Equilibrium concentration???

Question

At 25 degrees Celcius, K = 1.6*10^(-17) for the reaction:

N2(g) + 2O2(g) <-----> 2NO2(g)

Initially the concentration of N2 and O2 are : [N2]= 0.0320 mol/L and

[O2] = 0.00860 mol/L. What would be the equilibrium concentration of NO2??

Answer 1

[N2] = 0.0320-x

[O2] = 0.00860 - 2x

[NO2] = 2x

1.6 x 10^-17 = (2x)^2 / (0.0320-x)(0.00860-2x)^2

solve for x