JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
For the following reactions, equilibrium constants are given \(S\left( s \right) + {O_2}\left( g \right) \rightleftharpoons S{O_2}\left( g \right);{K_1} = {10^{52}}\) \(2S\left( s \right) + 3{O_2}\left( g \right) \rightleftharpoons 2S{O_3}\left( g \right);{K_2} = {10^{129}}\) The equilibrium constant for the reaction \(2S{O_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2S{O_3}\left( g \right)\) is
- A \(10^{77}\)
- B \(10^{25}\)
- C \(10^{181}\)
- D \(10^{154}\)
Answer & Solution
Correct Answer
(B) \(10^{25}\)
Step-by-step Solution
Detailed explanation
\(2S{O_2}(g) + {O_2}(g) \to 2S{O_3}(g)\) \({K_{eq}} = \frac{{{{[S{O_3}]}^2}}}{{[{O_2}]{{[S{O_2}]}^2}}}\) \( = \frac{{{K_2}}}{{{K_1}}} = \frac{{{{10}^{129}}}}{{{{10}^{104}}}} = {10^{25}}\)
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