Question

For the reaction below, Kc = 1.10 x 10⁻⁴. What is the equilibrium concentration of C if the reaction begins with 0.200 MA and 0.400 MB?

2 A (aq) + B (aq) ⇋ C (aq)

Answer 1

The equilibrium concentration of C is 1.76 x 10⁻⁵ M.

Step-by-step explanation:

Given that the equilibrium constant (Kc) for the reaction is 1.10 x 10⁻⁴, we can use this information to determine the equilibrium concentration of C. The equilibrium constant expression for this reaction is Kc = [C] / ([A]²[B]).

Let's assume that the equilibrium concentration of C is x M. Since the initial concentration of A is 0.200 M and the initial concentration of B is 0.400 M, we can set up the following equation:

1.10 x 10⁻⁴ = x / ((0.200)²(0.400))

Simplifying the equation, we have:

1.10 x 10⁻⁴ = x / (0.1600)

Multiplying both sides by 0.1600, we get:

x = (1.10 x 10⁻⁴)(0.1600)

x = 1.76 x 10⁻⁵ M