Michael S. answered • 13d
B.S. in Chemistry, Indiana University; Organic Chem Teaching Intern
[NO] = 0.015 M (1.5 × 10−2 M)
No ICE table needed — and that is the point of the problem
Read the wording carefully: "the concentrations at equilibrium are [N2] = 0.23 M and [O2] = 0.44 M." These are already equilibrium values, not starting amounts. So you can substitute them straight into the K expression and solve. Building an ICE table here would be wasted work and would invite errors.
The expression
N2(g) + O2(g) ⇌ 2 NO(g)
K = [NO]2 / ([N2][O2])
The coefficient 2 on NO becomes an exponent, not a multiplier. Writing 2[NO] instead of [NO]2 is the most common error on this problem.
Rearrange, then substitute
[NO]2 = K × [N2] × [O2]
[NO]2 = (2.1 × 10−3)(0.23)(0.44) = (2.1 × 10−3)(0.1012) = 2.13 × 10−4
[NO] = √(2.13 × 10−4) = 0.015 M
Do not forget that square root — solving for [NO]2 and reporting 2.13 × 10−4 is the other frequent slip.
Does the answer make sense?
K = 2.1 × 10−3 is much smaller than 1, which says the equilibrium lies well to the left — reactants dominate and very little product forms. And indeed 0.015 M NO is tiny next to 0.23 M and 0.44 M for the reactants. Consistent.
If your answer had come out larger than the reactant concentrations, that would contradict a small K, and you would know to check your algebra before submitting.
Worth knowing why this reaction matters: N2 and O2 sit together in the air all around you and barely react, exactly because K is so small at ordinary temperatures. But K climbs steeply with temperature, so inside a car engine or a lightning channel the equilibrium shifts far enough right to produce meaningful NO — which is where atmospheric nitrogen oxides come from.
