Question

A mixture initially contains A, B, and C in the following concentrations: [A] = 0.400M , [B] = 0.700M , and [C] = 0.700M . The following reaction occurs and equilibrium is established:

A+2B?C

At equilibrium, [A] = 0.210M and [C] = 0.890M . Calculate the value of the equilibrium constant, Kc.

Express your answer numerically.

Answer 1

aA + bB <---> cC + dD
([D]^d [C]^c) / ([A]^a [B]^b)

so u have A + 2B <---> C

put the concentrations of the products over the reactants:
[C] / ([A] [B]^2)
Remember that the coefficients of the original equation become exponents for the equilibrium expression.

You know the equilibrium values of A and C but not B. What you can do is make an ICE box.
Initial        : A = 0.4    | B = 0.7         | C = 0.7 |
Change   : - x             | - 2x               | + x |
Final         : 0.21         | B = ?            | C = 0.89 |

For the C (change) section, you use the ratio found in the original equation in this case it's 1 : 2 : 1. You put minus signs on the reactant side since they are being used up and u put plus signs on the product side since they are being made.

To find the equilibrium value for B, take any column let's say A and solve for x. You can see that 0.4 - x = 0.21

You will get x = 0.19 Look at the B column and you'll see 0.7 - 2x Plug in .19 for x so you end up with

0.7- 0.38 =0.32(which is the equilibrium value for B).

Replace the letters with the equilibrium values:
Kc = (0.89) / ((0.21)(0.38)^2) = 29.349 ........answer