Question

Calculate ${\Delta }_{\text{f}}{\text{G}}^{{}^{\circ }}$ for (NH₄Cl, s) at 310 K.

Given : ${\Delta }_f{H}^{\circ }\left(N{H}_{4}Cl,s\right)=-314.5kJ/mol;{\Delta }_r{C}_p=0$

$S_{N_2(g)}^{\circ }=192J{K}^{-1}mo{l}^{-1};S_{H_2(g)}^{\circ }=130.5J{K}^{-1}mo{l}^{-1}$

$S_{C{l}_2(g)}^{\circ }=233J{K}^{-1}mo{l}^{-1};S_{N{H}_{4}Cl(s)}^{\circ }=99.5J{K}^{-1}mo{l}^{-1}$

All given data at $300K$.

• A. $-\text{198.56 kJ/mol}$
• B. $-\text{426.7 kJ/mol}$
• C. $-\text{202.3 kJ/mol}$
• D. None of these

Answer 1

Answer: (A)

${\Delta }_f{S}^0\left(N{H}_{4}Cl,s\right)$ at $300K$

$={\text{S}}_{\text{N}{\left(\text{H}\right)}_4\text{C}\text{l}(\text{S})}^0-[\frac{1}{2}{\text{S}}_{{\left(\text{N}\right)}_2}^0+2{\text{S}}_{{\left(\text{H}\right)}_2}^0+\frac{1}{2}{\text{S}}_{\text{C}{\left(\text{l}\right)}_2}^0]$

$=-374\text{J}{\text{K}}^{-1}\text{m}\text{o}{\text{l}}^{-1}$

$\therefore {\Delta }_{\text{f}}{\text{C}}_{\text{p}}=0$

$\therefore {\Delta }_{\text{f}}{\text{S}}_{310}^0={\Delta }_{\text{r}}{\text{S}}_{300}^0$

$=-374\text{J}{\text{K}}^{-1}\text{m}\text{o}{\text{l}}^{-1}$

${\Delta }_{\text{f}}{\text{H}}_{310}^0={\Delta }_{\text{f}}{\text{H}}_{300}^0=-314.5$

${\Delta }_{\text{f}}{\text{G}}_{310}^0={\Delta }_{\text{f}}{\text{H}}^0-310{\Delta }_{\text{f}}{\text{S}}^0$

$=-314.5-\frac{310\left(-374\right)}{1000}$

$-198.56\text{k}\text{J}/\text{m}\text{o}\text{l}$