N2O4(g) 2 NO2(g)

…concentrations are ________constant

…forward and reverse rates are ________equal

At Equilibrium…

The Equilibrium Constant

…the equilibrium constant expression (Keq) is

Kc = [C]c[D]d

[A]a[B]b

[ ] is conc. in MK expressions do not include:

solids(s) or pure liquids(l)

K = [products][reactants]

For: aA + bB cC + dD

K > 1, the reaction is product-favored;more productat equilibrium.

K < 1, the reaction is reactant-favored;more reactantat equilibrium.

[P][R]

[P]

[R]

K of reverse rxn = 1/K

Kc = = 4.0 [NO2]2

[N2O4]

N2O4 2 NO2

Kc = = 1 .

(4.0)

[N2O4] [NO2]2

N2O42 NO2↔ ↔

K of multiplied reaction

= K^# (raised to power)

Kc = = (4.0)2[NO2]4

[N2O4]2

4 NO22 N2O4 ↔

Manipulating K

K of combined reactions = K1 x K2 …

A + C 3 B Kovr = (2.5)(60)

A B K1 = 2.5C 2 B K2 = 60

Reaction

Initial

Change

Equilibrium

Initial 0.100 M 0.200 M 0 M

Change –0.0935 –0.0935 +0.187

Equilibrium 0.0065 M 0.1065 M 0.187 M

RICE Tables organize info:(Number, Unit, Substance,…Time)

H2 I2 2 HI+

Calculate Kc at 448C.

Kc = [HI]2

[H2] [I2]= (0.187)2

(0.0065)(0.1065) = 51

Reaction

Initial 0

Change

Equilibrium

Initial 0.200 0 0

Change – 2x + x + x

Equilibrium 0.200 – 2x x x

RICE Tables: Use x if K is known(in terms of x)

2 NO N2 O2+

what are the equilibrium concentrations of NO , N2 , and O2 ?

Kc = [N2] [O2]

[NO]2

49 = x(0.200 – 2x)

x = 0.099

2.4 x 103 = (x)2

(0.200 – 2x)2

9.8 – 98x = x9.8 = 99x

[N2]eq = 0.099 M [O2]eq = 0.099 M[NO]eq = 0.0020 M

Equilibrium 0.200 – 2x x x

√√Avoid

polynomials by rooting

K

Q

ratef = raterR P

KQ

Q =[P][R]

K

Q

at equilibriumQ = K

= KQ =

[P]

[R]Q =

[P][R]

Q < K Q > Ktoo much R,shift faster

too much P,shift faster

ratef > raterR P

ratef < raterR P

Reaction Quotient, Q

- 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)

Le Châtelier’s Principle

more mol of gas (↑ngas)

no shift

(H + R P) (R P + H)(changes K)

(Ptotal)

in endo dir. to use up heatin exo dir. to make more heat

System at equilibrium disturbed by change (affecting collisions) will shift ( or ) to counteract the change.

Ksp = [X+]a [Y–]b

Solubility Product Constant (Ksp)XaYb(s) aX+(aq) + bY−(aq)

grams of solid (s) dissolved in 1 L (g/L)mol of solid (s) dissolved in 1 L (M)

product of conc.’s (M) of ions(aq) at equilibrium

solubility:molar solubility:Ksp :

XaYb aX+ + bY–

[XaYb] [X+] [Y–]molar solubility molar conc.’s of ions

Ksp = [X+]a[Y–]b

(Always solid reactant)

Ksp Calculations

PbBr2(s) Pb2+ + 2 Br–

0.010 M 0 M 0 M–0.010 +0.010 +0.020 0 M 0.010 M 0.020 M

If solubility (or molar solubility) is known, solve for Ksp . [PbBr2] is 0.010 M at 25oC .

HW p. 763 #48a

Ksp = (0.010)(0.020)2

Ksp = 4.0 x 10–61 PbBr2 dissociates into…

1 Pb2+ ion & 2 Br– ions

R ICE

Ksp = [Pb2+][Br–]2

(maximum that can dissolve)

(all dissolved = saturated) (any excess solid is irrelevant)

PbCl2(s) Pb2+ + 2 Cl–

x 0 M 0 M –x +x +2x 0 M x 2x

Ksp = [Pb2+][Cl–]2

Ksp = (x)(2x)2

Ksp = 4x3

[PbCl2] = 0.016 M [Pb2+] = 0.016 M [Cl–] = 0.032 M

1.6 x 10–5 = 4x3

3√4.0 x 10–6 = x

0.016 = x

If only Ksp is known, solve for x (M).Ksp for PbCl2 is 1.6 x 10–5 .

(molar solubility)

Ksp Calculations

R ICE

Common-Ion Effect (more Le Châtelier)

• adding common ion (product) shifts left (less soluble)

BaSO4(s) Ba2+(aq) + SO42−(aq)

BaSO4 would be least soluble in which of these 1.0 M aqueous solutions?

Na2SO4 BaCl2 Al2(SO4)3 NaNO3

most soluble?

Basic anions, more soluble in acidic solution.

Mg(OH)2(s) Mg2+(aq) + 2 OH−(aq)

H+

H+ NO Effect on:Cl– , Br–, I–,

NO3–, SO4

2–, ClO4–

Adding H+ would cause…shift , more soluble.

AgCl(s) Ag+(aq) + Cl−(aq)

NH3

Ag(NH3)2+

Forming complex ions…

…increases solubility

AgIO3(s) Ag+ + IO3– Ksp = [Ag+][IO3

–]

Ksp = 3.1 x 10–8

HW p. 764 #62b (is Q > K ?)

Q = [Ag+][IO3–]

Q =

100 mL of 0.010 M AgNO3

10 mL of 0.015 M NaIO3

[Ag+] = ________(0.010 M)(100 mL) = M2(110 mL)

(mixing changes M and V)M1V1 = M2V2

0.0091 M

[IO3–] = ________

(0.015 M)(10 mL) = M2(110 mL)

0.0014 M

Will a Precipitate Form?

Q = (0.0091)(0.0014)

Q = 1.3 x 10–5

Q > K , so…rxn shifts leftprec. will form

BaSO4(s) Ba2+ + SO42–

HW p. 764 #66a

(BaSO4) (SrSO4)SrSO4(s) Sr2+ + SO4

2–

Ksp = [Sr2+][SO42–]

When Will a Precipitate Form?

Ksp = [Ba2+][SO42–]

1.1 x 10–10 = (0.010)(x) 3.2 x 10–7 = (0.010)(x)

x = 1.1 x 10–8

[SO42–] = 1.1 x 10–8 M

x = 3.2 x 10–5

[SO42–] = 3.2 x 10–5 M

#66b

Ba2+ will precipitate first b/c… less SO42– is

needed to reach equilibrium (Ksp).