Chemical Equilibrium
AP Chemistry, LFHS
Chemical Equilibrium- the point when the
rate(speed) of a reverse process equals the
rate(speed) of its forward process.
All reactions will reach equilibrium. The time it takes to reach
equilibrium can range from a fraction of a second, to thousands of
years.
The amount of time a reaction takes to reach equilibrium is only
dependent on the Kinetics of the rxn.
Once a reaction reaches equilibrium, it will remain at equilibrium, until
a stress is applied by its surroundings to disrupt it.
Once equilibrium is reached the conc. of both
reactants and products remains constant
N2(g) + 3 H2(g)  2 NH3(g)
• http://phet.colorado.edu/en/simulation/reversible-reactions
Le Chatelier’s Principle-when a stress is
applied to a system at equilibrium, the
system will shift to re-establish equilibrium.
• Shift- an increase in speed of a forward or reverse reaction.
• To determine direction of shift:
• 1) The system will shift to consume added species
• 2) The system will shift to replace removed species
• 3) When a gas is compressed, the system will shift to the side with less moles of gas
• 4) A catalyst or inert substance is NOT a stress
• 5) Solids and liquids are always excluded, use only (aq) & (g)
Molarities partial pressures
Use LeChatlier’s Principle to determine the
shift by the system, if each stress is applied:
• Rxn:
N2(g)
Stress:
Add more NH3
Add H2
Cool container
Compress container
Add a catalyst
Add He gas
Heat container
Expand container
Remove some N2
Remove some NH3
+ 3 H2(g)  2 NH3(g)
System response:
DH = - 92.4 KJ/mol
• Rxn: CaCO3(s)
 CaO(s)
+
CO2(g) DH = + 556 KJ/mol
Stress:
System response:
Heat container
Add catalyst
Remove some CaCO3
Add some CO2
Expand container
Add some water(the water will dissolve some CO2)
Compress container
Add some CO2
Cool container
Add more CaO
Rxn: MgCl2(s)  Mg2+(aq)
+
2Cl-(aq) + Heat
Stress:
System response:
Heat container
Add catalyst
Remove some MgCl2
Add some Pb(NO3)2 to the solution
Expand container
Add some water(the water will dissolve some CO2)
Compress container
Add some Mg(NO3)2
Cool container
During Dynamic Equilibrium, the rate forward
process equals the rate of the reverse process.
• Write the rate law for the forward and reverse processes:
A(aq) + B(s)  C(aq)
Since the forward rate = reverse rate, set rate laws equal to each other.
Keq = Kf/Kr then
or
Keq = K1/K-1
Keq = [C]/[A]
(B is a solid)
SSSSSSSSSSSSSSOOOOOOOOOOOOOOOOOO…………………………….
Law of Mass Action
• The law of mass action- When a system reaches equilibrium, the ratio
of Products/Reactants is constant:
coeff.
If using conc.’s
[Products]
Kc = ----------------------coeff.
[Reactants]
Since Conc. a Press.
PProductscoeff.
OR:
If using partial press.’s
Kp =
---------------------coeff.
Reactants
P
• Write the equilibrium expressions Kc and Kp:
N2(g)
CaO(s)
+ 3 H2(g)
+ CO2(g)
 2 NH3(g)
 CaCO3(s)
4 NH3(g) + 7 O2(g)  4 NO2(g)
+
6 H2O(g)
Kp =
Dn
Kc(RT)
• Write the equilibrium expressions Kc and Kp:
2 NBr3(s)  N2(g) + 3 Br2(g)
At equilibrium: [N2] = 2.0M, 5.0 M. Solve for Kc
Solve for Kp
At equilibrium, PN2 = 1.2 atm. What is PBr2?
• Write the equilibrium expressions Kc and Kp:
What does the value of Keq mean?
Keq will tell us how much product compared to reactant
will form, when the rxn reaches equilibrium.
If Keq is < 1, the rxn favors reactants:
The reaction makes more reactant than product at equilibrium
If Keq = 1, the rxn favors neither products nor reactants:
Equal amounts of products and reactants will be present at equilibrium
If Keq > 1, the rxn favors products:
The reaction makes more products that reactants at equilibrium.
Manipulating Keq
1) When reversing rxn, inverse Keq
A + B  C
Keq
C  A + B
Keq’ = 1/Keq
Solve for Keq of reverse rxn:
H2(g) + Br2(g)  2 HBr(g)
Kc = 2.5 x 10-2
4 Fe(s)
2 NO2(g)
+
3O2(g)

 2 NO(g)
2 Fe2O3(s) Kp = 0.38
+
O2(g)
Kc = 14
• 2) When multiplying coefficients by X, raise Keq to X power:
Br2(g)  2 HBr(g)
H2(g) +
2 H2(g) +
2 Br2(g)  4 HBr(g)
4 Fe(s)
+
3O2(g)
2 Fe(s)
+
1.5 O2(g)

 2 NO(g)
6 NO(g)
+
3 O2(g)
Kc = __________
2 Fe2O3(s) Kp = 0.38

2 NO2(g)
Kc = 2.5 x 10-2
+
Fe2O3(s) Kp = __________
O2(g)
 6 NO2(g)
Kc = 14
Kc =___________
• 3) When combining Reactions, Multiply Keq’s:
A
C
(Target Eq.) A
NO2(g)
+
½ N2(g)
3/2 O2(g)
+
+
+
B  C
D  B + Q
D  Q
NO2(g)  N2O4(g)
N2O4(g)  N2(g) + O2(g)
K1
K2
K = K1 X K 2
Kp = 2.50
Kp = 3.00
2 NO2(g)  N2(g) + O2(g)
+ ½ O2(g)  NO2(g)
NO2(g)  ½ N2(g) + ½ O2(g)
+ 3/2 N2(g)  3 NO2(g)
Kp = _____
Kp = _____
Kp = _____
Kp = _____
Manipulating Equations w/ Energies and Keq’s
Manipulation
1) Reverse Rxn:
M + N  MN
MN  M + N
2) Multiply coeff. By X
3 M + 3 N  3 MN
Energies(DH, DS, DG)
change sign
DH = + 35 KJ/mol
DH = _________
Mult. Energy by X
DH = ________
Keq’s
inverse
Kc = 5
Kc = _______
Raise Kc to X power
Kc = _______
3) Apply Hess’s Law
MN  Z
M + N  Z
Add energies
DH = - 10 KJ/mol
DH = __________
Multiply Keq’s
Kc = 4
Kc = _______
Reaction Quotient- a calculation of law of
mass action using initial amounts.
coeff.
[Products]o
Qc = ----------------------coeff.
[Reactants]o
OR:
PProductsocoeff.
Qp =
----------------------
PReactants coeff.
o
Use Q to determine the how the system will achieve equilibrium:
H2O2(g)  H2(g)
+
O2(g)
The following conc. were found at equilibrium: [H2O2]= 1.50 M, [H2] = [O2] = 3.75 M
Solve for Kc
Does the reaction favor products or reactants?
If the following initial conc.’s were added, solve for Q and tell the direction of initial shift:
[H2O2]= 4.00 M, [H2] = [O2] = 1.75 M
[H2O2]= 0.500 M, [H2] = [O2] = 5.75 M
[H2O2]= 0.0600 M, [H2] = [O2] = .750 M
[H2O2]= 1.50 M, [H2] = [O2] = 0 M
Initial amounts given, how do I solve for equilibrium amounts? RICE Table!!!!
H2(g) +
Br2(g)

2 HBr(g) Kc = 25
1.25M H2 and 1.25 M Br2, 0.500 M HBr were placed in a container.
What is the amount of each present at equilibrium?
R
H2(g) + Br2(g)  2 HBr(g)
I
1.25M 1.25M
0.500 (solve for Q, Q<K, Shift )
C
-X
-X
+2X
E
1.25-X
1.25-X
0.500 +2X (find % dissociation, 5% or less rule)
Set up Equilibrium Expression & solve for X(look for perfect squares):
• 0.00213 moles of CO and H2O are placed in an evacuated 0.335 Liter container.
What is the concentration of all species at equilibrium?
CO(g) + H2O(g)  CO2(g) + H2(g)
Kc = 16
R
I
C
E
CO(g)
0.00636M
+
H2O(g)
Keq =
+
H2(g)
0.00636M
0
0
-X
+X
+X
+X
+X
-X
0.00636-X
 CO2(g)
0.00636-X
[CO2][H2]
[CO][H2O]
(Find Q, Q<K, shift )