Question

Calculate

$\Delta G^0$

the equilibrium constant for the formation of $N{O}_2$ from NO and $O_2$ at 298 K
$NO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\leftrightarrow N{O}_2\left(g\right)$

Where:
${\Delta }_f{G}^{\circ }\left(N{O}_2\right)=52.0kJ/mol{\Delta }_f{G}^{\circ }\left(NO\right)=87.0kJ/mol{\Delta }_f{G}^{\circ }\left(O_2\right)=0kJ/mol$

Answer 1

Fort the given reaction,
$\Delta G^{\circ }=\Delta G^{\circ }$ (Products ) - $\Delta G^{\circ }$ (Reactants)
$\Delta G^{\circ }=52.0-\{87.0+0\}=-35.0kJmo{l}^{-1}\left(or\right)-35\times {10}^{3}Jmo{l}^{-1}$

We know that,
$\Delta G^{\circ }=RTlog{K}_c$
$\Delta G^{\circ }=2.303RTlog{K}_c$
$K_c=\frac{-35.0\times {10}^3}{-2.303\times 8.314\times 298}=6.134\therefore K_c=antilog\left(6.134\right)=1.36\times {10}^6$
Hence, the equilibrium constant for the given reaction $K_c$ is $1.36\times {10}^6$