For the chemical reaction 3X_{(g)} + Y_{(g)} | vedclass

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(A) The given reaction is $3X_{(g)} + Y_{(g)} \rightleftharpoons X_3Y_{(g)}$.The change in the number of gaseous moles is $\Delta n_g = 1 - (3 + 1) =

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temperature and pressure

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(A) The given reaction is $3X_{(g)} + Y_{(g)} ightleftharpoons X_3Y_{(g)}$.The change in the number of gaseous moles is $\Delta n_g = 1 - (3 + 1) = -3$.Since $\Delta n_g 
eq 0$,the position of equilibrium is affected by a change in pressure according to Le Chatelier's principle.The equilibrium constant $K_p$ is a function of temperature,so a change in temperature will shift the equilibrium position.$A$ catalyst only increases the rate of both forward and backward reactions equally and does not affect the equilibrium position or the amount of product formed.Therefore,the amount of $X_3Y$ at equilibrium is affected by both temperature and pressure.

For the chemical reaction $3X_{(g)} + Y_{(g)} \to X_3Y_{(g)}$,the amount of $X_3Y$ at equilibrium is affected by

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