Question

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.500 M , [B] = 0.800 M , and [C] = 0.350 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.330 M and [C] = 0.520 M . Calculate the value of the equilibrium constant, Kc.

Answer 1

Answer:

The value of the equilibrium constant is 7.447.

Explanation:

A  +   2B  ⇌  C

Initially

0.500 M   0.800 M   0.350 M

At equilibrium :

(0.500-x) M (0.800-2x)M     (0.350+x)M

Concentration at equilibrium:

[A] = 0.330 M = (0.500-x)M

x = 0.17 M

[B] = (0.800-2x)M  = 0.800 - 2  0.17 M = 0.46 M

[C] = 0.520 M

The expression of an equilibrium constant will be give as:

$K_c=\frac{[C]}{[A][B]^2}$

$K_c=\frac{0.520M}{0.330 M\times (0.46)^2}=7.447$

The value of the equilibrium constant is 7.447.