Chemistry - Chemical Equilibrium Question with Solution | TestHub

(A) correct

(B) $\ell P_s=\ell n{P}_L$

$17.2\frac{-919}{\mathrm{T}}=15.2\frac{-700.5}{\mathrm{T}}$

$\frac{218.5}{\mathrm{T}}=2.0$

$\mathrm{T}=\frac{218.5}{2.0}=109.25$ (Correct)

(C) $\ln {\mathrm{P}}_{\mathrm{s}}=17.2-\frac{919}{109.25}=17.2-8.41$

$\ln {\mathrm{P}}_{\mathrm{s}}=8.79$

${\mathrm{P}}_{\mathrm{s}}={\mathrm{e}}^{8.79}=6750$

Or $\ell {\mathrm{nP}}_{\mathrm{L}}=15.2-\frac{700.5}{109.25}=15.2-6.41$

$\ell {\mathrm{nP}}_{\mathrm{L}}=8.79$

${\mathrm{P}}_{\mathrm{L}}={\mathrm{e}}^{8.79}=6750$

${\mathrm{P}}_{\mathrm{S}}={\mathrm{P}}_{\mathrm{L}}=\frac{6750}{750}=9$ bar (incorrect)

(D) $\Delta {\mathrm{G}}^{\circ }=\Delta {\mathrm{H}}^{\circ }-\mathrm{T}\Delta {\mathrm{S}}^{\circ }$

$-\mathrm{RT}\ell {\mathrm{nP}}_{\mathrm{s}}=\Delta {\mathrm{H}}_{\text{sub }}^{\circ }-\mathrm{T}\Delta {\mathrm{S}}^{\circ }$

$\ell {\mathrm{nP}}_{\mathrm{s}}=-\frac{\Delta {\mathrm{H}}_{\text{sub }}^{\mathrm{o}}}{\mathrm{RT}}+\frac{\Delta {\mathrm{S}}^{\circ }}{\mathrm{R}}$

$\ell {\mathrm{nP}}_{\mathrm{s}}=\frac{\Delta {\mathrm{S}}^{\circ }}{\mathrm{R}}-\frac{\Delta {\mathrm{H}}_{\text{sub }}^{\mathrm{o}}}{\mathrm{RT}}$

Similarly $\ell \mathrm{n}{\mathrm{P}}_{\mathrm{L}}=\frac{\Delta {\mathrm{S}}^{\circ }}{\mathrm{R}}-\frac{\Delta {\mathrm{H}}_{\text{vap }}^{\circ }}{\mathrm{RT}}$

$\frac{\Delta {\mathrm{H}}_{\text{sub }}^{\mathrm{o}}}{\mathrm{R}}-\frac{\Delta {\mathrm{H}}_{\text{vap }}^{\mathrm{o}}}{\mathrm{R}}=919-700.5$

$\Delta {\mathrm{H}}_{\text{sub }}^{\mathrm{o}}-\Delta {\mathrm{H}}_{\text{vap }}^{\mathrm{o}}=218.5\mathrm{R}$ (Correct)