To solve this problem, we will use the reaction quotient Q and the equilibrium constant K to determine the effect of changing the volume on the equilibrium position and the concentration of each species at equilibrium.First, let's write the balanced chemical equation and the expression for the reaction quotient Q:2NO g + O2 g 2NO2 g Q = [NO2]^2 / [NO]^2 * [O2] Initially, the concentrations of the species are:[NO] = 0.50 mol / 1.0 L = 0.50 M[O2] = 0.40 mol / 1.0 L = 0.40 M[NO2] = 0.10 mol / 1.0 L = 0.10 MNow, let's calculate the initial reaction quotient Q:Q_initial = 0.10 ^2 / 0.50 ^2 * 0.40 = 0.0100 / 0.100 = 0.10When the volume is contracted to 0.50 L, the new concentrations of the species are:[NO] = 0.50 mol / 0.50 L = 1.00 M[O2] = 0.40 mol / 0.50 L = 0.80 M[NO2] = 0.10 mol / 0.50 L = 0.20 MNow, let's calculate the new reaction quotient Q:Q_new = 0.20 ^2 / 1.00 ^2 * 0.80 = 0.0400 / 0.800 = 0.05Since Q_new < Q_initial, the reaction will shift to the right to reach equilibrium. This means that the concentration of NO and O2 will decrease, and the concentration of NO2 will increase.Let's denote the change in concentration as x:2NO g + O2 g 2NO2 g -2x -x +2xAt equilibrium, the concentrations will be:[NO] = 1.00 - 2x[O2] = 0.80 - x[NO2] = 0.20 + 2xSince Q_new < Q_initial, we can assume that the equilibrium constant K is close to Q_initial:K 0.10Now, let's write the expression for K using the equilibrium concentrations:K = [NO2]^2 / [NO]^2 * [O2] 0.10 = 0.20 + 2x ^2 / 1.00 - 2x ^2 * 0.80 - x This is a quadratic equation in x, which can be solved either by factoring or by using the quadratic formula. Solving for x, we get:x 0.042Now, we can find the equilibrium concentrations:[NO] = 1.00 - 2x 1.00 - 2 0.042 0.916 M[O2] = 0.80 - x 0.80 - 0.042 0.758 M[NO2] = 0.20 + 2x 0.20 + 2 0.042 0.284 MSo, the equilibrium concentrations after contracting the volume to 0.50 L are approximately:[NO] = 0.916 M[O2] = 0.758 M[NO2] = 0.284 M