Posted by **K** on Wednesday, May 25, 2011 at 10:40pm.

Determine Kc for the following reaction:

1/2N2(g) + 1/2O2(g)+ 1/2Br(g) <-> NOBr(g)

from the following information (at 298K)

2No(g) <-> N2(g) + O2(g) Kc = 2.1x10^30

NO(g) + 1/2Br2(g) <-> NOBr(g) Kc = 1.4

- Chemistry -
**DrBob222**, Thursday, May 26, 2011 at 12:47am
Take equn 1, divide by 2, and reverse it. That makes Kc of 2.1E30 = k' [1/(sqrt 2.1E30)]. That is sqrt because you took 1/2 the equation and the reciprocal because you reversed it.

Then add eqn 2 and check to see that it is the equation you want. K for the final rxn is k'*k2 = 1.4/(sqrt 2.1E30) = ??

- Chemistry -
**K**, Thursday, May 26, 2011 at 1:02am
Thank you so much. I got 9.7 x 10^-16. The answer is correct. I was just confused because why did we only substitute the k values for 2 compounds in the equation rather then all 4?

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