Compounds A, B, and C react according to the following equation. At 100 $\degree$C a mixture of these gases at equilibrium showed that [A] = 0.855 M, [B] = 1.23 M, and [C] = 1.75 M. What is the value of K_c for this reaction?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the given reaction is: K_c = [C] / ([A]^[d] * [B]^[e]) From the balanced chemical equation, we can see that the reaction involves 2 moles of A and 3 moles of B in order to produce 1 mole of C. Therefore, the reaction quotient (Qc) at equilibrium can be calculated as: Qc = [C]^1 / ([A]^2 * [B]^3) = (1.75 M) / (0.855 M)^2 * (1.23 M)^3 = 3.240 Comparing Qc to K_c , we find that the equilibrium lies to the left since Qc < K_c . Therefore, to reach equilibrium, the concentrations of A and B will decrease while the concentration of C will increase. Since we now know the values of Qc and the concentrations of A, B, and C at equilibrium, we can substitute these values into the equilibrium constant expression and solve for K_c : K_c = [C] / ([A]^2 * [B]^3) = (1.75 M) / (0.855 M)^2 * (1.23 M)^3 = 3.24 Therefore, the value of K_c for this reaction is 3.24. The correct answer choice is (D).

Compounds A, B, and C React According to the Following °\degree°

Compounds A, B, and C React According to the Following $\degree$