Question

The following reaction is performed at 298K
$2NO\left(g\right)+O_2\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
The standard free energy of formation of NO(g) is 86.6kJ/mol at 298K. 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. $R\left(298\right)In(1.6\times {10}^{12})-86600$
• B. $86600+R\left(298\right)In(1.6\times {10}^{12})$
• C. $86600-\frac{In(1.6\times {10}^{12})}{R\left(298\right)}$
• D. $0.5[2\times 86600-R(298\left)In\right(1.6\times {10}^{12}\left)\right]$

Answer 1

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

For the given reaction,
$2NO\left(g\right)+O_2\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
$Given,\Delta G_f^{\circ }\left(N\right))=86.6kJ/mol$
$\Delta G_f^{\circ }\left(N{O}_2\right)=?$
$K_p=1.6\times {10}^{12}$
Now, we have,
$\Delta G_f^{\circ }=2\Delta G_{f\left(N{O}_2\right)}^{\circ }-[2\Delta G_{f\left(NO\right)}^{\circ }+\Delta G_{f\left(O_2\right)}^{\circ }]$
$=-RTIn{K}_p=2\Delta G_{f\left(N{O}_2\right)}^{\circ }-[2\times 86,000+0]$
$\Delta G_{f\left(N{O}_2\right)}^{\circ }=\frac{1}{2}[2\times 86600-R\times 298In(1.6\times {10}^{12}\left)\right]$
$\Delta G_{f\left(N{O}_2\right)}^{\circ }=0.5[2\times 86,600-R\times (298\left)In\right(1.6\times {10}^{12}\left)\right]$