Predict the effect of decreasing the container volume on the amounts of each reactant and product in the following reactions:

(a) ${\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{HCl}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

(b) $\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}$

Le-Chatelier's principle states that, whenever a system at equilibrium is disturbed, the system will undergo reactions and try to cancel that effect and retain equilibrium. Changes in concentration of any component, temperature, pressure, or volume are all examples of disturbances.

(a)

Considering the given reaction,

${\mathbf{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{Cl}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{HCl}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

We have a total of 2 mol of gas on the reactant side and 2 mol of gas on the product side.

As the volume of the system decreases, the pressure rises, leaving less "space" for our gas molecules. In this case, the reaction will produce less gas, which can only happen if the reaction shifts to the side with the fewest gas molecules.

There are an equal number of gas molecules on both the reactant and product sides. There will be no change in the amount of the reaction components because the reaction will not shift.

(b)

Considering the given reaction,

$\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{2}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}$

We have a total of 3 mol of gas on the reactant side, but no gas molecules on the product side.

As the volume of the system decreases, the pressure rises, limiting the amount of "space" available for our gas molecules. In this case, the reaction will produce less gas, which can only be achieved by shifting the reaction to the side with fewer gas molecules.

There are no gas molecules on the product side. The reaction will shift to the right, with the amount of product increasing and the amount of reactant decreasing.

Therefore, both reactants ${\mathit{H}}_{\mathbf{2}}\left(g\right)$and ${\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$will decrease, and ${\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}$will increase.