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Q.

The equilibrium constant K p for the following reaction at 191oC is 1.24. what is Kc.? B(s)+32 F2(g)⇌BF3(g)

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a

6.7

b

0.61

c

8.30

d

7.6

answer is D.

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Detailed Solution

kp=kc·(RT)Δng ∆ng=1-32=-12 ; T=191°c=464k 1.24=Kc(0.0821×464)-1/2 Kc=7.6

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The equilibrium constant K p for the following reaction at 191oC is 1.24. what is Kc.? B(s)+32 F2(g)⇌BF3(g)