17-1

17 Chemical Equilibria

Consider the following reaction:

aA + bB ---> cC + dD

As written is suggests that reactants A + B will be used up informing products C + D. However, what we learned in the sectionon thermodynamics is that a reaction might not occur to completionto leave only C + D. Instead it can form any proportion of A and Bto C and D, including remaining primarily as the products.

17-2

Consider a system in which every compound is present:

A better way of considering the reaction is to recognize that from astarting point with A and B placed in a closed thermodynamicsystem, a reaction proceeds to the point in time at which fourcompounds, A, B, C and D come to exist simultaneously. This isshown in the next graph.

17-3

17-4

Extracting kinetics from the plot:

This plot provides us with enormous information about both thekinetics and the thermodynamics of the reaction. From a kineticperspectives we know that the reaction proceeds at a rate determinedby

- 1 d[A] 1 d[C]rate = =

a dt c dt

It changes rapidly at first and then levels off later in time.

17-5

Kinetics when nothing seems to be happening:

What is of greater interest in this chapter is the rate of reaction at thepoint where there appears to be no change in concentration ofreactants and products. In fact, at that point, we can consider tworeactions to be occurring, a forward reaction

aA + bB ----> c C + dDand a reverse reaction

cC + dD ----> aA + bB

17-6

Deriving the Equilibrium Constant, Keq, from Rate Constant, k.

To simplify things, consider reactions following simple one stepbimolecular mechanisms. Then we have rate laws of the followingform:

Rate = kf[A][B] forward reactionand

Rate = kr[C][D] reverse reaction

17-7

A little bit of math and presto: Keq

Obviously from the plot, at the point where all of the curves flattenout, the rates at which A, B, C and D are formed are the same. Thuswe can equate the two rate laws

kf[A][B] = kr [C][D]

kf [C][D]and rearranging Kc = __ = ______

kr [A][B]

17-8

The constant we will use the rest of the semester.

The ratio, Kc, is a very important value know as the equilibriumconstant. How important is it? Coming to understand and applythis constant will be the primary focus of THE REST OF THECOURSE. We will come to learn that though Kc is derived from thekinetics of the reaction, it is an important thermodynamic constantwhich is directly related to the most important of thermodynamicstate functions, the Gibbs free energy, ∆Go.

17-9

Generalizing the equilibrium constant, Kc.

It can be shown that no matter what the mechanistic pathway of areaction, the equilibrium constant, Kc, is defined as

aA + bB ⇔ cC + dD

[C]c[D]d

Kc = [A]a[B]b

for any reactants and products AT EQUILIBRIUM.

17-10

Example. What is the equilibrium expression for the followingreaction:

H2(g) + I2 (g) ⇔ 2HI (g)

[HI]2

Kc = [H2][I2]

17-11

A new way to describe all chemical reactions,the equilibrium expression:

Note that every reaction ultimately arrives at an equilibrium; i.e., thepoint at which the forward and reverse reactions have equal rates sothat there is no change in the overall concentration of species presentin the system. Thus every reaction has an equilibrium constant, Kc.A few hundred equilibrium constants for the work we will do inChapters 18-20 are found in Appendixes F, H and I of Davis.

17-12

The magnitude of Keq values

Equilibrium constants can vary enormously in magnitude. Very largeand very small numbers are possible:

Cu++ + S= ⇔ CuS Kc = 1.1 x 1035

HPO4-2 ⇔ H+ + PO4

-3 Kc = 3.5 x 10-13

Note that we can rewrite an equilibrium expression as the reversereaction and the new Kc

’ is calculated simply as the reciprocal of Kc.

17-13

Example. The equilibrium expression for the formation of CuS iswritten below:

[CuS]Cu++ + S= ⇔ CuS(s) Kf = 1.1 x 1035 =

[Cu++] [S=]

From this information, determine the equilibrium constant for thedissociation of CuS. The dissociation of CuS is written as

[Cu++] [S=]CuS(s) ⇔ Cu++ + S= Kd = = 1/Kf = 9 x 10-36

[CuS]

17-14

A few points to make about Kc:

1. Temperature dependence. The value is obtained for a specifictemperature. This is not surprising since K is a thermodynamicconstant and all thermodynamic constants are strongly dependentupon T. Once again, a standard temperature has to be selected forrecording values in the Appendix. The typical temperature used is25oC. Later we will learn to use the van’t Hoff equation to convertbetween Kc values at different temperatures.

17-15

2. In performing equilibrium calculations:

i. Pure solids and pure liquids are assigned an activity of 1

ii. For ideal solutions, the activity of samples dissolved in solution is given in molarity

iii. For ideal gases, the activity of samples is given as partial pressures in atm.

17-16

3. Equilibrium constants are dimensionless

17-17

.

4. Looking at K to determine reaction spontaneity.

If the value for K>1, the formation of products is favored overreactants. If the value for K<1, the formation of reactants overproducts is favored.

17-18

Example. Calculation of equilibrium constant.

One liter of an equilibrium mixture is found to contain 0.172 molesof PCl3, 0.086 moles of chlorine and 0.028 moles of PCl5. CalculateKc for the reaction.

PCl5 ⇔ PCl3 + Cl2

[PCl3][Cl2] (0.172mole/1l)(0.086mole/1l)Kc = = = 0.53 [PCl5] (0.028mole/1l)

17-19

Example. Calculation of equilibrium constant.

The decomposition of PCl5 at a different temperature involvesintroduction of one mole of PCl5 to an evacuated 1 liter container. Atequilibrium, 0.6 moles of PCl3 is present in the container. CalculateKc.

PCl5 ⇔ PCl3 + Cl2

initial 1M 0M 0M

change -0.6M +.6M +.6M

equilibrium 0.4M 0.6M 0.6M

(0.6M)(0.6M)Kc = = 0.9

(0.4M)

17-20

Example. Calculation of equilibrium constant.

In an evacuated 1 liter container, 0.8 moles of N2 and 0.9 moles of H2

are placed. At equilibrium, 0.2 mole of NH3 is present. Calculate Kc.N2 + 3H2 ⇔ 2NH3

initial 0.8M 0.9M 0

change -0.1M -0.3M +0.2M

equilibrium 0.7M 0.6M 0.2M

(0.2M)2

Kc = = 0.26 (0.7M)(0.6M)3

17-21

Relating the Reaction Quotient, Q, to Kc.

The calculation of Kc is accomplished by ratioing the equilibriumconcentration of products to reactant. However, nothing stops usfrom making a calculation of the ratio of products to reactants at anynon-equilibrium position along the reaction path. The ratio ofproducts to reactants is generally called the mass action equation, Q,or in Davis, the reaction quotient.

[C]c[D]d

Q = [A]a[B]b

Recall that at equilibrium, Q ------> K

17-22

Some rules for relating Q to K

We can calculate the value of Q anywhere along the course of thereaction. Note that in the reaction of A and B forming C and D, thevalue of Q starts off very small (reactants in denominator) comparedto the final K value. Q increases until the point that the forward andreverse reaction rates are equal, and the system is at equilibrium; thenKc can be calculated.

We can establish the following rules for describing where thereaction is relative to equilibrium by comparing the value of Q to K.

Q < K forward reaction dominates (example above)Q = K system at equilibriumQ > K reverse reaction dominates

17-23

17-24

Example. Calculating Q.

The equilibrium constant for the following reaction, Kc, is 49 at450oC. If 0.22 mole of H2, 0.22 mole of I2 and 0.666 mole of HIwere put into an evacuated container, would the system be atequilibrium? If not, in what direction must the reaction proceed toreach equilibrium?

H2 + I2 ⇔ 2HI (assume one liter)

[HI]2 (0.666M)2

Q = = = 9.2[H2][I2] (0.22M)(0.22M)

Q<K ∴ rxn proceeds to right to form more [HI]

17-25

Example. The equilibrium constant, Kc, is 71.00 for the reactionbelow. If 0.6 mole of SO2 and 0.2 mole NO2 are placed into anevacuated 2.0 liter container and allowed to reach equilibrium, whatwill be the concentration of each compound at equilibrium?

SO2 + NO2 ⇔ SO3 + NO

Initial 0.3M 0.1M 0M 0M

Change -xM -xM +xM +xM

Equilibrium (0.3-x)M (0.1-x)M xM xM

x2 x2

Kc = 71 = = (0.3-x)(0.1-x) 0.03-.4x+x2

0 = 70x2-28.4x+2.13 [NO2]=0.0007Msolve quadratic to get [SO2] = 0.2M

x = .0993 [NO] = [SO3] = 0.1M

17-26

Example. The equilibrium constant is 49 for the reaction below. If1.00 mole of HI is placed into a 1.0 liter container and allowed toreach equilibrium, what will be the equilibrium concentration of eachcompound?

H2 + I2 ⇔ 2HI

Initial 0M 0M 1M

Change +xM +xM -2xM

Equilibrium xM xM (1-2x)M

(1-2x)2 use quadratic equation [H2] = 0.11MKc = = 49 x = 0.11 [I2] = 0.11M

x2 [HI] = 0.78M

17-27

Stress and LeChatlier’s Principle:

Stress is a big deal, in real life and in chemical equilibria. What dowe do in life when things change? We do what we can to relieve thestress by moving away from it. We apply the same notion tochemical equilibria. In a system at equilibrium, when a changeoccurs a stress is imposed on the system. The system then respondsby adopting new equilibrium conditions that relieve the stress.

This is the definition of LeChatlier’s Principle: If a change inconditions occurs to a system at equilibrium, the system responds torelieve the stress and reach a new state of equilibrium.

What are some of the ways stress is introduced to a system?

1. Changes in concentration of system components.

2. Change in system temperature.

3. Change in volume or pressure (in gaseous systems.)

17-28

Case 1. LeChatlier’s Principle: Changes in Concentration

Consider the general reaction:

A + B ⇔ C + D

Suppose we add additional reactant A or B to the system. This meanswe have more reactant than is necessary to establish equilibrium andthe system shifts to reduce the amount of A and B. This results in anincrease in the amount of C and D. The counter to this is also true. Ifthe system changes by adding C and D to the system, the equilibriumshifts to the left forming more A and B.

Looking at the equilibrium constant equation gives us a mathematicaljustification for the direction of the change.

[C] [D]Kc = ___________

[A] [B]Note for example that if we increase the amount of A, then tomaintain the equality, the equilibrium must shift so that the amount ofC and D go up as well. What happens to the amount of B? It mustgo down.

17-29

Some General Rules for LeChatelier and Concentration Changes.

From a chemical perspective, we have an equilibrium system withrelative concentrations of compounds defined by Kc. We impose astress on the system by changing a concentration and create a non-equilibrium Q value for the system. However, because of the non-zero free energy now in the system, Q will in time return to K as ∆G---> 0.

We can make a little table telling us how equilibrium concentrationsadjust to compensate for changes in concentration.

A + B ⇔ C + Dcase 1: Add A B decreases C increases D increasescase 2: A decreases Add B C increases D increasescase 3: A increases B increases Add C D decreasescase 4: A increases B increases C decreases Add D

17-30

Example: LeChatlier’s Principle applied to a change inconcentration.

A 2 liter vessel contains an equilibrium mixture of 1.2 mole ofCOCl2, 0.6 mole of CO and 0.2 mole of Cl2. First calculate theequilibrium constant for the reaction:

CO + Cl2 ⇔ COCl2

initialequilibrium: 0.3M 0.1M 0.6M

[COCl2]Kc = = 20

[CO] [Cl2]

17-31

An additional 0.8 mole of Cl2 is added to the vessel. Calculate themolar concentration of CO, Cl2 and COCl2 when the new equilibriumis established.

CO + Cl2 ⇔ COCl2

Original equilibrium 0.3M 0.1M 0.6M

Stress added 0M +0.4M 0M

New initial conc. 0.3M 0.5M 0.6M

Change (0.3-x)M (0.5-x)M (0.6+x)M

New equilibrium 0.121M 0.321M 0.779M

(0.6)1) After stress Q = = 4

(0.3)(0.5)

Q<K so rxn proceeds forward

(0.6+x)(0.6+x)2) K = 20 = = (0.3-x)(0.5-x) 0.15-0.8x+x2

solve quadratic, 0 = 20x2 - 17x + 2.4, to get x = 0.179

17-32

Case 2. LeChatelier’s Principle: Changes in Temperature.

Consider the following exothermic reaction.

A + B ⇔ C + D + heat (∆H is -)

This means that heat is being generated by the reaction. If we addadditional heat to the system, the stress of the increased temperaturewill make the system respond to try and to cool things down. Thismeans the system reaction will shift to the left to reduce the heatgenerated by the exothermic process.

17-33

We can use similar sorts of reasoning to establish the following rules:

exothermic rxn temperatureincreases

reaction shifts left

exothermic rxn temperaturedecreases

reaction shifts right

endothermic rxn temperatureincreases

reaction shifts right

endothermic rxn temperaturedecreases

reaction shifts left

17-34

Examples: How will an increase in temperature affect each of thefollowing reactions?

2NO2(g) ⇔ N2O4 + heatExothermic reaction so equilibrium shifts to the left to reduce thetemp. of the system.

H2 (g) + Cl2 (g) ⇔ 2HCl (g) +92 kJExothermic reaction so equilibrium shifts to the left to reduce thetemp. of the system.

H2 (g) + I2 (g) ⇔ 2HI ∆H = +25kJEndothermic reaction so equilibrium shifts to the right to reduce thetemp. of the system.

17-35

Case 3. LeChatlier’s Principle: Changes in Volume and Pressure(application to reactions involving gases)

First note from the ideal gas law that P and V are inversely related. Ifwe reduce the volume of a system, this has the affect of increasingthe concentration of a gas in the system. The system will thusrespond by shifting the reaction in the direction to reduce theconcentration of gases in the system. This means the reaction shiftstoward the side that has the fewest moles of gas at equilibrium

Similarly reasoning yields the following:

V decreases (P, [ ] increase) shift reaction toward side with smallernumber of moles of gas

V increases (P, [ ] decrease) shift reaction toward side with largernumber of moles of gas

Note that if we assume an ideal gas, then if there are an equal numberof moles of gas on either side of the reaction, then pressure or volumechanges have no affect on the equilibrium.

17-36

Examples.

How will an increase in pressure (concentration) resulting from adecrease in volume affect the following equilibria?

H2 (g) + I2 (g) ⇔ 2HI (g)∆n = 2-2 = 0

therefore, no change in equilibrium

4NH3 (g) + 5O2 (g) ⇔ 4NO (g) + 6H2O (g)∆n = 10-9 = 1,

therefore, shifts to the left to reduce moles of gas.

PCl3 (g) + Cl2 (g) ⇔ PCl5 (g)∆n = 1-2 = -1,

therefore, shifts to the right to reduce moles of gas

2H2 (g) + O2 (g) ⇔ 2H2O (g)∆n = -1

therefore, shifts to the right to reduce moles of gas

17-37

Equilibrium Constant, Kp, Expressed for Gases:

Taking a look at the ideal gas law:

PV = nRT

The equation suggests a way to relate partial pressure, P, of a gas inthe system to concentration, [ ] by

[ ] = n/V = P/RT

Substituting this relationship into the general equilibrium equationyields:

aA (g) + bB(g) ⇔ cC (g) + dD (g)

(PC)c (PD)d

Kp = ____________________

(PA)a (PB)b

The relationship between Kc and Kp follows from the ideal gas law:

Kp = Kc(RT)∆n where ∆n =(ngas prod - ngas react)

17-38

Example:

For the reaction below with a Kc = 20 at 25oC, what is Kp?

CO + Cl2 ⇔ COCl2

∆n = 1-2 = -1

Kp = Kc(RT)∆n = 20[(0.082atml/molK)(298K)]-1

= 0.82

17-39

Example: A calculation interchanging Kc and Kp

Kc is 49 for the following reaction. If 1.0 mole H2 and 1 mole I2 areallowed to reach equilibrium in a 3 liter vessel, how many moles of I2

are unreacted at equilibrium?

H2 (g) + I2 (g) ⇔ 2HI (g)

initial 0.33M 0.33M 0M

change 0.33-xM 0.33-xM +2xM

equilibrium 0.073M 0.073M 0.514M

[HI]2 4x2 45x2-32.34x+5.34 = 0Kc = 49 = = solve quadratic

[H2][I2] 0.01089-0.66x+x2 x = 0.257

3(0.257) = 0.219 moles of reacted I2

17-40

Now what are the equilibrium pressures of H2, I2 and HI?

∆n = 0, therefore, Kc = Kp

[HI]2 PHI2 moles I2 = moles of H2, therefore,

______ = ______ [I2][H2] PI2 PH2

(2.2(PI2) )2

0.264(PI22) = PHI

2(0.0053) 49 = PI2

2

So PI2 = 0.044 = PH2

And 2.2(PI2) = PHI so PHI = 0.0968

17-41

Relating Kc to Thermodynamic Quantities through Go:

Recall that in the last chapter we introduced the idea of the freeenergy of the system, ∆Go, under standard conditions. ∆Go describesthe free energy change in the system that accompanies the conversionof all reactants in their standard states (1M, 1 atm) to products in theirstandard states (1M, 1 atm).

We now define the free energy, ∆G, which is the free energy of areaction at concentrations other than 1M or 1 atm. It is found from∆Go by correcting for non-standard conditions using the reactionquotient, Q:

∆G = ∆Go + RT ln Q

17-42

The special case of thermodynamics at equilibrium.

Remember that at equilibrium, ∆G = 0 and Q = K. This means theabove equation reduces to:

∆Go = -RT ln K (at equilibrium)

Among other things this relationship tells us that we can calculate theequilibrium constant, Kc, from the thermodynamic constants inAppendix K.

17-43

Examples: Converting between Kc and Go

Case 1. 2C graphite + 3H2 (g) + 12 O2 (g) D C2H5OH

∆G fo= -168.6 kKJ

mol

Kc = exp ( ∆Go

RT )

Kc = exp ( −168.6x103

(−8.3)(298K ))

Kc = 3.5 x 1029

So a spontaneous reaction has a K > 1

17-44

Case 2. N2(g) D 2N(g)

∆G fo = 456 kKJ

mol

Kc = exp ( −456x103

(8.3)(298))

Kc = 8.6 x 10 –81

So a non-spontaneous reaction has K < 1. Reaction is spontaneous inopposite direction.

17-45

What generalization can be made about ∆Go and Kc?

If ∆Go < 0 then K>1 products favored over reactantIf ∆Go > 0 then K< 1 reactants favored over products

Isn’t it exciting to see that chemical equilibrium is another model fordescribing thermodynamics? In this case we don’t measure thechange in energy, instead we measure the equilibrium concentrations.

17-46

End-of chapter equations and the Famouns Scientists whoinvented them:

It is the end of the chapter and it is time to throw a hard equation atyou. Just to let you know that this is a pretty common occurrence,consider:

end of chapterequation name

if weknow this:

then for thefollowing change:

we get a newphysical constant

Clausius-Clapeyron

∆Hvap ∆T P(vapor pressure)

Arrhenius Ea ∆T rate constantvan’t Hoff ∆H ∆T equilibrium

constant

17-47

The van’t Hoff Equation: Finding a new Kc at a new T:

Of interest in Chapter 17 is the bottom equation: the van’t Hoffequation that allows us to determine Kc for a chemical system at anytemperature we desire.

KT2 ∆Ho 1 1ln = . - KT1 R T1 T2

Notice that this is the quantitative way to look at LeChatelier’sprinciple applied to changes in temperature.

17-48

Example: Using the van’t Hoff equation to find a new K.

For the reaction below,2NO2 ⇔ 2NO + O2

∆Ho = 114 kJ/mole and Kp = 4.3 x 10-13 at 25oC.

So what is Kp at 250oC?

Insert T2 = 523K, R = 8.312 J/mol K. ∆Ho = 114 kJ/mole and Kp =4.3 x 10-13 at 25oC.

KT2 = KT1 [exp ( ∆H

R ( 1

T1

−1

T2

))]

exp (1n ( KT2

KT1

)) = exp [ ∆H

R ( 1

T1

- 1

T2

) ]

KT2 = (4.3 x 10-13) exp (114x103

8.314 ( 1

298−

1

523))

So at 250oC, KT2 = 1.7 x 10-4

Is this consistent with what you know from LeChatelier?