Question

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.650 M, [B] = 1.35 M, and [C] = 0.300 M. The following reaction occurs and equilibrium is established: A+2B<->C
At equilibrium, [A] = 0.550 M and [B] = 0.400 M. Calculate the value of the equilibrium constant, Kc

Answer 1

Answer: The value of $K_c$ for given reaction is 0.465

Explanation:

We are given:

Initial concentration of A = 0.650 M

Initial concentration of B = 1.35 M

Initial concentration of C = 0.300 M

Equilibrium concentration of A = 0.550 M

Equilibrium concentration of B = 0.400 M

For the given chemical equation:

                           $A+2B\rightarrow C$

Initial:                0.65     1.35     0.30

At eqllm:        0.65-x   1.35-2x   0.30+x

Evaluating the value of 'x'

$0.650-x=0.550\\\\x=0.100$

So, equilibrium concentration of B = 1.35 - 2x = [1.35 - 2(0.100)] = 1.15 M

Equilibrium concentration of C = (0.30 + x) = (0.300 + 0.100) = 0.400 M

The expression of $K_c$ for above equation follows:

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

Putting values in above equation, we get:

$K_c=\frac{0.400}{0.650\times (1.15)^2}\\\\K_c=0.465$

Hence, the value of $K_c$ for given reaction is 0.465