
to

00:04:21.963 --> 00:04:23.150
0.4

00:04:23.150 --> 00:04:24.177
and notice that

00:04:24.177 --> 00:04:25.010
Kc

00:04:25.010 --> 00:04:25.843
is also

00:04:25.843 --> 00:04:26.676
equal to

00:04:26.676 --> 00:04:27.509
0.4.

00:04:27.509 --> 00:04:31.100
So at this moment in
time, Qc is equal to Kc,

00:04:31.100 --> 00:04:34.940
which tells us the
reaction is at equilibrium.

00:04:34.940 --> 00:04:37.170
Next, we think about the
stress that was applied

00:04:37.170 --> 00:04:38.900
to the reaction at equilibrium.

00:04:38.900 --> 00:04:41.070
We decreased the volume.

00:04:41.070 --> 00:04:42.410
And if we look at the volumes here,

00:04:42.410 --> 00:04:46.180
we're going from 1.0 liters to 0.5 liters.

00:04:46.180 --> 00:04:49.080
So we're decreasing the
volume by a factor of two,

00:04:49.080 --> 00:04:51.520
which would cause an
increase in the pressure

00:04:51.520 --> 00:04:53.850
by a factor of two.

00:04:53.850 --> 00:04:57.010
And changing the volume would
change the concentration.

00:04:57.010 --> 00:05:00.167
So instead of 0.4, divided by 1.0,

00:05:00.167 --> 00:05:01.243
it'd be 0.4

00:05:01.243 --> 00:05:02.440
divided by

00:05:02.440 --> 00:05:03.600
0.5,

00:05:03.600 --> 00:05:05.084
which is equal to

00:05:05.084 --> 00:05:06.210
0.8

00:05:06.210 --> 00:05:07.043
molar.

00:05:07.043 --> 00:05:09.890
So the concentration has doubled.

00:05:09.890 --> 00:05:13.710
So if we calculate Q for our
second particular diagram,

00:05:13.710 --> 00:05:16.970
we plug in the concentration
of C, which is 0.8 molar.

00:05:16.970 --> 00:05:17.910
So Qc

00:05:17.910 --> 00:05:19.363
is equal to

00:05:19.363 --> 00:05:20.450
0.8,

00:05:20.450 --> 00:05:21.440
K is equal

00:05:21.440 --> 00:05:22.530
to 0.4.

00:05:22.530 --> 00:05:24.580
So Q is not equal to K,

00:05:24.580 --> 00:05:27.730
therefore the reaction
is not at equilibrium

00:05:27.730 --> 00:05:29.357
for our second particular diagram.

00:05:29.357 --> 00:05:34.040
Let me write in here, not at equilibrium.

00:05:34.040 --> 00:05:34.873
In this case,

00:05:34.873 --> 00:05:35.890
Qc

00:05:35.890 --> 00:05:37.690
is greater than Kc,

00:05:37.690 --> 00:05:39.850
which tells us we have too many products

00:05:39.850 --> 00:05:41.340
and not enough reactants.

00:05:41.340 --> 00:05:43.860
Therefore the net
reaction goes to the left

00:05:43.860 --> 00:05:45.580
to get rid of some of the products

00:05:45.580 --> 00:05:48.950
and to increase the amount of reactants.

00:05:48.950 --> 00:05:50.890
The net reaction keeps going to the left

00:05:50.890 --> 00:05:53.440
until we reach equilibrium again.

00:05:53.440 --> 00:05:55.330
So if we calculate the concentration of C

00:05:55.330 --> 00:05:57.100
in the third particular diagram,

00:05:57.100 --> 00:05:58.870
here there are only two particles.

00:05:58.870 --> 00:05:59.930
So that'd be

00:05:59.930 --> 00:06:00.763
0.2

00:06:00.763 --> 00:06:01.750
moles

00:06:01.750 --> 00:06:04.780
divided by a volume of 0.5 liters.

00:06:04.780 --> 00:06:05.690
So 0.2

00:06:05.690 --> 00:06:07.150
divided by 0.5

00:06:07.150 --> 00:06:08.660
is equal to

00:06:08.660 --> 00:06:09.493
0.4

00:06:09.493 --> 00:06:10.326
molar.

00:06:10.326 --> 00:06:11.230
So Qc

00:06:11.230 --> 00:06:12.063
is equal

00:06:12.063 --> 00:06:13.100
to 0.4.

00:06:13.100 --> 00:06:17.083
And since Qc is equal
to 0.4, Kc is still 0.4.

00:06:18.650 --> 00:06:22.560
So Qc is equal to Kc and
we're at equilibrium.

00:06:22.560 --> 00:06:26.180
So equilibrium has been re-established.

00:06:26.180 --> 00:06:27.900
Since the reaction is at equilibrium

00:06:27.900 --> 00:06:29.940
in a third particular diagram,

00:06:29.940 --> 00:06:31.700
the net reaction stops going to the left

00:06:31.700 --> 00:06:35.020
and the concentration
of C remains constant.

00:06:35.020 --> 00:06:37.150
Let's apply what we've
learned to another reaction,

00:06:37.150 --> 00:06:41.360
the synthesis of ammonia from
nitrogen gas and hydrogen gas.

00:06:41.360 --> 00:06:43.940
If we have a mixture of
these gases at equilibrium,

00:06:43.940 --> 00:06:45.230
and we introduce a stress,

00:06:45.230 --> 00:06:47.770
the system like a decrease in volume,

00:06:47.770 --> 00:06:49.560
the decrease in the
volume of the container

00:06:49.560 --> 00:06:52.150
would cause an increase in the pressure.

00:06:52.150 --> 00:06:54.080
And according to Le Chatelier's principle,

00:06:54.080 --> 00:06:55.940
the net reaction is
gonna go in the direction

00:06:55.940 --> 00:06:59.280
that relieves the stress that
was placed on the system.

00:06:59.280 --> 00:07:02.120
So if the stress is increased pressure,

00:07:02.120 --> 00:07:03.360
the net reaction says,

00:07:03.360 --> 00:07:06.350
I wanna move in the direction
that decreases that pressure.

00:07:06.350 --> 00:07:09.320
So the net reaction moves to the right,

00:07:09.320 --> 00:07:11.900
because there are four
moles of gas on the left

00:07:11.900 --> 00:07:13.970
and only two moles of gas on the right.

00:07:13.970 --> 00:07:15.790
And by moving to the right,

00:07:15.790 --> 00:07:18.020
that goes from four moles
of gas to two moles of gas,

00:07:18.020 --> 00:07:20.010
which decreases the amount of gas

00:07:20.010 --> 00:07:23.220
and causes a decrease in the pressure.

00:07:23.220 --> 00:07:26.370
If we had a mixture of
these gases at equilibrium,

00:07:26.370 --> 00:07:29.530
and we increased the volume
and increase in the volume

00:07:29.530 --> 00:07:32.190
would cause a decrease in the pressure.

00:07:32.190 --> 00:07:35.440
So the stress this time,
is decreased pressure.

00:07:35.440 --> 00:07:37.780
To relieve the stress, the net reaction,

00:07:37.780 --> 00:07:41.020
wants to move in the direction
that increases the pressure.

00:07:41.020 --> 00:07:43.860
Therefore, the net reaction
is going to move to the left

00:07:43.860 --> 00:07:45.140
because if it moves to the left,

00:07:45.140 --> 00:07:47.160
we're going from two
moles of gas in the right

00:07:47.160 --> 00:07:49.080
to four moles of gas on the left.

00:07:49.080 --> 00:07:52.000
So that's an increase in the moles of gas

00:07:52.000 --> 00:07:53.800
and increase in the amount of gas,

00:07:53.800 --> 00:07:56.730
causes an increase in the pressure.

00:07:56.730 --> 00:07:57.780
Now let's see what happens,

00:07:57.780 --> 00:07:59.584
when we have equal amounts of moles of gas

00:07:59.584 --> 00:08:01.930
on both sides of the equation.

00:08:01.930 --> 00:08:04.630
For example, for the
hypothetical reaction,

00:08:04.630 --> 00:08:06.530
where gas A turns into gas B,

00:08:06.530 --> 00:08:09.200
there's one mole of gas
on the reactant side,

00:08:09.200 --> 00:08:12.230
and there's one mole of
gas on the product side,

00:08:12.230 --> 00:08:15.460
and let's use particular
diagrams and reaction quotients

00:08:15.460 --> 00:08:17.080
to understand what's going on here.

00:08:17.080 --> 00:08:19.490
So for our first particular diagram,

00:08:19.490 --> 00:08:21.200
here's the Qc expression.

00:08:21.200 --> 00:08:23.520
It's equal to the concentration
of B to the first power,

00:08:23.520 --> 00:08:26.270
divided by the concentration
of A to the first power

00:08:26.270 --> 00:08:28.850
and the concentration
of B since B is blue,

00:08:28.850 --> 00:08:32.700
there's two blue spheres in
this first particular diagram.

00:08:32.700 --> 00:08:36.040
So two times 0.1 moles
is 0.2 moles of blue

00:08:36.040 --> 00:08:38.430
divided by a volume of one liter.

00:08:38.430 --> 00:08:43.350
Therefore the concentration
of B is equal to 0.2.

00:08:43.350 --> 00:08:46.700
The concentration of A would
also be equal to 0.2 molar

00:08:46.700 --> 00:08:50.040
because there's two
particles of A in here.

00:08:50.040 --> 00:08:52.360
So 0.2 divided by 0.2

00:08:52.360 --> 00:08:53.560
is equal to

00:08:53.560 --> 00:08:54.730
one.

00:08:54.730 --> 00:08:58.360
And since K is equal to
one for this reaction,

00:08:58.360 --> 00:09:01.240
Kc is equal to one at 25 degrees Celsius.

00:09:01.240 --> 00:09:04.950
Qc is equal to Kc at this moment in time.

00:09:04.950 --> 00:09:07.940
Therefore the reaction is at equilibrium

00:09:07.940 --> 00:09:10.220
in this first particular diagram.

00:09:10.220 --> 00:09:13.010
If you introduce a stress to
our system at equilibrium,

00:09:13.010 --> 00:09:16.080
let's decrease the volume
by a factor of two.

00:09:16.080 --> 00:09:20.050
So we're going from one,
one liter to 0.5 liters.

00:09:20.050 --> 00:09:23.190
That's gonna increase the
pressure by a factor of two.

00:09:23.190 --> 00:09:26.350
And it's also going to
double the concentrations.

00:09:26.350 --> 00:09:28.130
So the concentration of B

00:09:28.130 --> 00:09:29.310
is now

00:09:29.310 --> 00:09:30.143
0.4

00:09:30.143 --> 00:09:30.976
molar

00:09:30.976 --> 00:09:34.620
and the concentration
of A is also 0.4 molar.

00:09:34.620 --> 00:09:36.660
So 0.4 divided by 0.4

00:09:36.660 --> 00:09:37.493
is equal

00:09:37.493 --> 00:09:38.326
to

00:09:38.326 --> 00:09:39.159
one.

00:09:39.159 --> 00:09:40.310
So Qc

00:09:40.310 --> 00:09:41.250
is still

00:09:41.250 --> 00:09:42.540
equal to Kc.

00:09:42.540 --> 00:09:46.160
The reaction is still at equilibrium.

00:09:46.160 --> 00:09:49.150
Since changing the volume
didn't change the value for Q,

00:09:49.150 --> 00:09:51.320
Qc is still equal to Kc.

00:09:51.320 --> 00:09:53.960
And since there's no change,
there's no net reaction

00:09:53.960 --> 00:09:57.080
to the left and there's no
net reaction to the right.

00:09:57.080 --> 00:10:00.080
Therefore changing the volume
on a reaction at equilibrium,

00:10:00.080 --> 00:10:03.060
when there are equal amounts
of moles of gas on both sides,

00:10:03.060 --> 00:10:08.060
has no effect on the composition
of the equilibrium mixture.

00:10:09.000 --> 00:10:10.980
Let's go back to the
reaction for the synthesis

00:10:10.980 --> 00:10:14.470
of ammonia gas from nitrogen
gas and hydrogen gas.

00:10:14.470 --> 00:10:17.050
And let's say this
reaction is at equilibrium.

00:10:17.050 --> 00:10:19.170
And we add in some helium gas

00:10:19.170 --> 00:10:22.010
to the reaction mixture at equilibrium.

00:10:22.010 --> 00:10:25.340
Helium is an inert gas,
which means it doesn't react

00:10:25.340 --> 00:10:27.770
with any of the gases
that we have present.

00:10:27.770 --> 00:10:28.740
It's tempting to say,

00:10:28.740 --> 00:10:32.720
adding this inert gas would
increase the total pressure

00:10:32.720 --> 00:10:36.430
and therefore the net reaction
would shift to the right

00:10:36.430 --> 00:10:39.100
to get rid of that increased pressure.

00:10:39.100 --> 00:10:42.250
However, notice that helium is
not a part of the expression

00:10:42.250 --> 00:10:45.500
for Qc and therefore after you add it,

00:10:45.500 --> 00:10:48.630
you're not actually changing
any of these concentrations.

00:10:48.630 --> 00:10:53.630
And so Qc is still equal to Kc
after the addition of helium.

00:10:53.890 --> 00:10:55.700
And since Qc is equal to Kc,

00:10:55.700 --> 00:10:57.820
the reaction is still at equilibrium

00:10:57.820 --> 00:11:00.450
and there's no shift either direction.

00:11:00.450 --> 00:11:03.220
Therefore adding an
inert gas to a reaction,

00:11:03.220 --> 00:11:05.780
make sure at equilibrium, has no effect

00:11:05.780 --> 00:11:08.753
on the composition of
the reaction mixture.


Le Chȃtelier’s principle: Changing concentration

WEBVTT
Kind: captions
Language: en

00:00:00.930 --> 00:00:02.710
- [Instructor] Le
Chatelier's principle says,

00:00:02.710 --> 00:00:06.420
if a stress is applied to a
reaction mixture at equilibrium,

00:00:06.420 --> 00:00:08.770
the net reaction goes in the direction

00:00:08.770 --> 00:00:10.920
that relieves the stress.

00:00:10.920 --> 00:00:13.770
Change in the concentration
of a reactant or product

00:00:13.770 --> 00:00:18.010
is one way to place a stress
on a reaction at equilibrium.

00:00:18.010 --> 00:00:21.220
For example, let's consider
the hypothetical reaction

00:00:21.220 --> 00:00:24.770
where gas A turns into gas B.

00:00:24.770 --> 00:00:27.760
And let's say the reaction
is at equilibrium.

00:00:27.760 --> 00:00:29.590
And we suddenly introduce a stress

00:00:29.590 --> 00:00:34.590
such as we increase the
concentration of reactant A.

00:00:34.800 --> 00:00:37.083
According to Le Chatelier's principle,

00:00:38.200 --> 00:00:40.180
the net reaction is
gonna go in the direction

00:00:40.180 --> 00:00:41.520
that relieves the stress.

00:00:41.520 --> 00:00:44.390
And since we increase
the concentration of A,

00:00:44.390 --> 00:00:46.970
the net reaction is gonna go to the right

00:00:46.970 --> 00:00:50.190
to decrease the concentration of A.

00:00:50.190 --> 00:00:51.680
Let's use some particular diagrams

00:00:51.680 --> 00:00:53.060
so we can get into the details

00:00:53.060 --> 00:00:55.440
of how the reaction goes to the right.

00:00:55.440 --> 00:00:58.330
So we're gonna symbolize
gas A by red particles

00:00:58.330 --> 00:01:00.970
and gas B by blue particles.

00:01:00.970 --> 00:01:02.980
And for this hypothetical reaction,

00:01:02.980 --> 00:01:05.740
the equilibrium constant is equal to three

00:01:05.740 --> 00:01:09.070
at 25 degrees Celsius.

00:01:09.070 --> 00:01:11.480
Let's start by writing
the reaction quotient.

00:01:11.480 --> 00:01:16.480
Qc is equal to, and we get that
from our balanced equation.

00:01:16.490 --> 00:01:20.200
That would be the concentration
of B to the first power

00:01:20.200 --> 00:01:25.200
divided by the concentration
of A, also to the first power.

00:01:28.050 --> 00:01:30.520
Let's calculate the
concentrations of B and A

00:01:30.520 --> 00:01:33.010
from our first particular diagram.

00:01:33.010 --> 00:01:35.670
So B is represented by the blue spheres

00:01:35.670 --> 00:01:37.650
and there are three blue spheres.

00:01:37.650 --> 00:01:40.960
If each particle represents
0.1 moles of a substance,

00:01:40.960 --> 00:01:44.310
and the volume of the
container is one litter,

00:01:44.310 --> 00:01:46.040
since we have three particles,

00:01:46.040 --> 00:01:49.630
that'd be three times
0.1, which is 0.3 moles

00:01:49.630 --> 00:01:53.260
divided by a volume of
one litter is 0.3 molar..

00:01:53.260 --> 00:01:57.590
So the concentration of B is 0.3 molar.

00:01:57.590 --> 00:02:00.430
For A, we have one
particles, that's 0.1 mole

00:02:00.430 --> 00:02:03.530
divided by one litter, which is 0.1 molar.

00:02:03.530 --> 00:02:07.128
So the concentration of A is 0.1 molar.

00:02:07.128 --> 00:02:09.810
And 0.3 divided by 0.1 is equal to three.

00:02:09.810 --> 00:02:14.400
So Qc at this moment in
time is equal to three.

00:02:14.400 --> 00:02:16.490
Notice we could have just
counted our particles,

00:02:16.490 --> 00:02:20.590
three blues and one red
and said three over one.

00:02:20.590 --> 00:02:22.810
That would have been a little bit faster.

00:02:22.810 --> 00:02:27.780
So Qc is equal to three and
Kc is also equal to three.

00:02:27.780 --> 00:02:29.351
So I should have written a C in here.

00:02:29.351 --> 00:02:34.160
So when Qc is equal to Kc, the
reaction is at equilibrium.

00:02:34.160 --> 00:02:36.730
So in this first particular diagram here

00:02:36.730 --> 00:02:41.730
where Qc is equal to Kc, the
reactions are at equilibrium.

00:02:41.890 --> 00:02:43.480
Next, we're gonna introduce a stress

00:02:43.480 --> 00:02:45.500
to our reaction at equilibrium.

00:02:45.500 --> 00:02:48.930
We're going to increase
the concentration of A.

00:02:48.930 --> 00:02:51.790
So here, we're gonna
add four particles of A

00:02:51.790 --> 00:02:55.810
to the reaction mixture at equilibrium.

00:02:55.810 --> 00:02:58.540
The second particulate diagram
shows what the reaction

00:02:58.540 --> 00:03:02.010
looks like right after we
add those four red particles.

00:03:02.010 --> 00:03:05.100
So we started with one red
particle and we added four.

00:03:05.100 --> 00:03:07.610
So now there's a total
of five red particles.

00:03:07.610 --> 00:03:10.910
And we still have the
same three blue particles

00:03:10.910 --> 00:03:13.710
that we had in the first
particular diagram.

00:03:13.710 --> 00:03:17.200
Let's calculate Qc at this moment in time.

00:03:17.200 --> 00:03:20.300
So just after we introduced the stress.

00:03:20.300 --> 00:03:23.250
Since there are three blue
particles and five red particles,

00:03:23.250 --> 00:03:27.633
Qc is equal to three divided
by five, which is equal to 0.6.

00:03:29.470 --> 00:03:33.690
Since Qc is equal to 0.6,
and Kc is equal to three,

00:03:33.690 --> 00:03:38.241
at this moment in time,
Qc is less than Kc.

00:03:38.241 --> 00:03:41.810
So there are too many reactants
and not enough products.

00:03:41.810 --> 00:03:46.080
Therefore, the net reaction
is going to go to the right

00:03:46.080 --> 00:03:49.100
and we're going to decrease
in the amount of A,

00:03:49.100 --> 00:03:52.870
and we're gonna increase
in the amount of B.

00:03:52.870 --> 00:03:54.749
The third particular
diagram shows what happens

00:03:54.749 --> 00:03:58.120
after the net reaction moves to the right.

00:03:58.120 --> 00:04:00.250
So we said, we're gonna
decrease the amount of A

00:04:00.250 --> 00:04:02.880
and increase in the amount of B.

00:04:02.880 --> 00:04:04.910
We're going from three blues

00:04:04.910 --> 00:04:08.957
in the second particular diagram
to six blues in the third.

00:04:08.957 --> 00:04:13.957
And we're going from five
reds to only two reds.

00:04:14.940 --> 00:04:19.940
Therefore, three reds must
have turned into blues

00:04:20.110 --> 00:04:23.000
to get the third particular
diagram on the right.

00:04:23.000 --> 00:04:27.540
And if we calculate Qc for
our third particular diagram,

00:04:27.540 --> 00:04:30.980
it'd be equal to six divided by two,

00:04:30.980 --> 00:04:33.280
which is equal to three.

00:04:33.280 --> 00:04:37.780
So at this moment in
time, Qc is equal to Kc.

00:04:37.780 --> 00:04:39.090
They're both equal to three.

00:04:39.090 --> 00:04:41.827
So equilibrium has been reestablished

00:04:41.827 --> 00:04:44.940
in the third particular diagram.

00:04:44.940 --> 00:04:47.640
It isn't always necessary
to calculate Q values

00:04:47.640 --> 00:04:51.730
when doing a Le Chatelier's
changing concentration problem.

00:04:51.730 --> 00:04:54.040
However, for this hypothetical reaction,

00:04:54.040 --> 00:04:57.440
it's useful to calculate
Q values to understand

00:04:57.440 --> 00:04:59.760
that we're starting at equilibrium

00:04:59.760 --> 00:05:01.520
and then a stress is introduced

00:05:01.520 --> 00:05:04.080
such as changing the concentration
of a reaction or product.

00:05:04.080 --> 00:05:08.400
And that means the reaction
is no longer at equilibrium.

00:05:08.400 --> 00:05:10.460
Le Chatelier's principle
allows us to predict

00:05:10.460 --> 00:05:14.130
which direction the net reaction will go

00:05:14.130 --> 00:05:16.320
or we could also use Q to predict

00:05:16.320 --> 00:05:18.150
the direction of the net reaction.

00:05:18.150 --> 00:05:21.850
The net reaction will continue
going in that new direction

00:05:21.850 --> 00:05:23.970
until Q is equal to K again

00:05:23.970 --> 00:05:27.220
and equilibrium has been reestablished,

00:05:27.220 --> 00:05:28.530
let's look at another reaction.

00:05:28.530 --> 00:05:30.570
This is the synthesis of ammonia

00:05:30.570 --> 00:05:34.060
from nitrogen gas and hydrogen gas.

00:05:34.060 --> 00:05:36.800
And let's see the reaction
is at equilibrium.

00:05:36.800 --> 00:05:38.810
So let's also look at this on a graph

00:05:38.810 --> 00:05:41.320
of concentration versus time.

00:05:41.320 --> 00:05:44.580
At equilibrium, the concentrations
of reactants and products

00:05:44.580 --> 00:05:48.040
are constant, which is why we
see these straight lines here

00:05:48.040 --> 00:05:51.983
for the concentration of
hydrogen, ammonia, and nitrogen.

00:05:52.830 --> 00:05:56.480
And let's introduce a stress
to the system at equilibrium.

00:05:56.480 --> 00:05:58.710
So right now we are at equilibrium

00:05:58.710 --> 00:06:00.980
and all the concentrations are constant.

00:06:00.980 --> 00:06:05.160
And let's increase the
concentration of hydrogen.

00:06:05.160 --> 00:06:08.020
So we can see that on our graph.

00:06:08.020 --> 00:06:10.530
So right here, there's a sudden increase

00:06:10.530 --> 00:06:13.670
in the concentration of hydrogen.

00:06:13.670 --> 00:06:16.870
Adding hydrogen means that
Q is no longer equal to K

00:06:16.870 --> 00:06:19.870
and therefore the reaction
is not at equilibrium.

00:06:19.870 --> 00:06:21.880
So let's go ahead and write over here,

00:06:21.880 --> 00:06:24.173
Now we're not at equilibrium.

00:06:25.050 --> 00:06:27.600
And Le Chatelier's principle
allows us to predict

00:06:27.600 --> 00:06:30.560
which direction the
net reaction will move.

00:06:30.560 --> 00:06:33.130
So since we added a stress,

00:06:33.130 --> 00:06:36.600
the stress being increased
concentration of hydrogen,

00:06:36.600 --> 00:06:38.860
the net reaction is
gonna move to the right

00:06:38.860 --> 00:06:42.670
to get rid of some of that
hydrogen that was added.

00:06:42.670 --> 00:06:44.460
And when the reaction goes to the right,

00:06:44.460 --> 00:06:46.850
the amount of ammonia will increase.

00:06:46.850 --> 00:06:49.620
And that's what we can see right
here on this red line here,

00:06:49.620 --> 00:06:52.170
the amount of ammonia is increasing.

00:06:52.170 --> 00:06:54.340
And the amount of ammonia increases

00:06:54.340 --> 00:06:57.760
because nitrogen and hydrogen
are reacting to form ammonia.

00:06:57.760 --> 00:06:59.900
Therefore, the amount
of nitrogen and hydrogen

00:06:59.900 --> 00:07:00.851
will decrease.

00:07:00.851 --> 00:07:03.560
Here we can see the amount
of hydrogen is decreasing.

00:07:03.560 --> 00:07:05.711
And down here, we can see
the amount of nitrogen

00:07:05.711 --> 00:07:07.800
is decreasing.

00:07:07.800 --> 00:07:09.710
The reaction will continue
to go to the right

00:07:09.710 --> 00:07:12.060
until equilibrium is reestablished.

00:07:12.060 --> 00:07:14.330
And that happens at the
second dotted line here.

00:07:14.330 --> 00:07:16.250
And we know that because we can see

00:07:16.250 --> 00:07:19.830
all of these concentrations
are now constant.

00:07:19.830 --> 00:07:23.870
So the reaction has reached equilibrium.

00:07:23.870 --> 00:07:26.720
So far we've only talked about
change in the concentration

00:07:26.720 --> 00:07:28.600
of a reactant so for example,

00:07:28.600 --> 00:07:31.632
if we increase the
concentration of hydrogen,

00:07:31.632 --> 00:07:34.260
the net reaction goes to the right.

00:07:34.260 --> 00:07:36.360
We could also say shifts to the right.

00:07:36.360 --> 00:07:38.530
So for a reaction at equilibrium,

00:07:38.530 --> 00:07:40.730
if you increase the
concentration of reactants,

00:07:40.730 --> 00:07:42.520
such as the concentration of hydrogen

00:07:42.520 --> 00:07:45.750
or the concentration of nitrogen,

00:07:45.750 --> 00:07:49.070
the reaction will shift to the right

00:07:49.070 --> 00:07:52.970
to decrease the amount of
one of those reactants.

00:07:52.970 --> 00:07:54.840
And if our reaction is at equilibrium

00:07:54.840 --> 00:07:57.700
and we were to increase
the amount of our product,

00:07:57.700 --> 00:07:59.348
increase the amount of ammonia,

00:07:59.348 --> 00:08:00.940
so this time the stress place

00:08:00.940 --> 00:08:03.130
is increased concentration of a product,

00:08:03.130 --> 00:08:05.310
Le Chatelier's principle
says the net reaction

00:08:05.310 --> 00:08:08.560
is going to move in the direction
that decreases the stress.

00:08:08.560 --> 00:08:11.640
So in this case, the net
reaction would go to the left

00:08:11.640 --> 00:08:13.833
to decrease the amount of ammonia.

00:08:14.730 --> 00:08:16.460
And if our reaction is at equilibrium

00:08:16.460 --> 00:08:21.460
and we were to decrease the
concentration of our product,

00:08:21.490 --> 00:08:24.520
the net reaction would shift to the right

00:08:24.520 --> 00:08:27.110
to make more of the product.

00:08:27.110 --> 00:08:29.930
Or if we decrease the concentration
of one of our reactants,

00:08:29.930 --> 00:08:32.240
let's say of nitrogen,

00:08:32.240 --> 00:08:34.600
in this case the reaction
will shift to the left

00:08:34.600 --> 00:08:36.493
to make more of our reactant.


Worked example: Using the reaction quotient to find equilibrium partial pressures

WEBVTT
Kind: captions
Language: en

00:00:00.568 --> 00:00:02.932
- [Tutor] For the
reaction of iron two oxide

00:00:02.932 --> 00:00:07.463
plus carbon monoxide goes to
solid iron and carbon dioxide.

00:00:07.463 --> 00:00:12.463
The equilibrium constant Kp is
equal to 0.26 at 1000 Kelvin.

00:00:12.818 --> 00:00:15.929
Our goal is to find the
equilibrium partial pressures

00:00:15.929 --> 00:00:19.531
of our two gasses, carbon
monoxide and carbon dioxide.

00:00:19.531 --> 00:00:21.232
If the initial partial pressures

00:00:21.232 --> 00:00:24.210
are 0.80 atmospheres for carbon monoxide

00:00:24.210 --> 00:00:28.140
and 0.40 atmospheres for carbon dioxide,

00:00:28.140 --> 00:00:30.131
we can use the reaction quotient Q,

00:00:30.131 --> 00:00:32.833
to predict which direction
that reaction will go

00:00:32.833 --> 00:00:34.819
to reach equilibrium.

00:00:34.819 --> 00:00:36.569
So let's calculate Qp

00:00:37.648 --> 00:00:39.221
and Qp

00:00:39.221 --> 00:00:40.370
is equal to,

00:00:40.370 --> 00:00:42.139
first we think about our products

00:00:42.139 --> 00:00:44.885
and we leave solids out of
equilibrium expressions,

00:00:44.885 --> 00:00:47.787
and therefore we also leave it
out of our expression for Qp.

00:00:47.787 --> 00:00:49.381
So we're gonna leave out iron.

00:00:49.381 --> 00:00:52.818
We're going to include carbon
dioxide since it's a gas.

00:00:52.818 --> 00:00:56.115
So we're gonna write the partial
pressure of carbon dioxide,

00:00:56.115 --> 00:00:57.830
and since we have a coefficient of one

00:00:57.830 --> 00:00:59.229
in front of carbon dioxide,

00:00:59.229 --> 00:01:01.685
it's the partial pressure
raise to the first power

00:01:01.685 --> 00:01:04.140
divided by, next we look at our reactants,

00:01:04.140 --> 00:01:06.709
and we have a solid, so
we're gonna leave that out.

00:01:06.709 --> 00:01:09.231
So we have another gas, carbon monoxide.

00:01:09.231 --> 00:01:12.090
So this will be the partial
pressure of carbon monoxide

00:01:12.090 --> 00:01:13.460
raised to the first power,

00:01:13.460 --> 00:01:15.878
since there's also a coefficient of one.

00:01:15.878 --> 00:01:17.749
Next, we plug in our partial pressures

00:01:17.749 --> 00:01:19.218
at this moment in time.

00:01:19.218 --> 00:01:21.593
The partial pressure of carbon dioxide

00:01:21.593 --> 00:01:22.467
is

00:01:22.467 --> 00:01:23.300
0.40,

00:01:24.158 --> 00:01:26.631
and the partial pressure
of carbon monoxide

00:01:26.631 --> 00:01:27.464
is

00:01:27.464 --> 00:01:28.297
0.80

00:01:28.297 --> 00:01:29.635
atmospheres.

00:01:29.635 --> 00:01:32.146
So we plug those into
our expression for Qp

00:01:32.146 --> 00:01:33.108
and 0.40

00:01:33.108 --> 00:01:34.679
divided by 0.80

00:01:34.679 --> 00:01:36.236
is equal to

00:01:36.236 --> 00:01:37.750
0.50.

00:01:37.750 --> 00:01:39.390
So Qp

00:01:39.390 --> 00:01:43.210
at this moment in time is equal to 0.50.

00:01:43.210 --> 00:01:46.730
Since QP is not equal to
Kp at this moment in time,

00:01:46.730 --> 00:01:49.569
the reaction is not at equilibrium.

00:01:49.569 --> 00:01:54.120
So Qp is equal to 0.50
and Kp is equal to 0.26.

00:01:54.120 --> 00:01:54.953
So Qp

00:01:55.850 --> 00:01:57.199
is greater

00:01:57.199 --> 00:01:58.548
than Kp.

00:01:58.548 --> 00:02:00.986
And when Qp is greater than Kp,

00:02:00.986 --> 00:02:04.655
there are too many products
and not enough reactants.

00:02:04.655 --> 00:02:09.467
Therefore the net reaction
is going to move to the left.

00:02:09.467 --> 00:02:12.373
Next let's fill out our I.C.E
table for this reaction.

00:02:12.373 --> 00:02:15.679
I stands for the initial
partial pressure in atmospheres.

00:02:15.679 --> 00:02:18.369
And so the initial partial
pressure of carbon monoxide

00:02:18.369 --> 00:02:20.276
was 0.80 atmospheres.

00:02:20.276 --> 00:02:22.717
And the initial partial
pressure of carbon dioxide

00:02:22.717 --> 00:02:25.068
is 0.40 atmospheres.

00:02:25.068 --> 00:02:26.656
C stands for change.

00:02:26.656 --> 00:02:30.017
And E stands for the
equilibrium partial pressure.

00:02:30.017 --> 00:02:32.609
Calculating Qp allowed us to realize

00:02:32.609 --> 00:02:35.306
that the net reaction moves to the left.

00:02:35.306 --> 00:02:37.416
And if the net reaction moves to the left,

00:02:37.416 --> 00:02:39.999
we're going to lose some carbon dioxide

00:02:39.999 --> 00:02:43.569
and we're going to gain
some carbon monoxide.

00:02:43.569 --> 00:02:46.506
So first let's think about carbon dioxide.

00:02:46.506 --> 00:02:47.695
We're gonna lose some of it,

00:02:47.695 --> 00:02:50.146
but we don't know how much and therefore

00:02:50.146 --> 00:02:52.169
that's gonna be represented by X.

00:02:52.169 --> 00:02:55.767
So we're gonna write minus
X here for carbon dioxide.

00:02:55.767 --> 00:02:57.689
And since the coefficient is a one

00:02:57.689 --> 00:02:59.457
in front of carbon dioxide,

00:02:59.457 --> 00:03:02.177
and it's also one in
front of carbon monoxide,

00:03:02.177 --> 00:03:04.536
if we lose X for carbon dioxide,

00:03:04.536 --> 00:03:08.376
we're going to gain X for carbon monoxide.

00:03:08.376 --> 00:03:10.745
Therefore the equilibrium partial pressure

00:03:10.745 --> 00:03:12.619
for carbon monoxide

00:03:12.619 --> 00:03:13.452
would be

00:03:13.452 --> 00:03:14.926
0.80

00:03:14.926 --> 00:03:15.759
plus X.

00:03:16.629 --> 00:03:19.956
And the equilibrium partial
pressure for carbon dioxide

00:03:19.956 --> 00:03:20.789
would be

00:03:20.789 --> 00:03:21.622
0.40

00:03:22.646 --> 00:03:23.479
minus X.

00:03:24.417 --> 00:03:27.725
Our next step is to write an
equilibrium constant expression

00:03:27.725 --> 00:03:29.235
for this reaction.

00:03:29.235 --> 00:03:34.235
So Kp is equal to the partial
pressure of carbon dioxide

00:03:34.543 --> 00:03:38.845
divided by the partial
pressure of carbon monoxide.

00:03:38.845 --> 00:03:41.965
So the expressions for
Kp and Qp look the same,

00:03:41.965 --> 00:03:43.994
but the difference is for Kp,

00:03:43.994 --> 00:03:46.954
it would be the equilibrium
partial pressures only.

00:03:46.954 --> 00:03:48.185
Whereas for Qp,

00:03:48.185 --> 00:03:51.373
it's the partial pressures
at any moment in time.

00:03:51.373 --> 00:03:52.434
And since for Kp,

00:03:52.434 --> 00:03:55.535
we're talking about the
equilibrium partial pressures,

00:03:55.535 --> 00:03:58.591
we can take those directly
from our I.C.E table

00:03:58.591 --> 00:03:59.754
and plug them in.

00:03:59.754 --> 00:04:03.496
So we can plug in the equilibrium
partial pressure of CO2

00:04:03.496 --> 00:04:08.496
and the equilibrium partial
pressure of carbon monoxide.

00:04:08.833 --> 00:04:12.046
Here we can see our two
equilibrium partial pressures

00:04:12.046 --> 00:04:14.622
plugged into our Kp expression

00:04:14.622 --> 00:04:17.114
and also the equilibrium constant Kp

00:04:17.114 --> 00:04:19.634
is equal to 0.26 for this reaction,

00:04:19.634 --> 00:04:21.625
so that's plugged in as well.

00:04:21.625 --> 00:04:23.697
Our next step is to solve for X.

00:04:23.697 --> 00:04:27.736
So we multiply both sides by 0.80 plus X,

00:04:27.736 --> 00:04:31.115
and we get this and then
we do some more algebra

00:04:31.115 --> 00:04:35.536
and we get down to 1.26
X is equal to 0.192.

00:04:35.536 --> 00:04:36.369
So

00:04:36.369 --> 00:04:37.202
0.192

00:04:37.202 --> 00:04:38.035
divided by

00:04:38.035 --> 00:04:39.502
1.26

00:04:39.502 --> 00:04:40.335
is equal to

00:04:40.335 --> 00:04:41.470
0.15.

00:04:41.470 --> 00:04:42.303
So X

00:04:42.303 --> 00:04:43.564
is equal to

00:04:43.564 --> 00:04:44.397
0.

00:04:44.397 --> 00:04:45.230
1

00:04:45.230 --> 00:04:46.776
5.

00:04:46.776 --> 00:04:49.189
Now that we know that X is equal to 0.15,

00:04:49.189 --> 00:04:50.879
we can go back to our I.C.E table

00:04:50.879 --> 00:04:54.028
and solve for the equilibrium
partial pressures.

00:04:54.028 --> 00:04:55.720
So for carbon monoxide,

00:04:55.720 --> 00:04:59.760
the equilibrium partial
pressure is 0.80 plus X.

00:04:59.760 --> 00:05:01.240
So that's equal to

00:05:01.240 --> 00:05:02.348
0.80

00:05:02.348 --> 00:05:03.480
plus

00:05:03.480 --> 00:05:04.313
0.

00:05:04.313 --> 00:05:05.146
1

00:05:05.146 --> 00:05:05.979
5,

00:05:05.979 --> 00:05:07.185
which is equal to

00:05:07.185 --> 00:05:08.018
0.

00:05:08.018 --> 00:05:08.851
9

00:05:08.851 --> 00:05:09.684
5

00:05:09.684 --> 00:05:10.517
atmospheres.

00:05:10.517 --> 00:05:12.325
So that's the equilibrium partial pressure

00:05:12.325 --> 00:05:14.313
of carbon monoxide.

00:05:14.313 --> 00:05:15.733
For carbon dioxide,

00:05:15.733 --> 00:05:19.730
the equilibrium partial
pressure is 0.40 minus X.

00:05:19.730 --> 00:05:20.563
So

00:05:20.563 --> 00:05:21.529
0.40

00:05:21.529 --> 00:05:22.809
minus

00:05:22.809 --> 00:05:23.642
0.15

00:05:24.499 --> 00:05:25.796
is equal to

00:05:25.796 --> 00:05:26.629
0.

00:05:26.629 --> 00:05:27.462
2

00:05:27.462 --> 00:05:28.295
5

00:05:28.295 --> 00:05:29.128
atmospheres.

00:05:29.128 --> 00:05:31.135
So that's the equilibrium partial pressure

00:05:31.135 --> 00:05:33.394
for carbon dioxide.

00:05:33.394 --> 00:05:34.227
Finally,

00:05:34.227 --> 00:05:36.223
we can use the reaction quotient Qp

00:05:36.223 --> 00:05:38.183
to make sure that these two answers,

00:05:38.183 --> 00:05:41.393
for equilibrium partial
pressures are correct.

00:05:41.393 --> 00:05:44.539
So we can write that Qp is equal to

00:05:44.539 --> 00:05:47.363
the partial pressure of CO2

00:05:47.363 --> 00:05:50.753
divided by the partial pressure of CO.

00:05:50.753 --> 00:05:52.144
And we can plug in those,

00:05:52.144 --> 00:05:54.335
those equilibrium partial pressures.

00:05:54.335 --> 00:05:55.603
So this would be

00:05:55.603 --> 00:05:57.561
0.25 atmospheres,

00:05:57.561 --> 00:06:01.133
was the equilibrium partial
pressure of carbon dioxide

00:06:01.133 --> 00:06:04.334
and 0.95 was the
equilibrium partial pressure

00:06:04.334 --> 00:06:05.913
of carbon monoxide.

00:06:05.913 --> 00:06:07.913
And 0.25 divided by 0.95

00:06:08.807 --> 00:06:09.987
is equal to

00:06:09.987 --> 00:06:10.820
0.

00:06:10.820 --> 00:06:11.653
2

00:06:11.653 --> 00:06:12.928
6.

00:06:12.928 --> 00:06:13.761
And

00:06:13.761 --> 00:06:14.797
since Kp

00:06:14.797 --> 00:06:16.770
is also equal to

00:06:16.770 --> 00:06:17.603
0.26

00:06:18.487 --> 00:06:19.991
at this moment in time,

00:06:19.991 --> 00:06:20.985
Qp

00:06:20.985 --> 00:06:22.010
is equal

00:06:22.010 --> 00:06:23.075
to Kp

00:06:23.075 --> 00:06:25.963
and the reaction is at equilibrium.

00:06:25.963 --> 00:06:28.076
Therefore we know we have the correct

00:06:28.076 --> 00:06:30.300
equilibrium partial pressures.