For the reaction below, Kc = 1.10 × 10⁻⁴. Note Kc is sometimes called K. What is the equilibrium concentration of C if the reaction begins with 0.200 M A and 0.800 M B? 2 A (aq) + B (aq) ⇌ C (aq)

Question

For the reaction below, Kc = 1.10 × 10⁻⁴. Note Kc is sometimes called K. What is the equilibrium concentration of C if the reaction begins with 0.200 M A and 0.800 M B? 2 A (aq) + B (aq) ⇌ C (aq)

Answer 1

For the reaction 2 A (aq) + B (aq) ⇌ C (aq), the equilibrium constant expression, Kc, is:

Kc = [C]/([A]^2[B])

We are given the value of Kc as 1.10 × 10⁻⁴, and the initial concentrations of A and B as 0.200 M and 0.800 M, respectively.

Let the equilibrium concentration of C be represented by x M. At equilibrium, the concentrations of A and B will each decrease by 2x M (since 2 moles of A are consumed for every 1 mole of C produced, and 1 mole of B is consumed for every 1 mole of C produced). Therefore, the equilibrium concentrations of A and B will be 0.200 - 2x M and 0.800 - 2x M, respectively.

Substituting these equilibrium concentrations into the Kc expression, we get:

1.10 × 10⁻⁴ = x/[(0.200 - 2x)²(0.800 - 2x)]

Simplifying and solving for x, we get:

x = 0.00513 M

Therefore, the equilibrium concentration of C is 0.00513 M.