Question

$K_P$ for the reaction $N_2+{2H}_2\rightleftharpoons {2NH}_3$ is $1.6\times {10}^{-4}{atm}^{-2}$ at ${400}^{0}C$. What will be $K_P$ at ${500}^{0}C$? The heat of reaction at this temperature is $-25.14$ kcal?

• A. $K_{P_2}=1.428\times {10}^{-5}{atm}^{-2}$
• B. $K_{P_2}=15.55\times {10}^{-5}{atm}^{-2}$
• C. $K_{P_2}=12\times {10}^{-4}{atm}^{-2}$
• D. $K_{P_2}=7.88\times {10}^{-5}{atm}^{-2}$

Answer 1

The correct option is A $K_{P_2}=1.428\times {10}^{-5}{atm}^{-2}$

Vant Hoff equation for variation of $K_P$ with temperature can be written as
$2.303log\frac{K_{P_2}}{K_{P_1}}=\frac{△H[T_2-T_1]}{R\times T_2{T}_1}$
$K_{P_1}=1.6\times {10}^{-4};T_1=673K$
$K_{P_2}=?;T_2=773K$

$△H=-25.14\times {10}^{3}cal$
$R=2cal$
$\therefore 2.303log\frac{K_{P_2}}{1.6\times {10}^{-4}}=\frac{-25.14\times {10}^3}{2}\left[\frac{773-673}{773\times 673}\right]$
$K_{P_2}=1.428\times {10}^{-5}{atm}^{-2}$