JEE Mains · Chemistry · STD 11 - 6.1. Equilibrium - 1 (chemical Equilibrium)
The equilibrium constant for the reaction \(\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})\) is \(\mathrm{K}_{\mathrm{C}}=4.9 \times 10^{-2}\). The value of \(\mathrm{K}_{\mathrm{C}}\) for the reaction given below is \(2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})\) is _______.
- A \(4.9\)
- B \(41.6\)
- C \(49\)
- D \(416\)
Answer & Solution
Correct Answer
(D) \(416\)
Step-by-step Solution
Detailed explanation
\(\mathrm{K}_{\mathrm{C}}^{\prime}=\left(\frac{1}{\mathrm{~K}_{\mathrm{C}}}\right)^2=\left(\frac{1}{4.9 \times 10^{-2}}\right)^2\) \(\mathrm{~K}_{\mathrm{C}}^{\prime}=416.49\)
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