To solve this problem, we need to use the reaction quotient Q and the equilibrium constant K for the reaction. The reaction quotient is given by the formula:Q = [NH3]^2 / [N2] * [H2]^3 At equilibrium, Q = K. Let's assume that the initial concentrations of N2, H2, and NH3 are [N2]0, [H2]0, and [NH3]0, respectively. We are not given the initial concentrations, but we can express them in terms of moles n and volume V :[N2]0 = n_N2 / 2 L[H2]0 = n_H2 / 2 L[NH3]0 = n_NH3 / 2 LWhen the volume is decreased to 1 L, the concentrations will change:[N2]1 = n_N2 / 1 L[H2]1 = n_H2 / 1 L[NH3]1 = n_NH3 / 1 LSince the pressure remains constant, the number of moles of each gas will not change. Now, let's consider how the change in volume affects the reaction quotient:Q = [NH3]1 ^2 / [N2]1 * [H2]1^3 Since Q must equal K at equilibrium, we can set up the following equation: [NH3]0 ^2 / [N2]0 * [H2]0^3 = [NH3]1 ^2 / [N2]1 * [H2]1^3 Now, we can substitute the expressions for the concentrations in terms of moles and volume: n_NH3 / 2 L ^2 / n_N2 / 2 L * n_H2 / 2 L ^3 = n_NH3 / 1 L ^2 / n_N2 / 1 L * n_H2 / 1 L ^3 Simplifying the equation, we get: n_NH3 ^2 / n_N2 * n_H2 ^3 = 8 * n_NH3 ^2 / n_N2 * n_H2 ^3 Since the equilibrium constant K does not change, the reaction will shift to maintain the same value of K. In this case, the reaction will shift towards the side with fewer moles of gas to counteract the decrease in volume. The reaction will shift towards the formation of NH3, which means that the concentrations of N2 and H2 will decrease, and the concentration of NH3 will increase.However, without the initial concentrations or the value of K, we cannot determine the exact new concentrations of each gas at equilibrium.