Question

With the following information, determine standard Gibb's free energy of formation of $N_2{O}_4$(g).
(i) $\frac{1}{2}{N}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to NO\left(g\right)$ $\Delta G^0$ = 86.6 kJ
(ii) $NO\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to \frac{1}{2}{N}_2\left(g\right)$ $\Delta G^0$ = 34.82 kJ
(iii) $2N{O}_2\left(g\right)\to N_2{O}_4\left(g\right)$ $\Delta G^0$ = 5.77 kJ

Answer 1

(i) $\frac{1}{2}{N}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)⟶NO\left(g\right)$ $\Delta G^o=86.6kJ$

(ii) $NO\left(g\right)+\frac{1}{2}{O}_2⟶N{O}_2\left(g\right)$ $\Delta G^o=34.82kJ$

(iii) $2N{O}_2\left(g\right)⟶N_2{O}_4\left(g\right)$ $\Delta G^o=5.77kJ$

(i) $\times 2$,

$\Rightarrow N_2\left(g\right)+O_2\left(g\right)⟶2NO$ $\Delta G^o=173.2kJ⟶\left(iv\right)$

(ii) $\times 2$,

$\Rightarrow 2NO\left(g\right)+O_2⟶2N{O}_2$ $\Delta G^o=69.64kJ⟶\left(v\right)$

(iii)+(iv)+(v)

$N_2+2O_2⟶N_2{O}_4$

$\therefore \Delta G^o=173.2{+}_{6}9.64+5.77$

$=248.61kJ$ .