Question

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.350 M , [B] = 0.650 M , and [C] = 0.300 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.220 M and [C] = 0.430 M . Calculate the value of the equilibrium constant, Kc.

Answer 1

Answer: The value of equilibrium constant, $K_c$ for the given reaction is 12.85.

Explanation:

For the given chemical equation:

                         $A+2B\rightleftharpoons C$

At t = 0      0.350M     0.650M      0.300M

At $t=t_{eq}$  (0.350 - x)    (0.650 - 2x)     (0.300 + x)

We are given:

Equilibrium concentration of A = 0.220 M

Forming an equation for concentration of A at equilibrium:

$0.350-x=0.220\\x=0.130$

Thus, the concentration of B at equilibrium becomes = $0.650-(2\times 0.130)=0.390M$

Equilibrium concentration of C = 0.430 M

The expression of $K_c$ for the given chemical equation is:

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

Putting values in above equation:

$K_c=\frac{0.430}{0.220\times (0.390)^2}\\\\K_c=12.85$

Hence, the value of equilibrium constant, $K_c$ for the given reaction is 12.85.