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

(a) $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{O}\mathit{H}\mathbf{\left(}\mathbf{I}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{O}\mathit{H}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

(b) $\mathit{C}{\mathit{H}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathit{H}\mathit{C}\mathit{N}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

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,

$\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{O}\mathit{H}\mathbf{\left(}\mathbf{I}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{O}\mathit{H}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

We have no gas substance on the reactant side and 1 mol of gas on the product side.

As the volume of the system increases, the pressure of the system decreases, allowing more "space" for our gas molecules. In this case, the reaction will produce more gas, which will only happen if the reaction shifts to the side with more gas molecules.

There are more gas molecules on the product side than on the reactant side. The reaction will shift to the right, with the amount of product increasing and the amount of reactant decreasing.

Therefore, $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{O}\mathit{H}\mathbf{\left(}\mathbf{I}\mathbf{\right)}$will decrease, and $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{O}\mathit{H}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$will increase.

(b)

Considering the given reaction,

$\mathit{C}{\mathit{H}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathit{H}\mathit{C}\mathit{N}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

A total of 2 mol of gas is present on the reactant side, and a total of 4 mol of gas is present on the product side.

As the volume of the system increases, the pressure decreases, allowing more "space" for our gas molecules. In this case, the reaction will produce more gas, which can only happen if the reaction is shifted to the side with more gas molecules.

In comparison to the reactant side, the product side has more gas molecules. As the reaction progresses to the right, the amount of product produced will increase while the amount of reactant consumed will decrease.

Therefore, both the reactants $\mathit{C}{\mathit{H}}_{\mathbf{4}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$and $\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$will decrease, and both products $\mathit{H}\mathit{C}\mathit{N}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$and ${\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$will increase.