To solve this problem, we will use an ICE Initial, Change, Equilibrium table to keep track of the concentrations of each species at different stages of the reaction. Assuming the volume of the flask is 1 L, the initial concentrations are as follows:[A] = 0.10 M[B] = 0.20 M[C] = 0 M[D] = 0 MNow, let's denote the change in concentration of each species as x. At equilibrium, the concentrations will be:[A] = 0.10 - x[B] = 0.20 - x[C] = x[D] = xWe can now use the equilibrium constant expression to set up an equation:Kc = [C][D] / [A][B] = 2.0 x 10^-3Substitute the equilibrium concentrations into the expression: 2.0 x 10^-3 = x * x / 0.10 - x * 0.20 - x Now, we need to solve for x. Since Kc is small, we can assume that x is much smaller than 0.10 and 0.20, so we can simplify the equation: 2.0 x 10^-3 x^2 / 0.10 * 0.20 x^2 2.0 x 10^-3 * 0.10 * 0.20 x^2 4.0 x 10^-5x 4.0 x 10^-5 x 2.0 x 10^-3 MNow we can find the equilibrium concentrations of each species:[A] = 0.10 - x 0.10 - 2.0 x 10^-3 0.098 M[B] = 0.20 - x 0.20 - 2.0 x 10^-3 0.198 M[C] = x 2.0 x 10^-3 M[D] = x 2.0 x 10^-3 MSo, the equilibrium concentrations are approximately:[A] = 0.098 M[B] = 0.198 M[C] = 2.0 x 10^-3 M[D] = 2.0 x 10^-3 M