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## Q. 14.2

Consider the chemical equation and equilibrium constant for the synthesis of ammonia at 25 °C:
$N_{2}(g) + 3 H_{2}(g) \xrightleftharpoons[]{} 2 NH_{3}(g)$     K = 3.7 × 10$^{8}$
Calculate the equilibrium constant for the following reaction at 25 °C:
$NH_{3}(g) \xrightleftharpoons[]{} \frac{1}{2} N_{2}(g) + \frac{3}{2} H_{2}(g)$       K′ = ?

## Verified Solution

You want to manipulate the given reaction and value of K to obtain the desired reaction and value of K. The given reaction is the reverse of the desired reaction, and its coefficients are twice those of the desired reaction.

 Begin by reversing the given reaction and taking the inverse of the value of K. $N_{2}(g) + 3 H_{2}(g) → 2 NH_{3}(g)$      K = 3.7 × 10$^{8}$ $2NH_{3} (g)\xrightleftharpoons[]{}N_{2}(g)+3H_{2}(g)$  $K_{rev}=\frac{1}{3.7\times10^{8}}$ Next, multiply the reaction by$\frac{1}{2}$ and raise the equilibrium constant to the $\frac{1}{2}$ power. $NH_{3}(g) \xrightleftharpoons[]{} \frac{1}{2} N_{2}(g) + \frac{3}{2} H_{2}(g)$ K′ = K$^{1/2}_{reverse}=(\frac{1}{3.7\times100^{8}})^{1/2}$ Calculate the value of K′. K′ = 5.2 ×10$^{25}$