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

(a) ${\mathbf{F}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{2}\mathbf{}\mathbf{F}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

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

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,

${\mathit{F}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{}\mathbf{\rightleftharpoons }\mathbf{}\mathbf{}\mathbf{2}\mathbf{}\mathit{F}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

On the reactant side, the moles of gas are $1mol.$

In the product side, the moles of gas are $2mol$.

Because the equilibrium position shifts towards the product side as the container volume increases, the concentration of the product increases and the concentration of the reactant (F) decreases.

Therefore, ${\mathit{F}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$will decrease and $\mathit{F}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$will increase.

(b)

Considering the given reaction,

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

The number of moles of gas in the reactant: 2 mol

On the product side, the moles of gas are 4 mol.

The concentration of the product $\mathbf{(}{\mathit{C}}_{\mathbf{2}}{\mathit{H}}_2$and ${\mathit{H}}_{\mathbf{2}}\mathbf{)}$increases and the concentration of the reactant $\left(C{H}_4\right)$decreases because the equilibrium position shift towards the product side on increasing the volume of the container.

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