Question

The equilibrium constant $K_p$ for the reaction
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$
is $1.64\times {10}^{-4}$ at ${400}^{o}C$ and $0.144\times {10}^{-4}$ at ${500}^{o}C$. Calculate the mean heat of formation of $1$ mole of ${NH}_3$ from its elements in this temperature range.

Answer 1

The equilibrium constant $K_p$ for the reaction
$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2{NH}_3\left(g\right)$
is $1.64\times {10}^{-4}$ at ${400}^{o}C$ and $0.144\times {10}^{-4}$ at ${500}^{o}C$.
The following equation is used to calculate the mean heat of formation of $1$ mole of ${NH}_3$ from its elements in this temperature range.

$\log \frac{K_2}{K_1}=\frac{\Delta H}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$
$\log \frac{0.144\times {10}^{-4}}{1.64\times {10}^{-4}}=\frac{\Delta H}{2.303\times 1.987\times {10}^{-3}kcal/mol/K}[\frac{1}{(400+273.15)K}-\frac{1}{(500+273.15)K}]$
$-1.0565=0.042\Delta H$
$\Delta H=-25.15kcal{mol}^{-1}$
This is heat of formation of $2$ moles of ${NH}_3$ from its elements in this temperature range.
The mean heat of formation of $1$ mole of ${NH}_3$ from its elements in this temperature range will be
$\Delta H=\frac{-25.15kcal{mol}^{-1}}{2}$
$\Delta H=-12.57kcal{mol}^{-1}$