Question

What will be the equilibrium constant at ${127}^{0}C$ if equilibrium constant at ${27}^{0}C$ is 4 for reaction $N_2$ $+$ $3H_2$ $\rightleftharpoons$ $2N{H}_3$; $△H$ = 46.06 kJ ?

• A. $4\times {10}^{-2}$
• B. $2\times {10}^{-3}$
• C. ${10}^2$
• D. $4\times {10}^2$

Answer 1

Answer: (D)

$4\times {10}^2$

$\log \frac{K_1}{K_2}=\frac{\Delta H}{R}(\frac{1}{T_2}-\frac{1}{T_1})$

$\log \left(H\right)-\log K_2=\frac{\frac{-46.06KJ}{mol}\frac{{10}^{3}J}{KJ}}{\frac{8.314J}{molK}}(\frac{1}{400K}-\frac{1}{300K})$

$\therefore K_{{127}^{o}C}\simeq 4.0\times {10}^{-2}$