The equilibrium constant for the reaction SO_3(g) leftharpoons SO_2(g) + 1/2O_2(g) is K_c = 4.9 × 10^-2 . The value of K_c for the reaction 2SO_2(g) + - Chemistry MCQ AIEEE 2006 2026 | ExamTest.live

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The equilibrium constant for the reaction SO_{3} (g) ⇌ SO_{2} (g) + \frac{1}{2} O_{2} (g) is K_{c} = 4.9 \times 1 0^{- 2} . The value of K_{c} for the reaction 2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g) will be:

Solution & Step-by-step Explanation

The target reaction is obtained by:Reversing the original reaction (takes 1/ K_{c}).Multiplying the coefficients by 2 (takes the square). K_{c}^{'} = (\frac{1}{K _{c}})^{2} = \frac{1}{( 4.9 \times 1 0 ^{- 2} ) ^{2}} K_{c}^{'} = \frac{1}{24.01 \times 1 0 ^{- 4}} = \frac{10000}{24.01} \approx 416.49

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The equilibrium constant for the reaction SO_{3} (g) ⇌ SO_{2} (g) + \frac{1}{2} O_{2} (g) is K_{c} = 4.9 \times 1 0^{- 2} . The value of K_{c} for the reaction 2 SO_{2} (g) + O_{2} (g) ⇌ 2 SO_{3} (g) will be:

416

2.40 \times 1 0^{- 3}

9.8 \times 1 0^{- 2}

4.9 \times 1 0^{- 2}

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