Question

Before any reaction occurs, the concentration of A in the reaction below is 0.049 M. What is the equilibrium constant if the concentration of A at equilibrium is 0.0196 M? A(g)⇌B(g)+2C(g) Select the correct answer below:

Kc=5.2×10−3
Kc=0.088
Kc=11
Kc=190

Answer 1

The equilibrium constant (Kc) for the given reaction A(g) ⇌ B(g) + 2C(g) is approximately 190, calculated using the initial and equilibrium concentrations of A.

Step-by-step explanation:

To calculate the equilibrium constant (Kc) for the reaction A(g) ⇌ B(g) + 2C(g), we need to use the initial and equilibrium concentrations of A. Initially, we have [A] = 0.049 M and at equilibrium [A] = 0.0196 M.

The change in concentration of A, which we'll call 'x', is 0.049 M - 0.0196 M = 0.0294 M. According to the stoichiometry of the reaction, for every 1 mol of A that reacts, 1 mol of B is formed and 2 mol of C are formed.

Thus, the equilibrium concentrations of B and C are [B] = x and [C] = 2x respectively.

Now, the expression for Kc is:

Kc = ([B][C]^2) / [A] = (x)(2x)^2 / (0.0196)

Substituting the value of x = 0.0294 M into this expression, we get:

Kc = (0.0294)(2 * 0.0294)^2 / 0.0196

Upon calculating, the value of Kc is found to be approximately 190.

Therefore, the correct answer is Kc = 190.