Question

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.600 M , [B] = 1.30 M , and [C] = 0.500 M . The following reaction occurs and equilibrium is established: A+2B⇌C At equilibrium, [A] = 0.410 M and [C] = 0.690 M . Calculate the value of the equilibrium constant, Kc.

Answer 1

Explanation:

Chemical equation for the given reaction is as follows.

            $A + 2B \rightarrow C$

Initial :    0.6       1.30           0.5

Final :   (0.6 - x)  (1.30 - 2x)  (0.5 + x)

It is given that at equilibrium, [A] = 0.410 M.

So,    x = (0.600 - 0.410)

           = 0.19 M

Hence, [B] will also be calculated as follows.

       [B] = $1.30 - (2 \times 0.19)$

             = 0.92

Now, we will calculate the value of equilibrium constant as follows.

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

                  = $\frac{0.690}{0.410 \times (0.92)^{2}}$      

                  = $\frac{0.690}{0.347}$

                  = 1.988

Thus, we can conclude that the value of the equilibrium constant, $K_{c}$ is 1.988.