What happens when an inert gas is added to an equilibrium keeping volume unchanged?

When an inert gas is added to a system at equilibrium with the volume kept unchanged, we need to analyze the effect on the equilibrium. An inert gas does not react with any of the reactants or products. Therefore, it does not participate in the chemical reaction.

The equilibrium constant for a reaction, denoted as $$ K_c $$, is expressed in terms of concentrations. For a general reaction:

the equilibrium constant is:

where $$[A]$$, $$[B]$$, $$[C]$$, and $$[D]$$ are the molar concentrations of the species. Concentration is defined as the number of moles per unit volume, so $$[A] = \frac{n_A}{V}$$, and similarly for others.

When an inert gas is added at constant volume:

• The volume $$ V $$ remains the same.
• The number of moles of the reactants and products (e.g., $$ n_A $$, $$ n_B $$, $$ n_C $$, $$ n_D $$) does not change because the inert gas does not react with them.

Therefore, the concentrations of the reactants and products remain unchanged. For example:

and similarly for $$ B $$, $$ C $$, and $$ D $$.

Since the concentrations are unchanged, the reaction quotient $$ Q_c $$ (which has the same expression as $$ K_c $$) remains equal to $$ K_c $$:

Because $$ Q_c = K_c $$, the system remains at equilibrium. There is no shift in the equilibrium position towards reactants or products.

Although the total pressure increases due to the addition of the inert gas, pressure changes only affect the equilibrium if the volume changes (which alters concentrations) or if the reaction involves a change in the number of moles of gas. Here, volume is constant, and the inert gas does not alter the mole numbers of the reacting species, so the equilibrium is unaffected.

Hence, the correct answer is Option D.