Consider the following equilibrium:

Consider the following equilibrium:
2NO(g) + O2(g) 2NO2(g)

Given that we start with 0.037M NO(g) and 0.037M O2(g) AND end with 0.015M NO2, what is the equilibrium constant for this reaction and what are the equilibrium concentrations for NO and O2?
Kc = M-1
[NO]eq = M
[O2]eq = M

please help!

ratio reacts: 2NO(g) & O2(g) ---> 2NO2(g
notice that they react 2 mol & 1 mol --> 2mol
so NO reacts to make NO2 in a 1 to 1 ratio
when 1/2 as much O2 reacts

initially it was 0.037M & 0.037M --> none
change was - 0.015 & -1/2 (0.015) --> + 0.015
resulting in: 0.022 M & 0.0295 M --> 0.015M

so, your answers are
NO]eq = 0.022 M
[O2]eq = 0.0295 M


Kc = [NO2]^2 / [NO]^2 [O2]

Kc = [0.015]^2 / [0.022]^2 [0.0295]

Kc = (0.000225) / (0.000484) [0.0295]

Kc = 2.25 e-4 / 1.43 e-5

Kc = 15.7 M-1
yoiu might be expected to round to 2 sig figs, which would be
K2 = 16 M-1