Question

Given Kc values:

N2(g)+1/2 O2(g)<->N2O(g) Kc=2.7* 10^-18

N2O4(g)<->2NO2(g)
Kc=4.6*10^-3

1/2N2(g)+O2(g)<->NO2(g)
kc=4.1*10^-9

What is the Kc value for:
2N2O(g)+3O2(g)<->2N2O4(g)

Answer 1

[NO2]^2/[N2O4] = 4.6*10^-3
[N2O4]/[NO2]^2 = 217.4
[N2O4]^2 = 4.73*10^4 [NO2]^4
= 4.73*10^4 *[N2]^2[O2]^4

Now use the fact that
[N2O]^2 = [N2]^2*[O2]*7.29*10^-36
[N2]^2*[O2] = 1.37*10^35 [N2O]^2
[N2O4]^2 = 4.73*10^4*1.37*10^35 [N2O]^2*[O2]^3
= 6.49*10^39 [N2O]^2*[O2]^3
Kc = 6.49*10^39

check my math

Answer 2

The answer is wrong.

Kc (Equilibrium Constant) is always <1.

Answer 3

The answer may be wrong due to a math error I might have made, but Kc can certainly be greater than 1.

This is essentially an algebra problem. Using the concentration ratios you have been given, calculate the Kc for the indicated reaction.