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(D) The standard Gibbs free energy change is given by the formula: $\Delta_{r} G^{\ominus} = -RT \ln K_{P}$.Given $R = 8.314 \ J \ K^{-1} \ mol^{-1} =

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$163$

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(D) The standard Gibbs free energy change is given by the formula: $\Delta_{r} G^{\ominus} = -RT \ln K_{P}$.Given $R = 8.314 \ J \ K^{-1} \ mol^{-1} = 8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,and $K_{P} = 2.47 	imes 10^{-29}$.Substituting the values:$\Delta_{r} G^{\ominus} = -(8.314 	imes 10^{-3} \ kJ \ K^{-1} \ mol^{-1}) 	imes (298 \ K) 	imes \ln(2.47 	imes 10^{-29})$.$\ln(2.47 	imes 10^{-29}) = \ln(2.47) + \ln(10^{-29}) \approx 0.904 - 66.776 = -65.872$.$\Delta_{r} G^{\ominus} = -8.314 	imes 10^{-3} 	imes 298 	imes (-65.872) \approx 163.29 \ kJ$.Rounding to the nearest integer,we get $163 \ kJ$.

Consider the following reaction at $298 \ K$.
$\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)} ; K_{P} = 2.47 \times 10^{-29}$.
$\Delta_{r} G^{\ominus}$ for the reaction is $ . . . . . . \ kJ$. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

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Consider the following reaction at $298 \ K$.$\frac{3}{2} O_{2(g)} \rightleftharpoons O_{3(g)} ; K_{P} = 2.47 \times 10^{-29}$.$\Delta_{r} G^{\ominus}$ for the reaction is $ . . . . . . \ kJ$. (Given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)