Chem 128 Notes

General Chemistry II Chemistry 128

Lecture Notes

Darin J. Ulness Department of Chemistry

Concordia College, Moorhead MN

Spring 2007

CONTENTS

1 Calculus in a Nutshell 7Rates of Change . . . . . . . . . . . . . . . . . . . . 7Slope . . . . . . . . . . . . . . . . . . . . . . . . . . . 8The Anti-Derivative . . . . . . . . . . . . . . . . . . 9Group Work . . . . . . . . . . . . . . . . . . . . . . . 9Homework . . . . . . . . . . . . . . . . . . . . . . . . 9

2 Differential Rate Laws 11The Reaction Rate . . . . . . . . . . . . . . . . . . . 11Rate Laws . . . . . . . . . . . . . . . . . . . . . . . . 12Elementary Reactions and Molecularity . . . . . . . . 13Group Work . . . . . . . . . . . . . . . . . . . . . . . 14Homework . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Determination of Rate Laws 15Differential methods based on the rate law . . . . . . 15Group Work . . . . . . . . . . . . . . . . . . . . . . . 18Homework . . . . . . . . . . . . . . . . . . . . . . . . 18

4 Integrated Rate Laws 19Linearizing the integrated rate laws . . . . . . . . . . 19Group Work . . . . . . . . . . . . . . . . . . . . . . . 21Homework . . . . . . . . . . . . . . . . . . . . . . . . 21

5 The Arrhenious Equation 23Temperature Dependence of the Rate Constant . . . 23Application of the Arrhenious Equation . . . . . . . . 24Catalysts . . . . . . . . . . . . . . . . . . . . . . . . 25Group Work . . . . . . . . . . . . . . . . . . . . . . . 25Homework . . . . . . . . . . . . . . . . . . . . . . . . 26

6 Kinetics and Equilibrium 27Group Work . . . . . . . . . . . . . . . . . . . . . . . 28Homework . . . . . . . . . . . . . . . . . . . . . . . . 28

CONTENTS 1

7 Equilibrium and the Equilibrium Constant 29The Equilibrium Constant . . . . . . . . . . . . . . . 29Group Work . . . . . . . . . . . . . . . . . . . . . . . 31Homework . . . . . . . . . . . . . . . . . . . . . . . . 31

8 ICE Charts 33Group Work . . . . . . . . . . . . . . . . . . . . . . . 35Homework . . . . . . . . . . . . . . . . . . . . . . . . 35

9 The Reaction Quotient and LeChatelier’s Prin-ciple 37The Reaction Quotient . . . . . . . . . . . . . . . . . 37LeChatelier’s Principle . . . . . . . . . . . . . . . . . 38Group Work . . . . . . . . . . . . . . . . . . . . . . . 39Homework . . . . . . . . . . . . . . . . . . . . . . . . 40

10 Review for Exam 1 41

11 Acids and Bases 43Brønstad Acids and Bases . . . . . . . . . . . . . . . 43Conjugate Acid-Base Pairs . . . . . . . . . . . . . . . 43Group Work . . . . . . . . . . . . . . . . . . . . . . . 44Homework . . . . . . . . . . . . . . . . . . . . . . . . 45

12 Water and pH 47The pH Scale . . . . . . . . . . . . . . . . . . . . . . 48The pH of Strong Acids and Bases . . . . . . . . . . 48Group Work . . . . . . . . . . . . . . . . . . . . . . . 49Homework . . . . . . . . . . . . . . . . . . . . . . . . 49

13 Acid and Base Equilibria 51Group Work . . . . . . . . . . . . . . . . . . . . . . . 54Homework . . . . . . . . . . . . . . . . . . . . . . . . 54

14 Reactions with Weak Acids and Weak Bases 55Reaction Between a Monoprotic Weak Acid and a

Strong Base . . . . . . . . . . . . . . . . . . . . 55Reaction Between a PolyproticWeak Acid and a Strong

Base . . . . . . . . . . . . . . . . . . . . . . . . 57Reaction Between a Weak Acid and a Weak Base . . 58Work Group . . . . . . . . . . . . . . . . . . . . . . . 59Homework . . . . . . . . . . . . . . . . . . . . . . . . 59

2 CONTENTS

15 The Common Ion Effect and Buffers 61Buffer Solutions . . . . . . . . . . . . . . . . . . . . . 62Group Work . . . . . . . . . . . . . . . . . . . . . . . 64Homework . . . . . . . . . . . . . . . . . . . . . . . . 65

16 More on Buffers: The Henderson-Hasselbalch Equa-tion 67Group Work . . . . . . . . . . . . . . . . . . . . . . . 68Homework . . . . . . . . . . . . . . . . . . . . . . . . 69

17 Titrations Part I 71Titration of a Strong Acid with a Strong Base . . . . 71Titration of a Weak Acid with a Strong Base . . . . 73Group Work . . . . . . . . . . . . . . . . . . . . . . . 75Homework . . . . . . . . . . . . . . . . . . . . . . . . 75

18 Titrations Part II 77Group Work . . . . . . . . . . . . . . . . . . . . . . . 79Homework . . . . . . . . . . . . . . . . . . . . . . . . 79

19 Solubility of Weak Electrolytes 81Group Work . . . . . . . . . . . . . . . . . . . . . . . 82Homework . . . . . . . . . . . . . . . . . . . . . . . . 83


21 Review of the First Law, Enthalpy, and Hess’Law 87The First Law of Thermodynamics . . . . . . . . . . 87Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . 88Hess’ Law . . . . . . . . . . . . . . . . . . . . . . . . 90Group Work . . . . . . . . . . . . . . . . . . . . . . . 91Homework . . . . . . . . . . . . . . . . . . . . . . . . 91

22 Entropy 93Microstates and Configurations . . . . . . . . . . . . 93Group Work . . . . . . . . . . . . . . . . . . . . . . . 94Homework . . . . . . . . . . . . . . . . . . . . . . . . 94

23 The Second and Third Laws 95Calculating Entropy . . . . . . . . . . . . . . . . . . 96Group Work . . . . . . . . . . . . . . . . . . . . . . . 96Homework . . . . . . . . . . . . . . . . . . . . . . . . 97

CONTENTS 3

24 Gibbs Free Energy 99The Second Law . . . . . . . . . . . . . . . . . . . . 99Gibbs Free Energy . . . . . . . . . . . . . . . . . . . 100Group Work . . . . . . . . . . . . . . . . . . . . . . . 101Homework . . . . . . . . . . . . . . . . . . . . . . . . 102

25 Gibbs Free Energy and Equilibrium 103The difference between 4rxnG and 4rxnG

° . . . . . 104Group Work . . . . . . . . . . . . . . . . . . . . . . . 106Homework . . . . . . . . . . . . . . . . . . . . . . . . 106

26 Review of Concepts Surrounding Gibbs Free En-ergy 107

27 Oxidation—Reduction Reactions I 109Balancing redox reactions . . . . . . . . . . . . . . . 110Group Work . . . . . . . . . . . . . . . . . . . . . . . 110Homework . . . . . . . . . . . . . . . . . . . . . . . . 110

28 Oxidation—Reduction Reactions II 111Group Work . . . . . . . . . . . . . . . . . . . . . . . 111Homework . . . . . . . . . . . . . . . . . . . . . . . . 111

29 Electrochemical Cells 113Cell diagrams . . . . . . . . . . . . . . . . . . . . . . 114Standard Cell potentials . . . . . . . . . . . . . . . . 114Group Work . . . . . . . . . . . . . . . . . . . . . . . 115Homework . . . . . . . . . . . . . . . . . . . . . . . . 115

30 The Nernst Equation 117Group Work . . . . . . . . . . . . . . . . . . . . . . . 118Homework . . . . . . . . . . . . . . . . . . . . . . . . 119


32 Review of Molecular Geometry 123Charge Distribution in Covalent Compounds . . . . . 123VSEPR Model . . . . . . . . . . . . . . . . . . . . . 124Molecular Dipole Moments . . . . . . . . . . . . . . . 126Group Work . . . . . . . . . . . . . . . . . . . . . . . 126Homework . . . . . . . . . . . . . . . . . . . . . . . . 127

33 Intermolecular Forces 129Polarizability and Induced Dipoles . . . . . . . . . . 129Intermolecular Forces . . . . . . . . . . . . . . . . . . 130Group Work . . . . . . . . . . . . . . . . . . . . . . . 133

4 CONTENTS

Homework . . . . . . . . . . . . . . . . . . . . . . . . 133

34 Hydrogen Bonding and Water 135Water . . . . . . . . . . . . . . . . . . . . . . . . . . 137Group Work . . . . . . . . . . . . . . . . . . . . . . . 137Homework . . . . . . . . . . . . . . . . . . . . . . . . 137

35 Phase Diagrams 139Group Work . . . . . . . . . . . . . . . . . . . . . . . 140Homework . . . . . . . . . . . . . . . . . . . . . . . . 140

36 Homogeneous Solutions 141Liquid-solid solutions . . . . . . . . . . . . . . . . . . 141Liquid-liquid solutions . . . . . . . . . . . . . . . . . 142Liquid-gas solutions . . . . . . . . . . . . . . . . . . . 142Group Work . . . . . . . . . . . . . . . . . . . . . . . 143Homework . . . . . . . . . . . . . . . . . . . . . . . . 143

37 Properties of Solutions 145Henry’s Law . . . . . . . . . . . . . . . . . . . . . . . 145Raoult’s Law . . . . . . . . . . . . . . . . . . . . . . 146Colligative properties . . . . . . . . . . . . . . . . . . 147Group Work . . . . . . . . . . . . . . . . . . . . . . . 148Homework . . . . . . . . . . . . . . . . . . . . . . . . 149

38 Review for Final Exam 151

CONTENTS 5

Material for Exam I

6 CONTENTS

1 CALCULUS IN A NUTSHELL

It is so much easier to talk about chemical kinetics if we havejust a few ideas from calculus.

The following is all the calculus we need to know.

Rates of Change

When considering chemical kinetics we will be interested in howfast a reaction proceeds.

Experimentally this means we are in some way monitoring howthe reaction is changing in time.

We might be watching a reactant disappear as time goes on orwe might be watching a reactant appear.

We are measuring a rate of change of the reaction.

Mathematically the rate of change is called a derivative.

The concept of a derivative is one of two central concepts incalculus.

Let’s say we have some function that changes with time: f(t).We are interested in how it changes with time.

We could consider two times and compare the function at eachof these times.

4f = f(t2)− f(t1).

The rate of change might then be given by4t

4f

4t,

CALCULUS IN A NUTSHELL 7

where 4t = t2 − t1.

Unfortunately this only works if the rate of change is constantduring 4t.

Obviously if 4t is big then this might not be the case.

One of the central ideas of calculus is to consider the limit when4t gets infinitesimally small. We then write dt.

If dt is sufficiently small then

df

dt

is a good measure of the instantaneous rate of change.

In calculus this is called the derivative of the function f withrespect to t.

Since the derivative is an instantaneous rate of change it holdsfor any function not just those that have a constant rate ofchange.

Slope

If we graph the function f then the derivative has a very nicegraphical interpretation.

The derivative at any time t is equal to the slope of the functionat t.


The Anti-Derivative

As mentioned above, the derivative is one of two fundamentalconcepts in calculus.

The other is the anti-derivative, which is more commonly calledthe integral.

The integral is the answer to the question what function hasthis derivative?

Group Work

1. If a function is decreasing with time is it’s derivative pos-itive or negative.

2. If a derivative is always zero what must be true about itsanti-derivative?

3. Use a plot to show that

df

dt=4f

4t

only for a straight line.

4. If y = mx+ b for a straight line, what is

dy

dx?

Homework

Reading: Evaluation book, 14.1

CALCULUS IN A NUTSHELL 9


2 DIFFERENTIAL RATE LAWS

The Reaction Rate

Consider the reaction

aA+ bB → cC + dD. (2.1)

We can monitor the rate of this reaction in a number of ways.

We can watch either A or B disappear or we can watch eitherC or D appear.

That is, we can monitor the derivative of the concentration ofA, B, C or D with respect to time:

d[I]

dt,

where [I] is the concentration of A, B, C or D.

If we inspect the stoichiometry of the above reaction we seethat a moles of A react with b moles of B to make c moles ofC and d moles of D.

Therefore A and B do not disappear at the same rate if a 6= band C and D do not appear at the same rate if c 6= d.

But, since the stoichiometry is fixed for a given reaction, onecan easily adjust the derivative of the concentration with re-spect to time to give a reaction rate, v, that is consistent re-gardless of which reactant or product is monitored.

The rate (or velocity) of a reaction is

v =1

νI

d[I]

dt, (2.2)

DIFFERENTIAL RATE LAWS 11

where νI is the stoichiometric factor with the understandingthat it is negative for reactants and positive for products.

Thus for the reaction given above

v =1

−ad[A]

dt=1

−bd[B]

dt=1

c

d[C]

dt=1

d

d[D]

dt(2.3)

Rate Laws

One of the primary goals of chemical kinetics is to determinehow the rate of a reaction goes. That is to find the rate law forthe reaction.

Kinetics is very much an empirical science that depends muchmore heavily on experimental results than on a strong theoret-ical foundation.

As a result, rate laws must be experimentally determined ratherthan theoretically predicted.

One very important and very common class of rate laws areones in which the rate depends on the concentration of reac-tants present.

Again considering the reaction

aA+ bB → cC + dD,

the rate laws of this type would be

v = k [A]nA [B]nB , (2.4)

where nA and nB are constants, usually integers and k is therate constant for the reaction.

A differential rate law for this reaction is then

1

−ad[A]

dt= [A]nA [B]nB (2.5)

The nA and nB powers on the concentrations are called ordersof the reaction.


Thus one would say this reaction is of order nA in A, of ordernB in B, and of overall order nA + nB.

It is very important to note that nA and nB generally do nothave any relation to the stoichiometric factors a and b.

Elementary Reactions and Molecularity

In general an overall reaction is made up of so called elementaryreactions

Reactant → Product overall rxn (2.6)

Reactant → Intermediates → Product

Note that we shall use an equal sign when talking about theoverall reaction and arrows when talking about the elementaryreactions

ExampleLet

2A+ 2B → C +D (2.7)

be the overall reaction. One possible set of elementary stepscould be

elementary rxn molecularityA+A→ A0 BimolecularA0 → A00 Unimolecular

A00 + 2B → C +D Trimolecular

.

The rate laws for elementary reactions can be determined fromthe stoichiometry

molecularity elementary rxn rate lawUnimolecular A → Product v = k[A]Bimolecular A+A → Product v = k[A]2

Bimolecular A+B → Product v = k[A][B]Trimolecular A+A+A → Product v = k[A]3

Trimolecular A+A+B → Product v = k[A]2[B]Trimolecular A+B + C → Product v = k[A][B][C]

.

Conversely, rate laws for overall reactions can not be deter-mined by stoichiometry.

DIFFERENTIAL RATE LAWS 13

One of the reasons why one can not write a rate law directlyfrom the stoichiometry of an overall reaction has to do with theidea of the rate determining step. The rate determining step isthe one in the series of elementary reactions that has the lowestrate constant k.

Group Work

1. Write the rate for the following reaction three times, eachtime monitoring a different reactant or product.

2H2 +O2 → 2H2O

2. If the above reaction is elementary, write a differentialrate law for it.

Homework


Problems: Evaluation book Chapter 14: 3, 5, 9, 11, 12, 24,26, 27, 28, 51, 63, 64


3 DETERMINATION OF RATE LAWS

Concentrations [I] are measured not rates. To obtain the ratefrom the concentration we must examine how the concentrationchanges with time

1

νI

d[I]

dt

That is we must measure [I] as a function of time and find therate of change of this concentration curve.

The rates of chemical reactions vary enormously from sub-seconds to years. Consequently no one experimental techniquecan be used.

• For slow reactions (hrs/days) almost any technique formeasuring the concentration can be used.

• For medium reactions (min) either a continuous monitor-ing technique or a stopping technique can be used

— A stopping technique used rapid cooling or destruc-tion of the catalysts to stop a reaction at a givenpoint.

• Very fast (sec/subsec) reactions cause problems becausethe reaction goes faster than one can mix the reactants.

Differential methods based on the rate law

Methods based directly on the rate law rely on the determina-tion of the time derivative of the concentration.

1. Method of initial velocities

• for v = k[A]x[B]y rate laws.

DETERMINATION OF RATE LAWS 15

• initially v0 = kaxby where a and b are the initialconcentrations of A and B respectively

• taking the log of both sides gives ln v0 = ln[kaxby] =ln k + x ln a+ y ln b

• a and b can be varied independently so both x andy can be determined.

• usually the experiment is run so the order can beeasily determined by simple inspection.

• problems(a) if the concentration drops very sharply(b) if there is an induction period

2. Method of isolation

• for v = k[A]x[B]y rate laws

• flood with, say, A so v ≈ kax[B]y

• repeat for the other reactants

Example: Method of Initial Rates


aA+ bB → cC (3.1)

and the following experimental data

Run [A]0 [B]0 [A]95%I 0.1 M 0.1 M 20 minII 0.1 M 0.2 M 10 minII 0.2 M 0.1 M 10 min

The data from run I gives an initial rate of

vI =0.05× 0.1 M20 min

= 0.00025 M/min (3.2)

Likewise from runs II and III

vII =0.05× 0.1 M10 min

= 0.00050 M/min (3.3)

and

vIII =0.05× 0.2 M10 min

= 0.00100 M/min (3.4)

Now, by comparison we can obtain the order.


Comparing runs I and II we see the rate has doubled.

Thus doubling the concentration of B doubles the rate.

Therefore the order dependence on [B] must be one.

Comparing runs I and III we see the rate has doubled.

Thus doubling the concentration of A quadruples the rate.

Therefore the order dependence on [A] must be two.

So finally we can write the rate law as

v = k [A]2 [B]. (3.5)

Notice the orders have nothing to do with the stoichiometricfactors. They are equal to the stoichiometric factors only forelementary reactions.

Example: Method of Isolation

Again consider the reaction

aA+ bB → cC (3.6)

and the following experimental data.

Run [A]0 [B]0 obs. orderI 1.00 0.01 1II 0.01 1.00 2

From run I we see that the reaction depends on [B] to the firstpower

From run II we see that the reaction depends on [A] to thesecond power.

Therefore the overall rate law is

v = k [A]2 [B].

DETERMINATION OF RATE LAWS 17

Group Work

1. Sometimes the initial rate data is presented slightly dif-ferently. Consider the reaction

aA+ bB → cC

and the data

Run [A]0 [B]0 [A]10 minI 0.1 M 0.1 M 0.095 MII 0.1 M 0.2 M 0.090 MII 0.2 M 0.1 M 0.180 M

What is the overall rate law?

2. Consider the reaction

aA+ bB + cC → dD

and the following data

Run [A]0 [B]0 [C]0 [A]95%I 0.1 0.1 0.1 100 secII 0.1 0.1 0.2 50 secIII 0.1 0.2 0.1 100 secIV 0.2 0.1 0.1 100 sec

what is the overall rate law?

3. Explain in your own words the strategy of the method ofisolation.

Homework


Problems: Evaluation book Chapter 14: 14, 18, 48, 49, 53,54, 55, 56


4 INTEGRATED RATE LAWS

The previous lecture dealt with differential rate laws look di-rectly at the rate law which is a differential equation. Thedifferential equation is not solved.

We now solve the differential equations to yield what are calledthe integrated rate law.

The differential equations (rate law) and their solutions (inte-grated rate law) are simply listed here for a few rate laws.

type rate lawa) integrated rate lawa)

1st order 1νi

d[I]dt= k[I] [I] = [I0]e

νikt

2nd order 1νi

d[I]dt= k[I]2 1

[I]= 1

[I0]− νikt

nth orderb) 1νi

d[I]dt= k[I]n 1

[I]n−1 =1

[I0]n−1− (n− 1)νikt

enyzme 1νi

d[I]dt= k[I]

km+[I]km ln

[I0][I]+ ([I0]− [I]) = −νikt

a)[I] is the concentration of one of the reactants in an el-ementary reaction and νi is the stoichiometric factor for [I](n.b., νi is a negative number).

b)The order need not be an integer. For example n = 3/2 isa three-halves order rate law.

Linearizing the integrated rate laws

We see from the table above that the integrated rate law forfirst order behavior is

[I] = [I0]eνikt, (4.1)

which is of the form of an exponential.

INTEGRATED RATE LAWS 19

With a computer and a program like Microsoft Excel it iseasy to work with the data as is.

Sometimes, however, it is convenient to cast the data into linearform.

For a first order reaction this is done by taking the natural logof the integrated rate law

ln[I] = ln¡[I0]e

νikt¢

(4.2)

ln[I] = ln[I0] + ln¡eνikt

¢ln[I]|{z}y

= νikt|{z}mx

+ ln[I0]| {z }b

.

So, by plotting the natural log of experimental data versus time,first order reactions will appear as straight lines with a slopeequal to νik.

The second and higher order integrated rate laws given in thetable about are already in linearized form.

If a reaction is described by a second order rate law then whenone over the data is plotted one gets a straight line with slope−νik:

1

[I]|{z}y

= −νikt| {z }mx

+1

[I0]|{z}b

. (4.3)

The table lists the integrated rate laws for elementary reactions.

Over all reactions also follow these rate laws but now one cannot express the slopes of the linearized equations as a productof the rate constant and the stoichiometric factor.

Instead one uses an observed rate constant, kobs.

Thus overall reactions that appear to be first order follow thelinearized integrated rate law,

ln[I] = −kobst+ ln[I0] (4.4)

and second order reactions go as

1

[I]= kobst+

1

[I0](4.5)


Group Work

1. Identify “y”, “m”, “x” and “b” in the linearized nth orderintegrated rate law.

2. We have talked about first and second order reactions. Ifwe consider a zeroth order rate law, we must solve thedifferential equation

1

νi

d[I]

dt= k[I]0 = k

d[I]

dt= νik.

From what we learned in the first lecture, we can solvethis equation. The equation is asking us to find the func-tion [I] that has a constant derivative that is equal toνik.

(a) Given the slope interpretation of the derivative, whatis the most general type of function that has a con-stant slope for all time t?

(b) Given that [I] must equal [I]0 (a constant, equal tothe initial concentration) at t = 0

Homework

Reading: Evaluation book, 14.4, 14,5

Problems: Evaluation book Chapter 14: 10, 16, 20, 58, 59,61, 62

INTEGRATED RATE LAWS 21


5 THE ARRHENIOUS EQUATION

Temperature Dependence of the Rate Constant

We have learned that kinetics is an empirical science. That is,it is based more on experiment than on theory.

One experimental observations is that the rate of a reactiongenerally increases as the temperature increases.

In other words, the rate constant gets bigger with increasingtemperature.

An empirical rate constant that often captures the correct tem-perature dependence was proposed by Arrhenious:

k = Ae−EaRT (5.1)

where Ea is the Arrhenious activation energy and A is a con-stant called the frequency factor.

Notice that if one takes the natural log of the Arrhenious equa-tion one has

ln k = lnA− Ea

RT. (5.2)

If ln k is plotted versus 1Twe get a straight line whose slope is

EaRand intercept is lnA.

So one needs to know the rate constant at least two tempera-tures to determine the Arrhenious parameters.

THE ARRHENIOUS EQUATION 23

Application of the Arrhenious Equation

The best way to get a feel for the Arrhenious equation is towork a few examples.

Example 1: A reaction is found to go 1.5 times faster at 400K than at 300 K. What is the Arrhenious activation energy forthis reaction?

The reaction goes 1.5 times faster means k400 = 1.5k300. So

Ae−Ea400R = 1.5Ae−

Ea300R

e−Ea400R = 1.5e−

Ea300R .

Taking the natural log of both sides gives

− Ea

400R= ln 1.5− Ea

300R.

Solving for Ea gives

Ea

300R− Ea

400R= ln 1.5

Ea

µ1

300R− 1

400R

¶= ln 1.5

Ea =ln 1.5¡

1300R− 1

400R

¢Ea = 4045 J/mol

Example 2: The Arrhenious activation energy is 50.50 kJ/molfor some reaction. How much faster will the reaction go at 298K than at 273 K?

We want to find the ratio k298k273

. Thus,

k298k273

=Ae−

50500298R

Ae−50500273R

= e50500(1

273R− 1298R)

= 6.47.

So the reaction goes 6.47 times faster at 298 K than at 273 K.


Catalysts

The Arrhenious activation energy creates a barrier for reactionsto proceed.

The higher the energy barrier the slower the reaction will go.

We have learned that one way to speed up a reaction is to heatit.

Sometimes heating is not desirable.

There is an additional method for lowering the activation en-ergy barrier.

One can add a catalysts.

A catalyst is a species that participates in the reaction but isnot consumed by the reaction.

Reactants + Catylist → Products + Catylist (5.3)

Group Work

1. A certain reaction goes twice as fast at 1000 K than at300 K. What is the Arrhenious activation energy?

2. A certain reaction has an activation energy of 34 kJ/mol.How much faster will the reaction go at 300 K than at100 K?

3. Reaction I has a large Arrhenious activation energy andreaction II has a small one. Both reaction I and reactionII take the same amount of time to reach completion at298 K. Should you heat the reactions or cool the reactionsif you want reaction I to go to completion before reactionII?

THE ARRHENIOUS EQUATION 25

Homework

Reading: Evaluation book, 14.7, 14.8

Problems: Evaluation book Chapter 14: 131, 33, 37, 39, 41,68,. 69, 70, 71, 74, 75


6 KINETICS AND EQUILIBRIUM

Our next topic will be chemical equilibrium.

There is a relation between kinetics and equilibrium.

Last semester and up to this point we have been talking aboutreactions that go all the way to completion.

Most reactions, however, do not go to completion but reach anequilibrium with a certain amount of reactants still left over.

We can understand equilibrium from the standpoint of kineticsby not only considering the forward reaction,

aA+ aBkf→ cC + dD (6.1)

but also the reverse reaction,

cC + dDkr→ aA+ bB. (6.2)

These can be written as one reaction with a double arrow,

aA+ aBkfkr

cC + dD. (6.3)

The forward reaction rate depends on the concentration of re-actants,

vf = kf [A]nA[B]nB , (6.4)

while the rate of the backwards reaction depends on the con-centration of products

vr = kr[C]nC [D]nD . (6.5)

Equilibrium occurs when the reaction rate in both the forwardand reverse directions are equal,

vf = vr, (6.6)

or

kf [A]nA[B]nB = kr[C]

nC [D]nD (6.7)kfkr

=[C]nC [D]nD

[A]nA [B]nB= K

KINETICS AND EQUILIBRIUM 27

Group Work

1. Does the ratio of rate constants tell us how fast the reac-tion reaches equilibrium. Hint: consider the units.

Homework

Problems: Evaluation book Chapter 14: 78, 80, 81, 82, 84

28 KINETICS AND EQUILIBRIUM

7 EQUILIBRIUM AND THEEQUILIBRIUM CONSTANT

Last lecture we began to consider the fact that usually reactionsdo not go all the way to completion, but rather, they reach anequilibrium.

The topic of equilibrium is of essential importance to gen-eral chemistry and will be discussed for nearly the rest of thesemester.

The first topic in equilibrium we will discuss is the equilibriumconstant

The Equilibrium Constant

We already sort of saw the equilibrium constant last time.

For a given reaction,

aA+ bB cC + dD,

the equilibrium constant is defined as

K =[C]c [D]d

[A]a [B]b.

Notice here, unlike kinetics, we can rely on the stoichiometryof the reaction to give us the proper K values.

The equilibrium constant tells us how much of each of the reac-tants and products are present after the reaction reaches equi-librium.

EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT 29

Because of this, K provides a criteria for labeling a reaction asproduct favored or reactant favored.

Product favored reactions are those in which there are moreproducts around than reactants at equilibrium. Thus K > 1for product favored reactions.

Reactant favored reactions are those in which there are morereactants around than products at equilibrium. Thus K < 1for reactant favored reactions.

Sometimes concentrations are not the most convenient value touse for the equilibrium constant.

One example are gas phase reactions. Here partial pressuresare used in place of concentrations,

KP =P cCP

dD

P aAP

bB

.

Another case where one does not use concentration is whenreferring to the solvent. Consider the reaction

aA+ bB cC +H2O

that takes place in water.

Here water is the product and the solvent.

In these cases

K =[C]c

[A]a [B]b,

The concentration of water is not used.

One also does this when pure substances are used as reactantsor produces as products. Consider

aA(aq) + bB(s) cC(aq) + dD(aq).

Here

K =[C]c [D]d

[A]a,

30 EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT

Group Work

1. Write the equilibrium constant for the following reactions

(a)

H2CO3(aq) + H2O HCO−3(aq) + H3O+(aq)

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

(c)O2(g) + 2 H2(g) 2 H2O(l)

Homework

Reading: Kotz and Treichel, 16.1, 16.2, 16.3

Problems: Chapter 16: 1, 3, 5, 7, 8, 9, 10, 11

EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT 31

32 EQUILIBRIUM AND THE EQUILIBRIUM CONSTANT

8 ICE CHARTS

One of the most important things to be able to do with theequilibrium constant is to calculate the concentration of prod-ucts and reactants at equilibrium given an initial concentrationof reactants.

These calculations can be aided through the used of an ICEchart.

The “I” stands for initial concentration, the “C” stands forchange in concentration, and the “E” stands for equilibriumconcentration.

Use of ICE charts are best illustrated by way of examples.

Example 1: Consider the reaction

A(aq) + 2B(aq) C(aq), (8.1)

where K = 1.23 × 102. Initial concentrations of A and Bare each 1.00 M. Determine the concentration of reactants andproducts at equilibrium.We set-up our ICE chart

A B CI 1.00 1.00 0C −x −2x xE 1.00− x 1.00− 2x x

We now use the equilibrium constant,

K = 1.23× 102 = x

(1.00− x) (1.00− 2x)2(8.2)

We can rearrange this to give

492x3 − 984x2 + 616x− 123 = 0 (8.3)

Unfortunately we are now faced with a cubic equation to solve.

ICE CHARTS 33

This is hard to solve algebraically but you can use your calcu-lators to get three solutions: x = 0.459, x = 0.550, x = 0.992.

Only one of these solutions can be the real answer. We canfigure that out by inspection.

Since the final concentration of B is 1.00 − 2x, only the x =0.459 solution makes sense.

The other solutions give a negative concentration for B.

Thus the final equilibrium conditions are

[A] = 1.00− 0.459 = 0.541 M (8.4)

[B] = 1.00− 2× 0.459 = 0.082 M[C] = 0.459 M

Example 2: Consider the reaction

A(g) B(g) + C(g), (8.5)

where Kp = 4.23 × 10−1. The initial partial pressure of Ais 0.333 atm. Determine the concentration of reactants andproducts at equilibrium.First we set-up the ICE chart

A B CI 0.333 0 0C −x x xE 0.333− x x x

The equilibrium equation is

Kp = 4.23× 10−1 =(x) (x)

(0.333− x)(8.6)

This can be rearranged into

x2 + 0.423x− 0.141 = 0 (8.7)

This is a quadratic equation so we can easily get the solutionsvia the quadratic formula,

ax2 + bx+ c = 0

x = −b±√b2 − 4ac2a

(8.8)

34 ICE CHARTS

For this example,

x = −0.423±p0.4232 − 4 (0.141)2

x = −0.642, x = 0.219. (8.9)

Again, we need to determine the real solution by inspection.A negative x does not make physical sense so x = 0.219 is oursolution.

The final equilibrium conditions are

PA = 0.333− 0.219 = 0.114 atmPB = 0.219 atm (8.10)

PC = 0.219 atm

Group Work

1. Normally when we use the stoichiometry of a reaction wemust compare moles, but the ICE chart uses concentra-tion directly. Why does this work? Can you think ofsituations where it would not work?

2. Use the ideal gas law to derive the relationship betweenK and KP :

KP = K(RT )4n.

Homework

Reading: Kotz and Treichel, 16.4, 16.5


ICE CHARTS 35

36 ICE CHARTS

9 THE REACTION QUOTIENT ANDLECHATELIER’S PRINCIPLE

We have seen that the equilibrium constant is an importantquantity.

It turns out that one can extend the definition of the equilib-rium constant to non-equilibrium situations.

The Reaction Quotient

Under non-equilibrium situations the ratio of the concentra-tion (or partial pressures) of products to reactants is called thereaction quotient,

Q =[products][reactants]

. (9.1)

So, for the reaction,

aA+ bB cC + dD, (9.2)

the reaction quotient is

Q =[C]c [D]d

[A]a [B]b. (9.3)

The reaction quotient applies to the current situation of thereaction and thus changes over the course of the reaction.

Clearly, at equilibrium Q = K.

Beyond that, however, if Q < K the reaction will proceed inthe forward direction (make more products).

And, if Q > K the reaction will proceed in the reverse direction(make more reactants).

THE REACTION QUOTIENT AND LECHATELIER’S PRINCIPLE 37

LeChatelier’s Principle

One of the most important concepts in equilibrium is LeChate-lier’s principle which describe how systems behave when theyare taken out of equilibrium

LeChatelier’s principle states that when any factor that deter-mines the equilibrium condition of the system is changed, thesystem will respond in such a way to counteract and reduce theeffect of that change.


aA+ bB cC + dD, (9.4)

at equilibrium.

Now consider adding more reactant A or B.

Takes the system out of equilibrium to the reactant side of thereaction.

LeChatelier’s principle says the system will respond by makingmore product.

Conversely if more product is added the system is shifted outof equilibrium to the product side.

LeChatelier’s principle says the system will respond by goingin reverse to make more reactant.

We can capture LeChatelier’s principle mathematically by con-sidering the reaction quotient.

At equilibrium Q = K. If reactant is added then the denomi-nator of Q gets bigger and Q < K.

This Q < K condition says that the reaction will proceed for-ward.

Similarly, if product is added the system the numerator of Qgets better and Q > K.

38 THE REACTION QUOTIENT AND LECHATELIER’S PRINCIPLE

This Q > K condition says that the reaction will go in reverse.

Temperature can also play a role.

If a reaction is exothermic we can think of heat as one of theproducts.

Conversely if a reaction is endothermic we can think of heat asone of the reactants.

For an endothermic reaction at equilibrium, increasing the tem-perature will shift the reaction out of equilibrium in the reac-tant direction and the system will proceed towards products

For an exothermic reaction at equilibrium, increasing the tem-perature will shift the reaction out of equilibrium in the productdirection and the system will proceed towards reactants

Here too we can view this property in terms of the reactionquotient.

Now, however, it is the equilibrium constant that changes withtemperature.

For endothermic reactions K ↑ as T ↑ and for exothermic re-actions K ↓ as T ↑ .

So, if the system is at equilibrium at one temperature K = Q.

If the temperature changes then K changes while Q stays thesame.

Consequently, K 6= Q and the system will adjust accordingly.

Group Work

1. Let’s say an important drug was produced in a reaction

A+B Drug.

but the equilibrium constant was small. The reactantsA and B are both very expensive. Suggest a way to use

THE REACTION QUOTIENT AND LECHATELIER’S PRINCIPLE 39

LeChatelier’s principle to get the most drug for the leastamount of reactant.

2. Consider the above problem but now A is very expensivebut B is very cheap. Also assume that your suggestionfrom above was difficult to do in the lab for this drug.Use LeChatelier’s principle to get the most amount ofdrug for the least expense.

3. How would LeChatelier’s principle apply to gas phase re-action where the pressure is changed?

Homework


Problems: Chapter 16: 25, 26, 27, 28, 41, 43, 49

40 THE REACTION QUOTIENT AND LECHATELIER’S PRINCIPLE

10 REVIEW FOR EXAM 1

REVIEW FOR EXAM 1 41

Material for Exam II


11 ACIDS AND BASES

We are familiar with acids and bases and their reactions fromlast semester.

We will spend the next few lectures looking at acid—base reac-tions in the context of equilibrium

Brønstad Acids and Bases

Our first definition of an acid or base is what called the Brønstad-Lowry concept of an acid or base.

A Brønstad acid is any compound that can donate a proton

A Brønstad base is any compound that can accept a proton

Brønstad acids that can donate one proton, like HCl are calledmonoprotic acids.

Brønstad acids that can donate more than one proton, likeH2SO4 are called polyprotic acids.

Conjugate Acid-Base Pairs

Within the Brønstad-Lowry concept of an acid or base, an acid—base reaction involves the transfer of a proton from the acid tothe base,

HA + B− A− + HB (11.1)

Here HA in it is the acid and B− is the base.

ACIDS AND BASES 43

We could write the reaction in reverse,

A− + HB HA + B− (11.2)

Now A− is the base and HB is the acid.

So within the reaction HA and A− form a so-called conjugateacid-base pair. Likewise B− and HB form a conjugate pair.

For conjugate acid base pairs, the stronger one partner is, theweaker the other partner is.

So a strong acid has a corresponding weak conjugate base anda weak acid has a strong conjugate base.

For polyprotic acids we must consider each proton donationseparately,

H2A + B2− HA− + HB− A2− +H2 B.

Here the conjugate pairs are

H2A ⇐⇒ HA−

HA− ⇐⇒ A2−

HB− ⇐⇒ B2−

H2 B ⇐⇒ HB−

Group Work

1. Give the conjugate partner for

(a) HNO3

(b) H2CO3

(c) HCO−3(d) NH3

(e) H2O

2. Identify the conjugate acid-base pairs in the following re-actions

44 ACIDS AND BASES

(a)CH3CO2H + NH3 CH3CO−2 + NH

+4

(b)H2S + H2O HS− + H3O+

3. Amino acids have an interesting structure in that theyall have a region that acts like and acid and a regionthat acts like a base. The simplest amino acid is glycine,H2NCHCOOH. The nitrogen end can act like a Brønstadbase by accepting another proton whereas the oxygen endcan act like a Brønstad acid by giving up the proton at-tached to it. Write two separate reactions of glycine firstwith a Brønstad acid and then with a Brønstad base.

Homework


Problems: Chapter 17: 1, 2, 3, 5, 6, 7

ACIDS AND BASES 45

46 ACIDS AND BASES

12 WATER AND PH

We learned about the pH scale last semester and we saw thatit was a result of the fact that water is a weak electrolyte.

That is water under goes autoionization,

2H2O(l) H3O+(aq) +OH−(aq) (12.1)

We can write the equilibrium constant for this reaction as

K = Kw =£H3O+

¤ £OH−

¤. (12.2)

Notice that we do not include the concentration of H2O becauseit a considered a pure liquid.

At standard conditions of 25 ◦C,

Kw = 1.0× 10−14 =£H3O+

¤ £OH−

¤.

By the symmetry of the reaction, two water molecules createsone H3O+ and one OH−, then for pure water£

H3O+¤=£OH−

¤= 1.0× 10−7.

If we now consider the addition of an Brønstad acid to water,we see that the H3O+ goes up: Q > Kw.

By LeChatelier’s principle, the autoionization is driven back-wards,

H3O+(aq) +OH−(aq) → 2H2O(l),

The only source of OH− is from the original water. Thereforethe concentration of OH− goes down.

More precisely the autoionization reaction is driven backwardsuntil

£H3O+

¤ £OH−

¤= Kw.

A similar situation arrises for addition of a Brønstad base.

WATER AND PH 47

The pH Scale

The inverse relationship between£H3O+

¤and

£OH−

¤is seen in

the pH scale.

Consider taking the negative log of Kw,

− logKw = − log¡1.0× 10−14

¢= − log

¡£H3O+

¤ £OH−

¤¢14 = − log

£H3O+

¤| {z }pH

− log£OH−

¤| {z }pOH

14 = pH + pOH

The pH of Strong Acids and Bases

If one is dealing with strong acids and bases, determinationof pH is relatively easy based on the LeChatelier argumentsabove.

Example 1: What is the pH of a 2.3×10−3 M solution of HCl.

Since HCl is a strong acid, it fully dissociates to produce a2.3× 10−3 M solution of H3O+.

The pH is simply

pH = − log¡2.3× 10−3

¢= 2.63

Example 2: What is the pH of a 2.3 × 10−3 M solution ofNaOH.

Since NaOH is a strong base, it fully dissociates to produce a2.3× 10−3 M solution of H3O+.

The pOH is

pOH = − log¡2.3× 10−3

¢= 2.63

From the 14 =pH + pOH relationship

pH = 14− 2.63 = 11.36

48 WATER AND PH

Group Work

1. Identify the conjugate pairs in the autoionzation reactionfor water.

2. Considering the Kw data for different temperatures givenin the table below

◦C Kw

10 0.29× 10−1425 1.00× 10−1450 5.48× 10−14

(a) Is the autoionization reaction exothermic or endother-mic?

(b) What is the pH range at 50 ◦C?

3. What is the pH of a 4.56× 10−2 M solution of HNO3?

4. What is the pH of a 7.89 × 10−10 M solution of KOH?Does this answer make any sense? If not, why not?

Homework

Reading: Kotz and Treichel, 17.3,


WATER AND PH 49

50 WATER AND PH

13 ACID AND BASE EQUILIBRIA

Up until now we have been considering strong acids and basesin which the acid or base fully dissociates.

For weak acids and bases we must set-up an equilibrium ex-pression.

For a weak acid equilibrium reaction,

HA H+ +A−, (13.1)

the equilibrium expression is

Ka =

£H+¤ £A−¤

[HA](13.2)

where Ka is called the acid dissociation constant.

The larger the Ka the stronger the acid

Likewise for a weak base equilibrium reaction,

BOH B+ +OH−, (13.3)

the equilibrium expression is

Kb =

£OH−

¤ £B+¤

[BOH](13.4)

where Kb is called the base dissociation constant.

There is a relationship between Ka and Kb for conjugate acid—base pairs. The relationship is

KaKb = Kw. (13.5)

If the acid (or base) is polyprotic then there are multiple Ka

values (Ka1, Ka2, etc.) for each of the dissociation reactions.

ACID AND BASE EQUILIBRIA 51

Usually Ka1 À Ka2 À Ka3.

For polyprotic acids (or bases) we must consider separate equi-librium reactions

H2A H+ +HA− (13.6)

HA− H+ +A2−

where, for this case,

Ka1 =

£H+¤ £HA−

¤[H2A]

, (13.7)

Ka2 =

£H+¤ £A2−

¤£HA−

¤Example 1: Calculate the pH of a 0.012 M solution of niacin(HC 6H 4NO2). The Ka = 1.4× 10−5.

We first write the appropriate equilibrium reaction

HC6H4NO2 H+ +C6H4NO−2 . (13.8)

Next we setup an ICE chart,

[HC6H4NO2]£H+¤ £

C6H4NO−2¤

I 0.012 0 0C −x x xE 0.012− x x x

.

From Ka we get the equation,

Ka = 1.4× 10−5 =(x) (x)

(0.012− x)(13.9)

Rearranging and solving for x using the quadratic formula weget

x =£H+¤= 4.1× 10−5. (13.10)

Therefore the

pH = − log¡4.1× 10−5

¢= 3.39 (13.11)

Example 2: What is the pH of a 0.10 M solution of vitaminC (H 2C 6H 6O6). The Ka1 = 7.9×10−5 and Ka2 = 1.6×10−12.

In general this problem is more complicated because of the twoKa values. However, as will often be the case Ka1 À Ka2 soKa1 alone will determine the pH.


Like above we first write that appropriate reaction,

H2C6H6O6 H+ +HC6H6O−6

and produce an ICE chart

[H2C6H6O6]£H+¤ £

HC6H6O−6¤

I 0.10 0 0C −x x xE 0.10− x x x

The Ka1 equation gives

Ka1 = 7.9× 10−5 =(x) (x)

(0.10− x).

Rearranging and using the quadratic formula gives

x =£H+¤= 0.0028.

Therefore the

pH = − log (0.0028) = 2.55

Example 3: Calculate the concentration of C 6H 6O2−6 for the

case of example 2.

Now we do need the second dissociation reaction

HC6H6O−6 H+ +C6H6O2−6

and we setup and ICE chart where we use the

x =£HC6H6O−6

¤= 0.0028

result from example 2.

The ICE chart is£HC6H6O−6

¤ £H+¤ £

C6H6O2−6¤

I 0.0028 0 0C −y y yE 0.0028− y y y

.

Utilizing

Ka2 = 1.6× 10−12 =(y) (y)

(0.0028− y)

and solving to get

y =£C6H6O2−6

¤= 6.7× 10−8

ACID AND BASE EQUILIBRIA 53

Group Work

1. What is the pKa for niacin and what are the pKa1 andpK2 values for vitamin C?

2. What is the pH of a 0.10 M solution of the base, hydrazine(N2H2). The Kb = 1.7× 10−6.

3. What is the equilibrium concentration of SO2−3 if one ini-tially prepares a 0.100 M solution of H2SO3? The Ka val-ues are: Ka1 = 1.3× 10−2 and Ka2 = 6.3× 10−8.

Homework

Reading: Kotz and Treichel, 17.4

Problems: Chapter 17: 15, 17, 19, 21, 23, 25, 27, 29, 31


14 REACTIONS WITH WEAK ACIDSAND WEAK BASES

We have learned how to determine the pH of a weak acid orweak base solution.

We will now extend that to include calculating the pH after aacid—base reaction.

Reaction Between aMonoproticWeakAcid and a StrongBase

We first consider a reaction between a weak monoprotic weakacid with a strong base.

We will take the reactants to be present in stoichiometric ratios.We will not consider limiting reagent cases until later.

Since this is a reaction between a weak base and a strong basewe would expect the final pH to be basic.

Example: What is the final pH if 25 ml of a 0.10 M niacin(HC 6H 4NO2) is reacted with 25 ml of 0.10M NaOH. For niacinthe Ka = 1.4× 10−5.We first write a balanced reaction,

HC6H4NO2 +NaOH NaC6H4NO2 +H2O. (14.1)

It is convenient to express the net ionic equation,

HC6H4NO2 +OH− C6H4NO−2 +H2O. (14.2)

We know that sodium hydroxide is a strong base so it fullydissociates to produce OH−.

REACTIONS WITH WEAK ACIDS AND WEAK BASES 55

All of the OH− reacts with the H+ from niacin.

Now this leaves only C6H4NO−2 , but since this is the conjugatebase to a weak acid we must set-up a base dissociation reaction

C6H4NO−2 HC6H4NO2 +OH−.

The Kb for this reaction is obtain from the Ka for niacin andKw,

Kb =Kw

Ka= 7.14× 10−10. (14.3)

We will need to setup and ICE chart but first we must deter-mine the concentration of C6H4NO−2 .

To do this we see,

0.025 L HC6H4NO2 × 0.10 mol HC6H4NO2L

× 0.10 mol C6H4NO−20.10 mol HC6H4NO2

= 0.0025 mol C6H4NO−2

But we also had 25 mL from the NaOH solution so the totalvolume is now 50 mL thus,£C6H4NO−2

¤=0.0025 mol C6H4NO−2

0.050 L= 0.05 M C6H4NO−2

Using this, the ICE chart is£C6H4NO−2

¤[HC6H4NO2]

£OH−

¤I 0.05 0 0C −x x xE 0.05− x x x

.

So, we have

Kb = 7.14× 10−10 =(x) (x)

(0.05− x)

Rearranging and using the quadratic formula we get x = [OH] =5.97× 10−6. From this

pOH = − log¡5.97× 10−6

¢= 5.22.

Hence,pH = 14− 5.22 = 8.78.

The reaction between a weak base and a strong acid would bedone in a similar manner.

56 REACTIONS WITH WEAK ACIDS AND WEAK BASES

Reaction Between a Polyprotic Weak Acid and a StrongBase

The reaction of a weak polyprotic acid and a strong base is alittle more complicated but is nonetheless similar to what wehave done.

Example: What is the final pH of a reaction between 25 mlof a 0.10 M Vitamin C (H 2C 6H 6O6) solution and 25 mL ofa 0.20 M NaOH solution? The Ka1 = 7.9 × 10−5 and Ka2 =1.6× 10−12.

Like above we write a balanced reaction

H2C6H6O6 + 2NaOH Na2C6H6O6 + 2H2O

H2C6H6O6 + 2OH− C6H6O2−6 + 2H2O

All the NaOH reacted with the vitamin C to produce onlyC6H6O2−6 . The concentration of C6H6O2−6 is

0.025 L H2C6H6O6 × 0.10 mol H2C6H6O6L

× 0.10 mol C6H6O2−60.10 mol H2C6H6O6

= 0.0025 mol C6H6O2−6

The total volume after the reaction is 50 mL so,£C6H6O2−6

¤=0.0025 mol C6H6O2−6

0.050 L= 0.05 M C6H6O2−6 .

We need to now consider the weak base equilibrium of C6H6O2−6 .The Kb’s are obtained from the Ka’s,

Kb1 =Kw

Ka1= 1.2× 10−10

Kb2 =Kw

Ka2= 6.3× 10−3.

Since Kb2 À Kb1, we need only consider the reaction

C6H6O2−6 HC6H6O−6 +OH−

and the ICE chart is£C6H6O2−6

¤ £HC6H6O−6

¤ £OH−

¤I 0.05 0 0C −x x xE 0.05− x x x

.


From this

Kb2 = 6.3× 10−3 =(x) (x)

(0.05− x)

Rearranging and using the quadratic formula we get x =£OH−

¤=

0.015 andpOH = − log (0.015) = 1.82.

So, finally,pH = 14− 1.82 = 12.18

Reaction Between a Weak Acid and a Weak Base

The reaction between a weak acid and a weak base requires adifferent strategy since we can not exploit the full dissociationof the strong acid or base as we did above.

We can perform this calculation by using Hess’ law from lastsemester.

Example: What are the final concentrations of products if 25ml of a 0.10 M niacin solution (Ka = 1.4 × 10−5) is reactedwith 25 ml of a 0.10 M NH 3 solution (Kb = 1.8× 10−5)?

As before, we start with a balanced equation

HC6H4NO2 +NH3 C6H4NO−2 +NH+4 .

We now use Hess’ law to write

HC6H4NO2 H+ +C6H4NO−2 Ka = 1.4× 10−5

NH3 +H2O NH+4 +OH− Kb = 1.8× 10−5

H+ +OH− H2O 1/Kw = 1.0× 10+14

This gives the original reaction and the equilibrium constantfor this reaction is

K =KaKb

Kw= 2.5× 10+4

Upon mixing the volume of the solution is doubled so the initialconcentrations are half of what they are prior to mixing.


We can then setup an ICE chart as

[HC6H4NO2] [NH3]£C6H4NO−2

¤ £NH+4

¤I 0.05 0.05 0 0C −x −x x xE 0.05− x 0.05− x x x

Thus,

K = 2.5× 10+4 = (x) (x)

(0.05− x) (0.05− x)

Rearranging a solving for x gives x = 0.0496. So,

[HC6H4NO2] = 0.0004 M

[NH3] = 0.0004 M£C6H4NO−2

¤= 0.0496 M£

NH+4¤= 0.0496 M

Work Group

1. What is the final pH if 25 ml of a 0.10 M ammonia solu-tion is reacted with 25 ml of 0.10 M HCl? For ammoniathe Kb = 1.8× 10−5.

Homework

Reading: Kotz and Treichel, 17.5, 17.6, 17.7, 17.8, 17.9




15THE COMMON ION EFFECT ANDBUFFERS

It is important to consider an acid dissociation reaction (orbase reaction) in which a compound that contains one of theproduct ions is added

HA H+ +A− add H+ or A−.

By LeChatelier’s principle the reaction is shifted back towardsthe reactant.

This shift in the equilibrium by LeChatelier’s principle is calledthe common ion effect.

Example: How does the equilibrium shift when 10 mL of 0.010M HCl is added to 25 mL of a 0.10 M niacin (HC 6H 4NO2)solution (Ka = 1.4× 10−5)?

We first need to determine the initial concentrations of niacinand the, common ion, H+ upon mixing. The new volume is 35mL so,

0.025 L×0.1 mol HC6H4NO21 L

× 1

0.035 L= 0.071 M HC6H4NO2

and

0.01 L × 0.01 mol H+

1 L× 1

0.035 L= 0.0028 M H+

We now can setup an ICE chart,

[HC6H4NO2]£H+¤ £

C6H4NO−2¤

I 0.071 0.0028 0C −x x xE 0.071− x 0.0028 + x x

.

THE COMMON ION EFFECT AND BUFFERS 61

With the Ka value we can get the equilibrium concentrations,

Ka = 1.4× 10−5 =(0.028 + x) (x)

(0.071− x). (15.1)

Rearranging and using the quadratic formula we get x = 3.2×10−4. Hence

[HC6H4NO2] = 0.071− 3.2× 10−4 = 0.071£H+¤= 0.0028− 3.2× 10−4 = 0.0025£

C6H4NO−2¤= 3.2× 10−4

Buffer Solutions

An important application of the common ion effect occurs whenthe common ion is the conjugate pair to the original acid orbase.

These solutions are called pH buffer solutions (or buffers forshort) and they have the valuable property that one can add astrong acid or base to a buffer and not change the pH signifi-cantly.

The human body makes use of many buffers to control the pHof the extra- and intra- cellular environments.

Example, the acetate buffer (part 1): Calculate the pHof an acetate buffer that is prepared by adding 25 ml of a 0.10M acetic acid (HC 2H 3O2) solution to 15 mL of 0.10 M of asodium acetate (NaC 2H 3O2) solution (Ka = 1.7× 10−5).

We first determine the initial concentrations of acetic acid andacetate ion,

0.025 L × 0.10 mol HC2H3O21 L

× 1

0.040 mL= 0.063 M HC2H3O2

0.015 L × 0.10 mol C2H3O−2

1 L× 1

0.040 mL= 0.038 M C2H3O−2 ,

then setup an ICE chart

[HC2H3O2]£H+¤ £

C2H3O−2¤

I 0.063 0 0.038C −x x xE 0.063− x x 0.038 + x

.

62 THE COMMON ION EFFECT AND BUFFERS

Using,

Ka = 1.7× 10−5 =(x) (0.038 + x)

(0.063− x),

and the quadratic formula, we get x =£H+¤= 2.8× 10−5. So,

pH = − log¡2.8× 10−5

¢= 4.55.

Example, the acetate buffer (part 2): Calculate the pHif 10 ml of 0.01 M HCl is added to the above buffer. Thencompare this to the pH if the HCl was added to water.

First we figure out how many moles of H+ is coming from theHCl,

0.010 L × 0.010 mol H+

1 L= 0.0001 mol H+.

We take it that all the H+ reacts with C6H4NO−2 to makeHC2H3O2. Thus we must determine the new concentrationsof C6H4NO−2 and HC2H3O2.

From above we know the equilibrium concentrations of C6H4NO−2and HC2H3O2 before the HCl is added. These are

[HC2H3O2] = 0.063− 1.7× 10−5 = 0.063 M£C2H3O−2

¤= 0.038 + 1.7× 10−5 = 0.038 M

So the number of moles of each present are

0.063 mol HC2H3O21 L

× 0.040 L = 0.0025 mol HC2H3O2

and

0.038 mol HC2H3O21 L

× 0.040 L = 0.0015 mol C2H3O−2 .

After completely reacting with the 0.0005 mol of H+ from theHCl we have

0.0025 + 0.0001 = 0.0026 mol HC2H3O2

and0.0015− 0.0001 = 0.0014 mol HC2H3O2.

Now we need to get the new concentrations

0.0026 mol HC2H3O20.050 L

= 0.052 M HC2H3O2


and0.0014 mol C2H3O−2

0.050 L= 0.028 M C2H3O−2 .

Finally we now need to setup an ICE chart,

[HC2H3O2]£H+¤ £

C2H3O−2¤

I 0.052 0 0.028C −x x xE 0.052− x x 0.028 + x

.

Using

Ka = 1.7× 10−5 =(x) (0.028 + x)

(0.052− x)

and solving the quadratic equation we get x =£H+¤= 3.15×

10−5. So,pH = − log

¡3.15× 10−5

¢= 4.50

So a change in pH of only 0.05 units.

We now calculate the pH for the situation of adding the HClto pure water.

From above we see that 0.0001 mol of H+ are produced.

These moles are now in a total volume of 40 mL so,

£H+¤=0.0001 mol H+

0.040 mL= 0.0025 M

Hence a pH of

pH = − log (0.0025) = 2.60,

which is a change of 4.55− 2.60 = 1.95 units.

Group Work

1. Calculate the pH change when 15 ml of 0.0010 M NaOHis added to the buffer of the example in lecture.

2. What is the pH of an ammonium buffer that is made bymixing 50 mL of a 0.50 M NH3 solution with 50 mL of a0.30 M NH4Cl solution?


3. It will be important to be able to work the above buffercalculations in a different way so that we can prepare abuffer of a desired pH. Prepare 100 ml of a acetate bufferwith a pH of 4.25.

(a) First use the pH find£H+¤

(b) Now you know£H+¤and Ka. Solve for the ratio,£

C2H3O−2¤/ [HC2H3O2] . Any 100 ml solution that

has this ratio of concentrations will give the desiredpH.

Homework





16 MORE ON BUFFERS: THEHENDERSON-HASSELBALCH

EQUATION

As we learned from the third group work question last time, itis important to be able to make buffers of a desired pH.

If we consider the Ka equation for a buffer situation we have,

Ka =

£H+¤ £A−¤

[HA], (16.1)

for the buffer equilibrium,

HA H+ +A−. (16.2)

Now if we take of the negative log of the Ka equation, we get

− logKa = − logÃ£H+¤ £A−¤

[HA]

!

− logKa| {z }pKa

= − log£H+¤| {z }

pH

− logÃ£A−¤

[HA]

!

pKa = pH− logÃ£A−¤

[HA]

!

Solving for pH gives the Henderson—Hasselbalch equation:

pH = pKa + log

Ã£A−¤

[HA]

!.

We can use the Henderson—Hasselbalch equation to short-cutour buffer calculations.

Example 1: How would you make an acetate buffer that hasa pH of 4.25 (Ka = 1.7× 10−5)?

MORE ON BUFFERS: THE HENDERSON-HASSELBALCH EQUA-TION 67

ThepKa = − log 1.7× 10−5 = 4.77

Using the Henderson—Hasselbalch equation we have

4.25 = 4.77 + log

Ã £C2H3O−2

¤[HC2H3O2]

!

log

Ã £C2H3O−2

¤[HC2H3O2]

!= 4.25− 4.77 = −0.52Ã £

C2H3O−2¤

[HC2H3O2]

!= 10−0.52 = 0.30

This tells use that at equilibrium the ratio must be 0.30.

Technically we should back calculate from the equilibrium con-centrations to the initial concentrations.

However, for buffers the equilibrium concentrations of the acidand the conjugate base do not differ significantly from the start-ing concentrations.

Thus one way to make this buffer would be to mix, say, 50mL of a 0.2 M solution of HC2H3O2 with 50 mL of a 0.06 Msolution of NaC2H3O2.

Group Work

1. How would you make an acetate buffer that has a pH of5.00 (Ka = 1.7× 10−5)?

2. What would the pH of an acetate buffer made by com-bining 50 mL of a 1.00 M HC2H3O2 solution with 50 mLof a 1.25 M NaC2H3O2 solution?

3. It is a good idea to choose an acid base conjugate pairthat has a pKa that is near to the pH that one is tryingto make a buffer for. Why?

4. Let say you wanted to hold your pH to within 0.1 pH unitaround the pKa of acetic acid. What is the acceptable

range of the ratio [C2H3O

−2 ]

[HC2H3O2]?

68 MOREON BUFFERS: THE HENDERSON-HASSELBALCH EQUATION

Homework

Problems: Chapter 18: 11, 13, 15, 17, 18

MORE ON BUFFERS: THE HENDERSON-HASSELBALCH EQUA-TION 69

70 MOREON BUFFERS: THE HENDERSON-HASSELBALCH EQUATION

17 TITRATIONS PART I

Last semester in lab and in lecture we gained some experiencewith titrations.

We learned that titrations are important techniques for deter-mining the concentration of the analyte.

We now revisit titrations armed with our knowledge of acid—base equilibria.

Titration of a Strong Acid with a Strong Base

The simplest case of a titration reaction, and one we couldhandle last semester, is that involving a strong acid and a strongbase.

Since both the analyte and the titrant are strong electrolytesthey fully dissociate and therefore fully react.

Example: 25 mL of a 0.100 M HCl solution is titrated with0.100 M NaOH. Calculate the titration curve for this process.

This titration reaction is

HCl+NaOH NaCl+H2O (17.1)

The total number of moles of HCl is

0.025 L × 0.100 mol HCl1 L

= 0.0025 mol HCl (17.2)

This problem breaks into three parts, i) before the equivalencepoint, ii) the equivalence point and iii) after the equivalencepoint.

TITRATIONS PART I 71

Before the equivalence point NaOH is the limiting reagent.

If X L of NaOH is added then

0.100 M NaOH1 L

× X L= 0.1X mol NaOH (17.3)

have reacted with the HCl. Thus 0.0025− 0.1X moles of HClremains.

Then, £H+¤=

0.0025− 0.1X0.025 +X

(17.4)

pH = − logµ0.0025− 0.1X0.025 +X

¶At the equivalence point the HCl and NaOH exactly neutralizeone another and the pH= 7.00.

This occurs at

0.0025 mol HCl×1 mol NaOH1 mol HCl

× 1 L NaOH0.100 mol NaOH

= 0.025 L NaOH

(17.5)After the equivalence point HCl is the limiting reactant.

If Y L of NaOH is added then

Y L NaOH × 0.100 mol NaOH1 L

= 0.1Y mol NaOH (17.6)

have been added.

Of these moles 0.0025 mol react with the HCl leaving 0.1Y −0.0025 mol NaOH unreacted.

From this £OH−

¤=

0.1Y − 0.00250.025 + Y

(17.7)

pOH = − logµ0.1Y − 0.00250.025 + Y

¶Thus,

pH = 14−∙− log

µ0.1Y − 0.00250.025 + Y

¶¸(17.8)

The titration curve is shown in the figure below


0 10 20 30 40 500

2

4

6

8

10

12

14

pH

mL NaOH added

Titration of a Weak Acid with a Strong Base

This situation is complicated by the need to determine thedissociation equilibrium for the weak acid.

We have, however, learned enough to handle this problem.

Example: Calculate the titration curve when 25 mL of a 0.10M niacin (HC 6H 4NO2) solution, Ka = 1.4 × 10−5, is titratedwith 0.10 M NaOH.

As with strong acid, strong base titrations we need to split theproblem into three parts: (i) before the equivalence point, (ii)the equivalence point, and (iii) after the equivalence point.

The total number of moles of HCl is0.025 L × 0.100 mol HC6H4NO2

1 L= 0.0025 mol HC6H4NO2

(17.9)

Before the equivalence point NaOH is the limiting reagent.

Let’s say X mL of NaOH is delivered thus 0.1X mol of NaOHreacts fully with the niacin.


This leaves 0.0025−0.1X mol of niacin and 0.1X mol of C6H4NO−2

The concentrations acetic acid and acetate ion are

0.0025− 0.1X0.025 +X

and0.1X

0.025 +X

respectively.

Now we setup an ICE chart

[HC6H4NO2]£H+¤ £

C6H4NO−2¤

I 0.0025−0.1X0.025+X

0 0.1X0.025+X

C −x x xE 0.0025−0.1X

0.025+X− x x 0.1X

0.025+X+ x

Using in the Ka we can solve for x which gives the H+ concen-tration,

Ka = 1.4× 10−5 =¡

0.1X0.025+X

+ x¢(x)¡

0.0025−0.1X0.025+X

− x¢ .

At the equivalence point and equal amount of NaOH has fullyreacted with the niacin.

The total volume is now 50 mL and there are 0.0025 mol ofacetate ion, which is

0.0025

0.050= 0.05 M

We use this in an ICE chart for the conjugate base reaction

C6H4NO−2 +H2O HC6H4NO2 +OH−,£C6H4NO−2

¤[HC6H4NO2]

£OH−

¤I 0.05 0 0C −x x xE 0.05− x x x

Now since this is the conjugate base reaction we must use theKb which is

Kb =Kw

Ka= 7.1× 10−10.

So,

Kb = 7.1× 10−10 =(x) (x)

0.05− x.


Rearranging and using the quadratic formula gives x = [OH−] =6.0× 10−6.

The pH is then

pH = 14−¡− log

£6.0× 10−6

¤¢= 8.78

Notice that the pH at the equivalence point is not 7.0 as it isin the case of a strong acid strong base titration.

After the endpoint the niacin is the limiting reagent. So theproblem becomes exactly like the strong acid strong base case.

We figure out how much OH− is left over and get the pH fromthat.

Group Work

1. Calculate several values for the strong acid strong basetitration example

2. Calculate several values for the weak acid strong basetitration example

3. What is the pH at the equivalence point for the titrationof 25 mL of 0.10M NH3 with HCl.

Homework


Problems: Chapter 18: 23, 24, 25, 29, 79, 83 *Note* these arenot due until after the next lecture.



18 TITRATIONS PART II

With lecture and the group work problems we have tackledtitrations with monoprotic acids and bases.

Now we will complete the picture with the titration of weakpolyprotic acids (or bases) with strong bases (or acids)

Example : Calculate the titration curve when 25 ml of a 0.10M solution of vitamin C (H 2C 6H 6O6) is titrated with 0.10 MNaOH. The Ka1 = 7.9× 10−5 and Ka2 = 1.6× 10−12.

Now the titration process will experience two equivalence points.

Because Ka1 À Ka2 we take separate the process into twoparts: Reaction with the first equivalent of acid

H2C6H6O6 +NaOH NaHC6H6O6 +H2O

and the reaction with the second equivalent of acid

NaHC6H6O6 +NaOH Na2C6H6O6 +H2O

The problem is now like what we have done for monoproticacids so we will not go through every detail.

Instead, we will simply calculate the pH at each equivalencepoint.

The first equivalence point is controlled by theKa1 = 7.9×10−5reaction.

So we take the NaOH to fully react with the first proton ofvitamin C.

This occurs at 25 mL of NaOH added. Or,

0.025 L × 0.10 mol HC6H6O−6

1 L= 0.0025 mol HC6H6O−6 .

TITRATIONS PART II 77

Thus the concentration of HC6H6O−6 is

0.0025 mol HC6H6O−60.050 L

= 0.05 M HC6H6O−6

This HC6H6O−6 then goes back towards the acid via the conju-gate base reaction,

HC6H6O−6 +H2O H2C6H6O6 +OH−

So we need the Kb1 which is

Kb1 =Kw

Ka1= 1.27× 10−10

The ICE chart is£HC6H6O−6

¤ £OH−

¤[H2C6H6O6]

I 0.05 0 0C x x xE 0.05− x x x

Using the Kb1 equation and solving for x =£OH−

¤= 2.52 ×

10−6. Thus,

pH = 14−¡− log

£2.52× 10−6

¤¢= 8.40

Now, for the second equivalence point we repeat the procedure.

The second equivalence point occurs at the addition of a totalof 50 mL of NaOH.

So, £C6H6O−6

¤=0.0025 mol0.075 L

= 0.033 M.

Now we need the Kb2 which is

Kb2 =Kw

Ka2= 6.25× 10−3

The appropriate ICE chart for the reaction

C6H6O26 +H2O HC6H6O−6 +OH−

is £C6H6O−6

¤ £OH−

¤ £HC6H6O−6

¤I 0.033 0 0C −x x xE 0.033− x x x

Using the Kb2 equation and solving for x we get x =£OH−

¤=

0.0116. Thus

pH = 14− (− log [0.0116])12.06


Group Work

1. Work out several values of the titration curve for the ex-ample above.

Homework



TITRATIONS PART II 79


19 SOLUBILITY OF WEAKELECTROLYTES

We have been applying our ideas about equilibrium to acid basereactions.

But these ideas are general and apply to any reaction.

Here we consider application to the solubility of weak elec-trolytes.

Consider a general dissolution reaction

AncBna(s) ncAna+(aq) + naB

nc−(aq)

The equilibrium constant for this reaction is called the solubilityproduct,

Ksp =£Ana+

¤nc £Bnc−¤na .Remember the AncBna(s) is a solid so we do not include it inthe equilibrium constant.

For example of the solubility product for Fe2 (SO4)3 is

Ksp =£Fe3+

¤2 £SO2−4

¤3In the first semester we simply learned ionic compounds asstrong electrolytes, weak electrolytes and nonelectrolytes

The solubility product gives us a way of ranking how soluble acompound is.

For example Ca(OH)2 has aKsp of 5.5×10−5 whereas CaF2 hasa Ksp = 5.3× 10−10. We can say, quantitatively that Ca(OH)2is more soluble then CaF2.

Lets say we want to know the concentration of Ca2+ and OH−

upon putting solid Ca(OH)2 into a beaker.

SOLUBILITY OF WEAK ELECTROLYTES 81

We write the solubility reaction

Ca(OH)2 Ca2+ + 2OH−

andKsp = 5.5× 10−5 =

£Ca2+

¤ £OH−

¤2From the stoichiometry of the reaction 2x moles of OH− isproduced for every x moles of Ca2+ thus

5.5× 10−5 = (x) (2x)2

4x3 = 5.5× 10−5

x3 =5.5× 10−5

4

x =3

r5.5× 10−5

4x = 0.024

Thus£Ca2+

¤= x = 0.024 M and

£OH−

¤= 2x = 0.048 M.

We can take this further and determine the pH of the solution.

pH = 14− (− log 0.048)pH = 12.6

Group Work

1. Write the solubility product for

(a) copper II sulfide

(b) zinc II hydroxide

(c) lead II iodide

2. Put the solubility reaction we did in class into the ICEchart formalism.

3. The common ion effect occurs in solubility reactions aswell as acid-base reactions. How would the common ioneffect work here? Can you setup an ICE chart that willhandle the common ion effect?

82 SOLUBILITY OF WEAK ELECTROLYTES

Homework

Reading: Kotz and Treichel, 18.4, 18.5, 18.6, 18.7

Problems: Chapter 18: 35, 37, 39, 45, 47, 53, 55, 59, 65, 67

SOLUBILITY OF WEAK ELECTROLYTES 83

84 SOLUBILITY OF WEAK ELECTROLYTES



Material for Exam III


21 REVIEW OF THE FIRST LAW,ENTHALPY, AND HESS’ LAW

Last semester we tackled several ideas from thermodynamics.

Now, over the next several lectures, we will complete the pic-ture.

But first we need to review the key ideas from the first semester.

The First Law of Thermodynamics

The first law of thermodynamics is simply a statement of theconservation of energy for a system.

In words the first law states that a change in the total energyof a system is equal to the heat taken in by the system plus thework done on the system.

The first law is stated mathematically as

4E = q + w. (21.1)

Energy that enters or leaves a system is classified as either workor heat.

As system here means any macroscopic piece of matter such asa beaker of water or an iron bar.

Any work, w, done on the system is by convention taken to bea positive quantity

Any work done by the system is then negative.

REVIEW OF THE FIRST LAW, ENTHALPY, AND HESS’ LAW 87

Likewise any heat, q, taken in by the system is by conventiona positive quantity (endothermic)

Any heat given off by the system is negative (exothermic).

Enthalpy

Let us consider applying the first law to a gas in a piston

If we apply heat to the gas it expands against atmosphericpressure and in doing so it does work.

Since we are adding heat to the system, q > 0.

Since the system is doing work, w < 0.

88 REVIEW OF THE FIRST LAW, ENTHALPY, AND HESS’ LAW

The first law for this case says

4E = q(positive)+ w(negative) (21.2)

Therefore no matter what

4E < q (21.3)

So the change in total energy is less than the amount of heatenergy we added to the system. If we add 10 J of heat energyto the system the total energy does not change by 10 J

This is somewhat inconvenient so chemists have defined a newmeasure of energy called enthalpy (H)

Enthalpy is defined such that at constant pressure it is equalto the amount of heat energy taken in by the system

4H ≡ qP , (21.4)

where the subscript P reminds us that this is true only forconstant pressure. (We will always work at constant pressurein this course.)

Consider again the expanding gas example. Now if we add 10Jof heat to the gas then 4H = 10J

During a chemical reaction it is often rather convenient to mea-sure the heat given off by a reaction.

In fact, you will do this in lab.

If the heat is measured at constant pressure then it is simplythe change in enthalpy for the reaction:

4H = Hproducts −Hreactants (21.5)

If4H is positive that means that qP is positive and heat energyis taken in by the system.

Reactions with positive 4H are called endothermic reactions.

If 4H is negative that means that qP is negative and heatenergy is given off by the system.

Reactions with negative 4H are called exothermic reactions.


Hess’ Law

Both total energy, E, and enthalpy, H, are what are called statefunctions

A state function is a function that depends only on the partic-ular state that the system is in. It is completely independentof how the system arrived at that state.

For example consider heating in water from 300 K to 310 K.

You would calculate the same 4H or 4E regardless as towhether you heated directly from 300 K to 310 K or if youfirst cooled to 290 K then heated to 360 K and then cooled to310 K.

An important consequence of enthalpy being a state functionis Hess’ law which states that if a reaction is the sum of two ormore reactions, 4H for the overall reaction is the sum of the4H’s for the component reactions.

That is 4H is solely determined by

4H = Hproducts −Hreactants (21.6)

regardless of how the reactants became the products.

For example, 4H for the reaction

2C6H6 + 15O2 → 12CO2 + 6H2O (21.7)

is the same as the sum of the 4H’s for the following reactions

2C6H6 → 12C+ 6H2 (21.8)

12C+ 6H2 + 15O2 → 12CO2 + 6H2O (21.9)

Hess’s law is very handy because it would be impossible tomeasure enthalpy changes for every imaginable reaction wereit not for Hess’ law.


Group Work

1. Based on your experience, is 4H positive or negative forcombustion reactions?

2. Use Hess’ law to determine 4Hrxn for

4NH3 + 5O2 → 4NO + 6H2O

given the following data

N2 + O2 → 2NO; 4H = 180.6 kJ

N2 + 3H2 → 2NH3; 4H = −91.8 kJ2H2 +O2 → 2H2O; 4H = −483.7 kJ

3. Based on the previous problem what is 4Hrxn for

2NH3 +5

2O2 → 2NO + 3H2O

Homework




22 ENTROPY

Up to this point our discussion of thermodynamics has focussedon energy (either total energy or else enthalpy).

There is an equally important concept in thermodynamics calledentropy.

A simple way of thinking about entropy is that it provides ameasure of disorder in a system.

That is, for a given system, higher entropy means the system isdisordered whereas low entropy means the system is ordered.

This is idea of disorder is a good one and we should alwayskeep it in mind but it does not capture all of the sublties aboutentropy

Microstates and Configurations

One can get a feel for entropy by considering the possiblearrangements of the molecules in a system.

This requires a few definitions.

Ensemble: The collections of molecules in a macroscopic sys-tem

Microstate: One single arrangement of the ensemble

Configuration: Collection of indistinguishable microstates.Given the symbol W

ENTROPY 93

Entropy: Proportional to the number of configurations throughthe Boltzmann equation

S = k lnW, (22.1)

where k = 1.38× 10−23 J/K is the Boltzmann constant.

∗ ∗ ∗ WorkSheet on Microstate and Configuration ∗ ∗∗

Group Work

1. How many different ways are there to permute the let-ters in “CAT”? Call each permutation a microstate thencollect equivalent microstates into configurations. Howdoes the number of microstates compare to the numberof configurations?

2. Repeat the about for “ANN”

3. Multiply the Boltzmann constant time Avogadro’ss num-ber. Do you recognize the number?

4. The so-called thermal energy per molecule is kT . Verifythat it as proper units and calculate it for room temper-ature.

5. The so-called thermal energy per mole is RT . Verify thatit as proper units and calculate it for room temperature.

Homework


Problems: Chapter 19: 1,3

94 ENTROPY

23 THE SECOND AND THIRD LAWS

Entropy forms the basis for the second and third laws of ther-modynamics.

The mathematical statement of the second is for any sponta-neous process

T4S ≥ q. (23.1)

For isolated systems in which no heat energy can be exchangedwith the environment

T4S ≥ 0

4S ≥ 0. (23.2)

This leads to the important result that

4Suniverse = 4Ssys +4Senv ≥ 0 (23.3)

That is for every event the entropy of the universe increases.

Now it is important to realize that 4Ssys need not be greaterthan zero.

Natural processes can occur that lower the entropy of the sys-tem.

If this happens, however, the change in entropy of the environ-ment must be positive and outweigh that of the system.

The statement of the third law is a little beyond what we areable to handle so we shall only consider the consequence of thismathematical statement.

The third law puts entropy on an absolute scale like tempera-ture.

This is in contrast to energy which can not be put on an ab-solute scale so only energy differences have meaning.

THE SECOND AND THIRD LAWS 95

So we can not say a glass of water as and energy of so-and-so,but we can say it has an entropy of so-and-so.

Calculating Entropy

Like we learned for enthalpy, entropy is calculated in a similarmanner.

That is 4S is solely determined by

4S = Sproducts − Sreactants (23.4)

regardless of how the reactants became the products.

For example, 4S for the reaction

2C6H6 + 15O2 → 12CO2 + 6H2O (23.5)

is the same as the sum of the 4S’s for the following reactions

2C6H6 → 12C+ 6H2 (23.6)

12C+ 6H2 + 15O2 → 12CO2 + 6H2O (23.7)

We can also short cut the Hess’ law calculations by using theentropies of formation tables in Appendix L of the book.

4S° =X

S°f,products −X

S°f,reactants (23.8)

Group Work

1. Use either the entropy as a measure of disorder concept ofthe number of configurations concept to predict whetheror not the following spontaneous processes result in andincrease or decease in entropy

(a) melting of ice


(b) condensing of water vapor

(c) the reaction between hydrogen and oxygen to formwater

(d) dissolution of NaCl in water

2. Calculate 4rxnS° for

NH3 +1

2CO2

1

2NH2CONH2 +

1

2H2O

using the entropies of formation given in the table below

substance S°f¡

Jmol·K

¢NH3 193CO2 214

NH2CONH2 174H2O 70

3. What is 4rxnS° for

NH3 +CO2 NH2CONH2 +H2O

Homework



THE SECOND AND THIRD LAWS 97


24 GIBBS FREE ENERGY

We have now dealt with the two processes that drive all theprocesses of nature including chemical reactions: Energy andEntropy.

In most situations it is most convenient to use enthalpy as ourenergy.

Nature wants to minimize enthalpy

Nature also wants to maximize entropy

Often these driving factors are in opposition and nature mustadopt some compromise betweenminimizing enthalpy andmax-imizing energy.

For example consider water at room temperature and pressure.If only enthalpy was a factor than the water would be in thesolid state.

Conversely, if only entropy was a factor, the water would be avapor.

Since both factors contribute to the final state of water, it turnsout that water is a liquid under these conditions.

The Second Law

If we consider the second law

T4S ≥ q.

Now, for this course we are taking enthalpy to be heat energy(later in your career you will be more careful about this).

GIBBS FREE ENERGY 99

Hence,T4S ≥ 4H.

If we rearrange this equation we get

4H − T4S ≤ 0

The left hand side of this equation provides a very nice measureof spontaneity of a reaction.

We thus, define what is called Gibbs free energy as the lefthand side

4G = 4H − T4S.

If4G is negative the process will proceed, but if4G is positivethe process will reverse.

Finally, if 4G is zero the process is at equilibrium.

Gibbs Free Energy

Notice how 4G captures natures great compromise betweenenthalpy and entropy.

The definition of 4G also allows for another interpretation ofGibbs free energy.

The4H is the energy that the system can release or must takeon for the process to occur

The T4S has units of energy and can be interpreted as theenergy that is needed to produce the change in randomnessassociated 4S.

This energy is tied up and must be contained in the system.

So if we consider 4H as the total energy available and T4Sas the tied up in randomizing the system, then 4H − T4S isenergy free to do work. Hence the term free energy

Succinctly we say the Gibbs free energy is the energy availableto do work.


If 4G is negative then the free energy can be used to do work.

If 4G is positive then work must be done on the system to getthe process to go in the nonspontaneous direction.

There are several ways 4G can be negative.

If4G is negative because the reaction is very exothermic (4H <0) then we say the reaction is energy driven.

If 4G is negative because the reaction is the entropy increasethen we say the reaction is entropically driven.

Reactions can be energy driven, entropically driven, or both.

Group Work

1. Consider the following spontaneous processes and statewhether they are energy drive, entropically driven, orboth.

(a) melting of ice at room temperature

(b) condensing of steam at room temperature

(c) combustion of benzene into carbon dioxide and wa-ter.

(d) the rusting of iron

2. Consider the effect of temperature on 4G.

(a) At what temperatures is enthalpy the more impor-tant factor?

(b) At what temperatures is entropy the more importingfactor?

(c) A reaction that will not occur at 100K will occur at500K what must be true?

(d) A reaction that will occur at 100K will not occur at500K what must be true?

GIBBS FREE ENERGY 101

3. Like enthalpy and entropy, Gibbs free energy can be cal-culated using Hess’ law or free energies of formation. Cal-culate 4G for

C2H5OH+ 3O2 2CO2 + 3H2O,

givensubstance 4G°

f

¡kJmol

¢C2H5OH −174.8O2 0CO2 −394.4H2O −228.6

Homework




25 GIBBS FREE ENERGY ANDEQUILIBRIUM

We have seen that 4rxnG can tell us if we are at equilibriumand if we are not, it can tell us which way the reaction willproceed to get to equilibrium.

Thus there must be some connection between 4rxnG and boththe equilibrium constant K and the reaction quotient Q.

Indeed there is a connection, unfortunately you will need towait until later in your career to derive it from scratch.

So, here we will simply state the relation.

The equilibrium constant is related to 4rxnG° by

4rxnG° = −RT lnK (25.1)

Remember4rxnG° is the difference in free energies of the prod-

ucts and reactants under standard conditions (298 K, 1 atm,[i] = 1 M for all species)

This relation is very important because we can know K bycalculating 4rxnG

° for any reaction using the free energies offormation.

The 4rxnG for the reaction under any conditions is given byreferencing to 4rxnG

° and the reaction quotient,

4rxnG = 4rxnG° +RT lnQ. (25.2)

We will call this equation the 4G equation and it is the mostimportant equation in all of thermodynamics

GIBBS FREE ENERGY AND EQUILIBRIUM 103

The difference between 4rxnG and 4rxnG°

One point of confusion that often occurs is the distinction be-tween 4rxnG and 4rxnG

°.

4rxnG° is a constant for a given reaction that never changes

over the course of the reaction.

4rxnG is not constant, it changes from negative to zero as thereaction proceeds towards equilibrium.

Both 4rxnG and 4rxnG° tells us things about the character

of the reaction.

Since 4rxnG° is the difference in free energies of the products

and reactants under standard conditions, the value of 4rxnG°

tells us whether or not the reaction is product favored or reac-tant favored.

We see this because at standard condition all species present(both reactants and products) are 1 M.

If 4rxnG° < 0 then under these standard conditions the sys-

tem would move toward forming more products. Thus it isproduct favored.

If 4rxnG° > 0 then under these standard conditions the sys-

tem would move toward forming more reactants. Thus it isreactant favored.

Let’s see how this connects with the notion of product favoredor reactant favored that we got from K.

We said if K > 1, the reaction would be product favored. Con-versely if K < 1, then the reaction is reactant favored.

From

4rxnG° = −RT

positivez}|{ln

>1

K = negative (25.3)

4rxnG° = −RT lnK

<1|{z}negative

= positive

104 GIBBS FREE ENERGY AND EQUILIBRIUM

Turning to 4rxnG, this tells us how the reaction will go underthe current conditions of the reaction.

So if we consider the case where we mix reactants to performa reaction.

Initially there are no products and therefore Q = 0 and we have

4rxnG = 4rxnG° +RT

−∞z}|{ln 0 = −∞ (25.4)

So regardless of what 4rxnG°, even if very reactant favored,

initially there is a tremendous amount of thermodynamic forcefor the reaction to proceed.

Then some products are formed but before we reach equilibriumQ < K. Thus

4rxnG =

−RT lnKz }| {4rxnG

° +RT ln<K

Q

= −RT lnK +RT ln<K

Q

= RT lnQ

K|{z}<1| {z }

negative

= negative

and the reaction will continue to proceed towards products.

At equilibrium Q = K and

4rxnG =


° +RT ln=K

Q

= −RT lnK +RT lnK = 0

On the other side of equilibrium Q > K. Thus

4rxnG =


° +RT ln>K

Q

= −RT lnK +RT ln>K

Q

= RT lnQ

K|{z}>1| {z }

positive

= positive

and the reaction will reverse to form reactants.

GIBBS FREE ENERGY AND EQUILIBRIUM 105

Group Work

1. LeChatelier’s principle is a consequence of the 4G equa-tion. Use the4G equation to mathematically show LeChate-lier’ss principle. (Hint: the arguments will be similar towhat we just discussed at the end of lecture.

2. Earlier, when we doubled the stoichiometry of a reactionwe had to square the original K. Also when we turned areaction around when had to take one over the originalK. Use the 4G equation to show these relationships.

Homework



106 GIBBS FREE ENERGY AND EQUILIBRIUM

26 REVIEW OF CONCEPTSSURROUNDING GIBBS FREE

ENERGY

∗ ∗ ∗ Worksheet ∗ ∗∗

REVIEWOF CONCEPTS SURROUNDING GIBBS FREE ENERGY107

108REVIEWOF CONCEPTS SURROUNDING GIBBS FREE ENERGY

27 OXIDATION—REDUCTIONREACTIONS I

What distinguishes this type of reaction from the others is thefact that electrons are transferred during the reaction whichresults in a change in the oxidation numbers of the involvedcompounds.

We first need a few definitions.

• When a species increases its oxidation number during areaction it is oxidized.

• When a species decreases its oxidation number during areaction it is reduced.

• A species which has a tendency to be oxidized is called areducing agent because if it is oxidized the other reactantmust be reduced.

• A species which has a tendency to be reduced is called anoxidizing agent.

A general form for the oxidation—reduction reaction is

Ahigh +Dlow → Alow +Dhigh. (27.1)

Here a species A accepts an electron from a donor species D.A is reduced since its oxidation number decreases and D isoxidized since its oxidation number increases

For example an oxidation—reduction reaction between coppersulfate and zinc metal is written as

CuSO4(aq) + Zn(s) → ZnSO4(aq) +Cu(s). (27.2)

It is more illustrative to write the net ionic equation by can-celling the spectator sulfate ion

Cu2+(aq) + Zn0(s) → Zn2+(aq) +Cu

0(s). (27.3)

OXIDATION—REDUCTION REACTIONS I 109

Here we can clearly see that the oxidation number of the copperis reduced from 2+ to 0 whereas the oxidation number for zincis increased from 0 to 2+.

For this reaction two electrons are transferred from each zincatom to each copper ion to create the zinc ion and the copperneutral atom.

Zinc is oxidized and copper is reduced.

Zinc acts as a reducing agent and copper acts as an oxidizingagent

Balancing redox reactions

It is important for you to become proficient at balancing redoxreactions. This requires practice.

There are three cases one needs to consider. The first is simplybalancing redox reactions like the zinc—copper example above.The second is balancing redox reactions in acids and the thirdis balancing redox reactions in bases

∗∗∗ See Koltz & Treichel Section 20.1, Examples 1,2 and 3 ∗∗∗

Group Work

1. Work on handout.

Homework


Problems: Chapter 20: 1, 2, 3, 4, 5, 6 *Note* these are dueafter the next lecture.

110 OXIDATION—REDUCTION REACTIONS I

28 OXIDATION—REDUCTIONREACTIONS II

We will continue practicing balancing Redox reactions

Group Work

1. Work on handout.

Homework



OXIDATION—REDUCTION REACTIONS II 111

112 OXIDATION—REDUCTION REACTIONS II

29 ELECTROCHEMICAL CELLS

We now consider the example of the Daniell cell. The reaction

Zn(s) + Cu2+(aq)→ Cu(s) + Zn2+ (29.1)

will occur spontaneously when mixed in a beaker.

During the reaction electrons are transferred and the systemmoves to a lower energy state.

However, all of the chemical energy evolved is lost as heat andno work can be extracted.

In order to get usable energy from this reaction one must “sep-arate” the half reactions

Cu2+(aq) + 2e− → Cu(s) (29.2)

andZn(s)→ Zn2+ + 2e− (29.3)

into two solutions as shown below

ELECTROCHEMICAL CELLS 113

Cell diagrams

Conventions for writing down an electrochemical cell.

• list components in phases from electrode to electrode

• use a vertical line ( | ) to separate the phases

• use a double vertical line ( || ) to indicate the salt bridge

• start with the oxidation reaction on the left and then thereduction reaction on the right

Applying the conventions to the Daniell cell we write

Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s) (29.4)

to represent the cell.

Standard Cell potentials

The convention for the determining the potential of a cell isto take the REDUCTION potential for the “right” electrodeminus the REDUCTION potential for the “left” electrode.

εcell = εright − εleft (29.5)


For the Daniell cell we get:

left (oxidation) Zn → Zn2+ + 2e−red. pot.!!z }| {ε°Zn2+ |Zn −0.763 V

right (reduction) Cu2+ + 2e− → Cu ε°Cu2+ |Cu 0.337 V

cell reaction Zn(s) + Cu2+(aq)→ Cu(s) + Zn2+ ε°cell 1.100 V

Group Work

1. To get a redox to go in reverse one must apply an oppos-ing voltage in excess of the cell potential. This is how abattery recharger works. What voltage would need to beapplied to a Daniel cell to “recharge” it.

2. What is the standard cell potential for

3Mg(s) + 2Al3+(aq)→ 2Al(s) + 3Mg2+ (29.6)

3. How many electrons are transferred in the above reaction(be sure to consider the stoichiometry)?

Homework

Reading: Kotz and Treichel, 20.2, 20.3, 20.4, 20.5


ELECTROCHEMICAL CELLS 115


30 THE NERNST EQUATION

The cell potential allows one to determine the maximal work,−wmax the cell can produce. That is, the amount of work tomove n moles of electrons though a potential εcell.

−wmax = nFεcell (30.1)

where F = 96485 C/mol is Faraday’s constant.

Once we know the maximal work for the cell, we can determinethe free energy of the cell by

4rxnG = w0max = −nFεcell (30.2)

As also have the 4G equation,

4rxnG = 4rxnGª +RT lnQ, (30.3)

where Q = [products][reactants] is the reaction quotient (remember we are

not at equilibrium)

We can substitute 4rxnG = −nFεcell and 4rxnG° = −nFε°cell

into the above equation to give the Nernst equation

ε = ε° − RT

nFlnQ (30.4)

The Nernst equation is one of the most important equationsin all of electrochemistry since it provides a link to thermody-namics through Gibbs free energy via 4rxnG

° = −nFε° fromabove.

It is of interest to study the Nernst equation at equilibrium.At equilibrium Q = K and ε = 0 since the cell can do no work.The Nernst equation then becomes

0 = ε° − RT

nFlnK ⇒

4rxnG°z }| {−nFε° = −RT lnK

⇒ 4rxnG° = −RT lnK. (30.5)

THE NERNST EQUATION 117

As we expected from earlier.Example: What is ε for a Daniel cell at 300 K when theconcentration of Cu2+ is 0.100 M and the concentration of Zn2+

is 0.001 M?

Zn(s) + Cu2+(aq)→ Cu(s) + Zn2+ (30.6)

First we need ε° which we found last lecture to be 1.10 V.

Then we use the Nernst equation,

ε = 1.10− 8.314× 3002× 96485 ln

0.100

0.001= 1.04 V

The n = 2 because 2 electrons are transferred from the zinc tothe copper during the reaction.

Notice that even though the concentrations differ by a factorof 100, the difference in voltage from the standard potential isminimal.

Group Work

1. Explicitly derive the Nernst equation from the 4G equa-tion.

2. What is K for the Daniels cell reaction? Is it productfavored or reactant favored?

3. We can also use a differences in concentrations betweento half cells to do work. These systems are called concen-tration cells. Consider

Na(Hg)|NaCl(aq,M1)||NaCl(aq,M2)|Na(Hg) (30.7)

The double vertical lines ( || ) represent the salt bridge.

left (oxidation) Na(Hg) → Na+(aq,M1) + e− ε°Na+ |Na −2.714 V

right (reduction) Na+(aq,M2) + e− → Na(Hg) ε°Na+ |Na −2.714 V

cell reaction Na+(aq,M2) → Na+(aq,M1) ε°cell 0 V


The Nernst equation becomes

ε = 0− RT

n1↑Fln[Na]1[Na]2

= −RTFln[Na]1[Na]2

(30.8)

So we see that if [Na]1 6= [Na]2 then a potential will de-velop.

(a) What is the potential of a sodium concentration cellif the difference in concentration it a factor of 10?how about 100? how about 1000?

(b) What is the 4G for the above cases?

Homework


Problems: Chapter 20: 25, 26, 27, 28, 29, 31, 33, 35, 37, 39,45, 49

THE NERNST EQUATION 119




Additional Materialfor Exam IV


32 REVIEW OF MOLECULARGEOMETRY

We will be talking about the nature of intermolecular forces.

These forces are determined and influenced by the three dimen-sional geometry of the molecule.

Thus we need to review the molecular geometry material fromthe first semester.

Charge Distribution in Covalent Compounds

We know that in ionic compounds electrons move entirely fromone atom to another and the bond is formed due to the mutualattraction of the anion and the cation.

For covalent compounds the electron density is shared. How-ever, it is not necessarily shared evenly between all the atomsin the compound.

As a result some atoms in the compound carry what are calledpartial charges.

One way to characterize the charge distribution in compoundsand polyatomic ions is to introduce the idea of electronegativity,χ.

Electronegativity is a measure of the ability of an atom in acompound to attract electron density to itself.

∗ ∗ ∗ See Figs. 9.8, 9.9 and 9.11 of Kotz&Treichel ∗ ∗∗

REVIEW OF MOLECULAR GEOMETRY 123

For any given bonded pair the difference in electronegativitieswill determine just how equally the atoms share the bondingelectrons.As a rule of thumb,

• if the electronegativities for the bonding pair differ by lessthan 0.5 then they essentially equally share the electrons.This type of bond is called a nonpolar covalent bond.

• if the electronegativity difference is between 0.5 and 2.0then one atom attracts more electron density. This typeof bond is called a polar covalent bond. The atom with thehigher electronegativity carries a partial negative chargewhile its partner with lower electron density carries a pos-itive charge.

• if the electronegativity exceeds 2.0 the bond is an ionicbond.

VSEPR Model

Prior to discussing valence bond theory we need to learn some-thing about the shapes of molecules. A good approach to thisis given by the valence shell electron repulsion (VSEPR) model.

The primary principle of the VSEPRmodel is that bonds and/orlone pairs of the valence electrons on each element of a com-pound repel each other and try to be as far away from oneanother as possible.

Under the VSEPR model molecules can be classified accordingto there electronic geometry. The electronic geometry is thethree dimensional structure that the electron density–either abond or lone pair–adopts.

Electronic Geometries

• Linear: Two bonds or lone pairs

• Trigonal planar: Three bonds or lone pairs

• Tetrahedral: Four bonds or lone pairs

124 REVIEW OF MOLECULAR GEOMETRY

• Trigonal-bipyramidal: Five bonds or lone pairs

• Octahedral: Six bonds or lone pairs

Once the molecule has been assigned its electronic geometry,it then gets assigned its molecular geometry.

Molecular Geometries (electronic geometry in bold, molecularin italic)

• Linear

— Linear : Two bonds, no lone pairs.

• Trigonal planar

— Trigonal planar : Three bonds, no lone pairs.

— Bent: Two bonds, one lone pair.

• Tetrahedral

— Tetrahedral : Four bonds, no lone pairs.

— Trigonal Pyramidal : Three bonds, one lone pair.

— Bent: Two bonds, two lone pairs.

• Trigonal-bipyramidal

— Trigonal-bipyramidal : Five bonds, no lone pairs.

— Seesaw : Four bonds, one lone pair.

— T-shaped : Three bonds, two lone pairs.

— Linear : Two bonds, Three lone pairs.

• Octahedral

— Octahedral : Six bonds, no lone pairs.

— Square-pryamidal : Five bonds, one lone pair.

— Square-planar : Four bonds, two lone pairs.


Molecular Dipole Moments

Polar covalent bonds with in molecule can potentially lead toa net dipole moment for the molecule.

If all the polar bonds in a molecule cooperate in such a waythat one side of the overall molecule is partially negative andthe other side is partially positive then the molecule has a netdipole moment.

If all the polar bonds in a molecule cancel each other out thenthe molecule has no net dipole moment.

Group Work

1. Give the electronic and molecular geometries of the fol-lowing molecules. Also indicate the direction of the dipolemoment

(a) NH3

(b) H2CCl2

(c) BF3


Homework




33 INTERMOLECULAR FORCES

Polarizability and Induced Dipoles

Last time we reminded ourselves of the molecular dipole mo-ment.

The dipole moment is important in determined intermolecularforces and the physical properties of substances.

Another equally important and somewhat related property ofthe molecule is its polarization.

The polarization, α, is a measure of how the electron densityof the molecules responds to an external electric field.

An external electric field causes the electron cloud of a moleculeto become distorted.

This distortion means that, even for neutral molecules, therecan exits a dipole moment.

Unlike the permanent dipole moment that comes from appro-priate molecular geometry, this induced dipole is caused by thepresence of the electric field, ε,

μ = αε (33.1)

INTERMOLECULAR FORCES 129

Figure 1

Figure 2

Intermolecular Forces

There are several types of intermolecular forces that differ inthere make-up and in there interaction length.

The Coulombic Forces

Ion—Ion:The force between two ions is the strongest molecular force andit has the longest interaction length.

The force, F , falls off relatively slowly with distance, r,

F ∝ 1

r2(33.2)

Ion—dipole:This force is weaker than ion—ion interaction and it falls off

with distance quicker,

F ∝ 1

r3(33.3)


Figure 3

Figure 4

The Van der Waals Forces

Even when ions are not present one still sees intermolecularforces.

These forces are called Van der Waals forces.

The Van der Waals forces are weaker than the coulombic forces.They also have a shorter interaction length

dipole—dipole:The strongest of the Van der Waals forces with the longestinteraction length,

F ∝ 1

r4(33.4)

dipole—induced dipole:

When a molecule has a permanent dipole, that dipole createsan electric field that neighboring molecule feel.

Consequently, an induced dipole can be created in the neigh-boring dipole. As you might expect these forces are weaker


Figure 5

Figure 6

then dipole—dipole forces, having the form of

F ∝ 1

r5

induced dipole—induced dipole (called Loudon or dis-persive forces):

One might think that there would be no interaction betweenmolecules that have no permanent dipole.

In fact there is an attractive force that does arise in these sit-uations.

The electron cloud and hence the polarization of the moleculeis always fluctuating.

This random fluctuation spontaneously creates a transient di-pole moment which can then induce a dipole in a neighboringmolecule.

This induced dipole then strengthens the transient dipole byback induction.


The force is very weak and drops off very rapidly,

F ∝ 1

r6

Group Work

1. Give the principle type of intermolecular force for each ofthe following examples

(a) NaCl in water

(b) pure water

(c) “Salt bridges” in proteins

(d) methane dissolved in water

(e) pure CCl4

Homework





34 HYDROGEN BONDING ANDWATER

Last time we discussed intermolecular forces.

There is one very important and special type of intermolecularforce that we did not mention: hydrogen bonding.

Hydrogen bonding can occur in between some molecules thatcontain hydrogen.

If a hydrogen is bonded to an oxygen, nitrogen or fluorine atomin one molecule, it can hydrogen bond to an oxygen, nitrogenor fluorine atom in another molecule.

A hydrogen bond can be thought of as either an exceptionallystrong physical interaction or a very weak chemical bond.

Molecules that have a hydrogen bonded to an oxygen, nitrogenor fluorine atom can be hydrogen bond donors and acceptors.Molecules that have an oxygen, nitrogen or fluorine atom but nohydrogen attached to it can only be hydrogen bond acceptors.

Figure 1

HYDROGEN BONDING AND WATER 135

Figure 2

Figure 3

136 HYDROGEN BONDING AND WATER

Water

One of the most interesting, important and complicated sub-stances know is water.

We don’t always appreciate how unique water is because weare so familiar with it.

Almost all of the unique aspects of water are do to hydrogenbonds.

Some of the unique properties of water are

• An abnormally high melting point

• An abnormally high boiling point

• The fact that ice floats

• The Grotthuss mechanism

• A high heat capacity

• A high dielectric constant.

Group Work

1. Restate in your own words why hydrogen bonding is thereason for the unusual properties of water that we dis-cussed in lecture.

Homework


Problems: Chapter 13: 7, 8, 9, 10, 11, 13, 15, 17, 19, 21

HYDROGEN BONDING AND WATER 137

138 HYDROGEN BONDING AND WATER

35 PHASE DIAGRAMS

The phase change behavior of a substance can be expressedconveniently with a pressure versus temperature graph calleda phase diagram.

∗ ∗ ∗ HANDOUT ∗ ∗∗

The lines on the phase diagrams represent the precise temper-atures and pressures where an equilibrium between two phasesoccurs.

The solid—liquid line:

• crossing the line from solid to liquid represents melting

• crossing the line from liquid to solid represents freezing

The gas—liquid line:

• crossing the line from gas to liquid represents condensa-tion

• crossing the line from liquid to gas represents boiling

The solid—gas line:

• crossing the line from solid to gas represents sublimation

• crossing the line from gas to solid represents deposition

There are two special points on the phase diagram.

• The triple point

— The point where the solid-liquid line, solid-gas lineand gas-liquid line converge.

PHASE DIAGRAMS 139

— The single value of temperature and pressure in whichall three phases can co-exist.

• The critical point

— The termination of the gas-liquid line.

Group Work

1. How would the slope of the solid liquid line for the phasediagram for water compare to the generic one shown inthe handout.

2. Suggest a way to go from a liquid to a gas without boiling.

3. Is there a way to go from a solid to a liquid withoutmelting?

Homework

Reading: Kotz and Treichel, 13.6 (skim), 13.7 (skim), 13.8(skim), 13.9, 13. 10


140 PHASE DIAGRAMS

36 HOMOGENEOUS SOLUTIONS

Homogeneous solutions are homogeneous mixtures of a solutein a solvent.

There are several types of solutions based the nature of thesolute

• liquid-solid

• liquid-liquid

• liquid-gas

Whether or not a particular solute dissolves in a particularsolvent is determined by the intermolecular forces between themolecules.

The phrase “like dissolves like” captures the determining fac-tors in the dissolution process.

• Polar solutes dissolve more readily in polar solvents thanin non-polar solvents

• Non-polar solutes dissolve more readily in non-polar sol-vents than in polar solvents.

Liquid-solid solutions

The specific factors involved in the dissolution of a solid in asolvent are

• breaking the lattice structure of the solid

• the energy of interaction between the solute and solventmolecules

HOMOGENEOUS SOLUTIONS 141

• the energy of interaction between solvent molecules

• the net entropy gain or loss due to dissolution.

Liquid-liquid solutions

The specific factors involved in the mixing of liquids

• The interaction energy between the solute molecules

• The interaction energy between the solvent molecules

• The interaction energy between a solvent molecule and asolute molecule

• the net entropy gain or loss due to dissolution

We can draw a phase diagram that captures the behavior ofliquid-liquid mixtures.

Rather than a pressure versus temperature graph we now usetemperature versus mole fraction.

Liquid-gas solutions

The specific factors involved in the mixing of liquids

• The interaction energy between the gas molecule and thesolvent molecule

• The interaction energy between the solvent molecules

• The entropy loss by dissolving the gas in the liquid.

Notice that dissolving a gas in a liquid is entropically unfavor-able.

This means that gases dissolve better at low temperatures thanat high temperatures.


Figure 1

Group Work

1. An interesting thing happens when one put a salt likeNaCl in to a miscible mixture of acetonitrile and water.The solution separates into a water-rich and acetonitrile-rich phase. Why might this happen?

2. Related to the above there is a common “quick and dirty”technique for separating proteins from solutions calledsalting out. Here a salt is added to a protein—water so-lution to cause the protein to precipitate out as a solid.Why might this happen?

Homework



HOMOGENEOUS SOLUTIONS 143


37 PROPERTIES OF SOLUTIONS

We will now discuss some of the important properties of solu-tions.

If we consider the case for a solution in a closed container withsome headspace above the top of the liquid, there is a relation-ship between the make up of vapor phase and the solution.

Henry’s Law

First of all there is a relationship between the partial pressure ofthe solute in the vapor phase and the solubility in the solution.

PROPERTIES OF SOLUTIONS 145

This relation is called Henry’s law and is

Xi = kiPi (37.1)

where Xi is the mole fraction of the solute, Pi is the partialpressure and ki is the Henry’s law constant.

Henry’s law is related to LeChatelier’s principle.

If we consider the equilibrium establishing in the closed con-tainer, we have

solutesolution solutevapor

If we where to increase the partial pressure of the solute in thevapor phase we would expect, by LeChatelier’s principle, thatthe reaction above would go in the reverse direction.

We expect the mole fraction to go up.

If we look at Henry’s law, increasing the partial pressure makesthe mole increase.

Raoult’s Law

Another relationship between the vapor phase partial pressuresand the solution exist for the solvent.

According to Raoult’s law the partial pressure of the solvent inthe vapor phase above the solution is

Ps = XsP•s ,


where P •s is the vapor pressure, i.e, the partial pressure above

pure solvent.

Deviations from Raoult’s Law

Raoult’s law is a purely statistical law. It does not require anykind of interaction among the constituent particle making upthe solution.

Since, in reality, there are specific interactions between parti-cles, real solutions generally deviate from Raoult’s law.

The physical interpretation of deviation from Raoult’s law is

• positive deviation: the molecules prefer to be aroundthemselves rather than other types of molecules.

• negative deviation: the molecules prefer to be aroundother types of molecules than themselves.

• no deviation: the molecules have no preference.

Positive deviation from Raoult’s lawNegative deviation from Raoult’s law

Colligative properties

Interestingly, certain physical properties always change the sameway as a small amount of solute is added to a solvent.

The properties that do this are called colligative properties.

We will only discuss boiling point and freezing point.


When a small amount of solute is added to a solvent the boilingpoint is always initially elevated.

This phenomenon is called boiling point elevation.

Eventually as more solute is added the boiling point mightdecrease depending on the nature of the solute.

The initial boiling elevation, however, is independent of thenature of the solute.

As similar phenomenon occurs for freezing points.

But, in this case, the freezing point is lowered when a smallamount of solute is added.

This is called freezing point depression.

Again, as more solute is added the freezing point might beginto increase.

Group Work

1. What are the units on Henry’s law constant?

2. Sometimes Henry’s law is expressed in terms of concen-tration,

[i] = kiPi,

now what would the units of Henry’s law constant be?

3. For liquid—liquid mixtures it is not always clear what thesolute is and what the solvent is. As one liquid movesfrom a high mole fraction to a lowmole fraction, we wouldexpect the liquid to first obey Raoult’s law than to obeyHenry’s law. What must be true for about the relationbetween the vapor pressure and the Henry’s law constantfor the liquid to obey both laws at all mole fractions?


Homework


Problems: Chapter 14: 21, 23, 25, 27, 29, 33, 37, 39, 41, 47,51, 55



38 REVIEW FOR FINAL EXAM

REVIEW FOR FINAL EXAM 151

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