Main content

## Equilibrium

# Le Chȃtelier’s principle: Changing temperature

AP.Chem:

TRA‑7 (EU)

, TRA‑7.F (LO)

, TRA‑7.F.1 (EK)

, TRA‑8 (EU)

, TRA‑8.A (LO)

, TRA‑8.A.1 (EK)

, TRA‑8.A.2 (EK)

, TRA‑8.B (LO)

, TRA‑8.B.1 (EK)

, TRA‑8.B.2 (EK)

## Video transcript

- [Instructor] Le Chateliers's
Principle says if a stress is applied to a reaction at equilibrium, the net reaction goes in
the direction that relieves the stress. One possible stress is
to change the temperature of the reaction at equilibrium. As an example, let's look
at the hypothetical reaction where gas A turns into gas B. Delta H for this reaction
is less than zero, which tells us this is
an exothermic reaction. And for an exothermic
reaction, heat is given off, heat is released. Therefore we can go
ahead and write plus heat on the product side. Let's say that our hypothetical
reaction is at equilibrium, and then we change the temperature, so we're going to
increase the temperature. According to Le Chatelier's Principal, the net reaction is
gonna go in the direction that decreases the stress. And if we treat heat if
we treat like a product and we've increased the temperature, it says if we've increased the
amount of one of our products and therefore the net reaction
is going to shift to the left to decrease a product. Let's use particulate diagrams
and the reaction quotient Q to explain what's going on when
we increase the temperature on our reaction at equilibrium. The first particular
diagram shows the reaction at equilibrium, and let's
prove that by calculating QC at this moment in time. We can get the expression
for the reaction quotient, QC by looking at the balanced equation. So we have coefficients
of one in front of A and in front of B, therefore QC is equal to the concentration
of B to the first power divided by the concentration of A, also to the first power. To find the concentration
of B, we know that B is represented by blue spheres, so there are 1, 2, 3 blue spheres, And if each sphere represents
0.1 moles of a substance three times 0.1 is equal to 0.3 moles of B and at the volumes equal to 1.0 liter, 0.3 divided by 1.0 liter is 0.3 Molar. So the concentration of
B is equal to 0.3 Molar. There are also three particles of A therefore the concentration of A is also equal to 0.3 Molar. 0.3 divided by 0.3 is equal to one. So QC at this moment of
time is equal to one. KC for this reaction is equal
to one at 25 degrees Celsius. So QC is equal to KC. And when QC is equal to KC,
the reaction is at equilibrium. So for this first particular diagram, the reaction's at equilibrium. Next, we introduced the
stress to the reaction at equilibrium and the
stress is an increase in the temperature. In general, for an exothermic reaction, increasing the temperature
lowers the value for the equilibrium constant. So for this hypothetical
reaction at 25 degrees Celsius, KC is equal to one, but
since we've increased the temperature, the value
for the equilibrium constant is going to decrease. So let's say it goes 2.5, if
we increase the temperature to 30 degrees Celsius. So if we calculate QC for our
second particular diagram, we still have three blues and three reds, and the volume is still the same, therefore QC is still equal to one, but the difference is KC has now changed, so QC is not equal to KC, so we are not at equilibrium. And in this case, QC is greater than KC 'cause QC is equal to one
and KC is equal to 0.5. And when QC is greater than
KC, there are too many products and not enough reactants. And therefore the net
reaction goes to the left. When the net reaction goes to the left, we're going to have be turned into A. So we should see one blue
sphere turn into one red sphere. So if we have three blues and three reds, and one blue turns into a red,
that gives us only two blues and four reds now. So when we calculate QC for
our third particular diagram, the concentration of B would be 0.2 Molar, and the concentration
of A would be 0.4 Molar. So 0.2 divided by 0.4 is equal to 0.5. Well, KC is also equal to 0.5. Therefore QC is equal to KC, and the reaction is at equilibrium. And when a reaction is at equilibrium, the concentrations of reactants
and products are constant. Let's go back to our hypothetical
reaction at equilibrium, but this time we're going
to decrease the temperature. If we treat heat like a product
decrease in the temperature is like decreasing the amount
of one of the products. Therefore the net reaction
will go to the right to make more product. If we approach this problem
by thinking about the reaction quotient Q for an exothermic reaction, a decrease in temperature in
general causes an increase in the equilibrium constants. And if the equilibrium constant increases, then Q would be less than K. And when Q is less than K, the net reaction goes to the right. The net reaction would
continue to go to the right until Q is equal to K, and equilibrium has been re-established. Next let's look at an endothermic reaction where Delta H is greater than zero. When six water molecules
complex to a Cobalt two plus ion the resulting complex
ion is pink in color. And when for chloride anions
complex to a cobalt two plus ion, the resulting complex
ion is blue in color. When the pink ion reacts
with four chloride anions, the blue ion is formed. And since this reaction is endothermic, we can put heat on the reactant side. We're gonna use these
particular diagrams down here to help us understand what
happens to an endothermic reaction at equilibrium when
the temperature changes, however, these drawings aren't
designed to be completely accurate for this particular reaction. They're just to help us understand
what color we would see. For example, let's say that
this middle particulate diagram represents the reaction at equilibrium. And if there are decent
amounts of both the blue ion and the pink ion at equilibrium, the resulting equilibrium mixture, so this is an aqueous solution would appear to be purple or violet. If we were to increase the temperature for this endothermic reaction, we treat heat like a reactant. So increasing the temperature
is like increasing the amount of a reactant. And therefore the net reaction
will shift to the right to get rid of some of that reactant. Whether the net reaction
goes to the right, we're gonna increase in
the amount of blue ion, and we're going to decrease
in the amount of pink ion. Therefore looking at this
particular diagram on the right there are now more blue ions
than there are pink ions compared to our equilibrium
mixture in the middle. Therefore for this third
particular diagram, the resulting aqueous solution
is going to look blue. If we think about those
using Q in general, for an endothermic reaction,
an increase in temperature causes an increase in the
equilibrium constant K. And if K increases, then
the reaction quotient Q is less than K. And when Q is less than K, the net reaction goes to the right. Now let's go back to the
middle particular diagram. And so there are reactions at equilibrium and this time we're going
to decrease the temperature. If we treat heat as a reactant, and we decrease the temperature, it's like we're decreasing
one of our reactants. Therefore the net reaction
will shift to the left to make more of our reactant. And when that reaction goes to the left, we're gonna decrease in
the amount of the blue ion, and we're going to increase
in the amount of the pink ion. So when we compare the
middle particulate diagram to the one on the left
and the one that left, there's a lot more of the pink ion, then there is of the blue ion. Therefore the overall solution, the overall aqueous solution
is going to appear pink. If we think about what's happening using Q for an endothermic reaction in general, when you decrease the temperature, you decrease the equilibrium constant. And if the equilibrium constant decreases now, Q would be greater than K, which means too many products
and not enough reactance. Therefore the net reaction
would go to the left. Let's go back to our exothermic reaction. At this time let's pretend like
we're starting with only A, so we start with only A
and we have none of B. And our goal is to make
as much as we possibly can and to do it as fast as possible. One way to increase the rate of a reaction would be to increase the temperature. However, for an exothermic reaction, increasing the temperature
decreases the equilibrium constant K. And if you decrease the
equilibrium constant K, you would decrease the amount of B that you would have when
you reach equilibrium. So we can't run our hypothetical reaction at too high of a temperature
because that would decrease the equilibrium constant. So instead to speed up
the rate of the reaction, we could add a catalyst, let's look at a graph of
concentration of B versus time, and we're gonna start
with this blue curve here, which represents the hypothetical reaction without a catalyst, so
the uncatalyzed reaction. When time is equal to zero,
the concentration of B is zero because we start with only A. And as A turns into B the
concentration of B increases over time, and eventually
the concentration of B becomes constant. And when the concentration
of B becomes constant, the reaction reaches equilibrium. So this dotted line here
represents the concentration of B at equilibrium. The yellow line represents the reaction with a catalyst added. So once again, we're starting with only A, so when time is equal to zero, the concentration of B is equal to zero. And as time increases A turns into B, so the concentration of B increases and eventually the concentration
of B becomes constant. And the reaction reaches equilibrium. Notice that the reaction
reached equilibrium much faster with the addition of the catalyst than it did without the catalyst. So the addition of a
catalyst allows a reaction to reach equilibrium faster. However, notice that the final
equilibrium concentration of B is the same for both
the uncatalyzed reaction and the catalyzed reaction. Therefore, the addition of
a catalyst does not change the composition of the
equilibrium mixture. And that's because the catalyst speeds up both the forward reaction
and the reverse reaction, but the rates are still equal. And since the rates are equal, there's no change in the composition of the equilibrium mixture.