Final answer:
At equilibrium for the reaction A + B → C + D with initial moles in a ratio of 3:1, and upon the addition of more A, the reaction shifts to form more B and C equally. The concentration of D at equilibrium, being stoichiometrically related to B and C, can be determined from the change in moles of A and B.
Step-by-step explanation:
When a reaction A + B → C + D reaches equilibrium, the ratio of the concentrations of products to reactants is described by the equilibrium constant (Kc). In the given reaction with initial moles of B as n and A as 3n, and at equilibrium, the moles of C are equal to the moles of B. It's stated that additional A (0.50 moles) is added to the mixture, shifting the equilibrium towards the right to form more of B and C. The change in molar concentration is represented by a variable x, where an ICE table is used to interpret Initial, Change, and Equilibrium concentrations.
Given that the reaction proceeds with a mole ratio of 1:1 for A:B and that equilibrium shifts forward increasing both [B] and [C] by x, we can determine the equilibrium concentration of D when equilibrium is achieved. Since the reaction quotient (Qc) involves the concentrations of reactants and products and is derived from the stoichiometry, the moles of D would maintain a stoichiometric relationship to those of C and B, and hence can be deduced based on the information provided and the changes that occur when the reaction is shifted by the addition of A.