It is common to discuss the effect of pressure (of gases) on equilibrium but there is one HUGE problem with this. We cannot ever control pressure. Make sure you understand that. Pressure is always the dependent variable.
There is nothing you can do to directly change pressure.
You can change volume...which will change pressure.
You can change temperature...which will change pressure.
You can change the amount...which will change pressure.
But, you cannot directly change pressure.
So, when you are asked to consider the effect of pressure on equilibrium, this is a red herring. There is ALWAYS something else going on.
So let’s look at all of the ways that the pressure can be changed and see how they each affect equilibrium.
Change in temperature - this has already been addressed here.
Addition or removal of a gas involved in the reaction - this has already been addressed here.
Addition of a gas NOT involved in the reaction
Change in volume of the container
So there are really only two things to discuss.
Although “c” seems odd, it can be written as if it were a dilution problem. For instance:
The system 2 SO2 (g) + O2 (g) ⇄ 2 SO3 (g) is at equilibrium at 56oC and 1.50 atm. If the pressure is increased to 2.0 atm by addition of He to the container, what direction, if any, will the system shift to regain equilibrium?
However, this is NOT a dilution problem. When aqueous solutions are diluted the solute particles are less likely to collide due to the extra solvent molecules “in the way”. But gas particles are so far apart under normal circumstances that addition of another gas has no measurable effect on the number of collisions between the reactants.
So, addition of a gas not involved in the reaction is irrelevant and has no effect on the equilibrium.
In a gas, the particles move around independently, colliding with each other and the walls of the container. The temperature of the gas determines how fast the molecules move around, and consequently how often they collide.
If we decrease the size of the container but keep the number of particles the same and the temperature the same, the molecules will collide more often. This is simply a matter of crowding. The more tightly packed the molecules are, the more often they will collide and the more often reactions can occur.
Another way to think about this (really the same idea with different language) is that decreasing the volume of the container will increase the concentration of the gases, since concentration is moles/liters.
Let’s think about this reaction:
C2H4 (g) + Cl2 (g) ⇄ C2H4Cl2 (g)
When the volume of the container goes down, there are three consequences:
The concentration of ethene (C2H4) increases. Therefore the forward reaction will go faster.
The concentration of chlorine (Cl2) increases. Therefore the forward reaction will go faster.
The concentration of dichloro-ethane (C2H4Cl2) increases. Therefore the backward reaction will go faster.
Since there are two factors increasing the rate of the forward reaction and only one increasing the rate of the backward reaction, the forward reaction goes “faster-er” and the reaction will shift to the right.
Increasing the volume, would have the exact opposite effect. Larger volume means that ALL concentrations go down, therefore both reactions slow down, but because there are 2 factors on the left the forward reaction will be “slower-er” and the reaction will shift to the left.
So, if your question is just “WHAT direction will the reaction go if the volume is decreased?” the answer is that the reaction will shift toward the side with fewer gas particles.
If the question is “WHY does that shift occur?” you need to discuss the rate change for the forward AND backward reactions and then compare those two.
A last example
For the reaction H2 (g) + Cl2 (g) ⇄ 2 HCl (g) is unaffected by a change in volume, since both sides of the equation have 2 moles of gas particles.
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