Question

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.450 M , [B] = 1.40 M , and [C] = 0.650 M . The following reaction occurs and equilibrium is established: A+2B⇌C

At equilibrium, [A] = 0.330 M and [C] = 0.770 M . Calculate the value of the equilibrium constant, Kc.

Answer 1

Solution :-

ICE table for the reaction

   A         +     2B    ------ > C

0.450 M    1.40 M        0.650 M

-x                -2x                +x

0.330       1.40-2x           0.770

Lets find the value of x

0.450 – x = 0.330

0.450 – 0.330 = x

0.120 = x

Now lets calculate the equilibrium concentration of the B

B= 1.40 – 2x = 1.40 – (2*0.120) = 1.16 M

Kc = [C]/[A][B]^2

Kc = [0.770] /[0.330][1.16]^2

Kc = 1.734

So the equilibrium constant Kc = 1.73