TS EAMCET · Chemistry · Chemical Equilibrium
The equilibrium constant for the reaction \(\mathrm{SO}_2(g)+\frac{1}{2} \mathrm{O}_2(g) \rightleftharpoons \mathrm{SO}_3(g)\) is \(5 \times 10^{-2}\) atm. The equilibrium constant of the reaction \(2 \mathrm{SO}_3(g) \rightleftharpoons 2 \mathrm{SO}_2(g)+\mathrm{O}_2(g)\) would be
- A \(100 \mathrm{~atm}\)
- B \(200 \mathrm{~atm}\)
- C \(4 \times 10^2 \mathrm{~atm}\)
- D \(6.25 \times 10^4 \mathrm{~atm}\)
Answer & Solution
Correct Answer
(C) \(4 \times 10^2 \mathrm{~atm}\)
Step-by-step Solution
Detailed explanation
\begin{aligned} \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 & \rightleftharpoons \mathrm{SO}_3 \\ K_1 & =\frac{\left[\mathrm{SO}_3\right]}{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{1 / 2}} \\ 2 \mathrm{SO}_3 & \rightleftharpoons 2 \mathrm{SO}_2+\mathrm{O}_2 \\…
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