For the reaction,A_{(g)} + B_{(g)} rightlefth | vedclass

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(C) For the given reaction,$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_

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increasing of pressure

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(C) For the given reaction,$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$,the change in the number of moles of gaseous species is $\Delta n_g = (1 + 1) - (1 + 1) = 0$.Since $\Delta n_g = 0$,the pressure change does not affect the equilibrium position or the equilibrium constant.The equilibrium constant $K_c$ is a function of temperature only.Therefore,increasing the pressure will not disturb the equilibrium constant.

For the reaction,$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$; $\Delta H = Q \ kJ$,the equilibrium constant cannot be disturbed by

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