In the reaction A_{(g)} + B_{(g)} rightleftha | vedclass

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(A) According to Le Chatelier's principle,for a gaseous reaction,an increase in pressure shifts the equilibrium towards the side with fewer moles of g

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Decrease of pressure

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(A) According to Le Chatelier's principle,for a gaseous reaction,an increase in pressure shifts the equilibrium towards the side with fewer moles of gas. In the given reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$,the reactant side has $2$ moles of gas ($1$ mole of $A$ + $1$ mole of $B$) and the product side has $1$ mole of gas $(C)$. Since the product side has fewer moles,an increase in pressure will favour the forward reaction. Conversely,a decrease in pressure will favour the side with more moles of gas,which is the reactant side. Therefore,the backward reaction is favoured by a decrease in pressure.

In the reaction $A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}$,the backward reaction is favoured by

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