For the reaction PCl_{5(g)} rightleftharpoons | vedclass

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Question

(C) For the reaction,$PCl_{5(g)} \rightleftharpoons PCl_{3(g)}   Cl_{2(g)}$,the total number of moles of gaseous products is $2$ and reactants is $1$.

Vedclass · Accepted Answer

Introducing an inert gas at constant pressure

Vedclass · Answer

(C) For the reaction,$PCl_{5(g)} \rightleftharpoons PCl_{3(g)}   Cl_{2(g)}$,the total number of moles of gaseous products is $2$ and reactants is $1$.According to Le Chatelier's principle,for a reaction where the number of moles of gaseous products is greater than the number of moles of gaseous reactants $(\Delta n_g > 0)$:$1$. Introducing an inert gas at constant volume does not change the partial pressures of the reacting species,so the equilibrium remains unaffected.$2$. Introducing chlorine gas $(Cl_2)$ increases the concentration of products,which shifts the equilibrium in the backward direction.$3$. Introducing an inert gas at constant pressure increases the volume of the container. Since the number of moles of products is greater than reactants,the system shifts towards the side with more moles to counteract the increase in volume,thus favouring the forward reaction.$4$. Decreasing the volume of the container increases the total pressure,which shifts the equilibrium towards the side with fewer moles (backward direction).

For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the forward reaction at constant temperature is favoured by

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