Question

When $\text{NaN}{{\text{O}}_{3}}$ is heated in a closed vessel, ${{\text{O}}_{2}}$ is liberated and $\text{NaN}{{\text{O}}_{2}}$ is left behind. At equilibrium:
(A) addition of $\text{NaN}{{\text{O}}_{2}}$ favours reverse reaction
(B) addition of $\text{NaN}{{\text{O}}_{3}}$ favours forward reaction
(C) increasing temperature favours the forward reaction
(D) increasing pressure favours forward reaction.

Answer 1

Hint: The concept of solving this question is the le-chatelier principle. The le-chatelier principle is used to predict the behaviour of any closed system when the system changes pressure, concentration and temperature. Write the correctly balanced reaction of decomposition of sodium nitrate and the equilibrium expression.

Complete step by step solution:
The reaction of decomposition of sodium nitrate is $2\text{NaN}{{\text{O}}_{3}}\left( \text{s} \right)\rightleftharpoons 2\text{NaN}{{\text{O}}_{2}}\left( \text{s} \right)+{{\text{O}}_{2}}\left( \text{g} \right)$. The equilibrium expression will be $\left[ {{\text{O}}_{2}} \right]$. Pure solids are not included in the equilibrium expression. The $\vartriangle \text{n}$ of this reaction will be 1. As, $\vartriangle \text{n}$ is equal to moles of products subtracted by the moles of reactants.
Let us discuss the direction in which reaction will move if the above options undergo their changes:

S. No.	Options	In which direction will the reaction move	The reason for the movement of reaction in a particular direction
1.	addition of $\text{NaN}{{\text{O}}_{2}}$ favours reverse reaction	The reaction will be unaffected.	As sodium nitrate is solid. The le-chatelier principle does not apply to solid-solid equilibrium. This is because the activity of solids is 1. Pure solids are not included in the equilibrium expression. So, removing or adding solids from the system will not affect the direction of equilibrium.
2.	addition of $\text{NaN}{{\text{O}}_{3}}$ favours forward reaction	The reaction will be unaffected.	As sodium nitrate is solid. The le-chatelier principle does not apply to solid-solid equilibrium. This is because the activity of solids is 1. Pure solids are not included in the equilibrium expression.
3.	increasing temperature favours the forward reaction	The reaction will move forward.	The decomposition of sodium nitrate is an endothermic reaction. On increasing the temperature, more heat will be absorbed. This absorbed heat neutralizes the effect of increased temperature.
4.	increasing pressure favours the forward reaction	The reaction will move towards the left side or reverse the reaction.	As we know that when moles and temperature are constant, then, $\text{PV}$ is also constant. Increasing pressure will result in decreasing volume. Less volume of gas means fewer moles of the gas. The left-hand side has fewer moles of gaseous reactants.

When $\text{NaN}{{\text{O}}_{3}}$ is heated in a closed vessel, ${{\text{O}}_{2}}$ is liberated and $\text{NaN}{{\text{O}}_{2}}$ is left behind.

At equilibrium increasing temperature favours forward reaction, which is option ‘C’.

Note: The main point to note in this question and while writing the answer is that solids, as well as liquids, have activities are 1. In equilibrium expressions, only gaseous reactants and products are included. Take note that removing or adding solids from the system will not affect the direction of equilibrium.