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

Step-by-step 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 * 0.19)`

= 0.92

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

`K_(eq) = ([C])/([A][B]^(2))`

=
`(0.690)/(0.410 * (0.92)^(2))`

=
`(0.690)/(0.347)`

= 1.988

Thus, we can conclude that the value of the equilibrium constant,
`K_(c)` is 1.988.