Question

What are the value of $△G^{\circ }$ and the equilibrium constant for the formation of $N{O}_2$ from $NO$ and $O_2$ at 298 K. Given log $(1.365\times {10}^6)=6.14$
$NO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\rightleftharpoons N{O}_2\left(g\right)$
where $△G^{\circ }\left(N{O}_2\right)=52.0$ kJ/mol
$△G^{\circ }\left(NO\right)=87.0$ kJ/mol
$△G^{\circ }\left(O_2\right)=0$ kJ/mol

• A. 35 kJ mol${}^{-1}$ and $1.365\times {10}^6$
• B. -35 kJ mol${}^{-1}$ and $1.365\times {10}^{-6}$
• C. -350 kJ mol${}^{-1}$ and $1.3\times {10}^6$
• D. -35 kJ mol${}^{-1}$ and $1.365\times {10}^6$

Answer 1

The correct option is D -35 kJ mol${}^{-1}$ and $1.365\times {10}^6$

$△G^{\circ }=△G_{N{O}_2}^{\circ }-△G_{NO}^{\circ }=(52-87)kJmo{l}^{-1}=-35kJmo{l}^{-1}\because △G^0=-2.303RT\log K\log K=\frac{-35kJmo{l}^{-1}}{2.303\times -8.301J{K}^{-1}mo{l}^{-1}\times 298K}$
On solving $K=1.365\times {10}^6$