For the reaction CH_{4(g)} + H_{2(g)} longrig | vedclass

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Question

(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by: $\Delta G^{\circ} = -RT \ln K_p = -2.303 \ RT

Vedclass · Accepted Answer

$-100 \ kJ \ mol^{-1}$

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(B) The relationship between standard Gibbs free energy change and the equilibrium constant is given by: $\Delta G^{\circ} = -RT \ln K_p = -2.303 \ RT \log_{10} K_p$ Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,$K_p = 3.356 	imes 10^{17}$ $\Delta G^{\circ} = -2.303 	imes 8.314 \ J \ K^{-1} \ mol^{-1} 	imes 298 \ K 	imes \log_{10} (3.356 	imes 10^{17})$ $\log_{10} (3.356 	imes 10^{17}) = \log_{10} (3.356) + 17 \approx 0.526 + 17 = 17.526$ $\Delta G^{\circ} = -2.303 	imes 8.314 	imes 298 	imes 17.526 \ J \ mol^{-1}$ $\Delta G^{\circ} \approx -100,000 \ J \ mol^{-1} = -100 \ kJ \ mol^{-1}$

For the reaction $CH_{4(g)} + H_{2(g)} \longrightarrow C_2H_{6(g)}$,$K_p = 3.356 \times 10^{17}$,calculate $\Delta G^{\circ}$ for the reaction at $298 \ K$.

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