17.2 The equilibrium law

•17.2.1 Solve homogeneous equilibrium problems using the expression for Kc.

–The use of quadratic equations will NOT be assessed.

• In order to calculate the value of K, we need to know the concentration in mol/L or mol dm-3 of all reactants and products at equilibrium.

• Sometimes we do not immediately know all of the concentrations at equilibrium, but we are given enough information to determine all of the concentrations at equilibrium and then K.

For Example

0.100 mol/L Br2 and 0.100 mol/L F2

were initially mixed together and allowed to come to equilibrium according to the following reaction:

Br2(g) + F2(g) <-->2BrF(g)

At equilibrium the concentration of Br2 was found to be 0.00135 mol/L. Calculate the value of K.

Fill this in…

Br2(g) F2(g) 2BrF(g)

Initial Conc. 0.100M 0.100M 0

Change in conc.

Final Conc. 0.00135M

Should Look like:

Br2(g) F2(g) 2BrF(g)

Initial Conc. 0.100M 0.100M 0

Change in conc.

-0.09865M -0.09865M +[2 x 0.09865M] =0.1973M

Final Conc. 0.00135M 0.00135M 0.1973M

Good ol’ Stoichiometry

• We used stoichiometry (the relationship with the balanced chemical equation to help us solve for the final concentrations)

• Since Br2 and F2 were a 1:1 mole ratio and had the same initial concentration, they would have the same final.

• But BrF was a 1:2 mole ratio with Br2, so we had to double the change in concentration and therefore the final answer.

We’re not finished!

• Now we can solve for K because we found the final concentrations (at equilibrium)

• K =[BrF]2 / [Br2][F2]

=[0.1973M]2/ [0.00135M][0.00135M]

=2.15 x 104

Sample Problem 2 :

A closed container initially had a CO(g)concentration of 0.750 M and a H2O(g)concentration of 0.275 M. It was allowed to reach equilibrium. At equilibrium, analysis showed a CO2(g) concentration of 0.250 M.

The equation for the reaction is:CO(g) + H2O(g) <-->CO2(g) + H2(g)

Calculate KAnswer: K=5.00

SAMPLE PROBLEM 3A closed chemical system initially contained6.0 M SO2; 2.5 M NO2; and 1.0 M SO3.

Equilibrium was eventually reached for thereaction

SO2(g) + NO2(g)<--> SO3(g) + NO(g)

At equilibrium, the container was found to have 3.0 M SO3 present.

Calculate K.Answer: K=3.00

• A chemist puts 0.085 mol of N2 and 0.038 mol of O2 in a 1.0 L container, once equilibrium is reached its equilibrium constant is found to be 4.2 x10-8. What is the concentration of NO in the mixture at equilibrium?

• N2(g) + O2(g) < -- > 2NO(g)

answer

N2(g) O2(g) < -- > 2NO(g)

0.085 M 0.038 M 0

-x -x +2x

0.085-x 0.038-x 2x

Assumptions

• Normally, you would solve this using the quadratic equation, but IB has noted that their students will NOT need to use the quadratic equation.

• We’re going to do a short cut that we’ll check to make sure is okay later on.

• Instead of using 0.085-x, we’re going to ignore the x, same for 0.038-x (ignore the x) and solve for unknown

N2(g) O2(g) < -- > 2NO(g)

0.085 M 0.038 M 0

-x (ignore) -x (ignore) +2x

0.085 0.038 2x

• k = [2x]2 =4.2 x10-8

[0.085][0.038]

4x2 = (4.2x10-8) x [0.085] x [0.038]

4x2 = 1.357 x 10-10

x2 = 3.392 x 10-11

x = 5.82 x 10-6

[NO] = 2x = 2 x 5.82 x10-6 = 1.2 x10-5 M

• To show that the short cut was okay, check out your answer if the x wasn’t omitted.

• Ex: 0.085- 5.82 x 10-6 = 0.085 (when you only keep 2 sig figs) So it was okay to omit the x.

• In university you won’t be so lucky!!!

Additional practice:

Using quadratic: