Chemical EquilibriumChemical Equilibrium

Chapter 15Chapter 15

Chemical EquilibriumChemical Equilibrium Concentrations of all reactants and Concentrations of all reactants and

products remains constant with products remains constant with timetime

All reactions carried out in a closed All reactions carried out in a closed vessel will reach equilibrium:vessel will reach equilibrium: If little product is formed, If little product is formed,

equilibrium lies far to the left.equilibrium lies far to the left. If little reactant remains, If little reactant remains,

equilibrium lies far to the rightequilibrium lies far to the right

Reactions are reversibleReactions are reversible A + B C + D ( forward)A + B C + D ( forward) C + D A + B (reverse)C + D A + B (reverse) Initially there is only A and B so Initially there is only A and B so

only the forward reaction is only the forward reaction is possiblepossible

As C and D build up, the reverse As C and D build up, the reverse reaction speeds up while the reaction speeds up while the forward reaction slows down.forward reaction slows down.

Eventually the rates are equalEventually the rates are equal

Rea

ctio

n R

ate

Time

Forward Reaction

Reverse reaction

Equilibrium

What is equal at Equilibrium?What is equal at Equilibrium? Rates are equalRates are equal Concentrations are not.Concentrations are not. Rates are determined by Rates are determined by

concentrations and activation concentrations and activation energy.energy.

The concentrations do not The concentrations do not change at equilibrium.change at equilibrium.

The Haber ProcessThe Haber Process A process used to convert NA process used to convert N22 and and

HH22 into NH into NH33: N: N22 + 3H + 3H22 2NH 2NH33

Hydrogen is consumed at 3x the Hydrogen is consumed at 3x the rate of nitrogen.rate of nitrogen.

Ammonia is formed at 2x the rate Ammonia is formed at 2x the rate at which nitrogen is consumed.at which nitrogen is consumed.

Par

tial P

ress

ure

Time

H2

N2

NH3

Equilibrium Achieved

Law of Mass ActionLaw of Mass Action For any reaction For any reaction jjA + A + kkB B llC + C + mmDD K = [C]K = [C]ll[D][D]mm = [PRODUCTS] = [PRODUCTS]powerpower

[A] [A]jj[B][B]kk [REACTANTS][REACTANTS]powerpower

K is called the equilibrium K is called the equilibrium constant.constant.

is how we indicate a is how we indicate a reversible reactionreversible reaction

Playing with KPlaying with K If we write the reaction in If we write the reaction in

reverse.reverse. llC + C + mmD D jjA + A + kkB B Then the new equilibrium Then the new equilibrium

constant isconstant is K’ = [A]K’ = [A]jj[B][B]kk = 1/K = 1/K

[C][C]ll[D][D]mm

Playing with KPlaying with K If we multiply the equation by a If we multiply the equation by a

constantconstant njnjA + A + nknkB B nlnlC + C + nmnmDD Then the equilibrium constant isThen the equilibrium constant is

K’ =K’ = [A] [A]njnj[B][B]nknk

== ([A] ([A] j j[B][B]kk))nn

== K Knn

[C][C]nlnl[D][D]nm nm ([C]([C]ll[D][D]mm))nn

The units for KThe units for K Are determined by the various Are determined by the various

powers and units of concentrations.powers and units of concentrations. They depend on the reaction.They depend on the reaction.

Equilibrium Expression SummaryEquilibrium Expression Summary The equilibrium-constant depends only on The equilibrium-constant depends only on

the stoichiometry of the reaction, not its the stoichiometry of the reaction, not its mechanism.mechanism.

When the balanced equation for a reaction When the balanced equation for a reaction is multiplied by a factor, n, the equilibrium is multiplied by a factor, n, the equilibrium expression for the new reaction is the expression for the new reaction is the original expression raised to the nth poweroriginal expression raised to the nth power

KKnewnew = (K = (Koriginaloriginal))nn

For a reaction at a given temp., the value of K is For a reaction at a given temp., the value of K is constant regardless of the amounts of starting constant regardless of the amounts of starting materialsmaterials

Equilibrium ConstantEquilibrium Constant

One for each TemperatureOne for each Temperature

Calculate KCalculate K NN22 + 3H + 3H22 2NH 2NH33

Initial Initial At EquilibriumAt Equilibrium [N[N22]]00 =1.000 M [N =1.000 M [N22] = 0.921M] = 0.921M

[H[H22]]00 =1.000 M [H =1.000 M [H22] = 0.763M] = 0.763M

[NH[NH33]]00 =0 M [NH =0 M [NH33] = 0.157M] = 0.157M

Calculate KCalculate K NN22 + 3H + 3H22 2NH 2NH33

Initial Initial At Equilibrium At Equilibrium [N[N22]]00 = 0 M = 0 M [N [N22] = 0.399 M] = 0.399 M

[H[H22]]00 = 0 M = 0 M [H [H22] = 1.197 M] = 1.197 M

[NH[NH33]]00 = 1.000 M [NH = 1.000 M [NH33] = 0.157M] = 0.157M Think about which way this reaction Think about which way this reaction

is proceeding.is proceeding.

Equilibrium and PressureEquilibrium and Pressure Some reactions are gaseousSome reactions are gaseous PV = nRTPV = nRT P = (n/V)RTP = (n/V)RT P = CRTP = CRT C is a concentration in moles/LiterC is a concentration in moles/Liter C = P/RTC = P/RT

Equilibrium and PressureEquilibrium and Pressure 2SO2SO22(g) + O(g) + O22(g) 2SO(g) 2SO33(g)(g)

Kp =Kp = (P (PSO3SO3))22

(P(PSO2SO2))22 (P (PO2O2))

K =K = [SO [SO33]]22

[SO[SO22]]22 [O [O22]]

Equilibrium and PressureEquilibrium and Pressure

K =K = (P(PSO3SO3/RT)/RT)22

(P (PSO2SO2/RT)/RT)22(P(PO2O2/RT)/RT)

K =K = (P(PSO3SO3))22(1/RT)(1/RT)22

(P (PSO2SO2))22(P(PO2O2) (1/RT)) (1/RT)33

K = Kp K = Kp (1/RT)(1/RT)22 = Kp RT= Kp RT

(1/RT) (1/RT)33

Therefore K = KTherefore K = KppRTRT

General EquationGeneral Equation jjA + A + kkB B llC + C + mmDD

KKpp= (P= (PCC))l l (P(PDD))mm= = (C(CCCxRT)xRT)l l (C(CDDxRT)xRT)m m

(P(PAA))j j (P(PBB))k k (C(CAAxRT)xRT)jj(C(CBBxRT)xRT)kk

KKpp== (C(CCC))l l (C(CDD))mmx(RT)x(RT)l+ml+m

(C(CAA))jj(C(CBB))kkx(RT)x(RT)j+kj+k

KKp p == K (RT)K (RT)((l+m)-(j+k) = l+m)-(j+k) = K (RT)K (RT)nn

n=(l+m)-(j+k)=Change in moles n=(l+m)-(j+k)=Change in moles of gasof gas

Homogeneous EquilibriaHomogeneous Equilibria So far every example dealt with So far every example dealt with

reactants and products where all reactants and products where all were in the same phase.were in the same phase.

We can use K in terms of either We can use K in terms of either concentration or pressure.concentration or pressure.

Units depend on reaction.Units depend on reaction.

Heterogeneous EquilibriaHeterogeneous Equilibria If the reaction involves If the reaction involves purepure

solids or solids or purepure liquids the liquids the concentration of the solid or the concentration of the solid or the liquid doesn’t change.liquid doesn’t change.

As long as they are not used up As long as they are not used up we can leave them out of the we can leave them out of the equilibrium expression.equilibrium expression.

For example…For example…

For ExampleFor Example HH22(g) + I(g) + I22(s) 2HI(g)(s) 2HI(g)

Before we got… K = [HI]Before we got… K = [HI]22 [H[H22][I][I22]]

But the concentration of IBut the concentration of I2 2 does not does not

change because it is a pure solid, sochange because it is a pure solid, so K= [HI]K= [HI]2 2

[H[H22]]

Applications of the Equilibrium Applications of the Equilibrium Constant:Constant:

K > 1: K > 1: The concentrations of the products The concentrations of the products

are greater than the concentrations are greater than the concentrations of the reactants.of the reactants.

Equilibrium lies far to the right.Equilibrium lies far to the right. Generally, large, negative Generally, large, negative E.E.

Applications of the Equilibrium Applications of the Equilibrium Constant:Constant:

K < 1:K < 1: The concentrations of the reactants The concentrations of the reactants

are greater than the concentrations are greater than the concentrations of the products.of the products.

Equilibrium position is far to the leftEquilibrium position is far to the left

The time required to achieve The time required to achieve equilibrium:equilibrium: Related to reaction rate and Related to reaction rate and

activation energyactivation energy Not related to the magnitude of KNot related to the magnitude of K

In summary:In summary: E influences KE influences K EEaa influences rate influences rate

The Reaction QuotientThe Reaction Quotient Tells you the direction the reaction Tells you the direction the reaction

will go to reach equilibriumwill go to reach equilibrium Calculated the same as the Calculated the same as the

equilibrium constant, but for a equilibrium constant, but for a system not at equilibriumsystem not at equilibrium

Q = Q = [Products][Products]coefficientcoefficient

[Reactants] [Reactants] coefficientcoefficient

Compare value to equilibrium Compare value to equilibrium constantconstant

What Q tells usWhat Q tells us If Q<K If Q<K

Not enough products, reactants Not enough products, reactants are consumedare consumed

Shift to rightShift to right If Q>K If Q>K

Too many products, products are Too many products, products are consumedconsumed

Shift to leftShift to left If Q=K system is at equilibriumIf Q=K system is at equilibrium

Solving Equilibrium Solving Equilibrium ProblemsProblems

Type 1Type 1

ExampleExample for the reactionfor the reaction 2NOCl(g) 2NO(g) + Cl2NOCl(g) 2NO(g) + Cl22(g) (g) K = 1.55 x 10K = 1.55 x 10-5 -5 M at 35ºCM at 35ºC In an experiment 0.10 mol NOCl, In an experiment 0.10 mol NOCl,

0.0010 mol NO(g) and 0.00010 0.0010 mol NO(g) and 0.00010 mol Clmol Cl22 are mixed in 2.0 L flask. are mixed in 2.0 L flask.

Which direction will the reaction Which direction will the reaction proceed to reach equilibrium?proceed to reach equilibrium?

Solving Equilibrium ProblemsSolving Equilibrium Problems Given the starting concentrations Given the starting concentrations

and one equilibrium and one equilibrium concentration.concentration.

Use stoichiometry to figure out Use stoichiometry to figure out other concentrations and K.other concentrations and K.

Learn to create a table of initial Learn to create a table of initial and final conditions. (ICE Charts)and final conditions. (ICE Charts)

Consider the following reaction at Consider the following reaction at 600ºC600ºC

2SO2SO22(g) + O(g) + O22(g) 2SO(g) 2SO33(g)(g) In a certain experiment 2.00 mol of In a certain experiment 2.00 mol of

SOSO22, 1.50 mol of O, 1.50 mol of O22 and 3.00 mol of and 3.00 mol of SOSO33 were placed in a 1.00 L flask. At were placed in a 1.00 L flask. At equilibrium 3.50 mol of SOequilibrium 3.50 mol of SO33 were were found to be present. found to be present.

Calculate the equilibrium Calculate the equilibrium concentrations of Oconcentrations of O22 and SO and SO22, K and K, K and KPP

Use stoich (ICE chart) to solve for Use stoich (ICE chart) to solve for equilibrium concentrations of Reactantsequilibrium concentrations of Reactants

Consider the same reaction at 600ºCConsider the same reaction at 600ºC In a different experiment .250 mol In a different experiment .250 mol

OO22, .500 mol SO, .500 mol SO22 and .350 mol SO and .350 mol SO33 were placed in a 1.000 L container. were placed in a 1.000 L container. When the system reaches When the system reaches equilibrium 0.045 mol of Oequilibrium 0.045 mol of O22 are are present.present.

Calculate the final concentrations of Calculate the final concentrations of SOSO22 and SO and SO33 and K and Kp and K and Kp

What if you’re not given What if you’re not given equilibrium concentration?equilibrium concentration?

The size of K will determine what The size of K will determine what approach to take.approach to take.

First let’s look at the case of a First let’s look at the case of a LARGE value of K ( >100).LARGE value of K ( >100).

Allows us to make simplifying Allows us to make simplifying assumptions.assumptions.

ExampleExample HH22(g) + I(g) + I22(g) 2HI(g)(g) 2HI(g) K = 7.1 x 10K = 7.1 x 1022 at 25ºC at 25ºC Calculate the equilibrium Calculate the equilibrium

concentrations if a 5.00 L container concentrations if a 5.00 L container initially contains 15.9 g of Hinitially contains 15.9 g of H22 294 g I 294 g I22 . .

[H[H22]]00 = (15.9g/2.02)/5.00 L = 1.59 M = (15.9g/2.02)/5.00 L = 1.59 M [I[I22]]00 = (294g/253.8)/5.00L = 0.232 M = (294g/253.8)/5.00L = 0.232 M [HI][HI]00 = 0 = 0

Q= 0<K so more product will be Q= 0<K so more product will be formed.formed.

Assumption since K is large Assumption since K is large reaction will go to completion.reaction will go to completion.

Stoichiometry tells us IStoichiometry tells us I22 is is limiting so it will be smallest at limiting so it will be smallest at equilibrium (let it be x)equilibrium (let it be x)

Set up an I.C.E chartSet up an I.C.E chart

Choose X so it is small.Choose X so it is small. For IFor I22 the change in X must be X-.232 the change in X must be X-.232

MM Final must = initial + changeFinal must = initial + change

H2(g) I2(g) HI(g)

initial 1.59 M 0.232 M 0 M changefinal X

Using to stoichiometry we can findUsing to stoichiometry we can find Change in HChange in H2 2 = X-0.232 M= X-0.232 M

Change in HI = -twice change in HChange in HI = -twice change in H2 2 Change in HI = 0.464-2XChange in HI = 0.464-2X

H2(g) I2(g) HI(g) initial 1.59 M 0.232 M 0 M change X-0.232final X

Now we can determine the final Now we can determine the final concentrations by adding.concentrations by adding.

H2(g) I2(g) HI(g) initial 1.59 M 0.232 M 0 M

change X-0.232 X-0.232 0.464-2Xfinal X

Now plug these values into the Now plug these values into the equilibrium expressionequilibrium expression

K = K = (0.464-2X) (0.464-2X)22 = 7.1 x 10 = 7.1 x 1022 (1.328+X)(X) (1.328+X)(X)

Now you can see why we made sure Now you can see why we made sure X was small. X was small.

H2(g) I2(g) HI(g) initial 1.59 M 0.232 M 0 M

change X-0.232 X-0.232 0.464-2X final 1.36+X X 0.464-2X

Why we chose XWhy we chose X K = K = (0.464-2X) (0.464-2X)22 = 7.1 x 10 = 7.1 x 1022

(1.328+X)(X) (1.328+X)(X) Since X is going to be small, we can Since X is going to be small, we can

ignore it in relation to 0.464 and ignore it in relation to 0.464 and 1.3281.328

So we can rewrite the equationSo we can rewrite the equation 7.1 x 107.1 x 102 2 == (0.464) (0.464)22

(1.328)(X) (1.328)(X) Makes the algebra easyMakes the algebra easy

When we solve for X we get 2.2 x 10When we solve for X we get 2.2 x 10-4-4

So we can find the other concentrationsSo we can find the other concentrations II22 = = 2.2 x 102.2 x 10-4-4 M M

HH22 = 1.36 M = 1.36 M

HI = 0.464 MHI = 0.464 M

H2(g) I2(g) HI(g) initial 1.56 M 0.232 M 0 M

change X-0.232 X-0.232 0.464-2X final 1.36+X X 0.464-2X

Checking the assumptionChecking the assumption The rule of thumb is that if the value The rule of thumb is that if the value

of X is less than 5% of all the other of X is less than 5% of all the other concentrations, our assumption was concentrations, our assumption was valid.valid.

If not we would have had to use the If not we would have had to use the quadratic equation quadratic equation

More on this later.More on this later. Our assumption was valid.Our assumption was valid.

PracticePractice For the reaction ClFor the reaction Cl22 + O + O22

2ClO(g) K = 1562ClO(g) K = 156 In an experiment 0.100 mol ClO, In an experiment 0.100 mol ClO,

1.00 mol O1.00 mol O22 and 0.0100 mol Cl and 0.0100 mol Cl22 are are mixed in a 4.00 L flask.mixed in a 4.00 L flask.

If the reaction is not at equilibrium, If the reaction is not at equilibrium, which way will it shift?which way will it shift?

Calculate the equilibrium Calculate the equilibrium concentrations.concentrations.

Problems with small KProblems with small K

K< .01K< .01

Process is the sameProcess is the same Set up table of initial, change, Set up table of initial, change,

and final concentrations.and final concentrations. Choose X to be small.Choose X to be small. For this case it will be a product.For this case it will be a product. For a small K the product For a small K the product

concentration is small.concentration is small.

For exampleFor example For the reaction For the reaction

2NOCl 2NO +Cl2NOCl 2NO +Cl22 K= 1.6 x 10K= 1.6 x 10-5-5 If 1.20 mol NOCl, 0.45 mol of NO and If 1.20 mol NOCl, 0.45 mol of NO and

0.87 mol Cl0.87 mol Cl22 are mixed in a 1 L are mixed in a 1 L containercontainer

What are the equilibrium What are the equilibrium concentrationsconcentrations

Q = [NO]Q = [NO]22[Cl[Cl22] = (0.45)] = (0.45)22(0.87) = (0.87) = 0.15 0.15

MM [NOCl] [NOCl]22 (1.20) (1.20)22

Choose X to be smallChoose X to be small NO will be limitingNO will be limiting Choose NO to be XChoose NO to be X

2NOCl 2NO Cl2

Initial 1.20 0.45 0.87

Change

Final

Figure out change in Figure out change in NONO Change = final - initialChange = final - initial change = X-0.45change = X-0.45

2NOCl 2NO Cl2

Initial 1.20 0.45 0.87

Change

Final X

Now figure out the other changesNow figure out the other changes Use stoichiometryUse stoichiometry Change in ClChange in Cl22 is 1/2 change in NO is 1/2 change in NO Change in NOCl is - change in NO Change in NOCl is - change in NO

2NOCl 2NO Cl2

Initial 1.20 0.45 0.87

Change X-.45

Final X

Now we can determine final concentrations

Add

2NOCl 2NO Cl2

Initial 1.20 0.45 0.87

Change 0.45-X X-.45 0.5X - .225

Final X

Now we can write equilibrium constant K = (X)2(0.5X+0.645)

(1.65-X)2 Now we can test our assumption X is

small ignore it in + and -

2NOCl 2NO Cl2

Initial 1.20 0.45 0.87

Change 0.45-X X-.45 0.5X - .225

Final 1.65-X X 0.5X +0.645

K = (X)2(0.645) = 1.6 x 10-5 (1.65)2

X= 8.2 x 10-3 Figure out final concentrations

2NOCl 2NO Cl2

Initial 1.20 0.45 0.87

Change 0.45-X X-.45 0.5X - .225

Final 1.65-X X 0.5X +0.645

[NOCl] = 1.64 [Cl2] = 0.649

Check assumptions .0082/.649 = 1.2 % OKAY!!!

2NOCl 2NO Cl2

Initial 1.20 0.45 0.87

Change 0.45-X X-.45 0.5X - .225

Final 1.65-X X 0.5X +0.645

Practice ProblemPractice Problem For the reactionFor the reaction

2ClO(g) 2ClO(g) ClCl22 (g) + O (g) + O22 (g) (g) K = 6.4 x 10K = 6.4 x 10-3-3 In an experiment 0.100 mol ClO(g), In an experiment 0.100 mol ClO(g),

1.00 mol O1.00 mol O2 2 and 1.00 x 10and 1.00 x 10-2-2 mol mol ClCl22 are mixed in a 4.00 L container.are mixed in a 4.00 L container.

What are the equilibrium What are the equilibrium concentrations.concentrations.

Mid-range K’sMid-range K’s

.01<K<100.01<K<100

No SimplificationNo Simplification Choose X to be small.Choose X to be small. Can’t simplify so we will have to Can’t simplify so we will have to

solve the quadratic (we hope)solve the quadratic (we hope) HH22(g) + I(g) + I22 (g) 2HI(g) (g) 2HI(g) K=38.6K=38.6 What is the equilibrium What is the equilibrium

concentrations if 1.800 mol Hconcentrations if 1.800 mol H22, , 1.600 mol I1.600 mol I22 and 2.600 mol HI are and 2.600 mol HI are mixed in a 2.000 L container?mixed in a 2.000 L container?

Problems Involving PressureProblems Involving Pressure Solved exactly the same, with Solved exactly the same, with

same rules for choosing X same rules for choosing X depending on Kdepending on KPP

For the reaction NFor the reaction N22OO44(g) (g) 2NO2NO22(g) K(g) KPP = .131 atm. What are = .131 atm. What are the equilibrium pressures if a flask the equilibrium pressures if a flask initially contains 1.000 atm Ninitially contains 1.000 atm N22OO44??

Le Chatelier’s PrincipleLe Chatelier’s Principle If a stress is applied to a system If a stress is applied to a system

at equilibrium, the position of the at equilibrium, the position of the equilibrium will shift to reduce equilibrium will shift to reduce the stress.the stress.

3 Types of stress3 Types of stress

Change amounts of reactants Change amounts of reactants and/or productsand/or products

Adding product makes Q>KAdding product makes Q>K Removing reactant makes Q>KRemoving reactant makes Q>K Adding reactant makes Q<KAdding reactant makes Q<K Removing product makes Q<K Removing product makes Q<K Determine the effect on Q, will Determine the effect on Q, will

tell you the direction of shifttell you the direction of shift

Change PressureChange Pressure By changing volumeBy changing volume System will move in the direction System will move in the direction

that has the least moles of gas.that has the least moles of gas. Because partial pressures (and Because partial pressures (and

concentrations) change a new concentrations) change a new equilibrium must be reached.equilibrium must be reached.

System tries to minimize the System tries to minimize the moles of gas.moles of gas.

Change in PressureChange in Pressure By adding an inert gasBy adding an inert gas Partial pressures of reactants Partial pressures of reactants

and product are not changedand product are not changed No effect on equilibrium positionNo effect on equilibrium position

Change in TemperatureChange in Temperature Affects the Affects the ratesrates of both the of both the

forward and reverse reactions.forward and reverse reactions. Doesn’t just change the Doesn’t just change the

equilibrium position, changes the equilibrium position, changes the equilibrium constant.equilibrium constant.

The direction of the shift The direction of the shift depends on whether it is exo- or depends on whether it is exo- or endothermicendothermic

ExothermicExothermic H<0H<0 Releases heatReleases heat Think of heat as a productThink of heat as a product Raising temperature push toward Raising temperature push toward

reactants.reactants. Shifts to left.Shifts to left.

EndothermicEndothermic H>0H>0 Produces heatProduces heat Think of heat as a reactantThink of heat as a reactant Raising temperature push toward Raising temperature push toward

products.products. Shifts to right.Shifts to right.

What effect would a catalyst have What effect would a catalyst have on the equilibrium?on the equilibrium?

A catalyst can help the reaction A catalyst can help the reaction reach equilibrium faster, but it will reach equilibrium faster, but it will not shift the equilibrium one not shift the equilibrium one direction or the other.direction or the other.