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Current time:0:00Total duration:11:09

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.B (LO)

, TRA‑8.B.1 (EK)

- [Tutor] Le Chatelier's
principle says that if a stress is applied to a
reaction mixture at equilibrium, the net reaction goes in the direction that relieves the stress. One possible stress that we
could do is to change the volume on our reaction at equilibrium. Let's say we have a hypothetical
reaction, where solid A, which is symbolized by red
in our particular diagram, turns into solid B, which
is symbolized by a blue and also C, which is a gas and C will be symbolized by white spheres. The equilibrium constant
Kc for this reaction, is equal to 0.4 at 25 degrees Celsius. The first particular diagram shows the reaction at equilibrium, and we can see there's some
solid A and some solid B at the bottom of the container, and there's also some
gaseous particles of C. Let's introduce a stress, to our reaction mixture at equilibrium. Let's decrease the
volume of the container. So looking from the first
particular diagram to the second, we can see there's been a decrease in the volume of the container. That's gonna cause an
increase in the pressure because pressure comes
from these gas particles slamming into the sides of the container. And if we decrease the volume, now there's less distance
for these particles to travel before they slam into the
sides of the container, which means we increase
the collision frequency. And therefore the pressure increases. According to Le Chatelier's principle, the net reaction is gonna
go in the direction, that relieves the stress. So if the stress is increased pressure, the net reaction is going to
try to move in the direction that decreases the pressure. Looking at the equation for
this hypothetical reaction, the two solids aren't really
contributing to the pressure. So it's only gas C that
we need to worry about. And there's one mole of
gas C on the product side, and there are zero moles of
gas on the reactant side. So if the net reaction went to the right, we'd be going from zero moles
of gas to one mole of gas. So going to the right would
increase the moles of gas, which would increase the pressure. However, that's not what
the reaction wants to do. The goal of the reaction
is to relieve the stress and therefore the reaction
wants to decrease the pressure. So the reactions gonna go to the left, to get rid of some of that gas and decreasing the amount of
gas will decrease the pressure, therefore relieving the stress. If the net reaction moves to the left, we're gonna lose some of our products. So we're gonna decrease the amount of C and we're gonna decrease
in the amount of B and we're going to gain
some of our reactants. So we're gonna increase
in the amount of A, and we can see all that and
the third particulate diagram. So going from the second
particular diagram to the third particular diagram, we've decreased the amount of C, we've gone from four particles of C to only two particles of C. We've also decreased the amount of B. You can see that this blue, this blue solid has gotten smaller, and we've increased the amount of A. So A, is this, A, is this red sphere here, and you can see how it's gotten bigger from the second particulate
diagram to the third. And by going from four particles of C, to only two particles of C, we've decreased the amount of the gas, and therefore we've
decreased the pressure. To better understand what
happens to a reaction mixture at equilibrium when a
stress is placed on it, let's calculate the reaction quotient Q for these three particular diagrams, for the same hypothetical reaction. The expression for the
reaction quotient Qc, has the same form as the
equilibrium constant expression for Kc. So since solids are left out of the equilibrium constant expression, we only need to include the
concentration of the gas. And since there's a coefficient
of one in front of C, Qc is equal to the concentration of C raised to the first power. Since there are four particles of C in the first particular diagram, and if each particle
represents 0.1 moles of C, four times, 0.1 is equal to 0.4 moles of C. So to find the concentration of C, we take the moles and divide that by the volume of the
container, which is 1.0 liter. So 0.4 divided by 1.0 is equal to 0.4 molar. So that's the concentration of C in the first particular diagram. We plugged that into our expression for Q. So Qc is equal to 0.4 and notice that Kc is also equal to 0.4. So at this moment in
time, Qc is equal to Kc, which tells us the
reaction is at equilibrium. Next, we think about the
stress that was applied to the reaction at equilibrium. We decreased the volume. And if we look at the volumes here, we're going from 1.0 liters to 0.5 liters. So we're decreasing the
volume by a factor of two, which would cause an
increase in the pressure by a factor of two. And changing the volume would
change the concentration. So instead of 0.4, divided by 1.0, it'd be 0.4 divided by 0.5, which is equal to 0.8 molar. So the concentration has doubled. So if we calculate Q for our
second particular diagram, we plug in the concentration
of C, which is 0.8 molar. So Qc is equal to 0.8, K is equal to 0.4. So Q is not equal to K, therefore the reaction
is not at equilibrium for our second particular diagram. Let me write in here, not at equilibrium. In this case, Qc is greater than Kc, which tells us we have too many products and not enough reactants. Therefore the net
reaction goes to the left to get rid of some of the products and to increase the amount of reactants. The net reaction keeps going to the left until we reach equilibrium again. So if we calculate the concentration of C in the third particular diagram, here there are only two particles. So that'd be 0.2 moles divided by a volume of 0.5 liters. So 0.2 divided by 0.5 is equal to 0.4 molar. So Qc is equal to 0.4. And since Qc is equal
to 0.4, Kc is still 0.4. So Qc is equal to Kc and
we're at equilibrium. So equilibrium has been re-established. Since the reaction is at equilibrium in a third particular diagram, the net reaction stops going to the left and the concentration
of C remains constant. Let's apply what we've
learned to another reaction, the synthesis of ammonia from
nitrogen gas and hydrogen gas. If we have a mixture of
these gases at equilibrium, and we introduce a stress, the system like a decrease in volume, the decrease in the
volume of the container would cause an increase in the pressure. And according to Le Chatelier's principle, the net reaction is
gonna go in the direction that relieves the stress that
was placed on the system. So if the stress is increased pressure, the net reaction says, I wanna move in the direction
that decreases that pressure. So the net reaction moves to the right, because there are four
moles of gas on the left and only two moles of gas on the right. And by moving to the right, that goes from four moles
of gas to two moles of gas, which decreases the amount of gas and causes a decrease in the pressure. If we had a mixture of
these gases at equilibrium, and we increased the volume
and increase in the volume would cause a decrease in the pressure. So the stress this time,
is decreased pressure. To relieve the stress, the net reaction, wants to move in the direction
that increases the pressure. Therefore, the net reaction
is going to move to the left because if it moves to the left, we're going from two
moles of gas in the right to four moles of gas on the left. So that's an increase in the moles of gas and increase in the amount of gas, causes an increase in the pressure. Now let's see what happens, when we have equal amounts of moles of gas on both sides of the equation. For example, for the
hypothetical reaction, where gas A turns into gas B, there's one mole of gas
on the reactant side, and there's one mole of
gas on the product side, and let's use particular
diagrams and reaction quotients to understand what's going on here. So for our first particular diagram, here's the Qc expression. It's equal to the concentration
of B to the first power, divided by the concentration
of A to the first power and the concentration
of B since B is blue, there's two blue spheres in
this first particular diagram. So two times 0.1 moles
is 0.2 moles of blue divided by a volume of one liter. Therefore the concentration
of B is equal to 0.2. The concentration of A would
also be equal to 0.2 molar because there's two
particles of A in here. So 0.2 divided by 0.2 is equal to one. And since K is equal to
one for this reaction, Kc is equal to one at 25 degrees Celsius. Qc is equal to Kc at this moment in time. Therefore the reaction is at equilibrium in this first particular diagram. If you introduce a stress to
our system at equilibrium, let's decrease the volume
by a factor of two. So we're going from one,
one liter to 0.5 liters. That's gonna increase the
pressure by a factor of two. And it's also going to
double the concentrations. So the concentration of B is now 0.4 molar and the concentration
of A is also 0.4 molar. So 0.4 divided by 0.4 is equal to one. So Qc is still equal to Kc. The reaction is still at equilibrium. Since changing the volume
didn't change the value for Q, Qc is still equal to Kc. And since there's no change,
there's no net reaction to the left and there's no
net reaction to the right. Therefore changing the volume
on a reaction at equilibrium, when there are equal amounts
of moles of gas on both sides, has no effect on the composition
of the equilibrium mixture. Let's go back to the
reaction for the synthesis of ammonia gas from nitrogen
gas and hydrogen gas. And let's say this
reaction is at equilibrium. And we add in some helium gas to the reaction mixture at equilibrium. Helium is an inert gas,
which means it doesn't react with any of the gases
that we have present. It's tempting to say, adding this inert gas would
increase the total pressure and therefore the net reaction
would shift to the right to get rid of that increased pressure. However, notice that helium is
not a part of the expression for Qc and therefore after you add it, you're not actually changing
any of these concentrations. And so Qc is still equal to Kc
after the addition of helium. And since Qc is equal to Kc, the reaction is still at equilibrium and there's no shift either direction. Therefore adding an
inert gas to a reaction, make sure at equilibrium, has no effect on the composition of
the reaction mixture.