Question

Calculate $\Delta_{ f } G ^{\circ}$ for $\left( NH _4 Cl , s \right)$ at $310 K$.
Given: $\Delta_{ f } H ^{\circ}\left( NH _4 Cl , s \right)=-314.5 kJ / mol ; \Delta_{ r } C _{ p }=0$
$ \begin{array}{l} S _{ N _2( g )}^{\circ}=192 JK ^{-1} mol ^{-1} ; S _{ H _2( g )}^{\circ}=130.5 JK ^{-1} mol ^{-1} \\ S _{ Cl _2( g )}^{\circ}=233 JK ^{-1} mol ^{-1} ; S _{ NH _4 Cl ( s )}^{\circ}=99.5 JK ^{-1} mol ^{-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)

$\Delta_{ f } S ^0\left( NH _4 Cl , s \right)$ at $300 K$
$= S _{ NH _4 Cl ( S )}^0-\left[\frac{1}{2} S _{ N _2}^0+2 S _{ H _2}^0+\frac{1}{2} S _{ Cl _2}^0\right]$
$=-374 JK ^{-1} mol ^{-1}$
$\therefore \Delta_{ f } C _{ p }=0$
$\therefore \Delta_{ f } S _{310}^0=\Delta_{ r } S _{300}^0$
$=-374 JK ^{-1} mol ^{-1}$
$\Delta_{ f } H _{310}^0=\Delta_{ f } H _{300}^0=-314.5$
$\Delta_{ f } G _{310}^0=\Delta_{ f } H ^0-310 \Delta_{ f } S ^0$
$=-314.5-\frac{310(-374)}{1000}$
$-198.56 kJ / mol$