The equilibrium constant expression for the reaction 2BrF5 (g)
Br2 (g) + 5F2 (g) is
A) Kc = [Br2] [F2] / [BrF5]
B) Kc = [Br2] [F2]5 / [BrF5]2
C) Kc = [Br2] [F2]2 / [BrF5]5
D) Kc = [BrF5]2 / [Br2][F2]5
E) Kc = 2[BrF5]2 / ([Br2] × 5[F2]5)
Correct Answer:
Verified
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