Calculate the standard Gibbs free energy chan | vedclass

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Question

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

Vedclass · Accepted Answer

$162.74 \ kJ \ mol^{-1}$

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(A) The relationship between standard Gibbs free energy change $\Delta G^o$ and the equilibrium constant $K_p$ is given by the equation: $\Delta G^o = -RT \ln K_p$.Given values are $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,and $K_p = 3 	imes 10^{-29}$.Substituting these values into the equation:$\Delta G^o = -(8.314 \ J \ K^{-1} \ mol^{-1}) 	imes (298 \ K) 	imes \ln(3 	imes 10^{-29})$.$\Delta G^o = -2477.572 	imes (\ln 3 + \ln 10^{-29})$.$\Delta G^o = -2477.572 	imes (1.0986 - 66.779)$.$\Delta G^o = -2477.572 	imes (-65.6804) \approx 162735 \ J \ mol^{-1}$.Converting to $kJ \ mol^{-1}$,we get $\Delta G^o \approx 162.74 \ kJ \ mol^{-1}$.

Calculate the standard Gibbs free energy change $\Delta G^o$ at $298 \ K$ for the conversion of oxygen to ozone,given by the reaction: $\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)}$. The equilibrium constant $K_p$ for this conversion is $3 \times 10^{-29}$.

Similar Questions

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