The equilibrium constant for the reaction: SO 3 (g) ⇋ SO 2 (g) + 1 2 O 2 (g) is K c = 4.9…
Chemistry · JEE Main · NTA Exams — Equilibrium
The equilibrium constant for the reaction:
SO3(g) ⇋ SO2(g) + \(\frac{1}{2}\)O2(g) is Kc = 4.9 × 10-2. The value of for the reaction given below is
2SO2(g) + O2(g) ⇋ 2SO3(g) is (Jee Main 2024)
SO3(g) ⇋ SO2(g) + \(\frac{1}{2}\)O2(g) is Kc = 4.9 × 10-2. The value of for the reaction given below is
2SO2(g) + O2(g) ⇋ 2SO3(g) is (Jee Main 2024)
- 4.9
- 49
- 41.6
- 416
Answer
(D) 416
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