Ch. 15 EQUILIBRIUM&
Le CHÂTELIER
EQUILIBRIUM >Constant, Kc >Calculations, I.C.E. Tables RXN QUOTIENT, Qc
>Compare K - Q
Le CHÂTELIER’S PRINCIPLE
KINETICS RELATIONSHIP
USE [ ] & P
EQUILIBRIUM *State of balance *2 = & opp opposing forces occur @ same rate (still reacting)Chemistry *fwd & rev rxns @ same rate
*no in concentratons
N2O4(g) 2 NO2 (g) colorless brown
REVERSIBLE RXN rf = kf[N2O4] rr = kr[NO2]2
““unitless”unitless”
@ EQUILIBRIUM
set = & rearrange rf = kf[N2O4] rr = kr[NO2]2
kf[N2O4] = kr[NO2]2
[NO2]2/[N2O4] = kf /kr
= a constant; equilibrium constant; Kc
HABER PROCESS
3 H2(g) + N2(g) 2 NH3 (g)
LAW MASS ACTION
aA + bB cC + dD
[B]]A[
[D][C]
]reacts[
[pdts] K
ba
dc
y
xc Equilibrium Expression
Kc depends on………… not ………..nature of rxn how (mechanism)
value not depend on……… & ……..reactant product amounts
so, depends only on .… & …..rxn Temp
Le Châtelier
Equilibrium: balance between 2 opposing reactionsHow sensitive is this balance to changes in conditions?What can be done to change the equilibrium state?
If pdts. be withdrawn continuously, then reacting system can be kept constantly off-balance
More reactants used, more pdts. formedMost useful if 1 pdt. can 1. escape as gas 2. Condensed or frozen from gas phase as solid or liquid 3. Washed out of gas mixture by liquid spray which is especially soluble 4. Precipitated from gas or solution
Forward Rxn.: forward direction; reaction favored reactants to pdts.Reverse Rxn.: reverse direction; reaction favored pdts. to reactants
CaO (s) + 3 C (s) CaC2 (s) + CO (g)
Remove CO: reaction tipped toward CaC2 formation
TiCl4 (g) + O2 (g) TiO2 (s) + 2 Cl2 (g)
Production of calcium carbide
Production of titanium dioxide
TiO2 separates from gases as fine powdered solid thus, rxn. kept moving in forward direction
Synthesize Rxn.
CH3COOH + HOCH2CH3 CH3COOCH2CH3 + H20
Remove H2O: forward rxn. favored
Production of ammonia
N2 (g) + 3 H2 (g) 2 NH3 (g)
NH3 more soluble in water than H2 or N2
Wash out NH3 out of equilibrium mixture
Equilibrium usually temp. dependent Increase temp. both forward/reverse rxns. speed up. Why? molecules move faster, increase molecular collisions
Pressure
H2 (g) + I2 (g) 2 HI (g) + heat (given off)
Adding heat shifts rxn. to left
Temperature
Rxn shifts in direction of fewer gas molecules w/ P increase
Looking at rxn above:1. Which direction would rxn favor with an increase in P?
2. If rxn were at a temp at which iodine was a solid, which direction would rxn favor at an increase in P?
total 2 moles gas reactants -------> total 2 moles gas pdts; no change
1 mole gas reactant -------> 2 moles gas pdts; left (reverse) favored
Catalyst
No effect, but can increase speed which equilibrium is reached
If an external stress is applied to a system at chemical equilibrium, then theequilibrium pt. will change in such a way as to counteract (alleviate) the effectsof that stress.
Le Châtelier’s Principle
The amounts of reactants & pdts will shift in such a manner as to minimizethe stress
Problem Set #1
1. Write Kc for: CH4(g) + H2S(g) CS2 (g) + H2(g)
2. Write Kc for: Al(s) + HCl(aq) AlCl3 (aq) + H2(g)
S][H]CH[
]][H[CS K
4 2
224
422c
22(g)24 (g)(g)(g) HCSSHCH
[HCl][Al]
][H][AlCl
3 2 HCl(aq) 6 Al(s) 2
62
32
23
2(aq)3 (g)H
cK
AlCl
Ignore pure “solids & liquids”
Use Partial Pressure - atm 760 mmHg(torr) = 1 atm
Kc -----> Kp
HABER PROCESS 3 H2(g) + N2(g) 2 NH3 (g)
][H]N[
][NH K
322
23p
Kc Kp
convert using PV = nRT
[A]RT
RTV
n P
Molarity n/V RTV
n P
A
A
P = (n/V)RT
Relate Kp - Kc
Kp = Kc(RT)ngas
ngas = ( moles gas pdts) - ( moles gas reacts)
Problem Set #2
1. When does Kc = Kp? If ever!!
2. CH4(g) + H2S(g) CS2(g) + H2(g) Kc = 1.3*10-2 @ 475oC a) find ngas b) write Kp expression c) find Kp
when ( moles gas pdts) = ( moles gas reacts), then ngas= 0
a) 1 CH4(g) + 2 H2S(g) 1 CS2(g) + 4 H2(g) ngas = 5 - 3 = 2
b) Kp = (0.013)[(0.0821)(748)]2
c) Kp = (0.013)(3771.2863) = 49.03
EVALUATE K1- Kc = 4.56*109 2- Kc = 4.56*10-9
3- Kp = 49 4- Kp = 0.49
Gives info of mixture @ equilibrium
K = [pdt]/[react]
K >> 1 lies rgt; pdtsK = 1 50 - 50; = amtsK << 1 lies left; reacts
Evaluate List1- rgt; pdts2- left; reacts3- slightly more pdts than reacts
4. = amts
Figure 15.07ab
EQUILIBRIUM SYSTEMS
Systems look at:Reversible RXNTempP[ ]; 2 special rules
1: Driving Reversible RXN
state: rev rxn @ equilib - sys open; react/pdt allowed to escape - no longer @ equilib due to escaping particle - rxn shift to side that contains escaping particle Used to drive rev rxn to pdts that are wanted
Problem Set #2 HABER PROCESS3 H2(g) + N2(g) 2 NH3 (g)
5. If [H2] is increased a) Equilib shift direction? ____ b) [N2] will incr/decr? _____ c) [NH3] will incr/decr? ____
6. If [N2] is decreased a) Equilib shift direction? ____ b) [H2] will incr/decr? _____ c) [NH3] will incr/decr? ____
DecrIncr
IncrDecr
2: Temperature Changes
state: Energy shown as term in rxn @ equilib - follows Le Châtelier same manner as w/ [ ] - Endo add E, shifts equilib away from E side - Exo remove E, shifts equilib toward E sideE added by incr TempE removed by cooling
PS#2, cont.N2(g) + O2(g) +90 kJ 2 NO (g)
7. If temp is incr a) Equilib shift direction? ____ b) [O2] will incr/decr? _____ c) [NO] will incr/decr? ____
8. If [O2] is decreased a) Equilib shift direction? ____ b) [N2] will incr/decr? _____ c) Temp will incr/decr? ____
EnergyE added by incr Temp E produced, Temp incrE removed by cooling E used, Temp decr
DecrIncr
IncrIncr
3: Pressure Changes
state: affect sys w/ gases @ equilib - Incr P, shift to side w/ less total gas moles - Decr P, shift to side w/ more total gas moles
PS#32 NO2(g) N2O4 (g) + E
9. If temp is incr a) Equilib shift direction? ____ b) [NO2] will incr/decr? _____ c) [N2O4] will incr/decr? ____
10. If volume is decreased a) Equilib shift direction? ____ b) [NO2] will incr/decr? _____ c) [N2O4] will incr/decr? ____ d) Temp will incr/decr? ____
IncrDecr
DecrIncrIncr
4: Concentration Changes
2 Cases - add/remove solid or liquid @ equilib no shift - liquid solvent; add/remove no shift
I - Solids[consistent] by density add/remove, no in solid’s [ ]
11. NaCl(s) NaCl(aq)
add salt till soln saturated a) When is equilib reached? b) amt NaCl (s) incr/decr? _____ c) [NaCl(s)] incr/decr? _____ d) [NaCl(aq)] incr/decr? _____
Add more NaCl @ equilib Incr No No
Liquid SolventsLiq solvent list as term in eqn then +/- solvent no equlib
Subst dissolved in solvent, [liq solv] can .But, in practice nearly always negligible.
Solvent at such higher concentration than reaction substs.PS#3, cont12. Ca(NO3)2(s) + H2O(l) Ca+2(aq) + 2 NO3
-1(aq)
)aq()aq()s(-13
2 23 NO 2 Ca )NO(Ca O2H
a) Equilib shift direction, add water? ____ b) Equilib shift direction, add calcium compound? _____
No
DIRECTIONDIRECTIONN2(g) + O2(g) 2 NO(g) Kc = 1*10-30 @ 25oC
Write Kc and find for reverse
Kc = [NO]2/[[N2][O2]] = 1*10-30
Kc = [[N2][O2]]/[NO]2 = 1/1*10-30 = 1*10+30
EvaluateWhat is favored in this rxn?
What need to to favor NO?
N2 - O2
Incr Temp
PS #4
13. Haber Process @ 300oC, Kp = 4.34*10-3
Find Kp reverse?
14. For each: a) CaCO3(s) CaO(s) + CO2(g) (concen in M & atm)
Kc= Kp= b) CO2(g) + H2(g) CO(g) + H2O(l)
Kc= Kp=
c) Fe(s) + H2O(g) Fe3O4(s) + H2(g) (all concen listed in “atm”)
Kc= Kp =
1/(4.34*10-3 ) = 2.3*102
[CO2] [PCO2]
[CO]/[[CO2][H2]] [PCO]/[[PCO2][PH2 ]]
3 4 4 (PH2)4/(PH2O)4
PS #1- revisit2. Write Kc for: Al(s) + HCl(aq) AlCl3 (aq) + H2(g)
[HCl][Al]
][H][AlCl
3 2 HCl(aq) 6 Al(s) 2
62
32
23
2(aq)3 (g)H
cK
AlCl
Ignore pure “solids & liquids”
[HCl]
][H][AlCl
6
32
23cK
Diff. Phases: heterogeneous equilibrium
solids: same concen @ given temp; same # mols/L; no Vliquids: same applies therfore, we ignore pure solids & liquids
only concerned w/ those that will [ ] in rxn
Qc, Kc Expression Summary fig 15.9, pg 645
SUMMARY
1) Kc fwd is 1/Kc reverse2) Kc “unitless”3) Kc = [pdts]x/[reacts]y
ignore pure solids - liquids4) Kp = Kc(RT)ngas
5) Kc depends rxn & Temp6) K >> 1 K <<<<1 direction - evaluate
Calculate Kc2 methods
I. Know amts @ equilibrium [ ]/P @ spec Temp.
ex. Haber process analyze equil @ 472oC mixture: 7.38 atm H2, 1870 torr N2, & NH3; Ptot = 10.00 atm
Determine Kp
convert: N2 atm; find atm NH3; bal eqn
5-10*2.59 988.79
0.0256
(7.38))46.2(
(0.160)
)(P)P(
)(P K
3
2
3H22N
2NH3
p
II. I.C.E. TABLE SET UP -- find equil quantities & K
write balanced eqn, list given & unkn [ ] or P label conditions [initial, change, equlib]
ex. A mixture is analyzed to find [A] = 2.000*10-3 & [B]= 4.000*10-3 and allowed to react. The reaction is A(g) + 2 B(g) C(g) Find [ ] @ equilib
Kc = [C]/[[A][B]2] initial [A], [B], [C] need to find [all]
[A] + 2 [B] <------> [C]
initial change equilibrium
2.000*10-3 4.000*10-3 0.000 -x -2x +x0.002-x 0.004-2x +x
Can deduce from PDT amt [C]Now do mole conversion
[A] = [C] * mole ratio = (1.56*10-3)*(1A/1C) = -1.56*10-3 [A][B] = (1.56*10-3)*(2B/1C) = -3.12*10-3 [B]
[A] + 2 [B] <------> [C]
initial change equilibrium
2.000*10-3 4.000*10-3 0.000 -1.56*10-3 -3.12*10-3 +1.56*10-3
4.4*10-4 8.8*10-4 1.56*10-3
2c
[A][B]
[C] K
610*4.58
]10*][8.8010*[4.40
]10*[1.5624-4-
-3
PS #4, cont#15. find quantity X from know equil [ ] & K
2 NO (g) <----> N2 (g) + O2 (g) Kc = 4.6*10-1
[.976] [.781] X
222
]NO[
]O][N[Kc
0.561 ]N[
]NO[Kc]O[
2
22
#16. 0.500 mol ICl gas decomposes into two diatomic gases in a 5.00 L container.
1) construct a reaction table 2) concen @ equilb, Kc = 0.110Determine initial [ ]: ICl = 0.500 mol/2.00 L = 0.100 M since no rxn started; Cl2 & I2 = 0 mol
2 ICl (g) <------> Cl2 (g) + I2 (g)
initial change equilibrium
0.100 0 0 -2x +x +x0.100-2x +x +x
2
22
]ICl[
]I][Cl[Kc
2]x2100.0[
]x][x[ 110.0
2
2
]x2100.0[
]x[ 110.0
]x2100.0[
x 0.332
2 ICl <------> Cl2 + I2
initialchange
equilibrium
0.0332 = 1.664xx = 0.02
0.332[0.100-2x] = x
0.100 0.00 -2(0.02) +0.02 since Cl2 = I2
[0.04] [0.02] [0.02]
RXN QUOTIENT, Qc
K: equilibrium constantQ: rxn quotient
only 1 value @ equilb @ spec Tempvaries as rxn proceeds @ same spec Temp
Qc = Kc @ equilb
CH4 + Cl2 <-----> CH3Cl + HCl @ 1500 KPatm 0.13 0.035 0.24 0.47 Kp = 1.6*104
Find Qp, which direction? 25 0.035*0.13
0.47*0.24
QQ Q
24
33
ClCH
ClHClCHp
Qp < Kp ??
COMPARE Q <----> K
Q < K more pdts, fwd rxnQ = K no , equilibriumQ > K more reactant, rev rxn
now, values are “mol concen”; rxn in 250 mL flask
CH4 + Cl2 <-----> CH3Cl + HCl @ 1500 K [mol] 0.13 0.035 0.24 0.47 Kc = 1.6*104
Find Qc, which direction?
1) 0.25 L 2) find M, mols/L 3) use values Qc eqn solve & compare
EX. At 500oC, Kc = 1.6*10-2 for the rxn: 2 H2S(g) <----> 2 H2(g) + S2(g)
Calculate Kc for each: a) 0.5 S2(g) + H2(g) <----> H2S(g)
b) 5 H2S(g) <----> 5 H2(g) + 5/2 S2(g)
21
22 2
H S
S H
22 2
22
H S
H S
Solution: as reference: Kref =
a) reverse of original rxn by factor ½. Qc = (1/Kref)½
2/1
22
222
]SH[
]S[]H[
Kc = (1/1.6*10-2)½ = 7.9
b) original rxn by factor 5/2. Qc = (1/Kref)5/2 2/5
22
222
]SH[
]S[]H[
Kc = (1.6*10-2)5/2 = 3.2*10-5
PS #5 #17. H2 + Cl2 <-----> 2 HCl Kc = 7.6*108
Find Kc: 0.5 H2 + 0.5 Cl2 <-----> HCl
#18. Find Kc: 4/3 HCl <-----> 2/3 H2 + 2/3 Cl2
PS #5
40.585.025.02
21
22
210*2.8 10*7.6
ClH
HCl
ClH
HCl Kc .17#
6-32
83
43
223
22
32
22
2 fwdrev
10*1.4 10*7.6
1
HCl
ClH
ClH
HCl
1
K
1 K .18#
EX. Mixture of 5.00 volumes of N2 & 1.00 volume of O2 reaches equlibrium @ 900 K & 5.00 atm: N2(g) + O2(g) <----> 2 NO(g) Kp = 6.70*10-10
What is if partial pressure of NO?
Solution: construct I.C.E. table initial: 6 vol of gas @ 5 atm & 900 K. Vols are proportional to moles, so vol fraction = mole fraction
PN2 + PO2 = 5.00 atm PN2 = XN2Ptot = (5.00/6.00) = 4.17 atm PO2 = = (1.00/6.00) = 0.83 atm
N2 (g) + O2 (g) <------> 2 NO (g) Initial 4.17 0.83 0 Change -X -X +2X
Equilibrium 4.17-X 0.83-X 2X
Kp very small, assume [N2]eq = 4.17 - X = 4.17 & same O2 = 0.83
)83.0)(17.4(
(2X) 10*6.70
PP
P K
210-
O2N2
2NOp X = 2.41*10-5
NO: 2*(2.41*10-5) = 4.82*10-5 atm
assumption: [(2.41*10-5)/4.17]*100 = 0.0006% assumption < 5%
PS #5, cont#19. To obtain cleaner fuel from coal, a water-gas shift rxn is used. CO(g) + H2O(g) <-----> CO2(g) + H2(g)
@ equilb: [CO] = [H2O] = [H2] = 0.10 M & [CO2] = 0.40 M. 0.60 mol H2 is added to the 2.0-L vessel & new equilbr reached.What are new equilbr concentrations?
Solution: find Kc, find new initial [H2], construct I.C.E. table, find new [ ]s
4.0 0][0.10][0.1
0][0.40][0.1
O][CO][H
]][H[CO K
2
22c
CO(g) + H2O(g) <-----> CO2(g) + H2(g)
Initial 0.10 0.10 0.40 0.40 Change +X +X -X -X
Equilibrium 0.10+X 0.10+X 0.40-X 0.40-X
[H2] = 0.10 M + (0.60 mol/2.0-L) = 0.40 M
CO(g) + H2O(g) <-----> CO2(g) + H2(g)
Initial 0.10 0.10 0.40 0.40 Change +X +X -X -X
Equilibrium 0.10+X 0.10+X 0.40-X 0.40-X
2
2
2
22c
X)(0.10
X)-(0.40 4.0
X]X][0.10[0.10
X]-X][0.40-[0.40
O][CO][H
]][H[CO K
X0.10
X-0.40 2.0
)X10.0(
)X40.0( 0.4
2
2
2.0(0.10+X) = 0.40-X
X = 0.067
CO(g) + H2O(g) <-----> CO2(g) + H2(g)
Equilibr 0.10+0.067 0.40-0.067 [CO]=[H2O] = 0.167 M [CO2]=[H2]= 0.333 M
PS#6 #20. Phosgene (COCl2) that forms from CO & Cl2 at high temps. CO(g) + Cl2(g) <-----> COCl2(g)
0.350 mols of each reactant placed in 0.500-L flask @ 600 K.What are all [ ]s @ equilibrium? Kc = 4.95
CO(g) + Cl2(g) <-----> COCl2(g)
Initial 0.700 0.700 0.0 Change -X -X +X
Equilibrium 0.700-X 0.700-X X
Solution: find initial [CO & Cl2], construct I.C.E. table, find new [ ]s 0.350mol/0.5 L =0.700
0.490)1.400X-(X
(X) 4.95
X]X][0.700[0.700
[X]
][CO][Cl
][COCl K
22
2c
CO(g) + Cl2(g) <-----> COCl2(g)
Initial 0.700 0.700 0.0 Change -____ -____ +____
Equilib 0.700-___ 0.700-____ _____
a2
ac4bb- x
2
0.412 & 1.19 )4255.2(2
)4255.2)(95.4(4(-7.93)(-7.93)- x
2
4.95X2 - 7.93X + 2.4255 = 0
Check solutions for viability
1.19 value not possible, since 0.700 - X result in “-” value
Equilib 0.700-0.412 0.700-0.412 +0.412 [CO]=[Cl2] = _______ M _______ M = [COCl2]