Answer
$$[NO] = 3.5 \times 10^{-3} \space M$$
Work Step by Step
1. Calculate the concentrations:
$$Molarity(N_2) = \frac{0.20 \space mol}{1.0 \space L} = 0.20 \space M$$
$$Molarity(O_2) = \frac{0.15 \space mol}{1.0 \space L} = 0.15 \space M$$
2. The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ NO ] ^{ 2 }}{[ N_2 ][ O_2 ]}$$
3. Solve for the missing concentration:
$$ \sqrt[2]{K_C \times [ N_2 ][ O_2 ]}{} = [NO]$$
4. Evaluate the expression:
$$ [NO] = \sqrt[2]{(4.10 \times 10^{-4}) \times {( 0.20 )( 0.15 )}{}}$$
$$[NO] = 3.5 \times 10^{-3} \space M$$
