Question

When $NaN{O}_3$ is heated in a closed vessel oxygen is liberated and $NaN{O}_2$ is left behind. At equilibrium, which are correct?

• A. Addition of $NaN{O}_2$ favours reverse reaction.
• B. Addition of $NaN{O}_3$ favours forward reaction.
• C. Increasing temperature'favours forward reaction.
• D. Increasing pressure favours reverse reaction.

Answer 1

Answer: (C), (D)

Increasing temperature'favours forward reaction.
Increasing pressure favours reverse reaction.

$2NaN{O}_3\left(s\right)\to 2NaN{O}_2\left(s\right)+O_2\left(g\right)\uparrow$
$K_p=P_{O_2}$
(A) Addition of $NaN{O}_2$ has no effect as it is solid whose concentration does not appear in the equilibrium constant expression.
(B) Addition of $NaN{O}_3$ has no effect as it is solid whose concentration does not appear in the equilibrium constant expression.
(C) Increasing temperature favours forward reaction as the reaction is an endothermic reaction.
(D) Increasing pressure favours reverse reaction as the reaction proceeds with increase in the number of moles of gaseous species.
$\Delta n>0$