When the volume of the container is decreased to half the original volume, the initial concentrations of the reactants and products will change. Since the concentration is defined as the moles of solute per unit volume, when the volume is halved, the concentrations will double. So, the new initial concentrations will be:[A] = 0.4 M[B] = 0.6 M[C] = 0 M[D] = 0 MNow, let's assume that the reaction reaches a new equilibrium state after the volume change. We can represent the change in concentrations as follows:[A] = 0.4 - a*x[B] = 0.6 - b*x[C] = c*x[D] = d*xHere, x is the extent of the reaction, and a, b, c, and d are the stoichiometric coefficients of the reactants and products.The equilibrium constant K for the reaction remains the same, as it is only affected by temperature. So, we can write the expression for the equilibrium constant as:K = [C]^c * [D]^d / [A]â * [B]^b Now, we need to find the new equilibrium concentrations of [B]. Since we don't have the values of a, b, c, d, and K, we cannot provide a specific numerical value for the new equilibrium concentration of [B]. However, we can say that the new equilibrium concentration of [B] will be:[B] = 0.6 - b*xIf you have the values of a, b, c, d, and K, you can use the above expression to find the new equilibrium concentration of [B].