At 300 K,Delta_{r} G^{circ} for the reaction | vedclass

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(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by: $\Delta_{r} G^{\circ} = -RT \ln K_{eq}$.Given

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by: $\Delta_{r} G^{\circ} = -RT \ln K_{eq}$.Given values: $\Delta_{r} G^{\circ} = -11.5 \ kJ \ mol^{-1} = -11500 \ J \ mol^{-1}$,$T = 300 \ K$,and $R = 8.314 \ J \ mol^{-1} \ K^{-1}$.Substituting these values into the equation: $-11500 = -8.314 	imes 300 	imes \ln K_{eq}$.$\ln K_{eq} = \frac{11500}{8.314 	imes 300} \approx \frac{11500}{2494.2} \approx 4.61$.Since $\ln K_{eq} = 2.303 \log_{10} K_{eq}$,we have $2.303 \log_{10} K_{eq} \approx 4.61$.$\log_{10} K_{eq} \approx \frac{4.61}{2.303} \approx 2$.Therefore,$K_{eq} = 10^2 = 100$.

At $300 \ K$,$\Delta_{r} G^{\circ}$ for the reaction $A_{2(g)} \rightleftharpoons B_{2(g)}$ is $-11.5 \ kJ \ mol^{-1}$. The equilibrium constant at $300 \ K$ is approximately $(R=8.314 \ J \ mol^{-1} \ K^{-1})$.

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