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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).