Question

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 = ?

Answer 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 = ??

Answer 2

THANKS!

Answer 3

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

Answer 4

The previous answer is wrong by the way.

Answer 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.