CH15 Chemical Equilibrium

8/13/2019 CH15 Chemical Equilibrium

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Chapter 15:Chemical Equilibrium

Chemistry: The Molecular Nature

of Matter, 6E

Jespersen/Brady/Hyslop


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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 2

Dynamic Equilibrium in ChemicalSystems

Chemical equilibrium exists when

Rates of forward and reverse reactions are equal

Reaction appears to stop

[reactants] and [products] don't change overtime

Remain constant

Both forward and reverse reaction never cease Equilibrium signified by double arrows ( )

or equal sign (=)


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Dynamic Equilibrium

N2O4 2 NO2

Initially forward reaction rapid

As some reacts [N2O4] so rate forward

Initially Reverse reaction slow

No products

As NO2forms

Reverse rate

Ions collide more frequently as [ions]

Eventually rateforward= ratereverse

Equilibrium


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Dynamic Equilibrium

Almost all systems come to equilibrium

Where equilibrium lies depends on system

Some systems equilibrium hard to detect

Essentially no reactants or no products present

(Fig. 15.1)


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Reaction ReversibilityClosed system

Equilibrium can bereached from eitherdirection

Independent of whetherit starts with reactantsor products

Alwayshave the samecomposition atequilibrium under same

conditionsN

2O

4 2 NO

2


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For given overall system composition

Always reach same equilibrium concentrations

Whether equilibrium is approached from forwardor

reversedirection

N2O4 2 NO2

Reactants ProductsEquilibrium


7/97Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E 7

Equilibrium

Simple relationship among [reactants] and

[products] for any chemical system atequilibrium

Called = mass action expression

Derived from thermodynamics Forward reaction:A B Rate = kf[A] Reverse reaction:A B Rate = kr[B]

At equilibrium: A B kf[A] = kr[B] rate forward= rate reverse

rearranging:constant

[A]

[B]

r

f

k

k

kf

kr



Ex: H2(g)+ I2(g) 2HI(g) 440C

Expt

#InitialAmts

Equilm

Amts

Equilm

[M]

I 1.00molH2 0.222 molH2 0.0222 MH2

10 L 1.00mol I2

0.222 molI2

0.0222 MI2

0.00 mol HI 1.56 molHI 0.156 MHI

II 0.00 molH2 0.350 molH2 0.0350 MH2

10 L 0.100 mol I2 0.450 molI2 0.0450 MI2

3.50mol HI 2.80 molHI 0.280 MHI



Ex: H2(g)+ I2(g) 2HI(g) 440C

Expt#

Initial Amts EquilmAmts

Equilm[M]

III 0.0150molH2 0.150 molH2 0.0150 MH2

10 L 0.00 mol I2 0.135 molI2 0.0135 MI21.27 mol HI 1.00 molHI 0.100 MHI

IV 0.00 molH2 0.442 molH2 0.0442 MH2

10 L 0.00 mol I2 0.442 molI2 0.0442 MI2

4.00mol HI 3.11 molHI 0.311 MHI



Mass Action Expression (MAE)

Uses stoichiometric coefficients as exponent

for each reactant

For reaction: aA + bB cC + dD

Reaction quotient

Numerical value of mass action expression

Equals Q at any time, and

Equals K only when reaction is known to be atequilibrium

ba

dc

[B][A] [D][C]Q



Mass Action Expression

= same for all data sets at equilibrium]][I[H

[HI]

Q 22

2

]][I[H

[HI]Q

22

2

4.49)0222.0)(0222.0(

)156.0( 2

8.49)0450.0)(0350.0(

)280.0( 2

4.49)0135.0)(0150.0(

)100.0( 2

5.49)0442.0)(0442.0(

)311.0( 2

Equilibrium Concentrations(M)

Expt [H2

] [I2

] [HI]

I 0.0222 0.0222 0.156

II 0.0350 0.0450 0.280

III 0.0150 0.0135 0.100

IV 0.0442 0.0442 0.311

Average = 49.5



Equilibrium Law

For reaction at equilibrium write the following

Equilibrium Law (at 440 C)

Equilibrium constant= Kc= constant at given T

Use Kcsince usually working with concentrations inmol/L

For chemical equilibriumto exist in reactionmixture, reaction quotient Q must be equal toequilibrium constant, Kc

5.49]][I[H

[HI]

22

2

cK



Predicting Equilibrium Law

For general chemical reaction:

dD + eE fF + gG

Where D, E, F, and Grepresent chemical formulas

d, e, f, and gare coefficients

Mass action expression =

Note: Exponentsin mass action expression

are stoichiometric coefficientsin balancedequation.

Equilibrium lawis:

ed

gf

[E][D]

[G][F]

ed

gf

[E][D]

[G][F]cK



Predicting Equilibrium Law

Where only concentrations that satisfy this

equation are equilibriumconcentrations Numerator

Multiply [products] raised to their

stoichiometric coefficients Denominator

Multiply [reactants] raised to their

stoichiometric coefficients

is scientists conventiond

f

cK

]reactants[

]products[



Ex. Equilibrium Law

3 H2(g) + N2(g) 2 NH3(g)

Kc= 4.26 x 108at 25 C

What is equilibrium law?

8

23

2

23 1026.4][N][H

][NH cK



Learning Check

Write mass action expressions for the following:

2 NO2 (g) N2O4 (g)

2CO (g)+ O2 (g) 2 CO2 (g)

22

42

][NO

]O[NQ

][O[CO]

][COQ

22

22


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Manipulating Equations for ChemicalEquilibria

Various operations can be performed onequilibrium expressions

1. When directionof equation is reversed,

new equilibriumconstantis reciprocaloforiginal

A + B C + D

C +D A + B

c

c

K

K1

[C][D]

[A][B]

[A][B]

[C][D]cK



Ex. Manipulating Equilibria 1

1.When directionof equation is reversed, new

equilibriumconstantis reciprocalof original

3 H2(g) + N2(g) 2 NH3(g) at 25C

2 NH3(g) 3 H2(g)+ N2(g) at 25C

9

823

23

2 1035.21026.4

11

][NH

][N][H

c

cK

K

8

23

2

23 1026.4

][N][H][NH cK



Manipulating Equilibria 2

2.When coefficients in equation are

multiplied by a factor, equilibriumconstant is raisedto a powerequal to thatfactor.

A + B C + D

3A + 3B 3C + 3D

[A][B]

[C][D]cK

3

33

33

[A][B]

[C][D]

[A][B]

[C][D]

[A][B]

[C][D]

[B][A]

[D][C]cc KK




2. When coefficients in equation are multiplied by

factor, equilibrium constant is raisedto powerequal to that factor

3 H2(g) + N2(g) 2 NH3(g) at 25C

multiply by 3

9 H2(g)+ 3 N2(g) 6 NH3(g)

8

23

2

23 1026.4][N][H

][NH cK

3

3

2

9

2

63

][N][H

][NHcc KK


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3. When chemical equilibriaare added, their

equilibriumconstantsare multiplied

][CO][NO

]][CO[NO

][NO

][NO][NO

3

222

2

3

2 NO2(g) NO3(g)+ NO(g)

NO3(g)+ CO(g) NO2(g)+ CO2(g)

NO2(g)+ CO(g) NO(g)+ CO2(g)

22

3

][NO

][NO][NO1

cK

][CO][NO

]][CO[NO

322

2 cK

][CO][NO

][NO][CO

2

23

cK

Therefore 321 ccc KKK


][CO][NO

][NO][CO

2

2


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Learning CheckFor: N2(g) + 3 H2(g) 2 NH3(g)

Kc = 500at a particular temperature.What would be Kcfor following?

2 NH3(g) N2(g) + 3 H2(g)

N2(g) + 3/2 H2(g) NH3(g)

50011

cc

KK

22.4

0.002

50021

cc KK


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Equilibrium Constant, Kc Constant value equal to ratio of product

concentrations to reactant concentrationsraised to their respective exponents

Changes with temperature (vant HoffEquation)

Depends on solution concentrationsAssumes reactants and products are in

solution

d

f

c

K

]reactants[

]products[


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Equilibrium Constant, Kp Based on reactions in which substances are

gaseous

Assumes gas quantities are expressed inatmospheres in mass action expression

Use partial pressures for each gas in place ofconcentrations

Ex. N2

(g)

+ 3 H2

(g)

2 NH3

(g)

3

HN

2NH

22

3

PP

PPK


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How are Kpand Kc Related? Start with Ideal Gas Law

PV=nRT

Rearranging gives

Substituting P/RTfor molar concentrationinto Kcresults in pressure-based formula

n = moles of gas in productmoles ofgas in reactant

ncp RTKK

)(

MRTRTV

nP


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Learning Check

Consider the reaction: 2NO2(g) N2O4(g)

If Kp=0.480for the reaction at 25C, what isvalue of Kcat same temperature?

n = nproductsnreactants= 12 =1n

cp (RT)KK

1np

c)2980821.0(

480.0(RT)

KK

K

Kc= 11.7


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Jespersen/Brady/HyslopChemistry:TheMolecularNatureofMatter,6E

Your Turn!Consider the reactionA(g)+ 2B(g) 4C(g)

If the Kcfor the reaction is 0.99 at 25C, whatwould be the Kp?

A. 0.99

B. 2.0C. 24.

D. 2400

E. None of these

n=(4

3)=1

Kp= Kc(RT)n

Kp= 0.99*(0.082057*298.15)1

Kp= 24

29


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Homogeneous reaction/equilibrium

All reactants and products in same phase

Can mix freely

Heterogeneous reaction/equilibrium

Reactants and products in different phases

Cant mix freely

Solutions are expressed in M

Gases are expressed in M

Governed by Kc


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Heterogeneous Equilibria

2NaHCO3(s) Na2CO3(s)+ H2O(g)+ CO2(g)

Equilibrium Law =

Can write in simpler form

For any pure liquid or solid, ratio of moles tovolume of substance (M) is constant

Ex. 1 mol NaHCO3occupies 38.9 cm3

2 mol NaHCO3occupies 77.8 cm3

2)(3

)(2)(2)(32

][

]][][[

s

ggs

NaHCO

COOHCONaK

MM 7.25L0.0389

NaHCOmol1 3

MM 7.25L0.0778

NaHCOmol2 3


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Heterogeneous Equilibria2NaHCO3(s) Na2CO3(s)+ H2O(g)+ CO2(g)

Ratio (n/V) or M of NaHCO3is constant (25.7 mol/L)regardless of sample size

Likewise can show that molar concentration ofNa

2

CO3

solid is constant regardless of sample size

So concentrations of pure solids and liquidscan be incorporated into equilibrium constant,Kc

Equilibrium law for heterogeneous system writtenwithout concentrations of pure solids or liquids

]CO][OH[]NaHCO[

]CONa[)(2)(22

)(3

)(

32 ggs

s

c KK


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Learning Check

Write equilibrium laws for the following:

Ag+(aq)+ Cl(aq) AgCl(s)

H3PO4(aq)+ H2O() H3O+(aq)+ H2PO4

(aq)

]][Cl[Ag

1

cK

]PO[H

]PO][HO[H

43

423

cK


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Your Turn!Given the reaction:

3Ca2+(aq) + 2PO43(aq) Ca3(PO4)2(s)What is the mass action expression?

23

4

32

234

32243

234

32

243

234

32

][PO][Ca

]1[QD.

][PO][Ca])(POCa[QC.

]1[

][PO][CaQB.

])PO([Ca

]PO[][CaQA.

34


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Your Turn!Given the reaction:

3Ca2+(aq) + 2PO43(aq) Ca3(PO4)2(s)What is mass action expressionfor the reverse

reaction?

23432

234

32

243

234

32

243

234

32

]PO[]Ca[

]1[QD.

]PO[]Ca[

])(PO[CaQC.

]1[

]PO[]Ca[QB.

])(PO[Ca

]PO[]Ca[

QA.

35


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Interpreting KC

Large K (K>>1)

Means product rich mixture

Reaction goes far towardcompletion

Ex.

2SO2(g) + O2(g) 2SO3(g)

Kc= 7.0 1025at 25 C

1

100.7

][O][SO

][SO 25

2

2

2

23 cK


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Interpreting KC

Small K (K


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Interpreting KC

K1

Means product andreactant concentrations

close to equal Reaction goes only ~

halfway


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Size of K gives measure of howreaction proceeds

K >> 1 [products] >> [reactants]

K = 1 [products] = [reactants]

K


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Learning Check

Consider the reaction of 2NO2(g) N2O4(g)

If Kp= 0.480 at 25C, does the reaction favorproduct or reactant?

Kis small (K < 1)

Reaction favors reactantSince K is close to 1, significant amounts of

bothreactant and product are present

ilib i i i d Shif


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Equilibrium Positions and Shifts

Equilibrium positions

Combination of concentrations that allow Q = K

Infinite number of possible equilibrium positions

Le Chteliers principle System at equilibrium (Q = K) when upset by

disturbance (Q K) will shift to offset stress

System said toshift to rightwhen

forwardreaction is dominant (Q < K) System said toshift to leftwhen reverse

direction is dominant (Q > K)

R l ti hi B t Q d K


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Relationship Between Q and K

Q = K reaction at equilibrium

Q < K reactants products Too many reactants

Must convert some reactant to product to movereaction toward equilibrium

Q > K reactants products Too many products Must convert some product to reactant to move

reaction toward equilibrium

E l f L Cht li P i i l


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Examples of Le Chteliers Principle

Lets see how this works with changes in

1. Concentration

2. Pressure and volume

3. Temperature

4. Catalysts

5. Adding inert gas to system at constant

volume

1 Eff t f Ch i C t ti


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1. Effect of Change in ConcentrationCu(H2O)4

2+(aq)+ 4Cl(aq) CuCl42(aq)+ 4H2O

blue yellow Equilibrium mixture is blue-green

Add excess Cl(conc HCl) Equilibrium shifts to products

Makes more yellow CuCl42

Solution becomes green

4)()(

2

42

42)(

24

]Cl][O)Cu(H[

]OH][CuCl[

aqaq

aqcK

4)()(

242

)(24

42 ]Cl][O)Cu(H[

]CuCl[

]OH[ aqaq

aqcc

KK

1 Eff t f Ch i C t ti


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1. Effect of Change in Concentration

Cu(H2O)42+(aq)+ 4Cl(aq) CuCl4

2(aq)+ 4H2O

blue yellow

Add Ag+ Removes Cl: Ag+(aq)+ Cl(aq) AgCl(s) Equilibrium shifts to reactants

Makes more blue Cu(H2O)42+

Solution becomes bluer

Add H2O?

4)()(

242

)(24

]Cl][O)Cu(H[

]CuCl[

aqaq

aqcK

Effect of Change in Concentration


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Effect of Change in Concentration 2SO2(g) + O2(g) 2SO3(g)

Kc= 2.4 x 10-3 at 700 oC Which direction will the reaction move if

0.125 moles of O2is added to an equilibrium

mixture ?A. Towards the products

B. Towards the reactants

C. No change will occur

46

Eff t f Ch i C t ti


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Effect of Change in Concentration

When changing concentrations of reactants

or products Equilibrium shifts to removereactants or

products that have been added

Equilibrium shifts to replacereactants orproducts that have been removed

Eff t f P d V l Ch


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Effect of Pressure and Volume Changes

Consider gaseous system at constant Tand n

3H2(g) + N2(g) 2NH3(g) If reduce volume (V)

Expect Pressure to increase (P) To reduce pressure, look at each side of reaction Which has less moles of gas

Reactants = 3 + 1 = 4 moles gas

Products = 2 moles gas Reaction favors products (shifts to right)

3HN

2NH

22

3

PP

PKP

Eff t f P d V Ch


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Effect of P and V Changes

Consider gaseous system at constant Tand n

Ex. H2(g) + I2(g) 2 HI(g)

If pressure is increased, what is the effect onequilibrium?

nreactant= 1 + 1 = 2

nproduct= 2

Predict no change or shift in equilibrium

22 IH

2HI

PP

PKP

Eff t f P d V Ch


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2NaHSO3(s) NaSO3(s)+ H2O(g)+ SO2(g)

If you decrease volume of reaction, what is

the effect on equilibrium? Reactants: no moles gas = all solids

Products: 2 moles gas

V, causes P Reaction shifts to left (reactants), as this has

fewer moles of gas


22 SOOH PPKP



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Reducing volume of gaseous reaction

mixture causes reaction to decrease numberof molecules of gas, if it can

Increasing pressure

Moderate pressure changes have negligibleeffect on reactions involving only liquidsand/or solids

Substances are already almost incompressible Changes in V, P and [X] effect position of

equilibrium (Q), but not K


Effect of Change in Temperature


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Effect of Change in Temperature

Cu(H2O)42+(aq)+ 4Cl(aq) CuCl4

2(aq)+ 4H2O

blue yellow Reaction endothermic

Adding heat shifts equilibrium toward products

Cooling shifts equilibrium toward reactants 52

Ice

water

Boiling

water

Effect of Temperature Changes


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Effect of Temperature ChangesH2O(s) H2O() H=+6 kJ (at 0 C)

Energy + H2O(s) H2O()

Energy is reactant

Add heat, shift reaction right

3H2(g)+ N2(g) 2NH3(g) Hf=47.19 kJ

3 H2(g)+ N2(g) 2 NH3(g)+ energy

Energy is product Add heat, shift reaction left



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Tshifts reaction in direction that producesendothermic(heat absorbing) change

T shifts reaction in direction that producesexothermic(heat releasing) change

Changes in T change value of mass actionexpression at equilibrium, so K changed

K depends on T

T of exothermic reaction makes K smaller

More heat (product) forces equilibrium to reactants

T of endothermic reaction makes K larger

More heat (reactant) forces equilibrium to products

Catalysts And Equilibrium


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Catalysts And Equilibrium

Catalyst lowers Ea

for both forwardand reversereaction

Change in Eaaffects rates krand kfequally

Catalysts have noeffect onequilibrium

Effect of Adding Inert Gas


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Effect of Adding Inert Gas

Inert gas

One that does not react with components ofreaction

Ex. Argon, Helium, Neon, usually N2

Adding inert gas to reaction at fixedV(n andT), Pof all reactants and products

Since it doesnt react with anything

No change in concentrations of reactants orproducts

No net effect on reaction

How to Use Le Chteliers Principle


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How to Use Le Chtelier s Principle

1. Write mass action expression for reaction

2. Examine relationship between affectedconcentration and Q (direct or indirect)

3. Compare Q to K

If change makes Q > K, shifts Left

If change makes Q < K, shifts Right

If change has no effect on Q, no shift expected

Learning Check:


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Learning Check:

Consider:

H3PO4(aq)+ 3OH(aq) 3H2O()+ PO43(aq)

What will happen if PO43is removed?

Q is proportional to [PO43

] [PO4

3], Q

Q < K equilibrium shifts to right

]POH[]OH[

]PO[Q

433

34

Learning Check:


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Learning Check:The reaction

H3PO4(aq)

+ 3OH

(aq)

3H2O(aq)

+ PO43(aq)

is exothermic.

What will happen if system is cooled?

Since reaction is exothermic, heat is product

Heat is directly proportional to Q

T, Q

Q < K equilibrium shifts to right

heat

]POH[]OH[]PO[Q

433

34

Your Turn!


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Your Turn!The equilibrium between aqueous cobalt ion

and the chlorine ion is shown:[Co(H2O)6]

2+(aq) + 4Cl(aq) [Co(Cl)4]2(aq) + 6H2O()

pink blue

It is noted that heating a pink sample causes itto turn violet.

The reaction is:

A. endothermicB. exothermic

C. cannot tell from the given

information 60

Your Turn!


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Your Turn!

The following are equilibrium constants for the

reaction of acids in water, Ka. Which is themost efficient reaction?

A. Ka= 2.2103

B. Ka= 1.8105

C. Ka= 4.01010

D. Ka= 6.3103

61

Equilibrium Calculations


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Equilibrium Calculations

For gaseous reactions, use either KPor KC

For solution reactions, must use KC

Either way, two basic categories ofcalculations

1. Calculate Kfrom known equilibriumconcentrations or partial pressures

2. Calculate one or more equilibrium

concentrations or partial pressures usingknown KPor KC

Calculating KC from Equilibrium


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Calculating KCfrom EquilibriumConcentrations

When all concentrations at equilibrium areknown

Use mass action expression to relateconcentrations to KC

Two common types of calculationsA. Given equilibrium concentrations, calculate K

B. Given initial concentrations and one final

concentration Calculate equilibrium concentration of

all other species

Then calculate K

Calculating KC Given Equilibrium


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Calculating KCGiven EquilibriumConcentrations

Ex. 1 N2O4(g) 2NO2(g) If you place 0.0350 mol N2O4in 1 L flask at

equilibrium, what is KC?

[N2O4]eq= 0.0292 M [NO2]eq= 0.0116 M

][ ][ 42

2

2ONNOK

c ]0292.0[ ]0116.0[

2

cK

KC= 4.61 103

Your Turn!


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Your Turn!

For the reaction: 2A(aq)+ B(aq) 3C(aq)

the equilibrium concentrations are: A = 2.0 M,B = 1.0 M and C = 3.0 M. What is theexpected value of Kcat this temperature?

A. 14B. 0.15

C. 1.5

D. 6.75

[B][A]

[C]2

3cK

[1.0][2.0][3.0]2

3

cK

65

Calculating KCGiven Initial Concentrations


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g Cand One Final Concentration

Ex. 2 2SO2(g) + O2(g) 2SO3(g) 1.000 mol SO2and 1.000 mol O2are placed in

a 1.000 L flask at 1000 K. At equilibrium0.925 mol SO

3

has formed. Calculate KC

forthis reaction.

1stcalculate concentrations of each

Initial

Equilibrium

ML

mol

OSO 00.100.1

00.1

][][ 22

M

L

molSO 925.0

00.1

925.0][ 3

How to Solve:


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How to Solve:

Set up Concentration Table

Based on the following: Changes in concentration must be in same ratio

as coefficients of balanced equation

Set up table under balanced chemical equation

Initial concentrations Controlled by person running experiment

Changes in concentrations

Controlled by stoichiometry of reaction Equilibrium concentrations

EquilibriumConcentration

=Initial

Concentration

Change inConcentration

Next Set up Concentration Table


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Next Set up Concentration Table

2SO2(g) + O2(g) 2SO3(g)

Initial Conc. (M) 1.000 1.000 0.000

Changes in Conc. (M)

Equilibrium Conc. (M)

[SO2] consumed = Amt of SO3formed= [SO3] at equilibrium = 0.925 M

[O2] consumed = amt SO3formed

= 0.925/2 = 0.462 M[SO2] at equilibrium = 1.0000.975 = 0.075

[O2] at equilibrium = 1.000.462 = 0.538 M

0.925 0.462 +0.925

0.075 0.538 0.925

Ex 2


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Ex. 2

Finally calculate KCat 1000 K

][O][SO

][SO

22

2

23

c K

]538.0[]075.0[

]925.0[2

2

cK

Kc= 2.8 102 = 280

Summary of Concentration Table


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Used for most equilibrium calculations (Ch

15, 17, and 18)1. Equilibrium concentrations are only values

used in mass action expression

Values in last row of table

2. Initial value in table must be in units ofmol/L (M)

[X]initial= those present when reactionprepared

No reaction occurs until everything ismixed



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Summary of Concentration Table3. Changes in concentrations always occur in

same ratio as coefficients in balancedequation

4. In change row be sure all [reactants]change in same directions and all [products]change in opposite direction.

If [reactant]initial= 0, its change must be + ()because [reactant]finalnegative

If [reactants] , all entries for reactants in changerow should have minus sign and all entries forproducts should be positive

Calculate [X]equilibriumfrom Kc and [X]initial


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[ ]equilibrium c [ ]initial

When all concentrations but one are known

Use mass action expression to relate Kcand known concentrations to obtainmissing concentrations

Ex. 3 CH4(g)+ H2O(g) CO(g)+ 3H2(g)At 1500 C, Kc = 5.67. An equilibrium

mixture of gases had the following

concentrations: [CH4] = 0.400 M and[H2] = 0.800M and [CO] =0.300M.What is [H2O] at equilibrium ?

Calculate [X]equilibriumfrom Kc and [X]initial


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q

Ex. 3CH4(g)+ H2O(g) CO(g)+ 3H2(g)Kc = 5.67

[CH4] = 0.400 M; [H2] = 0.800M; [CO] =0.300M What is [H2O] at equilibrium?

First, set up equilibrium

Next, plug in equilibrium concentrations and Kc

O]][H[CH ][CO][H 24

3

2cK

[H2O] = 0.0678 M

27.2154.0

5.67)([0.400]800][0.300][0.O][H

3

2

cK][CH][CO][HO][H

4

3

22

Calculating [X]Equilibrium from Kc


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Calculating [X]Equilibriumfrom Kc

When InitialConcentrations Are Given

Write equilibrium law/mass action expression

Set up Concentration table

Allow reaction to proceed as expected,using x to represent change inconcentration

Substitute equilibrium terms from table into

mass action expression and solve

Calculate [X]equilibrium from [X]initial and KC


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Calculate [X]equilibriumfrom [X]initialand KC

Ex. 4 H2(g)+ I2(g) 2HI(g) at 425 C

KC= 55.64

If one mole each of H2and I2are placed in a0.500 L flask at 425 C, what are the

equilibrium concentrations of H2, I2and HI?

Step 1. Write Equilibrium Law

64.55]][[

][22

2

IH

HIKc

Ex. 4 Step 2. Concentration Table


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Ex. 4 Step 2.Concentration Table

Conc (M) H2(g) + I2(g) 2HI(g)

Initial 2.00 2.00 0.000

Change

Equilm

Initial [H2] = [I2] = 1.00 mol/0.500L =2.00M

Amt of H2consumed = Amt of I2consumed = x

Amt of HIformed = 2x

x +2xx

+2x2.00x2.00x

2

22

)00.2(

)2(

)00.2)(00.2(

)2(64.55

x

x

xx

x

Ex. 4 Step 3. Solve for x


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Both sides are squared so we can take

square root of both sides to simplify

2

2

)00.2(

)2(64.55

x

xK

)00.2(

2459.7

x

x

xx 2)00.2(459.7

xx 2459.7918.14

58.1459.9

918.14x

x459.9918.14

Ex. 4 Step 4. Equilibrium Concentrations


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Initial 2.00 2.00 0.00

Change

Equilm

[H2]equil= [I2]equil= 2.001.58 = 0.42 M

[HI]equil= 2x = 2(1.58) = 3.16

1.58 +3.161.58

+3.160.420.42



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[ ]equilibrium [ ]initial C

Ex. 5 H2(g)+ I2(g) 2HI(g) at 425 C

KC= 55.64

If one mole each of H2, I2and HIare placedin a 0.500 L flask at 425 C, what are the

equilibrium concentrations of H2, I2and HI?

Now have product as well as reactants initially

Step 1.Write Equilibrium Law

64.55]][[

][

22

2

IH

HIKc

Ex. 5 Step 2. Concentration Table


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p


Initial 2.00 2.00 2.00

Change

Equilm

x +2xx

2.00 + 2x2.00x2.00x

2

22

)00.2(

)200.2(

)00.2)(00.2(

)200.2(64.55

x

x

xx

x

2

2

)00.2(

)200.2(64.55

x

xK



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Ex. 5 Step 3.Solve for x

)00.2(

200.2459.7

x

x

xx 200.2)00.2(459.7

xx 200.2459.7918.14

37.1459.9

918.12

x

x459.9918.12 [H2]equil= [I2]equil= 2.00 x =2.00 1.37 = 0.63 M

[HI]equil= 2.00 + 2x= 2.00 + 2(1.37)

= 2.00 + 2.74

= 4.74 M

Your Turn!


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N2(g) + O2(g) 2NO(g)

Kc= 0.0123 at 3900o

CIf 0.25 moles of N2and O2are placed in a250 mL container, what are the equilibrium

concentrations of all species ?A. 0.0526 M, 0.947 M, 0.105 M

B. 0.947 M, 0.947 M, 0.105 M

C. 0.947 M 0.105 M, 0.0526 M D. 0.105 M, 0.105 M, 0.947 M

82

Your Turn! - Solution


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Conc (M) N2(g) +O2(g) 2NO(g) Initial 1.00 1.00 0.00

Change x x + 2x

Equil 1.00x1.00x + 2x

83

2 2

2

2

0.250 mol[N ] [O ] 1.00

0.250 L

(2 ) 20.0123 0.01231(1 )

0.0526 [NO] = 2 = 0.105

M

x xxx

x M x M



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[ ]equilibrium [ ]initial C

Ex. 6

CH3CO2H(aq)+ C2H5OH(aq) CH3CO2C2H5(aq)+acetic acid ethanol ethyl acetate H2O()

KC= 0.11

An aqueous solution of ethanol and aceticacid, each with initial concentration of 0.810M, is heated at 100 C. What are the

concentrations of acetic acid, ethanol andethyl acetate at equilibrium?

Ex. 6Step 1.Write Equilibrium Law


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p q

Need to find equilibrium values that satisfy

thisStep 2: Set up concentration table using x

for unknown

Initial concentrations Change in concentrations

Equilibrium concentrations

11.0H]COOH][CHH[C

]HCCO[CH

2352

5223 c

K

Ex. 6Step 2.Concentration Table


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p

(M) CH3CO2H(aq)+ C2H5OH(aq) CH3CO2C2H5(aq)+ H2O(l)

I 0.810 0.810 0.000

C

E

Amt of CH3CO2H consumed = Amt of C2H5OHconsumed =x

Amt of CH3CO2C2H5formed = + x

[CH3CO2H]eqand [C2H5OH ] = 0.810x [CH3CO2C2H5] = x

x +xx

+x0.810 x0.810 x

)8100)(8100(110

x.x.

x.



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p

Rearranging gives

Then put in form of quadratic equation

ax2+ bx+ c = 0

Solution for quadratic equation given by

xxx )62.16561.0(11.0 2

011.01782.007217.0 2 xxx

007217.01782.111.0 2 xx

a

acbbx

2

42



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p

This gives two roots: x = 10.6 and x =0.064

Only x = 0.064 is possible x = 10.6 is >> 0.810 initial concentrations

0.81010.6 = negative concentration,which is impossible

)11.0(2

)07217.0)(11.0(4)1782.1()1782.1( 2

x

22.0

164.11782.1

22.0

)032.0()388.1(1782.1

x

Ex. 6 Step 4. Equilibrium Concentrations


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(M) CH3CO2H(aq)+ C2H5OH(aq) CH3CO2C2H5(aq)+ H2O(l)

I 0.810 0.810 0.000

C

E

[CH3CO2C2H5]equil= x = 0.064 M

[CH3CO2H]equil= [C2H5OH]equil= 0.810 Mx= 0.810 M0.064 M= 0.746 M

0.064 +0.0640.064

+0.0640.7460.746



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When KCis very small

Ex. 7 2H2O(g) 2H2(g)+ O2(g)At 1000 C, KC= 7.3 10

18

If the initial H2O concentration is 0.100M,

what will the H2concentration be atequilibrium?

Step 1. Write Equilibrium Law

18

22

22

2 103.7O][H

][O][H cK

Ex. 7 Step 2.Concentration Table


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Conc (M) 2H2O(g) 2H2(g) + O2(g)

Initial 0.100 0.00 0.00Change

Equilm

2x +x+2x

+x+2x0.1002x

2

3

2

218

)2100.0(

4

)2100.0(

)2(103.7x

x

x

xx

Cubic equationtough to solve

Make approximation KCvery small, soxwill be very small

Assume we can neglectx

must prove valid later


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Ex. 7 Step 3.Solve for x


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Now take cubic root

xis very small

0.1002(2.6107) = 0.09999948

Which rounds to 0.100 (3 decimal places) [H2] = 2x = 2(2.610

7)= 5.2107M

2020

3 108.1

4

103.7

x

73 20 106.2108.1 x

Simplifications: When Can Youi i i l (C )?


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Ignore x in Binomial (Cix)? If equilibrium law gives very complicated

mathematical problems

And if Kis small

Change to reach equilibrium (xterm) is also small

Compare initial concentration Ciin binomial tovalue of K

Use proof to show that droppedxterm wassufficiently small

05.0termxdropped ?

iC

400K

Ci

Learning Check


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For the reaction 2A(g)B(g)given that Kp = 3.510

16at 25C, and we place 0.2

atm A into the container, what will be the pressure of Bat equilibrium?

2A B

I 0.2 0 atm

C 2x +x

E 0.22x x

2

NO

ON

P

PQ 42

2

16

)2.0(

x105.3

x = 1.41017

[B]= 1.41017 M

proof: 1.41017/0.2


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In the reaction shown, K = 1.8 105

HC2H3O2(aq) + H2O(l)H3O+(aq) + C2H3O2(aq)

If we start with 0.3M HC2H3O2, what will be theequilibrium concentration of C2H3O2

?

A. 0.3 MB. 0.002 M

C. 0.04 M

D. 0.5 M

96

Calculating KCGiven Initial Concentrationsd O Fi l C t ti


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and One Final Concentration

Ex. 2aH2(g)+ I2(g)2HI(g) @ 450 C Initially H2and I2concentrations are 0.200

mol each in 2.00L (= 0.100M); no HI ispresent

At equilibrium, HI concentration is 0.160M

Calculate KC

To do this we need to know 3 sets ofconcentrations: initial, change andequilibrium

CH15 Chemical Equilibrium

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