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 be withdrawn continuously, then reacting system can be kept constantly off-balance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

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]