When the volume of the container is suddenly decreased by a factor of 2, the concentrations of all the species reactants and products will double, since concentration is inversely proportional to volume. Let's denote the initial concentrations at equilibrium as [A], [B], [C], and [D]. After the volume change, the new concentrations will be 2[A], 2[B], 2[C], and 2[D].Now, the reaction will shift to re-establish equilibrium. Let's denote the changes in concentrations as x for A, y for B, z for C, and w for D. The stoichiometry of the reaction tells us that for every 2 moles of A consumed, 1 mole of B is consumed, 3 moles of C are produced, and 1 mole of D is produced. So, we can write the changes as:[A] = -2x[B] = -y[C] = 3z[D] = wThe new equilibrium concentrations will be:[A] = 2[A] - 2x[B] = 2[B] - y[C] = 2[C] + 3z[D] = 2[D] + wNow, we can write the new equilibrium expression using the equilibrium constant K:K = [C]^3[D] / [A]^2[B] 5.0 x 10^-3 = 2[C] + 3z ^3 2[D] + w / 2[A] - 2x ^2 2[B] - y Since K remains constant at the given temperature, the reaction will shift in such a way that the new concentrations of A, B, C, and D will satisfy the above equation. It is difficult to determine the exact changes in concentrations x, y, z, and w without more information about the initial concentrations. However, we can qualitatively predict the direction of the shift.Since the reaction has more moles of gas on the right side 3C + D = 4 moles than on the left side 2A + B = 3 moles , the reaction will shift towards the side with fewer moles of gas the reactants to counteract the increase in pressure caused by the decrease in volume. This means that the concentrations of A and B will increase, while the concentrations of C and D will decrease until a new equilibrium is established.