MCQ
For the reaction at $25\,^oC$ ${N_2}{O_{4(g)}}\,\, \rightleftharpoons \,\,2N{O_{2(g)}},$ if $\Delta {G^o}_f$ for $N_2O_4$ and $NO_2$ are $23.49$ and $12.39\,kcal$, then $K_p$ for the reaction is
- A$113332$
- B$11.33$
- C$1.133$
- ✓$0.113$
✓
Answer
Correct option: D.
$0.113$
d
$\Delta \mathrm{G}^{o}$ for change $=2 \times \Delta \mathrm{G}_{\mathrm{NO}_{2}}^{o}-\Delta \mathrm{G}_{\mathrm{N}_{2} \mathrm{O}_{4}}^{o}$
$\Delta \mathrm{G}^{o}$ for change $=2 \times \Delta \mathrm{G}_{\mathrm{NO}_{2}}^{o}-\Delta \mathrm{G}_{\mathrm{N}_{2} \mathrm{O}_{4}}^{o}$
$=2 \times 12.39-23.49=1.29 \,\mathrm{kcal}$
$\because $ $\Delta G^o= -2.303\, RT\, log \,K$
$\therefore 1.29 \times 10^{3}=-2.303 \times 1.987 \times 298 \times \log \mathrm{K}_{\mathrm{p}}$
$\mathrm{K}_{\mathrm{p}}=0.1132\, \mathrm{atm}$
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