Final answer:
To drive the reaction A + B to the right according to the law of mass action, increasing the concentration of reactants A and B will shift the equilibrium to the right, favoring the forward reaction. This is due to the reaction having a negative change in free energy and Q < K. The reaction follows a second-order rate law based on the concentrations of both A and B.
Step-by-step explanation:
According to the law of mass action, to drive the reaction A + B ⇌ to the right, we can alter conditions favourably. Increasing the concentration of reactants is one such condition that will shift the equilibrium to the product side. When we add 0.50 moles of A to the mixture, the system responds by shifting the equilibrium to counteract the change, leading to more products being formed, thus the forward reaction is favored. Furthermore, because the (ΔG < 0) and the reaction quotient (Q < K) due to Q being less than 1, the reaction will proceed spontaneously to the right.
The relationship between reactant concentrations and the reaction rate is captured by the rate law, and for the reaction A + B → C, the rate law is given by rate = k [A] [B], indicating that the reaction is first order with respect to both A and B. Therefore, the overall order of the reaction is 2, classifying it as a second-order reaction.