The following reaction is performed at 298 K 2 NO (g) + O_2 (g) rightharpoons 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? (KP = 1.6 × 1012)

0.5[2×86,600 – R(298) ln(1.6×1012)]

86600 - ln (1.6 \times 10^12) \over R (298)

Thermodynamics PYQ — JEE Main 2015 · Shift 0 · Q5 with Solution

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Correct answer: C

For the reaction $2NO(g)+O_2(g)\rightleftharpoons 2NO_2(g)$ we have $\Delta G^\circ_{\text{rxn}}=-RT\ln K_P$

Given:

So, $\Delta G^\circ_{\text{rxn}}=-R(298)\ln(1.6\times 10^{12})$

$\Delta G^\circ_{\text{rxn}}=\sum \nu \Delta G_f^\circ(\text{products})-\sum \nu \Delta G_f^\circ(\text{reactants})$

For the given reaction, $\Delta G^\circ_{\text{rxn}}=2\Delta G_f^\circ(NO_2)-\left[2\Delta G_f^\circ(NO)+\Delta G_f^\circ(O_2)\right]$

Since $O_2$ is an element in its standard state, $\Delta G_f^\circ(O_2)=0$

Also given, $\Delta G_f^\circ(NO)=86.6\,\text{kJ mol}^{-1}=86600\,\text{J mol}^{-1}$

Thus, $\Delta G^\circ_{\text{rxn}}=2\Delta G_f^\circ(NO_2)-2(86600)$

$2\Delta G_f^\circ(NO_2)-2(86600)=-R(298)\ln(1.6\times 10^{12})$

$2\Delta G_f^\circ(NO_2)=2(86600)-R(298)\ln(1.6\times 10^{12})$

$\Delta G_f^\circ(NO_2)=\frac{1}{2}\left[2\times 86600-R(298)\ln(1.6\times 10^{12})\right]$

This is exactly: $\boxed{0.5[2\times 86{,}600-R(298)\ln(1.6\times 10^{12})]}$

So the correct option is C.