Chemical Equilibrium — Medium JEE chemistry MCQ
For the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, if $K_p = 1.6 \times 10^{-4}$ at a certain temperature, then the value of $K_p$ for the reaction $\text{NH}_3(g) \rightleftharpoons \frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g)$ is:
- A. $6.25 \times 10^{3}$
- B. $79$
- C. $1.6 \times 10^{-4}$
- D. $2.5 \times 10^{-2}$
Solution
For the original reaction: $K_p = 1.6 \times 10^{-4}$
The new reaction is the reverse of the original, divided by 2.
When a reaction is reversed: $K'_p = \frac{1}{K_p}$
When coefficients are multiplied by $\frac{1}{2}$: $K''_p = (K'_p)^{1/2}$
So for the new reaction:
$$K_{p,new} = \left(\frac{1}{K_p}\right)^{1/2} = \left(\frac{1}{1.6 \times 10^{-4}}\right)^{1/2}$$
$$K_{p,new} = (6250)^{1/2} = 79.06 \approx 79$$