1. 13.1 CHEMICAL EQUILIBRIUM
Cato Maximilian Guldberg and his brother-in-law Peter Waage developed the Law of Mass Action

2. Chemical Equilibrium Reversible Reactions:
A chemical reaction in which the products
can react to re-form the reactants
Chemical Equilibrium:
When the rate of the forward reaction
equals the rate of the reverse reaction
and the concentration of products and
reactants remains unchanged
2HgO(s)  2Hg(l) + O2(g)
Arrows going both directions (  ) indicates equilibrium in a chemical equation

3. 2NO2(g)  2NO(g) + O2(g) Remember this from Chapter 12?
Why was it so important to measure reaction rate at the start of the reaction
(method of initial rates?)

4. 2NO2(g)  2NO(g) + O2(g)

5. jA + kB  lC + mD Law of Mass Action For the reaction:
Where K is the equilibrium constant, and is unitless

6. Product Favored Equilibrium
Large values for K signify the reaction is “product favored”
When equilibrium is achieved, most reactant has been converted to product

7. Reactant Favored Equilibrium
Small values for K signify the reaction is “reactant favored”
When equilibrium is achieved, very little reactant has been converted to product

8. Writing an Equilibrium Expression
Write the equilibrium expression for the reaction:
2NO2(g)  2NO(g) + O2(g)
K = ???

9. Conclusions about Equilibrium Expressions
The equilibrium expression for a reaction is the reciprocal for a reaction written in reverse
2NO2(g)  2NO(g) + O2(g)
2NO(g) + O2(g)  2NO2(g)

10. Conclusions about Equilibrium Expressions
When the balanced equation for a reaction is multiplied by a factor n, the equilibrium expression for the new reaction is the original expression, raised to the nth power.
2NO2(g)  2NO(g) + O2(g)
NO2(g)  NO(g) + ½O2(g)

11. Ex. N2(g) + O2(g) 2 NO(g) K = [NO]2 [N2][O2] =(1.1 x 10-5)
An equilibrium mixture N2, O2 and NO gases at 1500 K is determined to consist of 6.4 x 10-3 mol/L of N2, 1.7 x 10-3 mol/L O2, and 1.1 x 10-5 mol/L of NO. What is the equilibrium constant for the system at this temperature?
N2(g) + O2(g) NO(g)
K = [NO]2
[N2][O2]
=(1.1 x 10-5)
(6.4 x 10-3)(1.7 x 10-3 )
=1.1 x 10-5
Which direction is most favored?
K is small so not much product is formed

12. Equilibrium Expressions Involving Pressure
For the gas phase reaction:
3H2(g) + N2(g)  2NH3(g)

13. Heterogeneous Equilibria
The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present
Write the equilibrium expression for the reaction:
PCl5(s)  PCl3(l) + Cl2(g)
Pure
solid
Pure
liquid

14. jA + kB  lC + mD The Reaction Quotient
For some time, t, when the system is not at equilibrium, the reaction quotient, Q takes the place of K, the equilibrium constant, in the law of mass action.
jA + kB  lC + mD

15. Significance of the Reaction Quotient
If Q = K, the system is at equilibrium
If Q > K, the system shifts to the left, consuming products and forming reactants until equilibrium is achieved
If Q < K, the system shifts to the right, consuming reactants and forming products until equilibrium is achieved

16. Solving for Equilibrium Concentration
Consider this reaction at some temperature:
H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0
Assume you start with 8 molecules of H2O and 6 molecules of CO. How many molecules of H2O, CO, H2, and CO2 are present at equilibrium?
Here, we learn about “ICE” – the most important problem solving technique in the second semester. You will use it for the next 4 chapters!

17. Solving for Equilibrium Concentration
H2O(g) + CO(g)  H2(g) + CO2(g) K = 2.0
Step #1: We write the law of mass action for the reaction:

18. Solving for Equilibrium Concentration
Step #2: We “ICE” the problem, beginning with the Initial concentrations
H2O(g) + CO(g)  H2(g) + CO2(g)
Initial:
Change:
Equilibrium: 8 6 -x -x +x +x
8-x
6-x x x

19. Solving for Equilibrium Concentration
Step #3: We plug equilibrium concentrations into our equilibrium expression, and solve for x
H2O(g) + CO(g)  H2(g) + CO2(g)
Equilibrium:
8-x
6-x x
x = 4

20. Solving for Equilibrium Concentration
Step #4: Substitute x into our equilibrium concentrations to find the actual concentrations
H2O(g) + CO(g)  H2(g) + CO2(g)
Equilibrium:
8-x
6-x x
x = 4
Equilibrium:
8-4=4
6-4=2 4

21. Ex.
At a particular temperature, K = 1.00 x 102 for the reaction. H2(g) + I2(g) 2 HI(g) In an experiment, 1.00 mol H2, 1.00 mol I2, and 1.00 mol HI are introduced into a 1.00 L container. Calculate the equilibrium concentrations of all reactions and products?
Write the balanced equation and the equilibrium expression, omitting pure solids and liquids.
H2(g) +I2(g) 2HI(g) K=[HI]2
[H2][I2] I
C -x -x +2x
E 1.00-x x x
K = 100. = ( x)2
(1.00 – x)(1.00 – x)

22. Homework
Read p
P. 614 # 18, 20, 22,