Question

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 $298 \,K ?\left( K _{n}=1.6 \times 10^{12}\right)$

• A. $R (298) \, in \, (1.6 \times 10^{12}) - 86600$
• B. $86600 + R(298) \, in \, (1.6 \times 10^{12})$
• C. $86600 - \frac{In \, (1.6 \times 10^{12})}{R(298)}$
• D. $0.5[2 \times 86600 - R(298) \, In \, (1.6 \times 10^{12})]$

Answer 1

Answer: (D)

$0.5[2 \times 86600 - R(298) \, In \, (1.6 \times 10^{12})]$

Answer 2

$-\frac{ R \times 298 \,\ell\, n \,1.6 \times 10^{12}}{2}$ $=\Delta G _{ r }^{ O }=2 \Delta G _{ NO _{2}}^{0}-2 \Delta G _{ NO }^{0}$ $\Delta G _{ NO _{2}}^{0}=86.6 \times 10^{3}-\frac{298\, K \,\ell\, n 1.6 \times 10^{12}}{2}$