Question

When \[{\rm{NaN}}{{\rm{O}}_3}\] is heated in a closed vessel, oxygen is liberated and \[{\rm{NaN}}{{\rm{O}}_2}\] is left behind. At equilibrium, which is/are not correct?
This question has multiple correct options.
(A) Addition of \[{\rm{NaN}}{{\rm{O}}_2}\] favours reverse reaction
(B) Addition of \[{\rm{NaN}}{{\rm{O}}_3}\] favours forward reaction
(C) Increasing temperature favours forward reaction
(D) Increasing pressure favours reverse reaction

Answer 1

Hint: Apply the Le chatelier’s principle. According to Le chatelier’s principle, if a system under equilibrium is subjected to a change in temperature, pressure or concentration, then the equilibrium shifts in such a manner as to reduce or to counteract the effect of change.

Complete answer: or Complete step by step answer:
if the concentration of one of the reactants is increased, the system will try to minimize this increase in the concentration by favouring the forward reaction.
Similarly, if the concentration of one of the products is increased, the system will try to minimize this increase in the concentration by favouring the reverse reaction.
For a gaseous reaction involving different numbers of moles of gaseous reactants and gaseous products, the increase in pressure will shift the equilibrium to the side containing fewer moles of gaseous species.

When \[{\rm{NaN}}{{\rm{O}}_3}\] is heated in a closed vessel, oxygen is liberated and \[{\rm{NaN}}{{\rm{O}}_2}\] is left behind.
Write the balanced chemical reaction.

\[{\rm{2NaN}}{{\rm{O}}_3}\left( s \right) + {\rm{heat}} \rightleftharpoons 2{\rm{NaN}}{{\rm{O}}_2}\left( s \right) + {{\rm{O}}_2}\left( g \right)\]
In the above reaction, heat is written on the reactant side as the reaction is heated and heat energy is supplied to the reaction.
Addition of \[{\rm{NaN}}{{\rm{O}}_2}\] favours reverse reaction. This is a correct statement. If the concentration of one of the products is increased, the system will try to minimize this increase in the concentration by favouring the reverse reaction.
Addition of \[{\rm{NaN}}{{\rm{O}}_3}\] favours forward reaction. This is a correct statement. If the concentration of one of the reactants is increased, the system will try to minimize this increase in the concentration by favouring the forward reaction.
Increasing temperature favours forward reaction. This is a correct statement. Since the reaction is an endothermic reaction, heat appears as a reactant in the balanced thermochemical equation.

Increasing pressure favours reverse reaction. This is a correct statement. The number of moles of gaseous products are more than the number of moles of gaseous reactants. With increase in pressure will shift the equilibrium to the side containing fewer moles of gaseous species.

Hence, all the four options are correct answers.

Note:
Please apply the concept of Le Chateliar’s concept properly.
For a gaseous reaction involving different numbers of moles of gaseous reactants and gaseous products, the increase in pressure will shift the equilibrium to the side containing fewer moles of gaseous species, so as to minimize the effect of increase in pressure.
The increase in temperature of an endothermic reaction will favour the forward reaction, so as to minimize the effect of increase in temperature. An endothermic reaction is one, to which heat is supplied.