TS EAMCET · Chemistry · Chemical Equilibrium
If the equilibrium constant for the reaction, \(2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3\) is 64 at \(500 \mathrm{~K}\), then the equilibrium constant for the reaction \(\mathrm{SO}_3 \rightleftharpoons \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2\) at the same temperature is
- A \(8\)
- B \(\frac{1}{8}\)
- C \(32\)
- D \(\frac{1}{64}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{8}\)
Step-by-step Solution
Detailed explanation
Given, \(2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3, K=64\) \(\mathrm{SO}_3 \rightleftharpoons \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2, K^{\prime}=?\) \(\because\) New \((K)=[K]^{1 / n}\) i.e. \(K^{\prime}\) where, \(n=\) factor to the new equilibrium…
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