For the reaction S₂F₆ (g)⇌ 2SF₂ (g)+ F₂ (g),[SF₂] = 0.18 M,[F₂] = 0.54 M.The quilibrium constant is:

Name: For the reaction S 2 F 6 (g)⇌ 2SF 2 (g)+ F 2 (g),[SF 2 ] = 0.18 M,[F 2 ] = 0.54 M.The quilibrium constant is:
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Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is: Kc=([SF₂]2[F₂])/([S₂][F₆]) Substituting the given values, we get: Kc=([0.18]2[0.54])/([1][1]) Kc=0.1458≈0.15 (rounded to two significant figures) Therefore, the correct answer is choice B.

For the Reaction S2F6 (G)⇌ 2SF2 (G)+ F2 (G),[SF2] =

For the Reaction S₂F₆ (G)⇌ 2SF₂ (G)+ F₂ (G),[SF₂] =