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

(a) ${\mathit{C}}_{\mathbf{3}}{\mathit{H}}_{\mathbf{8}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{50}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{3}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}$

(b) $\mathbf{4}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{3}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{2}\mathbf{}{\mathit{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

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,

${\mathit{C}}_{\mathbf{3}}{\mathit{H}}_{\mathbf{8}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{50}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{3}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}$

We have a total of 6 mol of gas on the reactant side and 3 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.

The number of gas molecules on the product side is lower than on the reactant side. The reaction will shift to the right, producing more product and less reactant.

Therefore, both reactants ${\mathit{C}}_{\mathbf{3}}{\mathit{H}}_{\mathbf{8}}\left(g\right)$and ${\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$will decrease, and both $\mathit{C}{\mathit{O}}_{\mathbf{2}}$and ${\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}$will increase.

(b)

Considering the given reaction,

$\mathbf{4}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}{\mathbf{30}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{2}\mathbf{}{\mathit{N}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{+}\mathbf{6}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

We have a total of 7 mol of gas on the reactant side, and 9 mol of gas 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 fewer gas molecules on the reactant side than on the product side. The reaction will shift to the left, with the amount of product decreasing and the amount of reactant increasing.

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