Question

The following reaction is performed at 298 $K$
$2NO\left(g\right)+O_2\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
The standard free energy of formation of $NO\left(g\right)$ is 86.6 kJ/mol at 298 K. What is the standard free energy of formation of $N{O}_2\left(g\right)$ at 298 K? ($K_p=1.6\times {10}^{12}$)

• A. R(298)ln($1.6\times {10}^{12}$) - 86600
• B. 86600 + R(298)ln($1.6\times {10}^{12}$)
• C. 86600 - $\frac{ln(1.6\times {10}^{12})}{R\left(298\right)}$
• D. 0.5 [2 $\times$ 86600 - R(298)ln($1.6\times {10}^{12}$)

Answer 1

The correct option is D 0.5 [2 $\times$ 86600 - R(298)ln($1.6\times {10}^{12}$)

2$\Delta G_{f\left(N{O}_2\right)}^o-[2\Delta G_{f\left(NO\right)}^o+\Delta G_{f\left(O_2\right)}^o]$ = $\Delta G_r^o=-RTln{K}_p$

On rearranging, we get
$\Delta G_{f\left(NO\right)}^o=0.5[2\times$ 86600 - R(298)ln($1.6\times {10}^{12}$)