JEE PYQ: Chemical Equilibrium Question 16
Question 16 - 2020 (06 Sep Shift 2)
The value of $K_c$ is 64 at 800 K for the reaction $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$. The value of $K_c$ for the following reaction is:
$\text{NH}_3(g) \rightleftharpoons \frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g)$
(1) 1/64 (2) 8 (3) 1/4 (4) 1/8
Show Answer
Answer: (4) 1/8
Solution
Reverse and halve: $K_c’ = (1/K_c)^{1/2} = (1/64)^{1/2} = 1/8$.
BETA
