Question

Given the values of K shown below, determine the value of K for the reaction,

N2(g) + 2 O2(g) <=> 2 NO2(g).

N2(g) + O2(g) <=> 2 NO(g) K = 92

NO2(g) <=> NO + 1/2 O2(g) K = 36

the answer is 0.0710 please show work to how to get that answer

Answer 1

consider the given reactions

1) N2 + 02 ---> 2N0

the equilibrium constant is given by

K1 = [ N0]^2 / [ N2] [02] = 92

2) N02 ----> N0 + 0.5 02

in this case

the equilirium constant is given by

K2 = [N0] [02]^0.5 / [N02] = 36

3)

now consider the required reaction

N2 + 2 02 --> 2 N02

the equilibrium constant is given by

K3 = [N02]^2 / [N2] [02]^2

now

from all the above equations

we can see that

K3 = K1 / (K2)^2

given

K1 = 92

K2 = 36

so

using those values

we get

K3 = 92 / ( 36)^2

K3 = 0.071

so

the value of K for the required reaction is 0.071