Final answer:
The equilibrium constant, Kc, for the reaction A(g) + B(g) ⇌ 2 C(g) + D(g) can be determined by applying the stoichiometry of the reaction to the given initial and equilibrium concentrations. By calculating the changes in concentrations, we find that Kc = 0.68, corresponding to answer choice B.
Step-by-step explanation:
To answer the question regarding the equilibrium constant (Kc) of the reaction A (g) + B (g) ⇌ 2 C (g) + D (g), we first need to write the expression for Kc based on the balanced chemical equation. The equilibrium constant expression for this reaction would be Kc = [C]²[D]/[A][B]. Given that the equilibrium concentration of C is 0.75 M, we can find the changes in concentrations of A and B that result from the formation of C and D. If 2 moles of C are formed from 1 mole of A and 1 mole of B, the changes in their concentrations would be proportional to their stoichiometry. Initially, we have 1.1 M of A and 1.0 M of B. If x is the amount of A reacted, then 2x would be the amount of C formed. Given that the equilibrium concentration of C is 0.75 M, we can say that 2x = 0.75 M and hence x = 0.375 M.
Thus, at equilibrium, the concentration of A would be 1.1 M - 0.375 M = 0.725 M, and the concentration of B would be 1.0 M - 0.375 M = 0.625 M. The concentration of D will be the same as x, which is 0.375 M since it is produced in a 1:1 ratio with A. Therefore, the value of Kc would be (0.75)²(0.375)/(0.725)(0.625), and when calculated, you will find that Kc = 0.68. Therefore, the correct answer is B. Kc = 0.68.