Question

Calculate ${\Delta }_f{G}^{\circ }$ for $\left(N{H}_{4}Cl,s\right)$ at $310K$.
Given: ${\Delta }_f{H}^{\circ }\left(N{H}_{4}Cl,s\right)=-314.5kJ/mol;{\Delta }_r{C}_p=0$
$\begin{array}{l}{S}_{N_2(g)}^{\circ }=192J{K}^{-1}mo{l}^{-1};S_{H_2(g)}^{\circ }=130.5J{K}^{-1}mo{l}^{-1}\\ S_{C{l}_2(g)}^{\circ }=233J{K}^{-1}mo{l}^{-1};S_{N{H}_{4}Cl(s)}^{\circ }=99.5J{K}^{-1}mo{l}^{-1}\end{array}$

• 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 $300K$
$=S_{N{H}_{4}Cl(S)}^0-\left[\frac{1}{2}{S}_{N_2}^0+2S_{H_2}^0+\frac{1}{2}{S}_{C{l}_2}^0\right]$
$=-374J{K}^{-1}mo{l}^{-1}$
$\therefore {\Delta }_f{C}_p=0$
$\therefore {\Delta }_f{S}_{310}^0={\Delta }_r{S}_{300}^0$
$=-374J{K}^{-1}mo{l}^{-1}$
${\Delta }_f{H}_{310}^0={\Delta }_f{H}_{300}^0=-314.5$
${\Delta }_f{G}_{310}^0={\Delta }_f{H}^0-310{\Delta }_f{S}^0$
$=-314.5-\frac{310(-374)}{1000}$
$-198.56kJ/mol$