Question

Given the following reaction at equilibrium, the addition of inert gas at constant pressure would cause:

$S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons S{O}_3\left(g\right)$

• A. the formation of more of $S{O}_3$
• B. the formation of less of $S{O}_3$
• C. no affect on the equilibrium concentration of $S{O}_3$
• D. the system to move to a new equilibrium position which cannot be theoretically predicted

Answer 1

The correct option is B the formation of less of $S{O}_3$

For the reaction $S{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons S{O}_3\left(g\right)$, the forward reaction occurs with decrease in the number of moles from 1.5 to 1. Also, the reverse reaction occurs with an increase in the number of moles from 1 to 1.5.

When inert gas is added at constant pressure, the equilibrium will shift in the direction in which there is an increase in the number of moles of gases. Thus the addition of inert gas at constant pressure will increase the dissociation of $S{O}_3$.