The standard Gibbs energy change at 300 , K f | vedclass

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(D) Given: $\Delta G^{\circ} = 2494.2 \, J$,$T = 300 \, K$,$R = 8.314 \, J/K \cdot mol$.First,calculate the equilibrium constant $K_c$ using $\Delta G

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reverse direction because $Q > K_c$

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(D) Given: $\Delta G^{\circ} = 2494.2 \, J$,$T = 300 \, K$,$R = 8.314 \, J/K \cdot mol$.First,calculate the equilibrium constant $K_c$ using $\Delta G^{\circ} = -RT \ln K_c$.$2494.2 = -(8.314) 	imes 300 	imes \ln K_c$.$\ln K_c = -2494.2 / 2494.2 = -1$.$K_c = e^{-1} \approx 0.367$.Next,calculate the reaction quotient $Q$ for the given concentrations $[A] = 0.5, [B] = 2, [C] = 0.5$.$Q = ([B][C]) / [A]^2 = (2 	imes 0.5) / (0.5)^2 = 1 / 0.25 = 4$.Since $Q = 4$ and $K_c \approx 0.367$,we have $Q > K_c$.When $Q > K_c$,the reaction proceeds in the reverse direction to reach equilibrium.

The standard Gibbs energy change at $300 \, K$ for the reaction $2A \rightleftharpoons B + C$ is $2494.2 \, J$. At a given time,the composition of the reaction mixture is $[A] = 1/2, [B] = 2$ and $[C] = 1/2$. The reaction proceeds in the:

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