Keq, PLEASE HELP?

Question

A 1.00 litre flask contained 0.24 mol NO2 at 700 k which decomposed according to the following equation. When equilibrium was achieved, 0.14 mol NO was present. Calculate Keq.
2NO2(g) --> 2NO(g) + O2(g)

a. 9.6 x 10-3
b. 1.1 x 10-2
c. 9.8 x 10-2
d. 1.4 x 10-1
e. 5.7 x 103

Answer 1

Keq= [NO]^2[O2]/[NO2}^2

if .14 mol of NO was formed, then .14 mol of NO2 were used and .07 mol of O2 were formed. So [NO2]=(.24-.14)/1
[NO]=.14/1
[O2]=.07/1

just plug'em in