JEE PYQ: Chemical Equilibrium Question 23
Question 23 - 2019 (09 Jan Shift 2)
Consider the following reversible chemical reactions:
$\text{A}_2(g) + \text{B}_2(g) \xrightleftharpoons{K_1} 2\text{AB}(g)$ …… (1)
$6\text{AB}(g) \xrightleftharpoons{K_2} 3\text{A}_2(g) + 3\text{B}_2(g)$ ….. (2)
The relation between $K_1$ and $K_2$ is:
(1) $K_1 K_2 = 1/3$ (2) $K_2 = K_1^3$ (3) $K_2 = K_1^{-3}$ (4) $K_1 K_2 = 3$
Show Answer
Answer: (3) $K_2 = K_1^{-3}$
Solution
Reaction (2) is reverse of (1) raised to power 3. $K_2 = (1/K_1)^3 = K_1^{-3}$.
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