# Consider the following reactions and their equilibrium constants.

(NO(g) + 0.5Br2(g) <===> NOBr(g) Kp = 5.3

2NO(g) <===> N2(g) + O2(g) Kp = 2.1 x 1030

Use these equations and their equilibrium constants to determine the equilibrium constant for the following reaction:

N2(g) + O2(g) + Br2(g) <===> 2NOBr(g) Kp = ?

Question ID
538311

Created
April 28, 2011 3:51am UTC

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0

5

Views
4979

1. Write K expression for eqn 1.
(NOBr)/(NO)(Br2)^1/2 = 5.3.
Now square that.
(NOBr)^2/(NO)^2(Br2) = 5.3^2
Multiply that equation by Keq for equation #2 which is
(NOBr)^2/(N2)(O2) = 2.1 x 10^30

The (NO)^2 cancels and you are left with Keq for the reaction you want and K for the final reaction is just 5.3^2 x 2.1 x 10^30 = ??

538330

Created
April 28, 2011 4:14am UTC

Rating
-4

2. THANKS!

538337

Created
April 28, 2011 4:28am UTC

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0

3. Consider the following equilibrium:
2NO(g) N2(g) + O2(g); Keq = 2.1 × 1030

1378616

Created
March 31, 2016 5:16pm UTC

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0

4. The previous answer is wrong by the way.

1630387

Created
October 19, 2017 5:32pm UTC

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0

5. the final answer should be 5.3^2 x( ( 2.1 x 10^30 )^-1) ) since you are doing the second reaction in reverse, you have to take the inverse of that reactions Kp value.