What is the equilibrium concentration of NO in this reaction: N2(g) + O2(g) 2NO(g)?

Question

Considering, the equilibrium constant for the reaction below is 0.49, and at equilibrium, [O2] = 0.11 and [N2] = 0.15.

Answer 1

N(2)(g) + O(2)(g) <====> 2NO(g) ?

K = [NO]^2 / [N(2)] * [O(2)]
.49 = [NO]^2 / 0.11 * 0.15
[NO] = 0.090 M

Answer 2

[NO]^2/[N2][O2] = .49

or [NO] = sqrt(.49*.11*.15) = 8.99 x 10^-2