The following reaction is performed at 298 K.
$$2NO(g) + O_2(g) \rightleftharpoons 2NO_2(g)$$
The standard free energy of the formation of $$NO(g)$$ is 86.6 kJ mol$$^{-1}$$ at 298 K. What is the standard free energy of the formation of $$NO_2(g)$$ at 298 K? $$(K_P = 1.6 \times 10^{12})$$

For the gas-phase equilibrium

$$2NO(g)+O_2(g)\;\rightleftharpoons\;2NO_2(g)$$

the standard Gibbs free-energy change of the reaction is related to the equilibrium constant by the well-known formula

$$\Delta_rG^\circ=-RT\ln K_P.$$

Here $$R$$ is the universal gas constant and $$T$$ is the absolute temperature. We know $$T=298\ \text{K}$$ and $$K_P=1.6\times10^{12}.$$ Substituting these values we have

$$\Delta_rG^\circ=-R(298)\ln(1.6\times10^{12}).$$

Now we express $$\Delta_rG^\circ$$ in terms of the standard free energies of formation of the species participating in the reaction. The definition is

$$\Delta_rG^\circ=\sum\nu_i\Delta_fG_i^\circ,$$

where $$\nu_i$$ are the stoichiometric coefficients taken with sign convention (products positive, reactants negative). Applying this to the present reaction we obtain

$$\Delta_rG^\circ=2\bigl[\Delta_fG^\circ(NO_2)\bigr]-2\bigl[\Delta_fG^\circ(NO)\bigr]-1\bigl[\Delta_fG^\circ(O_2)\bigr].$$

Because $$O_2(g)$$ is in its standard state, $$\Delta_fG^\circ(O_2)=0.$$ Denoting the unknown standard free energy of formation of $$NO_2(g)$$ by $$x,$$ we can rewrite the above equality as

$$\Delta_rG^\circ=2x-2(86\,600\ \text{J mol}^{-1}).$$

We already have another expression for $$\Delta_rG^\circ,$$ so equating the two gives

$$-R(298)\ln(1.6\times10^{12})=2x-2(86\,600).$$

Rearranging to isolate $$x$$ yields

$$2x=2(86\,600)-R(298)\ln(1.6\times10^{12}),$$

and consequently

$$x=\tfrac12\Bigl[2(86\,600)-R(298)\ln(1.6\times10^{12})\Bigr].$$

This algebraic expression is exactly what is offered in Option A.

Hence, the correct answer is Option A.