Question 17 5 pts Consider the reaction 2 NO(g) + Br2(g) =2 NOBr(g) Kp = 0.0352...

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Question 17 5 pts Consider the reaction 2 NO(g) + Br2(g) =2 NOBr(g) Kp = 0.0352 (at 298K) What is the Kc for the reaction at

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Answer 1

Answer #1

Kp= Kc(RT)^∆n

Kc= Kp/(RT)^ ∆n

∆n = number of moles of product- number of moles of reactants

∆n = 2-3= -1

Kc= .0352/( .0821×298)^-1

Kc=.0321×.0821×298

Kc=.861

Answer A

Answer 2

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