Consider an equilibrium mixture of four chemicals (A, B, C, and D, all gases) reacting in a closed flask according to the equation: $$\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g)$$ a. You add more A to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. b. You have the original setup at equilibrium, and you add more D to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer.

According to Le Chatelier's principle, when more of the reactant A is added, the equilibrium will shift to minimize the disturbance. In this case, the system will move toward consuming more A by producing more C and D, and in the process, reactant B will also be consumed. This shift will occur until a new equilibrium is established. As a result, the concentration of A will decrease from its initial value, while the concentrations of C and D will increase. The concentration of B will decrease as well since it reacts with A. After the new equilibrium is established, the concentrations of A and B will be lower than their original values, while the concentrations of C and D will be higher than their original values.

In this scenario, more of product D is added to the system. Similar to the previous case, the system will adjust according to Le Chatelier's principle to minimize the disturbance. Adding more D will cause the equilibrium to shift in the opposite direction, moving toward the reactants to consume more D. Thus, more A and B will be produced, and in the process, C will also be consumed. As a result, the concentration of D will decrease from its initial value, while the concentrations of A and B will increase. The concentration of C will decrease as well since it reacts with D. After the new equilibrium is established, the concentrations of A and B will be higher than their original values, while the concentrations of C and D will be lower than their original values.