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A mixture initially contains A , B , and C in the following concentrations: [A] = 0.550 M , [B] = 1.05 M , and [C] = 0.550 M . The following reaction occurs and equilibrium is established: A+2B ⇌ C At equilibrium, [A] = 0.390 M and [C] = 0.710 M . Calculate the value of the equilibrium constant, Kc

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Final answer:

To calculate the equilibrium constant for the reaction A + 2B ⇌ C, changes in reactant and product concentrations were determined, and the Kc was calculated using the formula Kc = [C]^1 / ([A]^1[B]^2), resulting in a value of 2.58.

Step-by-step explanation:

The question is about calculating the equilibrium constant, Kc, for a given chemical reaction where the initial and equilibrium concentrations of the reactants and products are provided. The reaction is A + 2B ⇌ C. We are given initial concentrations and equilibrium concentrations for A and C, and we need to calculate the equilibrium concentration of B to determine Kc.

First, let's determine the change in concentration for A (ΔA). The initial concentration of A was 0.550 M and it changed to 0.390 M at equilibrium. So, ΔA = 0.550 M - 0.390 M = 0.160 M. Since A reacts with 2B to form C, the change in concentration of B (ΔB) will be twice that of A, so ΔB = 2 x 0.160 M = 0.320 M. The initial concentration of B was 1.05 M, so the equilibrium concentration of B ([B]) is the initial concentration minus the change, [B] = 1.05 M - 0.320 M = 0.730 M.

Now we can calculate Kc using the equilibrium concentrations: Kc = [C]^1 / ([A]^1[B]^2) = (0.710 M)^1 / (0.390 M)^1(0.730 M)^2 = 0.710 / (0.390 x 0.730 x 0.730) = 2.58. This is the value of the equilibrium constant for the given reaction at equilibrium.

User Armen Hovhannisian
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