Question

$K_p$ for the reaction, $N_2+3H_2\rightleftharpoons 2N{H}_3$ is $1.6\times {10}^{-4}at{m}^{-2}$ at ${400}^{\circ }C$. What will be $K_p$ at ${500}^{\circ }C$? Heat of reaction in this temperature range is $-25.14kcal$.

Answer 1

Using the relation
$\log \frac{K_{p2}}{K_{p1}}=\frac{\Delta H}{2.303R}\left[\frac{T_2-T_1}{T_2{T}_1}\right]$
given
$K_{p1}=1.6\times {10}^{-4}at{m}^{-2};\Delta H=-25.14kcal;R=2\times {10}^{-3}kcal{deg}^{-1}{mol}^{-1}$
$T_1=(400+273)K=673K;T_2=(500+273)K=773K$
$\log \frac{K_{p2}}{1.6\times {10}^{-4}}=\frac{-25.14}{2.303\times 2\times {10}^{-3}}\left[\frac{773-673}{773\times 673}\right]$
$\log K_{p2}=\log (1.6\times {10}^{-4})-\frac{25.14\times {10}^3\times 100}{2.303\times 2\times 773\times 673}$
$\log K_{p2}=-4.8450$
$K_{p2}=1.429\times {10}^{-5}{atm}^{-2}$