1. Brown, LeMay Ch 15 AP Chemistry Monta Vista High School
Chemical Equilibrium
Brown, LeMay Ch 15
AP Chemistry
Monta Vista High School

2. 15.1: Chemical Equilibrium
Occurs when opposing reactions are proceeding at the same rate
Forward rate = reverse rate of reaction
Ex:
Vapor pressure: rate of vaporization = rate of condensation
Saturated solution: rate of dissociation = rate of crystallization
Expressing concentrations:
Gases: partial pressures, PX
Solutes in liquids: molarity, [X]

3. Forward reaction: A → B Rate = kforward [A]
Reverse reaction: B → A Rate = kreverse [B] or
R = L•atm
mol•K
Forward reaction:
Reverse reaction:

4. http://www. kentchemistry. com/links/Kinetics/Equilibrium/equili brium
brium.swf ( Good link on Equilibrium) or

5. Kc= kf/kr, at equilibrium, if K> 1, then more products at equilib
Kc= kf/kr, at equilibrium, if K> 1, then more products at equilib. And if k<1, then reactants favored at equilb. K=1 (conc. Of reactants and products nearly same at equilibrium)
The magnitude of Kc gives us an indication of how far the reaction has proceeded toward the formation of products, when the equilibrium is achieved.
The larger the value of K, the further the reaction will have proceeded towards completion when equilibrium is reached (more products present at equilibrium).

6. Equilibrium
Equilibrium is dynamic. The forward and reverse rxns occur at the same rate.
There is a spontaneous tendency towards equilibrium. This does not mean that equilibrium will occur quickly, it simply means that there is always a drive TOWARD the equilibrium state. The amount of drive is measured as Free Energy (D G)
The driving force towards equilibrium diminishes as equilibrium is approached. Thus the appearance of products actually decreases the forward impetus of the reaction, making the free energy change less negative.

7. Equilibrium
The equilibrium position is the same at a given temperature, no matter from which direction it is approached.
It is possible to force an equilibrium one way or the other temporarily by altering the reaction conditions, but once this “stress” is removed, the system will return to its original equilibrium.

8. Figure 1: Reversible reactions
[A]0 or PA0 / RT
PX or [X]
Time →
[A] or PA / RT
Equilibrium is established
[B] or PB / RT

9. Reversible Reactions and Rate
Forward rate
Reaction Rate
Time
Equilibrium is established:
Forward rate = Backward rate
Backward rate
When equilibrium is achieved:
[A] ≠ [B] and kf/kr = Keq

10. 15.2: Law of Mass Action
Derived from rate laws by Guldberg and Waage (1864)
For a balanced chemical reaction in equilibrium:
a A + b B ↔ c C + d D
Equilibrium constant expression (Keq):
Cato Guldberg Peter Waage ( ) ( ) or
But Waage and Guldberg were also related through two marriages; Guldberg married his cousin Bodil Mathea Riddervold, daughter of cabinet minister Hans Riddervold, and the couple had three daughters. Waage married Bodil's sister, Johanne Christiane Tandberg Riddervold by whom he had five children, and after her death in 1869, he became Guldberg's brother-in-law a second time, in 1870, by marrying one of Guldberg's sisters, Mathilde Sofie Guldberg, by whom he had six children.
Keq is strictly based on stoichiometry of the reaction (is independent of the mechanism).
Units: Keq is considered dimensionless (no units)

11. Relating Kc and Kp Convert [A] into PA:
where Dn = = change in coefficents of products – reactants (gases only!) = (c+d) - (a+b)

12. Magnitude of Keq
Since Keq a [products]/[reactants], the magnitude of Keq predicts which reaction direction is favored:
If Keq > 1 then [products] > [reactants] and equilibrium “lies to the right”
If Keq < 1 then [products] < [reactants] and equilibrium “lies to the left”

13. Relationship Between Q and K
Reaction Quotient (Q): The particular ratio of concentration terms that we write for a particular reaction is called reaction quotient.
For a reaction, A B, Q= [B]/[A]
At equilibrium, Q= K
Reaction Direction: Comparing Q and K
Q<K, reaction proceeds to right, until equilibrium is achieved (or Q=K). Why?
Q>K, reaction proceeds to left, until Q=K. Why?

14. Value of K For the rxn, A>B, For the reverse rxn, B >A,
For the reaction,
2A > 2B
For the rxn,
A > C
C > B
K(rxn)= [B]/[A]
K= 1/K(rxn)
K= K(rxn)2
K (overall)= K1 X K2

15. 15.3: Types of Equilibria
Homogeneous: all components in same phase (usually g or aq)
N2 (g) + H2 (g) ↔ NH3 (g) 1 3 2
Fritz Haber (1868 – 1934)
German chemist, who received the Nobel Prize in Chemistry in 1918 for his development of synthetic ammonia, important for fertilizers and explosives. He is also credited as the "father of chemical warfare" for his work developing and deploying chlorine and other poison gases during World War I; this role is thought to have provoked his wife to commit suicide. Despite his contributions to the German war effort, Haber was forced to emigrate from Germany in 1933 by the Nazis because of his Jewish background; many of his relatives were killed by the Nazis in concentration camps, gassed by Zyklon B. Though he had converted from Judaism in an effort to become fully accepted, he was forced to emigrate from Germany by the Nazis in 1933 on account of his being Jewish in their eyes. He died in the process of emigration.
The Haber process now produces 500 million tons of nitrogen fertilizer per year, mostly in the form of anhydrous ammonia, ammonium nitrate, and urea. 1% of the world's annual energy supply is consumed in the Haber process (Science 297(1654), Sep 2002). That fertilizer is responsible for sustaining 40% of the Earth's population, as well as various deleterious environmental consequences.

16. CaCO3 (s) ↔ CaO (s) + CO2 (g)
Heterogeneous: reactants and products in different phases
CaCO3 (s) ↔ CaO (s) + CO2 (g)
Definition: What we use:
Concentrations of pure solids and pure liquids are not included in Keq expression because their concentrations do not vary, and are “already included” in Keq (see p. 548).
Even though the concentrations of the solids or liquids do not appear in the equilibrium expression, the substances must be present to achieve equilibrium.

17. 15.4: Calculating Equilibrium Constants
ICE tables are used to calculate equilibrium
concentrations of the reactants and products
from the initial concentrations.
Steps to use “ICE” table:
“I” = Tabulate known initial and equilibrium concentrations of all species in equilibrium expression
“C” = Determine the concentration change for the species where initial and equilibrium are known
Use stoichiometry to calculate concentration changes for all other species involved in equilibrium
“E” = Calculate the equilibrium concentrations

18. Sample Problem 1 on Equilibrium:
Ex: Enough ammonia is dissolved in 5.00 L of water at 25ºC to produce a solution that is M ammonia. The solution is then allowed to come to equilibrium. Analysis of the equilibrium mixture shows that [OH1-] is 4.64 x 10-4 M. Calculate Keq at 25ºC for the reaction:
NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)

19. NH3 (aq) + H2O (l) ↔ NH41+ (aq) + OH1- (aq)
Initial
Change
Equilibrium
NH3 (aq)
H2O (l)
NH41+ (aq)
OH1- (aq) X M
0 M
0 M X
- x
+ x
+ x X M
4.64 x 10-4 M
4.64 x 10-4 M
x = 4.64 x 10-4 M

20. Sample Problem 2 on Equilibrium
Ex: A L flask is filled with x 10-3 mol of H2 and x 10-2 mol of I2 at 448ºC. The value of Keq is What are the concentrations of each substance at equilibrium?
H2 (g) + I2 (g) ↔ 2 HI (g)

21. H2 (g) + I2 (g) ↔ 2 HI (g) 1.000x10-3 M 2.000x10-3 M 0 M
Initial
Change
Equilibrium
H2 (g)
I2 (g)
HI (g)
1.000x10-3 M
2.000x10-3 M
0 M
- x M
- x M
+ 2x M
(1.000x10-3 – x) M
(2.000x10-3 – x) M
2x M
4x2 = 1.33[x2 + (-3.000x10-3)x x10-6]
0 = -2.67x2 – 3.99x10-3x x10-6
Using quadratic eq’n: x = 5.00x10-4 or –1.99x10-3; x = 5.00x10-4
Then [H2]=5.00x10-4 M; [I2]=1.50x10-3 M; [HI]=1.00x10-3 M

22. 15.6: Le Châtelier’s Principle
If a system at equilibrium is disturbed by a change in:
Concentration of one of the components,
Pressure, or
Temperature
…the system will shift its equilibrium position to counteract the effect of the disturbance.
ry/flash/lechv17.swf
Henri Le Châtelier (1850 – 1936)
French, father was a prominent aluminum industrialist, had seven children, four girls and three boys.

23. 4 Changes that do not affect Keq:
Concentration
Upon addition of a reactant or product, equilibrium shifts to re-establish equilibrium by consuming part of the added substance.
Upon removal of reactant or product, equilibrium shifts to re-establish equilibrium by producing more of the removed substance.
Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l)
Add HCl, temporarily inc forward rate
Add H2O, temporarily inc reverse rate

24. 2. Volume, with a gas present (T is constant)
Upon a decrease in V (thereby increasing P), equilibrium shifts to reduce the number of moles of gas.
Upon an increase in V (thereby decreasing P), equilibrium shifts to produce more moles of gas.
Ex: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g)
If V of container is decreased, equilibrium shifts right.
XN2 and XH2 dec
XNH3 inc
Since PT also inc, KP remains constant.

25. 3. Pressure, but not Volume
Usually addition of a noble gas, p. 560
Avogadro’s law: adding more non-reacting particles “fills in” the empty space between particles.
In the mixture of red and blue gas particles, below, adding green particles does not stress the system, so there is no Le Châtelier shift.

26. 4. Catalysts
Lower the activation energy of both forward and reverse rxns, therefore increases both forward and reverse rxn rates.
Increase the rate at which equilibrium is achieved, but does not change the ratio of components of the equilibrium mixture (does not change the Keq)
Ea, uncatalyzed
Ea, catalyzed
Energy
Rxn coordinate

27. 1 Change that does affect Keq:
Temperature: consider “heat” as a part of the reaction
Upon an increase in T, endothermic reaction is favored (equilibrium shifts to “consume the extra heat”)
Upon a decrease in T, equilibrium shifts to produce more heat.
Effect on Keq
Exothermic equilibria: Reactants ↔ Products + heat
Inc T increases reverse reaction rate which decreases Keq
Endothermic equilibria: Reactants + heat ↔ Products
Inc T increases forward reaction rate increases Keq
Ex: Co(H2O)62+ (aq) + 4 Cl1- ↔ CoCl42- (aq) + 6 H2O (l); DH=+?
Inc T temporarily inc forward rate
Dec T temporarily inc reverse rate

28. Vant Hoff’s Equation
Vant Hoff’s equation shows mathematically how the equilibrium constant is affected by changes in temp.
ln K2 = -d H0 rxn (1 – 1)
K R T2 T1

29. Effect of Various Changes on Equilibrium
Disturbance
Net Direction of Rxn
Effect of Value of K
Concentration
Increase (reactant)
Towards formation of product
None
Decrease(reactant)
Towards formation of reactant
Increase (product)
Decrease (product)

30. Effect of Pressure on Equilb.
Increase P
(decrease V)
Towards formation of fewer moles of gas
None
Decrease P
(Increase V)
Towards formation of more moles of gas
( Add inert gas, no change in V)
None, concentrations unchanged

31. Effect of Temperature on Equilb
Increase T
Towards absorption of heat
Increases if endothermic
Decreases if exothermic
Decrease T
Towards release of heat
Increases if exothermic
Decreases if endothermic
Catalyst Added
None, forward and reverse equilibrium attained sooner
None