Question

For the reaction $2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$ The incorrect statement is :

• A. Decreasing the pressure will shift the equilibrium to the left
• B. Addition of inert gas at constant pressure will shift the equilibrium towards right
• C. Value of $\frac{K_P}{K_C}$ is less than one
• D. No change in state of equilibrium when inert gas is added at constant volume

Answer 1

The correct option is B Addition of inert gas at constant pressure will shift the equilibrium towards right

$2S{O}_2\left(g\right)+O_2\left(g\right)\rightleftharpoons 2S{O}_3\left(g\right)$
$\left(A\right)K_P=K_C(RT{)}^{\Delta ng}$
$\Delta n_g=2-(2+1)=-1$
$\frac{K_P}{K_C}=\frac{1}{RT}<1$

So, this is correct statement.
(B) Addition of inert gas at constant pressure will shift the equilibrium towards more no. of gaseous moles or towards left. So, given statement is incorrect.
(C) When inert gas is added at constant volume, there is no change in state of equilibrium. So, given statement is correct.
(D) On decreasing the pressure, the equilibrium will shift to the more no. of gaseous moles or towards left. So, given statement is also correct. Therefore, the incorrect statement is option ‘B’