For equilibrium systems containing gases, a change in the volume of the system produces a shift in equilibrium. To understand this section, recall that the pressure of a gas is inversely proportional to the volume of the container.

If the volume of the container is decreased, the pressure is increased. This causes an equilibrium system to shift toward the side of the reaction that has the lowest number of moles of gas molecules. This shift reduces the total number of gas molecules in the system and, therefore, reduces the pressure.

Conversely, if the volume of the container is increased, the pressure is decreased. This causes an equilibrium system to shift towards the side of the reaction that has the highest number of moles of gas molecules. This shift increases the total number of gas molecules and, therefore, increases the pressure.

Note that volume (pressure) changes do not affect the value of K_c.

Example

In the equilibrium system below, 3 moles of gas are on the left and 2 moles of gas are on the right. If the pressure is increased by decreasing the container's volume, the equilibrium will shift to the right.

\( \mathrm { 2SO_2(g) + O_2(g) \leftrightharpoons 2SO_3(g) } \)

The graph below shows how the concentrations of the reactants and product change in response to a pressure change.

Read pages 694 in the textbook.

The collision-reaction theory can be used to explain how volume/pressure changes cause equilibrium shifts. If the volume of the container is decreased, the reacting gases are compresed into a smaller volume, thus increasing their concentration. As you learned in the previous section, increasing the concentration of reacting entities increases the number of collisions. As the number of collisions increase, the reaction rate increases. If we change the pressure of a gaseous equilibrium system by adding an inert gas, no shift occurs. The addition of a gas that does not participate in the reaction has no effect whatsoever on the concentrations of gases in the reaction, so the rates of forward and reverse reactions remain unchanged.