Question

When $NaN{O}_3$ is heated in a closed vessel, $O_2$ is liberated and $NaN{O}_2$ is left behind. At equilibrium:

• A. increased temperature favours forward reaction
• B. addition of $NaN{O}_2$ favours reverse reaction
• C. increased pressure favours forward reaction
• D. adding of $NaN{O}_2$ favours forward reaction

Answer 1

The correct option is B addition of $NaN{O}_2$ favours reverse reaction

$2NaN{O}_3\rightleftharpoons 2NaN{O}_2+O_2$

At equilibrium when $NaN{O}_2$ is added then according to Le Chatlier's principle the system will follow reverse reaction.