EQUILIBRIUMEQUILIBRIUM

Opposing reactions proceed at equal ratesConcs. of reactants & products do not

change over time

Examples:– vapor pressure above liquid– saturated solution

Now:equilibrium of chemical reactions

Reactions we discussed so far could be described without equilibrium.i.e. CH4 + 2 O2 CO2 + 2 H2O

But this is not true for many reactions.

VAPOR PRESSUREVAPOR PRESSUREan equilibrium process

time

rate

time

rate of evaporation

rate of condensation

dynamicequilibrium(rates equal)

DYNAMIC EQUILIBRIUMrate forward = rate backwardno net change although change is occuring

HABER PROCESSHABER PROCESSan equilibrium

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

Initial state: reactantsreactants only

Initial state:productsproducts only

Final state: ratio of products to reactantsis the same for both sets of conditions

EQUILIBRIUMEQUILIBRIUMEquilibrium reached when rate of forward reaction equals rate of reverse reaction

Example:N2O4(g) 2 NO2(g)

forwardN2O4 NO2 + NO2 Ratef = kf[N2O4]

reverseNO2 + NO2 N2O4 Rater = kr[NO2]2

Equilibrium reached when Ratef = Rater

or

kf[N2O4] = kr[NO2]2

[ ][ ]

22

2 4

NOequil. constant

N Of

r

kk

= =

Reorganizing:

EQUILIBRIUM CONSTANTEQUILIBRIUM CONSTANT

aA + bB cC + dD

Kc = [C]c [D]d

[A]a [B]b

Kp =PC

c PDd

PAa PB

b

law ofmass action

Kp = Kc (RT)Δn Δn = change in number of moles in going fromreactants to products

K has no units[ ] means mol/LP means atm

CO + 3 H2 CH4 + H2O 1200 K

forward & reverse reactions both proceed

–

–

–3

2

1

mol

es

time

H2

CO

CH4 or H2O

Kc = [CH4] [H2O][CO] [H2]3

start: 1 mol CO 3 mol H2

in 10.0 L volume

At equilibrium:0.613 mol CO1.839 mol H20.387 mol H2O & CH4

EQUILIBRIUMEQUILIBRIUM

CO + 3 H2 CH4 + H2O

Kp =PCH4 PH2O

PCO PH23

PV = nRT

P = nRT

V =nV

RT

Kp =[CH4]RT [H2O]RT[CO]RT ([H2]RT)3

Kp = [CH4] [H2O] (RT)2

[CO] [H2]3 (RT)4 = Kc

(RT)2

Kp = Kc (RT)Δn

Δn = (2) – (4) = – 2

reactants

products

T= 1200K

Kc = [CH4] [H2O][CO] [H2]3

(0.0387)(0.0387)

(0.0613)(0.1839)3= = 3.93

Kp = 3.93{(0.0821)(1200)}2 = 4.05 x 10–4

Calculate KCalculate KCC and Kand KPP

DIRECTION OF DIRECTION OF EQUILIBRIUMEQUILIBRIUM

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

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

Kc =[N2][O2]

[N2][O2]

[NO]2

[NO]2Kc =

Kc expressions are reciprocals

Same equilibrium -- can approachfrom either direction

EQUILIBIA FOREQUILIBIA FORRELATED REACTIONSRELATED REACTIONS

OLD EXAM QUESTION

At a certain T, Kc for the reactionH2 + I2 2 HI

is 16. At the same T, what is Kc forthe reaction

HI H212

12 + I2

A. 1/16B. 4C. 1/4D. 16E. None of these

MORE EQUILIBRIUM PROBLEMSMORE EQUILIBRIUM PROBLEMS

KC for the reaction 2 H2O2(g) 2 H2O(g) + O2(g) is given by

A. D.

B. E.

C.

[ ] [ ][ ]

22 2

22 2

H O O

H O

[ ] [ ][ ]2 2

2 2

H O OH O

[ ] [ ][ ]

12

2 2

2 2

H O OH O

[ ][ ] [ ]

22 2

22 2

H O

H O O

[ ][ ] [ ]

2 2

2 2

H OH O O

MORE EQUILIBRIUM PROBLEMSMORE EQUILIBRIUM PROBLEMS

For which of the following reactions does KC = KP?

I. CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

II. CO(g) + H2O(g) CO2(g) + H2(g)

III. C2H4(g) + 3 O2(g) 2 CO2(g) + 2 H2O(g)

A. I and II

B. I and III

C. II and III

D. I, II, and III

E. none of the above

GENERAL APPROACH TOGENERAL APPROACH TOCALCULATE ANCALCULATE AN

EQUILIBRIUM CONSTANT EQUILIBRIUM CONSTANT

1. Write the balanced equation2. Write the general form for K3. Set up a data table

- initial conditions- changes in concentrations- equilibrium concentrations

(may need algebraic unknown)4. Substitute into K expression and solve

EQUM CONSTANTEQUM CONSTANTFROM KNOWN PRESS.FROM KNOWN PRESS.

A mixture of H2 and N2 is added to a reaction vessel at 472 oC and allowed to reach equilibrium. The equilibrium mixture of gasses was analyzed and found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data, calculate the equilibrium constants KP and KC for the reaction

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

CALCULATION OFCALCULATION OFEQUM CONSTANTEQUM CONSTANT

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

[HI] [H2] [I2]Initial 0.10 0 0

Change ↓ ↑ ↑– 2x +x +x

ProblemStatement

2.0 L flask with 0.20 mol HIAt equm: [HI] = 0.078 M

[HI] = 0.10 – 2x = 0.078 Mso x = 0.011

[H2] = [I2] = 0.011 M

Kc = [H2] [I2]

[HI]2=

(0.011)(0.011)

(0.078)2= 0.020

Equm 0.10–2x x x

CALCULATION OFCALCULATION OFEQUM CONSTANTEQUM CONSTANT

N2O4(g) 2 NO2(g)

[N2O4] [NO2]Initial 0.200 0.000

Change ↓ ↑– x +2x

ProblemStatement

1.0 L flask with 0.200 mol N2O4

At equm: [N2O4] = 0.120 M

[N2O4] = 0.200 – x = 0.120 Mso x = 0.0800

[ ][ ]

( )( )

( )( )

2 2 22

2 4

2 0.1600.200

00

.213.120c

NO xN O

Kx

= = = =−

Equm 0.200 – x +2x

1.00 mol SO2(g) and 1.00 mol O2(g) are added to a 1.00 L vessel and react until equilibrium is reached. At equilibrium, the vessel has 0.919 mol SO3(g). KC = 238 for the reaction at 25°C.

What are the equilibrium concentrations of SO2(g) and O2(g)?

What is KP for the reaction?

EQUILIBRIUM PROBLEM #1EQUILIBRIUM PROBLEM #1

Enough glacial acetic acid (pure acetic acid) is added to pure water to make a 0.200 M solution. Some of the acid dissociates according to the reaction

HC2H3O2(aq) H+(aq) + C2H3O2–(aq)

At equilibrium, the concentration of H+ is found to be 1.88 * 10–3 M. Calculate KC for the reaction. Assume that the H+ conc. in pure water is zero.

EQUILIBRIUM PROBLEM #2EQUILIBRIUM PROBLEM #2

HETEROGENEOUSHETEROGENEOUSEQUILIBRIUMEQUILIBRIUM

Heterogeneous equm: reactants andproducts in more than one phase

What is Kc? [Fe3O4][H2]4

[Fe]3[H2O]4Kc' =

3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g)

?

What is [Fe] ? [Fe3O4] ?

They are constants

Put them into the equm constant

[H2]4

[H2O]4Kc' =

[Fe]3

[Fe3O4]

Correct Kc is:[H2]4

[H2O]4Kc =

Leave solids and pure liquids out of equmconstant expressions

HETEROGENEOUSHETEROGENEOUSEQUILIBRIUMEQUILIBRIUM

EXAMPLEEXAMPLE

A 1.0 L container holds 224 g of Fe and 5.00mol H2O(l). It is heated to 1000 K and reachesequm. 56.0 g Fe are left unreacted. What is Kc ?

3 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g)

Init 4.0 5.0 0 0

[H2]4

[H2O]4Kc = =

(4.0)4

(1.0)4= 256

Final 1.0 ? ? ?Change –3.0 –4.0 +1.0 +4.0Equm 1.0 1.0 1.0 4.0

HETEROGEN. EQUILIBRIUMHETEROGEN. EQUILIBRIUMEXAMPLE #2EXAMPLE #2

Consider the reactionCO(g) + NiO(s) CO2(g) + Ni(s)

140 g of CO(g) and 523 g of NiO(s) are added to a 1.00 L reaction vessel at 500 oC and allowed to reach equilibrium. Analysis of the reaction mixtures shows that the concentration of CO(g) at equilibrium is 0.24 M.

What mass of NiO(s) remains at equilibrium? What is the value of KC?

USING AN USING AN EQUILIBRIUMEQUILIBRIUMCONSTANTCONSTANT

(1) Interpret value – does equm favorreactants or products ?

(2) Predict direction of reaction fromgiven conditions – will movetoward equlibrium

Calculate Q, compare to K(3) Calculate equm concentrations

VALUE OF KVALUE OF Kcc

Reaction goes almost completely to products

Kc ≈[Pr][Rc] Kc ≈

“big”“little”

so K → large1

Kc ≈[Pr][Rc]

Reaction doesn’t go – almost no products

Kc ≈“little”“big”

so K → small2

Kc ≈[Pr][Rc]

Both reactants and products present

[Pr] ≈ [Rc] K ≈ 1

0.1 < K < 10

3

Example of small KExample of small K

PbI2(s) Pb+2(aq) + 2 I–(aq)

K = 7.1 x 10–9

Reactants dominateReactants dominate

yellow colorless colorless

Examples of large KExamples of large K

Cu+2(aq) + 4 NH3(aq) Cu(NH3)4+2(aq)

K = 5.0 x 1012

pale blue colorless purple

Ni+2(aq) + 6 NH3(aq) Ni(NH3)6+2(aq)

K = 5.5 x 108

Products dominateProducts dominate

pale green colorless dark blue

PREDICTING THE DIRECTIONPREDICTING THE DIRECTIONOF A REACTIONOF A REACTION

aA + bB cC + dD

Q = [C]c [D]d

[A]a [B]bReactionQuotient Q

The concentrations used in Q are NOTequilibrium concentrations

When Q = Kc system is at equm

When Q < Kc reaction needs to move to rightto produce more product

When Q > Kc reaction needs to move to leftto produce more reactant

CO(g) + H2O(g) CO2 (g) + H2(g)

Kc = [CO2] [H2][CO] [H2O]

= 4.0

Initial conditions:[CO] = [H2O] = [CO2] = [H2] = 0.02 M

Is the system at equilibrium? If not, which direction will it go?

For Q → K, need larger numerator &smaller denominator

=(0.02)(0.02)(0.02)(0.02)

= 1System is

not at equmQ < Kc

Q

CO(g) + H2O(g) CO2 (g) + H2(g)

reaction will go to right

PRODEDURE TO DETERMINEPRODEDURE TO DETERMINEEQUILIBRIUM CONCS EQUILIBRIUM CONCS

• write the balanced equation for reaction

• write the equm constant expression

• list the initial conditions

• calculate Q and determine the direction ofshift to reach equm

• define the change needed to reach equmfor each species

• define the equm concs for each species(initial – change)

• substitute equm concs into equm expressionsolve for unknown

• check equm concs found by substituting theminto the equm expression

CALCULATION OFCALCULATION OFEQUILIBRIUM CONCSEQUILIBRIUM CONCS

2 IBr(g) Br2(g) + I2(g)

Kc = [Br2] [I2]

[IBr]2= 2.5 x 10–3

ProblemStatement

0.05 mol of each gas is put in a1.0 L vessel. What are the equmconcs of all three gases?

[I2] = [Br2] = 0.05 – 0.043 = 0.007 M[IBr] = 0.05 + 2(0.043) = 0.136 M

Takesquareroot

(0.05 – x)(0.05 + 2x)

= 0.05

x = 0.043

Check the answer:

(0.007)(0.007)(0.136)2

= 2.5 x 10–3

[IBr] [Br2] [I2]Initial 0.05 0.05 0.05

Change ↑ ↓ ↓2x –x –x

(0.05 – x)(0.05 – x)

(0.05 + 2x)2= 2.5 x 10–3

2 IBr(g) Br2(g) + I2(g)

Equm 0.05+2x 0.05–x 0.05–x

H2(g) + F2(g) 2 HF(g)

= 1.15 x 102

3.0 mol of each gas was added to a 1.5 L vessel. What are the equilibrium concs of all species?

Kc = [HF]2

[H2] [F2]

CALCULATION OFCALCULATION OFEQUILIBRIUM CONCSEQUILIBRIUM CONCS

[H2] = [F2] = 2.0 – x = 0.47 M[HF] = 2.0 + 2x = 5.06 M

(5.06)2

(0.47) (0.47)= 1.15 x 102

Check the answer:

H2(g) + F2(g) 2 HF(g)

Initial concs: [H2] = [F2] = [HF] = 2.0 M

Q = (HF)2

(H2)(F2)=

(2.0)2

(2.0)(2.0)= 1.0 Q < Kc

H2(g) + F2(g) 2 HF(g)

initial 2.0 2.0 2.0

1.15 x 102 = (2.0 + 2x)2

(2.0 – x) (2.0 – x)

x = 1.53

change – x – x + 2xequm 2.0 – x 2.0 – x 2.0 + 2x

CALCULATION OFCALCULATION OFEQUILIBRIUM CONCS #3EQUILIBRIUM CONCS #3

A 0.50 L reaction vessel is filled with 0.100 mole of CO(g) and H2O(l) and is then filled with 1.00 mole of CO2(g) and H2(g). The vessel is then heated to 1200 K and the mixture allowed to reach equilibrium. KC = 4.0 for the reaction.

Calculate the equilibrium concentrations of all species. Calculate partial pressure of H2.

Left blank for scratch paper

PRACTISE THE SETPRACTISE THE SET--UPUP

Consider the reaction 2 NH3(g) N2(g) + 3 H2(g). Suppose that 3 moles of pure NH3 were placed in a 1.0 L reaction vessel and allowed to reach equilibrium. If 3X represents the concentration of H2(g) at equilibrium, which of the following represents the concentration of NH3 at equilibrium in moles per liter?

A. 3 – XB. 3 – 2XC. 3 + 2XD. XE. 2X

Write the expression that you would use to solve for the equilibrium concentrations if you knew KC.You do not need to solve this expression.

Left blank for scratch paper

OLD EXAM QUESTIONOLD EXAM QUESTIONConsider this equilibrium

H2(g) + Br2(g) 2 HBr(g)

A. 0.019 atmB. 0.090 atmC. 0.235 atmD. 0.445 atmE. 0.863 atm

Kp = 0.0388 at 500 K

1.00 atm of HBr is placed in a 5.00 L vessel at500 K and allowed to reach equilibrium. What isthe partial pressure of HBr at equilibrium?

Left blank for scratch paper