Chapter 14
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′ = ?
Step-by-Step
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} |