Question

The equilibrium constant $K_p$ for the reaction,
$N_2+3H_2\rightleftharpoons 2N{H}_3$ is $1.64\times 10^{-4}$ at $400^{\circ }C$ and $0.144\times 10^{-4}$ at $500^{\circ }C$. Then the mean heat of formation of $1$ mole of $N{H}_3$ from its elements in this temperature range______.

Answer 1

Answer: 12.57

We know that, $\log \frac{K_2}{K_1}=\frac{\Delta H}{2.303R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

$\log \frac{0.144}{1.64}=\frac{\Delta H}{2.303\times 1.987\times 10^{-3}}\left(\frac{1}{673}-\frac{1}{773}\right)$

$\Delta H=-25.14kcal$ for $2mole$

$=12.57kcalmo{l}^{-1}$