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Reaction, $A \rightleftharpoons B+3 C$ at $25^{\circ} \mathrm{C}$ is in equilibrium. If equilibrium constant and Gibbs' free energy for this reaction are $y$ and $x$ respectively, then the Gibbs' free energy for reaction, $\frac{1}{2} A \rightleftharpoons \frac{1}{2} B+\frac{3}{2} C$ is
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$x / 2$
$\begin{aligned} & \mathbf{} A \rightleftharpoons B+3 C \\ & K=\frac{[B][C]^3}{[A]} \\ & \Delta G^{\circ}=-R T \ln K=x\end{aligned}$
For the reaction, $\frac{1}{2} A \rightleftharpoons \frac{1}{2} B+\frac{3}{2} C$
$\begin{aligned} & K^{\prime}=\frac{[B]^{1 / 2}[C]^{3 / 2}}{[A]^{1 / 2}} \quad \text { or } \quad\left(K^{\prime}\right)^2=\frac{[B][C]^3}{[A]}=K \\ & \Delta G_2^{\circ}=-R T \ln K^{\prime}=-R T \ln \sqrt{K} \\ & =-\frac{1}{2} R T \ln K=\frac{x}{2}\end{aligned}$
For the reaction, $\frac{1}{2} A \rightleftharpoons \frac{1}{2} B+\frac{3}{2} C$
$\begin{aligned} & K^{\prime}=\frac{[B]^{1 / 2}[C]^{3 / 2}}{[A]^{1 / 2}} \quad \text { or } \quad\left(K^{\prime}\right)^2=\frac{[B][C]^3}{[A]}=K \\ & \Delta G_2^{\circ}=-R T \ln K^{\prime}=-R T \ln \sqrt{K} \\ & =-\frac{1}{2} R T \ln K=\frac{x}{2}\end{aligned}$
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