How can you change the volume of gas being produced by the reaction?

Many industrial gas-phase reactions are run at very high gas pressure created by compressing gases originally at room pressure to a much smaller volume. To see why this is done, let’s take a closer look at the effect of changes in volume on gas-phase reactions.
If all other factors remain constant, changing the volume occupied by a gas will change its concentration, and therefore, change the rate at which it reacts with other substances. For example, for the following reaction, decreasing the volume occupied by the gases by half will double their concentrations.
N2O4(g) Double arrow 2NO2(g)
Doubling the concentration of N2O4 doubles the forward rate of reaction. In contrast, because there are two moles of NO2 involved in the reverse reaction, doubling the concentration of NO2 leads to four times the rate of the reverse reaction. The first important point here is that changing the volume occupied by a gas-phase reaction system leads to change in both the forward and reverse reaction rates. The second important point is that the effect on these two rates may not be the same. If the effect on the rates is different, equilibrium will be disrupted, and the reaction will shift toward more products or more reactants. In our case, because the reverse rate is increased more than the forward rate, the system will shift toward more reactants.
In general, decreased volume and increased concentration will lead to an increase in both the forward and reverse rates, but it will cause a greater increase in the rate (forward or reverse) whose “reactants” have more moles of gas. (Remember, the “products” are the “reactants” of the reverse reaction.) Thus decreased volume for a gas-phase reaction will shift the system toward the side of the reaction with the fewest moles of gas. For example, decreased volume and therefore increased concentration of both reactants and products for the following reaction at equilibrium will shift the system toward more products.
CO(g) + Cl2(g) Double arrow COCl2(g)
2 moles Double arrow 1 mole
The decreased volume only disrupts the equilibrium if the moles of gaseous products and moles of gaseous reactants are unequal. If there are an equal number of moles of gaseous substances on both sides of the arrow, the change in volume has an equal effect on the concentrations of reactants and of products. Thus, it has an equal effect on the forward and reverse rates, and the system remains at equilibrium. For example, a change in volume does not disrupt the equilibrium for the reaction that forms hydrogen gas.
CO(g) + H2O(g) Double arrow CO2(g) + H2(g)
2 moles Double arrow 2 moles

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