Question

When $NaN{O}_3$ is heated in a closed vessel, oxygen is liberated and $NaN{O}_2$ is left behind.
The reaction is :
$2NaN{O}_3\left(s\right)\rightleftharpoons 2NaN{O}_2\left(s\right)+O_2\left(g\right)$
Which of the following is true at equilibrium?

• A. Addition of $NaN{O}_3$ favors backward reaction
• B. Addition of $NaN{O}_2$ favors forward reaction
• C. Increasing pressure favors reverse reaction.
• D. All of the above.

Answer 1

The correct option is C Increasing pressure favors reverse reaction.

$2NaN{O}_3\left(s\right)\rightleftharpoons 2NaN{O}_2\left(s\right)+O_2\left(g\right)$

In this reaction, solid species which are $NaN{O}_{2}andNaN{O}_3$ do not have any effect on addition or removal , because in the $K_{eq}$ expression , it is only dependant on concentration of $O_2$ which is in gas phase..
Here $\Delta n_g=1$ which is positive (>0)

For $\Delta n_g>0$ as the pressure is increased, $\frac{n_g}{V}\uparrow$, the reaction goes where $n_g$ decreases.
It decreases in the backward direction.
So , by increasing the pressure , reaction will be favoured reverse i.e. backward direction .
Option c is correct.