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(B) The standard free energy change for the reaction is given by $\Delta G^{\circ} = -RT \ln K_p$.For the reaction $2 NO_{(g)} + O_{2(g)} \rightleftha

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$0.5 [2 	imes 86600 - R(298) \ln (1.6 	imes 10^{12})]$

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(B) The standard free energy change for the reaction is given by $\Delta G^{\circ} = -RT \ln K_p$.For the reaction $2 NO_{(g)} + O_{2(g)} ightleftharpoons 2 NO_{2(g)}$,the standard free energy change is $\Delta G^{\circ} = 2 \Delta G^{\circ}_{f}(NO_2) - [2 \Delta G^{\circ}_{f}(NO) + \Delta G^{\circ}_{f}(O_2)]$.Since $\Delta G^{\circ}_{f}(O_2) = 0$,we have $\Delta G^{\circ} = 2 \Delta G^{\circ}_{f}(NO_2) - 2 \Delta G^{\circ}_{f}(NO)$.Given $\Delta G^{\circ}_{f}(NO) = 86.6 \, kJ/mol = 86600 \, J/mol$ and $T = 298 \, K$,we substitute these into the equation:$-R(298) \ln (1.6 	imes 10^{12}) = 2 \Delta G^{\circ}_{f}(NO_2) - 2(86600)$.Rearranging for $\Delta G^{\circ}_{f}(NO_2)$:$2 \Delta G^{\circ}_{f}(NO_2) = 2(86600) - R(298) \ln (1.6 	imes 10^{12})$.$\Delta G^{\circ}_{f}(NO_2) = 0.5 [2 	imes 86600 - R(298) \ln (1.6 	imes 10^{12})]$.

The following reaction is performed at $298 \, K$.
$2 NO_{(g)} + O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$
The standard free energy of formation of $NO_{(g)}$ is $86.6 \, kJ/mol$ at $298 \, K$. What is the standard free energy of formation of $NO_{2(g)}$ at $298 \, K$? $(K_p = 1.6 \times 10^{12})$

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The following reaction is performed at $298 \, K$.$2 NO_{(g)} + O_{2(g)} \rightleftharpoons 2 NO_{2(g)}$The standard free energy of formation of $NO_{(g)}$ is $86.6 \, kJ/mol$ at $298 \, K$. What is the standard free energy of formation of $NO_{2(g)}$ at $298 \, K$? $(K_p = 1.6 \times 10^{12})$