Final answer:
The equilibrium constant (Keq) for the reaction where moles of C are three times that of A at equilibrium is determined using the ICE table method. The correct answer is 3, which is option (b).
Step-by-step explanation:
The student's question is about determining the equilibrium constant (Keq) for a chemical reaction where initially moles of reagents A and B are equal, and at equilibrium, moles of C are three times that of A. The reaction is represented as A + B ⇋ C + D. To solve this problem, we can set up an ICE table (Initial, Change, Equilibrium) to calculate the concentrations of the reactants and products at equilibrium.
Let's assume the initial moles of A and B are both 'a'. Since A and B react in a 1:1 ratio, if 'x' is the amount of A that reacted to reach equilibrium, then 'x' is also the amount of B that reacted and 'x' moles of C and D would have formed. Given that at equilibrium, [C] = 3[A], we can say that 3x = 3([A] - x), meaning 3x = 3a - 3x, so x = a/2. Therefore, at equilibrium, [A] = a - x = a/2, [B] = a - x = a/2, [C] = 3x = 3a/2, and [D] = x = a/2.
The equilibrium constant expression for this reaction is Keq = ([C][D])/([A][B]). Plugging in the equilibrium concentrations, we get Keq = ((3a/2)(a/2))/((a/2)(a/2)) = 3. Thus, the equilibrium constant Keq is equal to 3, and the correct answer to the student's question is option (b) 3.