1. SUMMARY of Unit 8: Equilibrium
At Equilibrium…
…forward and reverse rates are ________
…concentrations are ________
equal
constant
N2O4(g) NO2(g)

2. The Equilibrium Constant
For: aA + bB
cC + dD
…the equilibrium constant expression (Keq) is
Kc =
[C]c[D]d
[A]a[B]b
K =
[products]
[reactants]
[ ] is conc. in M
K expressions do not include:
solids(s) or pure liquids(l)
K > 1, the reaction is product-favored;
more product
at equilibrium.
K < 1, the reaction is reactant-favored;
more reactant
at equilibrium.
[P]
[R]
[P]
[R]

3. ↔ ↔ ↔ Manipulating K K of reverse rxn = 1/K N2O4 2 NO2 N2O4 2 NO2
Kc = = 4.0
[NO2]2
[N2O4]
N2O4
2 NO2 ↔
Kc = =
(4.0)
[N2O4]
[NO2]2
N2O4
2 NO2
K of multiplied reaction = K^# (raised to power)
K of combined reactions = K1 x K2 … ↔
Kc = = (4.0)2
[NO2]4
[N2O4]2
4 NO2
2 N2O4
A  B K1 = 2.5
C  2 B K2 = 60
A + C  3 B Kovr = (2.5)(60)

4. RICE Tables organize info: (Number, Unit, Substance,…Time)
Initial
0.100 M
0.200 M
0 M
Change
–0.050
+0.100
Equilibrium
0.050 M
0.150 M
Reaction
Initial
Change
Equilibrium
H I2
2 HI +
If K is NOT known…use x’s.
[HI]2
[H2][I2]
Kc =

5. Q = Reaction Quotient, Q R P [P] [R] R P Q = [P] [R] Q = K K Q K K Q
ratef < rater
ratef = rater
ratef > rater
Q =
[P]
[R]
Q =
[P]
[R] Q
Q =
[P]
[R]
= K K Q K K Q
Q < K
Q = K
Q > K
too much R,
shift faster
at equilibrium
too much P,
shift  faster

6. Le Châtelier’s Principle
System at equilibrium disturbed by change (affecting collisions) will shift ( or ) to counteract the change.
add R or P:
remove R or P:
volume: ↓V shifts to
↑V shifts to
temp.
↑T shifts
↓T shifts
catalyst:
shift away faster (consume)
shift toward faster (replace)
fewer mol of gas (↓ngas)
(Ptotal)
more mol of gas (↑ngas)
(changes K)
(H + R  P) (R  P + H)
in endo dir. to absorb heat
in exo dir. to produce heat
no shift

7. Solubility Product Constant (Ksp)
XaYb(s) 
aX+(aq) + bY−(aq)
Ksp = [X+]a [Y–]b
(Always solid reactant)
grams of solid (s) dissolved in 1 L (g/L)
mol of solid (s) dissolved in 1 L (M)
product (mult.) of M’s of ions(aq) at equilibrium
solubility:
molar solubility:
Ksp: R I C E
XaYb  aX bY–
# 0 M M
–# +a# b#
0 M a# b#
Ksp = [X+]a[Y–]b

8. 1 PbBr2 dissociates into…
Ksp Calculations
HW p. 763 #48a
If solubility (or molar solubility) is known, solve for Ksp .
[PbBr2] is M at 25oC .
(maximum that can dissolve) R I C E
PbBr2(s)  Pb Br–
0.010 M M M –
0 M M M
Ksp = [Pb2+][Br–]2
(all dissolved = saturated)
(any excess solid is irrelevant)
Ksp = (0.010)(0.020)2
1 PbBr2 dissociates into…
1 Pb2+ ion & 2 Br– ions
Ksp = 4.0 x 10–6

9. Ksp Calculations If only Ksp is known, solve for x (M).
Ksp for PbCl2 is 1.6 x 10–5 . R I C E
PbCl2(s)  Pb Cl–
x M M
–x x x
0 M x x
Ksp = [Pb2+][Cl–]2
Ksp = (x)(2x)2
Ksp = 4x3
(molar solubility)
1.6 x 10–5 = 4x3
3√4.0 x 10–6 = x
= x
[PbCl2] = M
[Pb2+] = M
[Cl–] = M

10. Common-Ion Effect (on solubility)
adding common ion (add product) shifts left (less soluble) XY(s)  X+(aq) + Y–(aq)
Basic anions, more soluble in acid (H+).
H+ (remove product) shifts right (OH–, F–, etc.)
H+ NO Effect on:
Cl– , Br–, I–, NO3–, SO42–, ClO4–
Forming complex ions…
…increases solubility
by (removing product) shifts right
Ag(NH3)2+
NH3
AgCl(s)  Ag+(aq) + Cl−(aq)

11. Will a Precipitate Form?
(is Q > K ?)
HW p. 764 #62b
AgIO3(s)  Ag+ + IO3–
Ksp = [Ag+][IO3–]
Ksp = 3.1 x 10–8
100 mL of M AgNO3
10 mL of M NaIO3
Q = [Ag+][IO3–]
(mixing changes M and V)
M1V1 = M2V2
Q = (0.0091)(0.0014)
Q =
[Ag+] = ________
(0.010 M)(100 mL) = M2(110 mL) M
Q = 1.3 x 10–5
Q > K , so…
rxn shifts left
prec. will form
[IO3–] = ________
(0.015 M)(10 mL) = M2(110 mL) M

12. When Will a Precipitate Form?
HW p. 764 #66a
(BaSO4)
(SrSO4)
BaSO4(s)  Ba2+ + SO42–
SrSO4(s)  Sr2+ + SO42–
Ksp = [Ba2+][SO42–]
Ksp = [Sr2+][SO42–]
1.1 x 10–10 = (0.010)(x)
3.2 x 10–7 = (0.010)(x)
x = 1.1 x 10–8
x = 3.2 x 10–5
[SO42–] = 1.1 x 10–8 M
[SO42–] = 3.2 x 10–5 M
#66b
less SO42– is
needed to reach equilibrium (Ksp).
Ba2+ will precipitate first b/c…