Question

What will be the equilibrium constant at $127$. If equilibrium constant at $27$ is $4$ for reaction?

$N_2+{3H}_2\rightleftharpoons {2NH}_3$; $\Delta H=-46.06kJ$

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

Answer 1

The correct option is A $4\times {10}^{-2}$

$N_2+3H_2\rightleftharpoons 2{NH}_3;\Delta H=-46.06KJ$

Given $K$ at ${27}^{0}C$ is $4$

We know from Gibbs - Helomolt & equation

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

$K_1$ be at ${127}^0$ equilibrium constant

$K_2$ be at ${27}^{0}C$ equilibrium constant

$In\left(\frac{K_1}{4}\right)=-\left(\frac{-46.06\times {10}^3}{8.314}\right)(\frac{1}{400}-\frac{1}{300})$

$In\left(\frac{K_1}{4}\right)=-4.6167$

$\frac{K_1}{4}=98.85\times {10}^{-4}$

$K_1=395.41\times {10}^{-4}$

$\approx 4\times {10}^{-2}$