JEE PYQ: Chemical Equilibrium Question 21
Question 21 - 2019 (08 Apr Shift 2)
For the following reactions, equilibrium constants are given:
$\text{S}(s) + \text{O}_2(g) \rightleftharpoons \text{SO}_2(g)$; $K_1 = 10^{52}$
$2\text{S}(s) + 3\text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$; $K_2 = 10^{129}$
The equilibrium constant for the reaction, $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ is:
(1) $10^{154}$ (2) $10^{181}$ (3) $10^{25}$ (4) $10^{77}$
Show Answer
Answer: (3) $10^{25}$
Solution
(ii) $-$ 2(i): $K = 10^{129}/10^{104} = 10^{25}$.
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