At 300 K,for the reaction PCl_{5(g)} rightle | vedclass

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Question

(A) The relationship between standard Gibbs free energy change $\Delta G^0$ and equilibrium constant $K_p$ is given by the formula: $\Delta G^0 = -RT

Vedclass · Accepted Answer

$38.72 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The relationship between standard Gibbs free energy change $\Delta G^0$ and equilibrium constant $K_p$ is given by the formula: $\Delta G^0 = -RT \ln K_p$. Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$K_p = 1.8 	imes 10^{-7}$. Substituting the values: $\Delta G^0 = -(8.314 \ J \ K^{-1} \ mol^{-1}) 	imes (300 \ K) 	imes \ln(1.8 	imes 10^{-7})$. $\Delta G^0 = -2494.2 	imes (\ln(1.8) + \ln(10^{-7}))$. $\Delta G^0 = -2494.2 	imes (0.5878 - 16.118)$. $\Delta G^0 = -2494.2 	imes (-15.5302) \approx 38735 \ J \ mol^{-1}$. Converting to $kJ \ mol^{-1}$: $\Delta G^0 \approx 38.74 \ kJ \ mol^{-1}$. Thus,the closest option is $38.72 \ kJ \ mol^{-1}$.

At $300 \ K$,for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the equilibrium constant $K_p = 1.8 \times 10^{-7}$. Calculate its standard Gibbs free energy change $\Delta G^0$.

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