Equilibrium Unit 4 Chapters 17, 18, 19, 20. Chapter 17 Equilibrium – when two opposite reactions occur simultaneously and at the same rate Equilibrium

EquilibriumEquilibrium

Unit 4Unit 4

Chapters 17, 18, 19, 20Chapters 17, 18, 19, 20

Chapter 17Chapter 17

EquilibriumEquilibrium – when two opposite reactions – when two opposite reactions occur simultaneously and at the same rateoccur simultaneously and at the same rate

Dynamic equilibriumDynamic equilibrium – equilibrium can be – equilibrium can be shifted if one or more of the reactants is shifted if one or more of the reactants is changed in any way. The equilibrium will changed in any way. The equilibrium will “react”“react”

Chemistry ConnectionsChemistry Connections

Connection to KineticsConnection to Kinetics Equilibrium affects reaction rates, especially for Equilibrium affects reaction rates, especially for

multi-step reactions!multi-step reactions! Factors that affect equilibrium affect reaction ratesFactors that affect equilibrium affect reaction rates

Connection to ThermoConnection to Thermo If the reaction rate slows, the energy to be released If the reaction rate slows, the energy to be released

or absorbed may be released more slowly or or absorbed may be released more slowly or quickly….can be dangerous or take quickly….can be dangerous or take FOREVER!...when setting up a reaction MUST FOREVER!...when setting up a reaction MUST keep this in mind!keep this in mind!

Equilibrium Basics – Kinetics!Equilibrium Basics – Kinetics!

For a reversible reaction with a one step For a reversible reaction with a one step mechanism:mechanism:

aA + bB aA + bB cC + dD cC + dD

Rate = kRate = kff[A][A]aa[B][B]b = b = kkRR[C][C]cc[D][D]dd

Definition of equilibrium: RDefinition of equilibrium: Rff = R = Rrr

At EquilibriumAt Equilibrium

kkff[A][A]aa[B][B]b = b = kkRR[C][C]cc[D][D]dd

KKff == [C][C]cc[D][D]dd

kkR R [A][A]aa[B][B]bb

kkff / k / kRR = k = kcc

kkcc is the equilibrium constant (mass action expression) is the equilibrium constant (mass action expression) The constant can be determined by the ratio of the The constant can be determined by the ratio of the

constants or the concentrations!constants or the concentrations! Example 17-1, 17-2, 17-3Example 17-1, 17-2, 17-3

Equilibrium ConstantEquilibrium Constant

kkcc

Just as in kinetics, intermediates are not allowed, Just as in kinetics, intermediates are not allowed, so substitution may be necessary!so substitution may be necessary!

Must be found experimentally or by means of Must be found experimentally or by means of equilibrium concentrations from thermodynamic equilibrium concentrations from thermodynamic data.data.

Also varies with temperature, and constant at a Also varies with temperature, and constant at a given temperature (just like kinetic rate constants)given temperature (just like kinetic rate constants)

Independent of initial concentrationIndependent of initial concentration

Equilibrium ConstantEquilibrium Constant

Equilibrium constants are expressed as a ratio Equilibrium constants are expressed as a ratio of “activities” rather than concentrations in of “activities” rather than concentrations in order to be dimensionless.order to be dimensionless.

Activity:Activity: Pure liquid or solid = 1 (if it is pure, amt. does not Pure liquid or solid = 1 (if it is pure, amt. does not

affect)affect) Solution = the molar concentrationSolution = the molar concentration Ideal gas = partial pressure in atmIdeal gas = partial pressure in atm

Reaction QuotientReaction Quotient

Reaction quotientReaction quotient – calculated the same way – calculated the same way as kas kcc, but the reaction is NOT YET at , but the reaction is NOT YET at

equilibrium. (mass action expression)equilibrium. (mass action expression)

If reaction quotient is calculated, comparing it If reaction quotient is calculated, comparing it to kto kcc will allow prediction of the direction of will allow prediction of the direction of

the reaction.the reaction.

Comparing Quotient to KComparing Quotient to K

aA + bB aA + bB cC + dD cC + dD

[C][C]cc[D][D]d d = Q= Q

[A][A]aa[B][B]bb

When kWhen kcc > Q, reaction moves forward > Q, reaction moves forward

When kWhen kcc < Q, reaction moves reverse < Q, reaction moves reverse

Reaction QuotientReaction Quotient

Forward reaction, Forward reaction, [C][C]cc[D][D]d d = kc= kc

[A][A]aa[B][B]bb

Reverse reaction, Reverse reaction, [A][A]aa[B][B]bb = 1/kc = kc’= 1/kc = kc’

[C][C]cc[D][D]d d

Reversing the reaction, kc must be raised to -1 Reversing the reaction, kc must be raised to -1 power.power.

Manipulation of kc Multiply the equation by ½

Kc* = = = kc1/2

Whatever factor (n) is applied to the balanced equation, the new kc* is kc raised to the power of n

Example 17-4, 17-5

Dd

Cc

Bb

Aa

2222

22

22

][][

][][ba

dc

BA

DC 2

1

][][

][][

ba

dc

BA

DC

What is kc used for?What is kc used for?

Generally used to calculate concentrations at Generally used to calculate concentrations at equilibrium of various reactants and productsequilibrium of various reactants and products

Example 17-6Example 17-6 Example 17-7Example 17-7

Le Chatelier’s PrincipleLe Chatelier’s Principle

O la laO la la

Le Chatlier’s Principle (1884)Le Chatlier’s Principle (1884)

If a change in conditions (stress) is applied to a If a change in conditions (stress) is applied to a system at equilibrium, the system responds to system at equilibrium, the system responds to reduce the stress and reach a new state of reduce the stress and reach a new state of equilibrium.equilibrium.


Major stresses include:Major stresses include:

1.1. Change in ConcentrationChange in Concentration (or pressure for (or pressure for gasses)gasses)

PV = nRTPV = nRT

P = (n/V) RTP = (n/V) RT

n/V = molar concentration if V is in Ln/V = molar concentration if V is in L

At constant T, P At constant T, P concentration so it can be concentration so it can be used in its place in calculations.used in its place in calculations.

ExampleExample

HH22 (g) + I (g) + I22 (g) (g) 2HI (g) 2HI (g) What happens when you add HWhat happens when you add H22??

What happens when you remove HWhat happens when you remove H22?? What happens when you compress the container?What happens when you compress the container?

Example 17-10Example 17-10


2.2. Change in VolumeChange in Volume – same moles, but smaller V, – same moles, but smaller V, affects gas concentrations, but not liquid, solid, or affects gas concentrations, but not liquid, solid, or solution.solution.

2NO2NO22 (g) (g) N N22OO44 (g) (g)

What will happen if we decrease the volume of the What will happen if we decrease the volume of the container?container?

You can increase the pressure by pumping in an inert You can increase the pressure by pumping in an inert gas, which does not increase the concentration.gas, which does not increase the concentration.


3.3. Effect of Temperature ChangeEffect of Temperature Change – depends – depends on “thermicity” of reaction on “thermicity” of reaction

Thermicitiy is considering heat as a reactant or a Thermicitiy is considering heat as a reactant or a product and writing it into the equation.product and writing it into the equation.

C (s) + OC (s) + O22 (g) (g) CO CO22 (g) + 393.5kJ (g) + 393.5kJ

If I add heat, which way will equilibrium move?If I add heat, which way will equilibrium move?


4.4. Catalysts – Catalysts –


Catalysts decrease the activation energy of Catalysts decrease the activation energy of both the forward and reverse reaction, both the forward and reverse reaction, therefore, no net effect on equilibrium, only on therefore, no net effect on equilibrium, only on the rate of reaction.the rate of reaction.

Example Problem 17-8Example Problem 17-8 Example Problem 17-9Example Problem 17-9 Example Problem 17-10Example Problem 17-10

The Haber ProcessThe Haber Process

A Classic Example of EquilibriumA Classic Example of Equilibrium

Haber ProcessHaber Process

NN22 (g) + 3H (g) + 3H22 (g) (g) 2NH 2NH33 (g) + 92.22kJ (g) + 92.22kJ

Is the enthalpy (Is the enthalpy (H) Favorable?H) Favorable?

Is the entropy (Is the entropy (S) Favorable?S) Favorable?

Will this reaction occur spontaneously?Will this reaction occur spontaneously?

Example Problem 17-8Example Problem 17-8

Mole % NHMole % NH33 in Equilibrium Mixture in Equilibrium Mixture

ooCC KKcc 10atm10atm 100atm100atm 1000at1000atmm

2525 3.6e83.6e8 near 0near 0 near 0 near 0 near 0near 0

209209 650650 5151 8282 9898

467467 0.500.50 44 2525 8080

758758 0.0140.014 0.50.5 55 1313


Problems:Problems:NN22 (g) + 3H (g) + 3H22 (g) (g) 2NH 2NH33 (g) + 92.22kJ (g) + 92.22kJ

1.1. Reaction is very slow at low Reaction is very slow at low temperatures, where Kc is highesttemperatures, where Kc is highest

2.2. Increasing the temperature increases Increasing the temperature increases the rate, but decreases the yield the rate, but decreases the yield because heat is produced.because heat is produced.


Solutions:Solutions:1.1. Increase the Temperature (generally 450Increase the Temperature (generally 450ooC)C)

2.2. Increase the Pressure (generally 200-1000atm)Increase the Pressure (generally 200-1000atm)

3.3. Use excess ReactantsUse excess Reactants

4.4. Remove the Ammonia as it is producedRemove the Ammonia as it is produced

5.5. Use a catalyst (solid Iron Oxide, Potassium Use a catalyst (solid Iron Oxide, Potassium Oxide, and Aluminum Oxide)Oxide, and Aluminum Oxide)

Partial Pressures and the Partial Pressures and the Equilibrium ConstantEquilibrium Constant

kkpp

Partial PressuresPartial Pressures

For gas phase equilibria, the equilibrium For gas phase equilibria, the equilibrium constant (kconstant (keqeq or k or kcc) may be expressed in terms ) may be expressed in terms

of the partial pressure (kof the partial pressure (kpp))

Partial PressurePartial Pressure

2Cl2Cl22 (g) + 2H (g) + 2H22O (g) O (g) 4HCl (g) + O 4HCl (g) + O22 (g) (g)

22

22

24

][][

][][

ONCl

OHClkc

22

4

)()(

)()(

22

2

ONCl

OHCl

p PP

PPk

Partial PressurePartial Pressure

If Kc or Kp are given and the other is wanted, If Kc or Kp are given and the other is wanted, you can use the relationship between them to you can use the relationship between them to calculate (PV = nRT)calculate (PV = nRT)

Scary Math ProofScary Math Proof

4

5

22

4

22

4

1

1

22

2

22

2

RT

RTx

PP

PP

RT

P

RT

P

RT

P

RT

P

kOHCl

OHCl

OHCl

OHCl

c

Scary Math Proof ExplainedScary Math Proof Explained

Kc = Kp (1/RT) for this reactionKc = Kp (1/RT) for this reaction Kc R T = KpKc R T = Kp

For every reaction: For every reaction: Kc = Kp(RT)Kc = Kp(RT)--nn

Kp = Kc(RT)Kp = Kc(RT)nn

Example Problem 17-11, 17-12, 17-13, 17-14Example Problem 17-11, 17-12, 17-13, 17-14

Heterogeneous EquilibriaHeterogeneous Equilibria

Heterogeneous EquilibriaHeterogeneous Equilibria

Remember:Remember: The activities of pure solids and liquids are 1The activities of pure solids and liquids are 1 Solvents of very dilute solutions are treated as pure Solvents of very dilute solutions are treated as pure

liquids with activity of 1liquids with activity of 1

Heterogeneous Equilibria Heterogeneous Equilibria (Examples)(Examples)

Write Kc and Kp expressions:Write Kc and Kp expressions:1.1. CaCOCaCO33 (s) (s) CaO (s) + CO CaO (s) + CO22 (g) (g)

kc = [COkc = [CO22] kp = (P] kp = (PCO2CO2))

2.2. HH22O (l) + CaFO (l) + CaF22 (s) (s) Ca Ca2+2+ (aq) + 2F (aq) + 2F1-1- (aq) (aq)kc = [Ca2+][F-]2kc = [Ca2+][F-]2 kp = undefined kp = undefined

(no gas)(no gas)

3.3. 3Fe(s) + 4H3Fe(s) + 4H22O (g) O (g) Fe Fe33OO44 (s) + 4H (s) + 4H22 (g) (g)

kc = [Hkc = [H22]]44/[H/[H22O]O]44 kp = (P kp = (PH2H2))44/(P/(PH2OH2O))44

Relating Equilibrium to Relating Equilibrium to ThermodynamicsThermodynamics

Relation to ThermoRelation to Thermo

GGrxnrxn = = GGoorxnrxn + RTlnQ + RTlnQ

R = 8.314 J / k molR = 8.314 J / k mol

Can use (+ 2.303RT logQ)Can use (+ 2.303RT logQ)

At equilibrium, At equilibrium, Grxn = 0 and Q = kGrxn = 0 and Q = k

k can be related directly to k can be related directly to Grxn Grxn

Relation to ThermoRelation to Thermo

0 = 0 = GGoorxnrxn + RTlnk + RTlnk

GGoorxnrxn = -RTlnk = -RTlnk

K is the thermodynamic equilibrium constantK is the thermodynamic equilibrium constant kc where solutions are involvedkc where solutions are involved kp where gases are involvedkp where gases are involved

Finding K at different Finding K at different TemperaturesTemperatures

Van’t Hoff EquationVan’t Hoff Equation

Knowing Knowing HHoo and kc at T1 allows the and kc at T1 allows the ESTIMATION of the kc at T2ESTIMATION of the kc at T2

ln(kln(kT1T1/k/kT2T2) = ) = HHoo (T (T22-T-T11))

R (TR (T22TT11))

ln(kln(k22/k/k11) = () = (H/R) ((1/TH/R) ((1/T11) – (1/T) – (1/T22))))

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Equilibrium Unit 4 Chapters 17, 18, 19, 20. Chapter 17 Equilibrium – when two opposite reactions occur simultaneously and at the same rate Equilibrium