## Q. 13.14

Applying Le Châtelier’s Principle to Pressure and Volume Changes

Does the number of moles of reaction products increase, decrease, or remain the same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?

(a) $\mathrm{PCl}_5(g) \rightleftharpoons \mathrm{PCl}_3(g)+\mathrm{Cl}_2(g)$

(b) $\mathrm{CaO}(s)+\mathrm{CO}_2(g) \rightleftharpoons \mathrm{CaCO}_3(s)$

(c) $3 \mathrm{Fe}(s)+4 \mathrm{H}_2 \mathrm{O}(g) \rightleftharpoons \mathrm{Fe}_3 \mathrm{O}_4(s)+4 \mathrm{H}_2(g)$

STRATEGY

According to Le Châtelier’s principle, the stress of a decrease in pressure is relieved by net reaction in the direction that increases the number of moles of gas.

## Verified Solution

(a) Because the forward reaction converts 1 mol of gas to 2 mol of gas, net reaction will go from reactants to products, thus increasing the number of moles of $PCl_3$ and $Cl_2$.

(b) Because there is 1 mol of gas on the reactant side of the balanced equation and none on the product side, the stress of a decrease in pressure is relieved by net reaction from products to reactants. The number of moles of $CaCO_3$ therefore decreases.

(c) Because there are 4 mol of gas on both sides of the balanced equation, the composition of the equilibrium mixture is unaffected by a change in pressure. The number of moles of $Fe_3O_4$ and $H_2$ remains the same.