CHEM1020 Module 1 2011 Notes

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‘ ’

Module 1em ca nerget cs

Lecturers: Prof Bob Gilbert [email protected]

Dr Horst Schirra [email protected]

Prof Matt Trau [email protected]

School of Chemistry & Molecular BiosciencesCHEM1020 Sem 2 / 2011 – Module 1: Thermodynamics & Equilibrium

em ca nerge cs eac v y

Text: Chemistr . Blackman Bottle Schmid Mocerino and

Wille.

John Wiley. 2008

For the Electronic Course Profile – see Blackboard

(http://blackboard.elearning.uq.edu.au)


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• PASS (attendance very strongly recommended)



50% MCQ and SAQ exam at end of semester (must gain 40%or higher in final exam to gain a grade of 4 or higher).

20% practical work

12% 2 QUIZZES

18% IS-IT Chemistry Collaborative Research Task (begins inweek 2 – details via Bb)

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Collaborative Group Task (weeks 2-12):

• ou c oose a scenar o – represen g o a sc ence ssues.

• You will be working in Groups of 4

• Groups are assembled by one of three ways (either

choose your team members; opt into a group; or be placed

- .

• Full details & sign on will be available in week 2 through

‘ ’- .


What is PASS?

PASS provides one hour of mentored group study, whereyou are able to:

•Clarify concepts not fully understood

•Practice questions in exam format

•Revise your own learning

•Identify gaps in your own knowledge•Practice learning in the language of the discipline


•Be part of an interactive learning community!

CHEM1020 S2 2009Mean Grades by PASS Attendance

5.5

e

e a n G r a d

4.5 v e r a l l M

4


0 (295) 1‐‐4 (251) 5‐‐8 (238) 9+ (287)

PASS Session Attendance #



CHEM1020 S2 2009Grade Distribution: Effect of PASS Attendance

45%

50%

35%

40%

25%

30%

d e n t s

15%

20% S t

5%

10%

0%

1 2 3 4 5 6 7


0 (295) 1‐‐4 (251) 5‐‐8 (238) 9+ (287)

PASS Classes

begin in Week 3:

on ay u us


Strategies for learning quantitative chemistry

In module 1 …

• Reading before lectures

• In class model Qs

• JITTerbug quizzes (Feedback)

…

• In-class problem-solving

• Homework between lectures

In module 3 …

• In-class problem-solving


• Homework between lectures

Module 1: 12 Lectures 1. States of Matter . Chapters 6 and 7 (p202 – 281)

- Gases and Gas Laws

- Intermolecular Forces- Phase Changes

2. Thermochemistry . Chapter 8 (p281 – 331)- The First and Second Laws

- Defining Internal Energy and Enthalpy- Quantifying Enthalpy

- Defining Entropy- The Gibbs Function

3. Equilibrium . Chapter 9 (p332 – 372)- Chemical Equilibrium

- Equilibrium and Gibbs Free Energy- How systems at equilibrium respond to change

- Equilibrium calculations

= A Key Concept oo or e con

with page numbers!= A worked example in the text book you should look at


ALWAYS BRING A CALCULATOR + PAPER + PEN




evident in this image?

Think Back!

How can we explain why some substances exist as gases at room

temperature and atmospheric pressure, but others are liquids or

=> Forces that occur between individual atoms,mo ecu es or ons nt ermo ecu ar orces


et s eg n w t gases s nce t ey are t e eas est ….


(Elements in their pure state are mostly solids, with exceptions)

We must understand gases to understand and solve someimportant problems:

• Greenhouse Effect

• o u on, p o oc em ca smog

• Energy production

• Biochemistry of lungs

•

What are the characteristics of ases?




6.2, 203

A gas will expand to occupy all available space it is in and will

.

shape.

Atoms in a gas move independently of each other and move

randomly (Because the atoms in a gas are moving at random,

the gaseous state is the simplest to describe mathematically by

.

Gases exert a ressure . .


Gas Pressure

Gas molecules are in constant motion and collide with each other and the

container walls

they exert a pressure on the surroundings .

= =

The pressure is defined as: p =FORCE (N)

2

A

F

Gas molecules have mass(m) and velocity (v)

Gas molecules have mass(m) and velocity (v)

Opposing forceOpposing force

N

-2


2m

1 Nm-2 = 1 Pascal (Pa) = 1 J m-3

You are going to come across other units:

Atmosphere (atm)

1atm = 1.01325 105 Pa (101.325 kPa)= 760 mm Hg (= 760 Torr)

= 14.696 psi (pound per square inch)


‘ ’

Standard tem erature = 0 oC = 273 K

Standard pressure: Two definitions:

1 atmosphere = 100 000 Pa (IUPAC + Blackman Chemistry)

-1

SI units:

Temperature = KVolume = m3 (103 dm3 or 103 L)


and think about units!



The Ideal Gas E uation


6.2, 205

p = n ea as quat on

One of the most important equations in physical chemistry.

Describes how an ideal gas behaves under given conditions of

p = pressure (Pa or atm)

,

V = volume (m3 or L)

n = molar amount of gas (mol)

11

)(molKJ8.314 SIRR = gas constant

=


moatm.

Assum tions in the Ideal Gas Law?

6.2, 211-13

• The molecules are very small compared to the distance between

em = ey ave neg g e vo ume .

• The molecules do not ‘see’ each other (= no intermolecular

interactions.

• The molecules undergo elastic collisions with container walls (= no

energy is lost due to the collision).

• The molecules move in com letel random motion different s eeds

and direction).

=> e n on o an ea as:

If a as meets these conditions then it will obe


pV = nRT , and it is ‘Ideal’

6.2, 204 - 205

a oes e ea

Gas Equation tell us?

ow oes e pressure o a gas c ange w en we vary e

volume, temperature, or amount of gas (number of moles)?




11V p 11

T n

22FinallyV p

R 22n

2211 V p V p If one variable remains


2211 , .

Example

A sample of helium gas is held at constant temperature inside a cylinder

with a volume 0.57 L. The external pressure is 2.1 x 105 Pa. If the piston is

withdrawn until the gas volume is 2.55 L. Calculate the final pressure.

the no of moles of He & T are constant in this problem

p 1V 1 = p 2 V 2 and rearranging to: p 2 = p 1V 1

V 2

p 2 = (2.1 x 105 Pa) x (0.57 x 10-3 m3) = 0.47 x 105 Pa


2.55 x 10-3 m3

Units! 1 L = 1 x 10-3 m3

1. Watch out for units:

If R is 8.314 J K mol-1, p is Pa (Jm-3 or Nm-2) and V is m3

If R is 0.08206 L atm K-1 mol-1, p is atmos, V is L

2. A useful value (by substituting into the ideal gas law):

1 mole of gas occupies 22.4 L @ 273 K & 1 atm

or 24.5 L @ 298 K & 1 atm


A sample of chlorine gas occupies a volume of 946 mL at a pressure

of 726 mmHg. Calculate the pressure of the gas if the volume isreduced to 154 mL (assume temperature remains constant)

Solut ion: Since pV = nRT = constant, p1V 1 = p2V 2, hence

mL946

2

112

V

V p p

Hgmm104.46

mL154gmm

3

kPa101.3Hgmm760

Hgmm104.46 3

2 p or to convert to Pa:


kPa594=



Module 1: Lecture 2


6.2, 204

ases an so ute emperature

o e a e n ercep o eac ne es a - . , nown as a so ute zero .

0 K = -273.15 °CV

1

P2

P3Absolute Zero

T (°C)-273.15 °CAbsolute zero: the lowest temperature that can be. It is theoretical and hasnot been reached (Liquid Helium gets to 4 K). All calculations done using


negative values).

Kelvin: absolute temperature scale.

(Room Temp) 25 °C 298 K

0 K = -273.15 °C

° = .

Note: man of the e uations we

will use require the use of temperature

in Kelvins.

Also note: Temperature differences are the same in Kelvin or °C


Suppose we have a 12.2 L sample containing 0.50 mol oxygen gas at

pressure of 1 atm. If all this oxygen were converted to ozone (O3) at

same an , w a wou e e vo ume

2O3O

No. moles O3 produced = 32

Omol2Omol0.50

2mo

= 0.33 mol O3

Since V/n is constant,2

2

1

1

n

V

n

V

1

212

n

nV V

L8.1

L12.2mol0.50

mol0.33




Air Bags

Sodium azide decomposes rapidly

gas.

a 3 s a s + 2 g

Causes inflation in < 0.05s

How many litres of nitrogen would100 NaN roduce at room tem


www.howstuffworks.com

and pressure?

2NaN3 s 2Na s + 3N2 g


6.4, 214/215

Determination of molar mass M

Number of moles (n ) = mass/molecular weight = m/M R

pV = nRT <=> pV = m /M R RT <=> M R = mRT / pV

= m/V

Using this equation and n = m /MR, we can rearrange pV=nRT to give:

gas =

m

V =

p MR

RT


Worked examples in Blackman 6.4 p214/5

ne c eory o ases

The Ideal Gas Law provides a numerical route for predicting gas

behaviour based on macroscopic observations (changes inpressure and volume).

behaviour on a molecular basis?

Richard Feynman: “It’s all about atoms jiggling about!”




6.2, 204 - 205

a oes e ea

Gas Equation tell us?

ow oes e pressure o a gas c ange w en we vary e

volume, temperature, or amount of gas (number of moles)?


Descri tion of an ideal as from a molecular

viewpoint.

t . e d u /

velocity (v ) – therefore have kinetic

energy: / w w w . p i

E =1

mv2 t o n h t t p :

o

h n N o r

We assume that translational kinetic energy is the only internal ener ossessed b a as molecule. We i nore rotational


vibrational etc energy).

6.3, 208

Ekinetic =1

2mv2

Gas molecules are in constantrandom motion. The energy of amolecule is related to its speed.

There is a distribution of speeds, v , amongst a population

o gas mo ecu es.

l e c u l e s

Mean energy

e r o f M

N u

m


Relationshi between Kinetic Ener & Tem erature

Experiments show that

temperature increases

To find the average kinetic energy we add the individualmolecular kinetic energies and divide by the total number of


molecules



6.3, 211

Ekinetic =3RT

= kT3

(kinetic energy of one gas molecule)A

Multi l b N to et avera e kinetic ener er mole of as

=3 This expression is independent

ne c, mo ar2 of the identity of the gas!

Temperature: A measure of the average kinetic energy of a gas.

e tota net c energy s t e t erma energy.

Im ortant! Tem erature and thermal ener describe different


properties!!

–

kin

Molecules are in constant motion and collisions can occur betweenmo ecu es or w e oun ar es o e con a ner.

The number of collisions/unit area = Force/unit area = pressure

Elastic Collisions

x

x

e ore

after


Derivation (for your information only)

L

F

Area

Force p

2

Elastic Collisionsx

-x

before

x v v 2

x

Lt

Lv

L

mv

L

v v m

t

v m ma F x x

x i

222

a terx v t

Force of one particle hitting one side of the cube

mv mv mv F p i

i 112 2

3

2

2

2

26 sides to a cube => pressure of one particle

mv nN pV mv

nN p AA 1 2

2

This looks already like the ideal gas law

pressure from many particles:

E nN

V A 2 kin kin E mv mv E 21 22

RT nN A 32

3

kin

RT kT E

33


N A23 A. . .

Historically, it went the other way round…

mv nN pV V

mv nN p AA

3

1

3

22

This looks already like the ideal gas law

E nN

pV kin A 2

3 kin kin E mv mv E 2

2

1 22

RT n

pV nRT pV

kT RT pV

E kin

333

=> Using the ideal gas law to

define absolute tem erature inAA

terms of kinetic energy of molecules




Rates of as movement

6.3, 217

E =3RT

E = 1 mv2and2NA 2

So2

mv2

2NA

=

Solve for v (noting that m NA is the mass of 1 mole of

gas = M R)

R

M

RT v

3 _

=> The heavier a gas molecule, the slower its


diffusion and effusion.

•

molecules and hence the number of collisions per unit area

and increases pressure (constant V & n )

• Reducing the amount of gas molecules, n, decreases the

num er o co s ons per un area re uces pressure

(constant V & T ) or volume (constant p & T )

can be explained in terms of kinetic theory!


Module 1: Lecture 3


’

6.5, 219

.

‘The total ressure of a mixture of ases is the sum of the ressures that each gas would exert if it were alone in the container.’

T = 1 2 3……. ii =


www.kentchemistry.com/links/GasLaws/dalton.htm



If a sample contains gases A and B, the total pressure is given by:

pt =

nTRT

V

= RT (nA + nB)

= n RT + n RT

V

V V

= pA + pB


pA nART/V

pA

pT

=(nA + nB)RT /V

= nA

=pA + pB

(nA + nB)

= xA

xA is the mole fraction of A, xA: 0 1

pA = xApT

xA0 1

pB = xBpT


In general for component i: pi = xipT.

ues on:

A SCUBA diver is using EANx32, an enhanced

Nitrox gas mixture. This mixture contains 32%

Oxygen with 68% Nitrogen. If the cylinder is at

a pressure of 250 kPa what is the partial

pressure of Oxygen?

http://www.oceanearth.org/pastinfo.html

Mole fraction of oxygen = 0.32

p O2 = 0.32 x 250 kPa

= 80 kPa


Deep sea divers use helium instead of nitrogen (Heliox) – why?

What ou should be able to do at this sta e:

1. List the qualities that define an ideal gas.

2. Describe ideal gas behaviour in terms of the relationship, , .

3. Apply the ideal gas law pV = nRT

4. Explain how and why kinetic energy influences p, V and T

(microscopic).

5. Understand the difference between temperature and thermalenergy.

6. Describe how the partial pressures of individual gases in a


.



What about ‘Real Gases’?

The ideal gas law begins to

rea own a ow empera ures

or pressures.

Why?

Because gas molecules possess

a vo ume an can see eac

other (=intermolecular

interactions)

Solution?

Modify the ideal gas law to incorporate corrections for real gases


=> e.g. van der Waals Equation (2nd year)

6.7, 226

a es o a er – n ermo ecu ar orces

increased numbers of collisions (increase in p), intermolecular

forces become more relevant.

Intermolecular forces are the forces that occur between molecules.A chemical bond is not an intermolecular force (intramolecular


.

.

-

Three t es of molecular force de endin on molecule olarit :

1.Dispersion Forces.

2.Dipolar forces.

. .

Relative Strength:

Dispersion forces < Dipolar forces < Hydrogen Bonds


6.8, 231

-

Molecules with dipole moments can line up so that negative

an pos ve en s are c ose – ea s o an e ectrostat c attraction. Generally important over short distances.

A dipole on a molecule arises from the combination of molecular


shape and bond polarity (revise).



Why do non-polar atoms or molecules exist in condensed states

under some conditions?

For example helium exists as a liquid below about 4 K.


phys.kent.edu

6.8, 229

Transient changes in the charge

distribution around a nucleus or

molecule) can occur, and when this

happens near another atom, the

other atom.

This results in a weak and short-

lived interaction known as London dispersion force .


6.8, 231

Intermolecular Forces

Halogens are diatomic (F2) and interact

via DISPERSION forces.

More electrons = more easily

polarisable = stronger bond


6.8, 233

Hydrogen bonds are a special form of dipolar forces. Due toig y po ar on s e.g O-H wit an e ectron e icient

hydrogen, and the small size of hydrogen. Usually involve Hattached to O, N or F.

A network of H-bonding


.



7.1, 243

Surface Tension.

.

Molecules in a liquids experienceattractive forces from other molecules,but molecules at the surface experienceuneven forces surface tension.

The surface area stays as small as.

o es ve orces: a rac mo ecu es nthe liquid to one another.


Intermolecular Forces in Liquids.

Capillary action.

Adhesive forces: attractmolecules in the liquid to thewalls of the container.

Capillary action occurs whenthe adhesive forces are

encarta.msn.com

stronger than cohesive forcesand gravity.


Intermolecular Forces in Liquids.

7.1, 244

Vapour Pressure

Some mo ecu es are a e to escape rom t e sur aceof the liquid. These gaseous molecules exert a pressure.


Intermolecular Forces in Solids.

7.2, 245

In solids, intermolecular forces are so strong that there is noarge sca e movemen o mo ecu es.

Amorphous – no extended order

Solids

(e.g. glass, plastics)

Crystalline – highly ordered(e.g. diamond)

In liquids the boiling point tells us about strength of

intermolecular forces. In solids, it is the melting point.


See table 7.1, pg 246

7 2 246 7 2 247



7.2, 246

.

lattice positions (eg ice). Intermolecular forces are relatively much

weaker than the covalent intramolecular forces.

Sugar Dry IceNapthalene

The molecules are typically bound by dispersion, dipolar or H-bonds.


7.2, 247

But, there are other solids where bonding is very strong:

.

2. Metallic

3. Ionic solids.

Network > Metallic > Ionic


Bondin in Solids

7.2, 247

Network solids (highest melting point).

Network solids have a array of covalent bonds whichlinks each atom to its neighbouring atoms.

The extent of covalent bonding influences properties.


Diamond Graphite

sp 3

dispersion3550°C

sp 2

Graphite


encarta.msn.com

Buckyballs

7 2 248



7.2, 248

Bondin in Solids

Metallic solids.

Bonding in metals is characterised by non-directional covalent

bonding.

Electrons (covalent) are arranged around metal atoms in such away that they are mobile (non-directional).

Why are the electrons delocalised?

When two metal atoms interact, two molecular orbitals result. Withmany atoms, a large number of molecular orbitals result – all

.around different atomic molecular orbitals (Band Model).


Bondin in Solids

Metallic solids.

Metals have unique properties: high electrical and thermal conductivity,

malleability

Electrons are arranged around metal atoms in such a way that they

are mobile.


“electron sea” model

7.2, 249

Bondin in Solids

Ionic solids.

Ionic solids contain cations or anions that are heldtogether by opposing electrostatic forces.

These are strong interactions so meltingtemperatures are generally high.

NaCl Na+ + Cl- Fp = 801°CNa

NaCl

The stoichiometry of an ionic solid reflects that they

must be electrically charged.

CaF2 Ca2+

+ 2F-

CaF2


Module 1: Lecture 4


7 3 250



7.3, 250


a ac ors ec o u y

o u y maximum amount of solute that will dissolve in a given amount of

solvent at specific temperature (heterogeneous equilibrium)

Polar solvent/polar solute or nonpolar solvent/nonpolar solute favoured

eg hexane (C6H14) is better solvent for grease (C20H14) than methanol

(CH3OH)


Revision

Depends on bond polarity and molecular shape …..

nonpolar polar ionic

Blue = ositive char e


Red = negative chargeChpt 5.5

Atom Electrone ativit

A. C-CH 2.1

B. C-HC. H-Br

C 2.5Br 2.8

D. N-H

E. C-OCl 3.0

F. O-H.

O 3.5

S 2.5


.



Jud in olarit of molecules!

•The addition of a carbonyl

functional rou -C=Ointroduces a dipole.

•

localised in one region of

the molecule.

• Dipole-dipole interactions

can occur


Assumed

knowledge from

Senior Chem

Dispersion

Di ole-di oleIncreasing

hydrogen bonding (special dipole-dipole case)

H2O CCl4

Sucrose


Polar Nonpolar

- -

O

H

O

H

H H

CC

H

O

H

H

H H


Blackman Chapter 10 p 378

Consider 3 solvents:

Water (H2O)

Hexane (C6H14)

Carbon tetrachloride (CCl4)


http://jchemed.chem.wisc.edu/JCESoft/CCA/pirelli/pages/cca2like.html



Which of these vitamins do you predict to be soluble in water?

H3C CH3

CH3 CH3

CH

OH Hydrophobic

Fat soluble: A, D, E, K

Vitamin A

Like dissolves Like!

OH

Hydrophilic

Water soluble: B C

OOH


HO OHVitamin C

Will this molecule dissolve

.

2. Yes

3. I don’t know!


Solute-solvent interactions must overcome


p 374-

are attracted to the polar water molecules:

A nonpolar solid will only dissolve in a nonpolar solvent

because the dispersion forces are of comparable strength


p 380



ons n e crys a a ce can

interact with polar solvents

through formation of ion-dipolebonds.

In water, each ion is surrounded

by water molecules which form a

hydration shell and the ion is

solvated (‘aquated’)


p 379

Enthalpy diagram for a solid dissolving in a liquid

ahead:

enthalpy is a

function!


What we measure is the change represented by the purple arrowp 380

Chapter 8

e a y o pre c e o e rec on an e ex en

of spontaneous chemical and physical change under

.

‘Thermochemistry’

e s u y o energy c anges nvo ve w c em ca reac ons


http://www.youtube.com/watch?v=KTHiIwxcexI

y are we o ng ermoc em stry

un y ng ramewor or no on y un ers an ng s eam eng nes,but for understanding a diverse range of chemical reactions:

Why some reactions are spontaneousWhy gases expand and solutions mix

Why chemical reactions proceed towards equilibriumWhy proteins fold/unfold, form multimers

Refrigeration…..…..




What we will investigate:

The role of ene r gy in chemical reactions

The Fi r s t Law o f Ther m odynamics

The Second Law o f The r modynam ics

How to understand and manipulate thermochemical data to pred ic t how ene r gy w i l l change in chemical reactions.

How energy changes can p r ed i ct spon tane i t y .

m ax m u m w o r reaction.


What is Energy??

1. The capacity to do work.

E.g. Lifting an object.

Work done = force x distance

. .

Heat = the process of t r ans fe r o f t he r m a l ene r gy between two

o es or sys ems a eren empera ures. e cons er ea asbeing unable to do work.


=> Heat is a means of energy transfer, not energy itself!!


nergy anges n em ca eac ons

r ua y a c em ca reac ons a sor or re ease energy.

In order to understand this, we need to focus on a limited, well-

defined part of universe, called the system . Everything else is called

the surroundings .

The Surroundings +

System = Universe.


8.2, 284



Open

System

Matter

Energy

pen sys em:

can exchange matter and energy with the

surroundings.

ClosedMatter Closed system:

SystemEnergy

surroundings.

IsolatedSystem

Matter

Ener

Isolated system:

can exchange neither matter nor energy

with the surroundin s.


e proper es o a sys em a any one me s s a e

Module 1: Lecture 5


Chemical Reactions involve

8.2, 284

Energy Transfer

The system is usually the chemical

reactants and products. The system is ourframe of reference.

Energy is transferred to/from a system from/to the surroundings.

or w an ea q are e wo un amen a ways n

which energy is transferred to or from a system.


nergy os y sys em = energy ga ne y surroun ngs

The First Law of Thermodynamics

Energy cannot be created or destroyed, it

another.

=> Internal energy difference between reactants and products.


8.3, 285The Energy that is transferred comes from the system’s internal energy U



The Energy that is transferred comes from the system s internal energy, U .

n erna nergy an e rs aw oThermodynamics

The internal energy, U is the sum of all the energies – e.g. potential,kinetic – for all particles in the system.

.

The change in internal energy: U = U final – U initial

If U is positive – system gains energy

If U is ne ative – s stem loses ener

All energy must be released or gained from the surroundings:


U system = -U surroundings

CH + 2O CO + 2H O l

Exam le: combustion of methane:

Internal Energy system

CH4(g), 2O2(g)Uinitial

internal energy than

reactants

q

CO2(g), 2H2O(l)Ufinal

In the combustion of methane the system has lower energy at the end


– .

=> Exothermic reaction

Exam le: Combustion of methane ener O2HCOO2CH

CH4

CO2

Combustion of methane releases energy in form

O2

H2O

of heat (energy flows from system to

surroundings). => exothermic reaction.

q


reactants

Other reactions cause energy to flow from surroundings to system:

.

)2NO(energy)(O)(N 22 ggg e.g.

NO(g)Uinitial

4

O2

CO2

N2 NO

qq

H2ONO N2(g), O2(g)Ufinal

Endothermic: reactants have lower internal

energy than productcs


,

place to another

8.3, 285



In chemical reactions energy is exchanged with the surroundingsas either heat (q) or work (w).

With respect to chemical reactions, the First Law can be expressedin terms of w and q.

U = q + w

ere q = ea a e o sys em q = ea remove

w = work done on system (w<0 = work by system)


Positive work or heatmp es t at energy o

system increases.

Negative work or heat

implies that energy of

system decreases.


‘ ’

8.3, 285

There are many different types of work – electrical, mechanicaletc. At this stage we are interested in work associated withcontraction and expansion of a gas (“pV” work).

e.g. If we raise the temperature of a gas the volume increases.As the as ex ands it ushes back the surroundin s i.e. doeswork.

= - ext

= final – initial


Rationalisation:

Work = F x d = F x h (Change in height)

= p x A x h

= pV


8.3, 287



xamp e:A gas expands in volume from 0.1 m3 to 0.2 m3 against a

.

V p w ext Solution:

m0.1m0.2mN10100

4

3323

kJ10

.

Gas “does” -10 kJ of work.


(negative sign= work done by system)

‘ ’

Heat = the process of t r ans fe r o f t he r m a l ene r gy betweentwo bodies or systems at different temperatures. We consider

heat as being unable to do work.

Heat is a m e a n s of energy transfer, not energy itself!!

We cannot measure heat directly, but we can measure the

body to another:q = C T

Where T = Tfinal –Tinitial and C=heat capacity


=> Calorimetry: (Blackman p287-293 – your responsibility!)

U = q + w ,

At constant volume V is 0, so w = 0 and

U = q v

=> =

Energy as q

Internal energy changes by q U r can bemeasured using a


bomb calorimeter

8.3, 284

But, in the ma ority of cases reactions occur under conditions

.

of constant pressure (atmospheric).

= ,

U = qp – pV <=> qp = U +pV

Need a new thermodynamic property that is equivalent to theheat of reaction at constant pressure – ENTHALPY (H )

H = U + pV General definition

H = U + pVEnthalpy changes (e.g.

during a chemical reaction)at constant pressure


Enthalpy

E h i E d h i

8.3, 292



Enthalpy.

U = qp – pV Energy as PV work

qp = U + pV

H = U + VEnergy as q

=

Enthalpy changes by q

p

=> H = heat of reaction at constant pressure

Enthalpy change for a chemical reaction:


= products - reactants

Exothermic Endothermic

R , H P

, H

P E n t h a l p

RH > 0 (positive)H < 0 (negative)

E n t h a l p

Exothermic: Endothermic:Hproducts < Hreactants, H < 0 Hproducts > Hreactants, H > 0

The ‘coffee cup’ calorimetercan measure H directl .

(specific heat/heat capacity andcalorimetry: Blackman p287-293 –


chemlab.truman.edu

your responsibility!)

Internal Ener versus Enthal .

H = U + (pV)

,

(i.e. H ≈ U).

For a reaction invo ving an i ea gas:

H = U + (nRT)

At constant temperature:

H = U + RTn

= –


Internal Ener vs Enthal .

C2H5OH(l) + O2(g) CH3COOH(l) + H2O(l), H° = -492.7 kJ

What is the standard internal energy changefor this reaction at 298K ?

Solution:At constant temperature we know that:

U = H – RTn

In this case n = 0 mol – 1 mol = -1 mol

U = H – RT -1 mol = -492.7 x 103 J – (8.314 JK-1mol-1 x 298 K x -1 mol)

= -490.2 x 103 J


The difference = -2.5 kJ = work done by the system.

I t t E th l i t

I t t E th l i t



Im ortant Enthal oints.

1.Enthapy is an ex tens i ve property.If the amounts of reactants doubles, theenthalpy doubles. Implicit that a quoted H

.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

2CH4(g) + 4O2(g) 2CO2(g) + 4H2O (g) H = -1604 kJ

‘ ’


(it includes the enthalpy change of the reaction)


2. Enthalpy change of the rever se reac t i on is of s am e m a g n i t u d e but oppos i te s ign .

4 g 2 g 2 g 2 g = -

2 g 2 g 4 g 2 g =



3. Enthalpy change for a reaction depends onthe state of reactants and products.

CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H = -802 kJ

CH (g) + 2O (g) CO (g) + 2H O(l) H = -890 kJ

Difference is:


2 g 2 = -

Module 1: Lecture 6


State Functions Pro ert of State

8.x, xxx




State Functions Pro ert of State .

When a system changes state, the State Functionsare onl de endent on initial and final states.

The change of state is independent of the path

.

Work and heat are dependent on the way in which.

Functions.

10°C 10°C

na s a e n a s a en a s a e

heat freeze


http://itl.chem.ufl.edu/2045_s00/lectures/lec_7.html


The Final State is independent of the path taken

The distance between

remains the same (a

state function) but the

distance travelled to go

between them is

variable (not a state

function).


Corollary: If you do a round trip, nothing has happened! ( X = 0)



Analogy:

between two points on amap remains the same (a

s a e unc on u e


between them or the actual

work going up and down are


function .





Analogy:

between two points on amap remains the same (a

s a e unc on u e





function .








Analogy:

between two points on a

map remains the same (a

state function) but the





function .



important and useful?

The Final State is independent of the path

taken.

‘The Enthalpy change for a reaction is dependent

products and is independent of the path or numberof ste s taken’ Hess’s Law.

Enthalpy change = H(final) – H(initial)

= H(products) – H(reactants)


Germain Henri Hess (1802 – 1850)

Hess’s Law.

8.x, xxx

,

H 3 = -112kJ

H ( k J

)

2NO2(g ) 2NO2(g )H 2 = 180kJ

N2(g ), 2O2(g )

1 2 31

N2(g ), 2O2(g )

–

= H(products) – H(reactants)


Hess’s Law

.

Can use Hess’s Law to calculate H for many reactionsusing tabulated enthalpies.

Important tabulated values are Standard Enthalpies (or heats)of Formation.

En tha lpy o f Fo r ma t ion (Hf ): enthalpy change associatedwith formation of 1 mole of a substance from its elements (C +

2

But: depends on conditions.


Standard Enthal of Formation Standard Enthal of Formation



Standard Enthal of Formation.

Depends on state (s, l, g), temperature andpressure.

–at 1 atm and temperature of interest (usually 298K).

For compounds: Gases – 1 atmLiquids – pureSolutions: 1M

. . , .



Standard enthalpy of formation (H°f )

‘Enthalpy change that results when 1 mole of a compound is

formed from its elements, with all substances in standardstates’

6C(s) + 6H2(g) + 3O2(g) C6H12O6(s)

Not:

6CO2(g) + 6H2O(l) C6H12O6 (s) + 6O2(g)

.

H2(g) + O2(g) H2O2(l), will not happen.



The standard enthalpy of formation for an elementin its standard state is ZERO.

‘Elements are not formed, they just……are!’

O2(g) O2(g) H°f = 0


Note: H°f is in kJ/mol


Blackman, Appendix A, pageA-1 to A-3

How do we use it? This means we can mani ulate



How do we use it?

r calorimetry.

u : e ous, me consum ng an no very exc ng o a waysfeasible.

We can use thermodynamic tables to determine H rxn,

w ithout entering a laboratory.

‘If a reaction is carried out in a series of steps, the enthalpy

individual steps.’


This means we can mani ulatechemical reactions.

1. Take reactants apart to their constituent elements

2. Form products from the elements

Reactants ProductsHrxn

H°f (r))

H°f (p))

Constituent Elements


For a general (hypothetical) reaction:

Then:

° ° ° ° °rxn f f - f f

H°rxn = nH°

f (products) - nH°f (reactants)

Where n = coefficients in the balanced equation.


for combustion of ammonia in air:

3 2 2 2

H°rxn = [cH°

f (C) + dH°f (D)] - [aH°

f (A) + bH°f (B)]

H°rxn = [4H°

f (NO2(g)) + 6H°f (H2O(l))] - [4H°

f (NH3(g)) + 7H°f (O2(g))]

= 4 33.2 + 6 -285.8 – 4 -46.1 – 0


= -1396 kJ Is this a sensible answer?

A ain: im ortant thin s to remember



’ Example 8.9, page 300 of Blackman.


A ain: im ortant thin s to rememberwhen manipulating reactions.

1. If you reverse a reaction be sure to change the sign on H.

CuO(s) + H2(g) Cu(s) + H2O(l) H = -130.6 kJ

Cu(s) + H2O(l) CuO(s) + H2(g) H = +130.6 kJ

[You will come across this concept again in Module 4 (electrochemistry)]

2. If you multiply by a coefficient, multiply H by the same

coefficient (H is extensive).

CuO(s) + H2(g) Cu(s) + H2O(l) H = -130.6 kJ

-


2 2 - .

Exam le:

Two forms of carbon are graphite and. ,

calculate H for the conversion of graphite to diamond.

www.encarta.msn.com

C(graphite) + O2(g) CO2(g) H = -394 kJ

C(diamond) +O2(g) CO2(g) H = -396 kJ


xamp e:

Solution:

Reverse the second reaction and sum the two

C(graphite) + O2(g) CO2(g) H = -394 kJ

CO2(g) C(diamond) +O2(g) H = +396 kJ

C ra hite C diamond H = +2 kJ

graphite to diamond (endothermic process)


Exam le:

Example:



Ethanol in petrol

With petrol approaching $2/litre it is cheaper to buy ethanol

. ,effect does it have on fuel efficiency?

Relevant data:

1

188

52

molkJ251)](HC[

molkJ278)](OHHC[

l H

l H

f

f

1

2

1

2

molkJ286)](OH[

molkJ5.393)](CO[

l H

g H

f

f

How much heat is roduced by burnin one ram of ethanol,


compared to one gram of octane?

p

The combustion of octane:

Ethanol in petrol

On a mass basis:


Example: Ethanol in petrol

The combustion of ethanol:

On a mass basis:


RACQ - ‘Blends of up to 10 percent ethanol in petrol (E10) have been declared suitable for use in most –but not all – post-1986 petrol-engine vehicles in Australia. E10 increases fuel consumption between 2.6


and 5 per cent relative to unblended petrol’.

So: it’s cheaper and potentially more environmentally friendly but you use more of it………..

Enthal and Phase Chan es.

7.3, 251

What you should be able to do at this stage:



Heat is also required for a phase change (Module 2)..

dependent on the strength of the intermolecular forces.

solid liquid (heat in) fusH

liquid gas va H

Criticalpoint

solid gas subHSolid Liquid

s s u r e ( a t m )

liquid solid (heat out) solidH = -fusH

= -

Gas P r e

Triple

point

condensation

vap

gas solid depositionH = -subHTemperature (°C)


Note: you may want to go back and review this slide and intermolecular forces, once you have encountered phasechanges.

1. Define and explain the First Law of Thermodynamics and its relation to

chemical reactions.

2. Explain what is meant by the terms System, Surroundings and the Universe

with respect to chemical reactions.

3. Explain how energy is transferred in chemical reactions.

4. Explain what Internal energy (U ) and Enthalpy (H ) are and how they are

re a e .

5. How is ‘work’ defined in chemical reactions (with relation to enthalpy)?

. .

7. Explain what a State Function is and how Hess’s law exploits it.

‘ ’. .9. Use thermochemical data to predict the enthalpy change for a reaction.


. xp a n w a an xo erm c reac on an an n o erm c reac on mean n

relation to enthalpy changes.

Module 1: Lecture 7


What we will investi ate:

.

The role of energy in chemical reactions

First Law of Thermodynamics

How to understand and manipulatethermochemical data to redict how enerwill change in chemical reactions.

How energy changes can predict spontaneity.


What determines whether or not a

A S ontaneous Process:



process will be spontaneous?

We observe that many exothermic reactions aren ee spon aneous.

+ - -2 - .

u : so are some en o erm c reac ons.

= °2

2

.


Lets look at some spontaneous processes

A hot object cooling

Not spontaneous

Hot Coldspontaneous

Hotter

o o ec a ways coo s, a o o ec w never spon aneous y

absorb heat from colder surroundings to heat up.


A S ontaneous Process:A hot object cooling

Hot ColdA natural change

What is different between the hot cube and the cold cube?

ENERGY

– the hot cube is characterised by molecules having

temperature).


Think: where has this energy gone with cooling?

“SPONTANEOUS”

occurs w ou“outside intervention”

Thermodynamics tell us about the directiono a process – o e spee o e process

e.g. n pr nc p e a amon s ouchange spontaneously to graphite- owever: ery s ow process


– .



feeling or native tendency without external

.

There is nothing in the definition of ‘spontaneous’ that describes how FAST the process is.

ne cs e s us ow as a reac on s o u e .Thermodynamics tells us if it will proceed at all.

Remember this!


e.g.

• A Ball rolls down a hill never spontaneously up a hill

• A gas fills a container uniformly

• Heat flow always from hot to cold object – never the reverse

• o ,

> 0 oC ice spontaneously melts

WHY?


No: The meltin of ice occurs s ontaneouslabove 0oC – Endothermic process

Looking ahead: The driving force forspontaneity is an “increase” in the

“entro ” of the universe


8.4, 304

o w a s n ropy

n ropy s a measure o nergy s r u on

um er o arrangemen savailable to a system

Nature spontaneouslyproceeds towards statesthat have the highest

probability of existing.


Entro is a measure of Ener distribution

8.4, 305


8.4, 305



High-energy molecules Low-energy moleculesLet’s bring 3 high-energy molecules in contact with 3 low-

energy molecules => What is going to happen??




Number of arrangements available to a system

0 energy units transferred 1 energy unit transferred 2 energy units transferred 3 energy units transferred


Nature spontaneously proceeds towards states that have the highest probability of existing.


8.4, 305



Macrostates

Number of arrangements available to a system



Nature spontaneously proceeds towards states that have the highest probability of existing.


8.4, 305



Macrostates

Microstates = Number of arrangements available to a system



Nature spontaneously proceeds towards macrostates that have the highest probability ofexisting = the highest number of microstates.

Lets think in terms of ‘Probabilit ’

8.4, 305

This is im ortant:

8.4, 306



p=5% p=45% p=45% p=5%

Which do you think is the most ‘probable’ state?

Spontaneous processes proceed from macrostates of low probability

to macrostates of higher probability


‘Spon t aneous p rocesses tend t o d i sperse energy ’ .

‘Spon t aneous p rocesses tend t o d i sperse energy ’ .

We can now introduce a quantity that is a measure for how

–


.

8.4, 307

c e s s i b

l e

e l s

b e r o f a c

e n e r g y l e

N u m

The distribution of an identical amount of energy is greater in the larger volume.

This is important:


,factor when considering entropy!!

Second Law of Thermod namics8.4, 309

increase in the Entropy of the Universe.

Suniverse = Ssys + Ssurr

If Suniv = positive (>0) – Entropy of the universe increases.

If Suniv = negative (<0) – process is spontaneous in the

opposite direction

If Suniv = 0 – equilibrium has been reached.


8.4, 309

Our two Law s of Thermodynamics



The Second Law of Thermod namics.

Nature spontaneously proceeds

diffusion of energy and thus have


Our two Law s of Thermodynamics.

Ener is conserved.

Entropy in maximised.


We are interested in Entro chan es.

Entropy is a state function, so

S = Sfinal – Sinitial

Similarly to Enthalpy, we have to take into account the system.

Where the surroundings + system = universe.

Suniv = entropy change in the universe

=sys

Ssurr = entropy change in the surroundings


Interplay of S and S determines thesign of Suniv

Suniverse = Ssystem + Ssurr > 0

Zumdahl, 2006

Note: one of the entropy changes may be negative, but sum must be positive.


If Suniv < 0, process will not be spontaneous as written, but is spontaneous in thereverse direction.

Entropy changes and real life.

A thermod namic caveat.



How do we end up withcomplex systems that have

no a arent entro y – i.e.how do we end up with asystem having localised


Large and positive

Negative


Transferrin the Second Law to ‘Life’ is not a ood idea!Remember these laws were essentially developed for chemicalreactions (steam trains and reactions in test tubes) where the

= ……..

i.e. You can ask yourself if life violates the Second Law but’

D. R. Brooks and E. O. Wiley.. , .

Thorne K. S. et al.–


– .QB843,B55.B59

8.4, 309

To determine spontaneity (=Suniv), we need to know

Ssys and Ssurr .

How does the entropy of the system and the surroundingsc ange ur ng a process

The only way we can exchange energy between system andthe surroundings is via work (w) or heat (q).

=> We need to find a t h e r m o d y m a n i c definition of entropy,


factor.

ENTROPY is:

q r S = J K-1

(So heat flow is the important factor in determining entropy.)

It all started with steam engines…


Optional reading: Carnot Heat Engines

The Carnot C cleIn the days of steam engines

The Carnot C cle



In the days of steam engines…

http://commons.wikimedia.org/wiki/File:Tower_bridge_steam_engine.jpg


http://cache-media.britannica.com/eb-media/11/71511-004-321E1907.jpg http://www.todayinsci.com/H/Hornblower_Jonathan/HornblowerJonathanPatent1298.htm

Cyclic process with 4 steps:1. Isothermal expansion at T1 (hot)2. Adiabatic expansion (T1 → T2

3. Isothermal compression at T2 (cold)

4. Adiabatic compression (T2 → T1)

We can look at Work , Heat and efficienc for this circular rocess:

= Model for any cyclic process!!

q1=-wtot + -q2 -wtot = q1+q2

=> = -w = +

T1 t t

Also: = (T1-T2)/T1 (derivation: RED BOX CHALLENGE!)

system

q1

-wtot

=(q1+q2)/q1 = (T1-T2)/T1 1+ q2 /q1 = 1 – T2 /T1

=> q1 /T1 + q2 /T2 = 0 for this processT2

-q2


=> q/ T = state function!! => Entropy!!!

We can define a standard entropy (S°)

Notice enthalpy as H

But entropy as just S

y

To understand that we need to focus onthermodynamic entropy = Entropy as it relates to

the occupancy and probability of energy levels with


re ation to t erma energy – i.e. eat.

°definition of Entropy:

The thermodynamic definition of ENTROPY is:

q r -

T

At absolute zero T=0 so heat cannot be transferredso = eore ca y


increases until we arrive at S° (298 K, 1 atm)

8.4, 308




Consider a perfect crystal of HCl at 0 K:

Entropy of such a crystal is+- +- +- +- +-

,way to achieve perfect

order

+- +- +- +- +-

+-+- +- +- +-

crystal at 0 K is zero.

Disorder appears as crystal is warmed, leading to increasein entropy. Values of S° are absolute entropies (cf H , U ,


etc)

Module 1: Lecture 8


Example: Calculate S° for the synthesis of ammonia from

2 an 2:

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

S°rxn = 2 S°(NH3) – [S°(N2) + 3 S°(H2)]

= 2 * 192.5 – [191.6 + 3 * 130.7]= -198.7 JK-1


Note: S° for elements is not equal to zero

Srxn < 0 for the Haber process: N2(g) + 3H2(g) 2NH3(g) ‘Rules’ for Entro Chan es



rea er n ropy Less Entropy


Reduction of ways in which energy can be distributed

(1) S° for elements is not equal to zero

(2) Reactions that consume or generate gases can have

.Could predict that S° for Haber reaction < 0, since thereare 4 moles of gas on the left and 2 on the right.

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

I2(g) 2I(g) S0 > 04Fe(s) + 3O2(g) 2Fe2O3(s) S0 < 0

Example: Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)


1. Prediction: the roblem of rust

+

4 mol Fe (s)

3 mol O2 (g)

2 mol Fe2O3 (s)

S < 0


2. Prediction: Com lex molecules

Al2O3(s) + 3H2(g) 2Al(s) + 3H2O(g)

As 3 mol hydrogen forms 3 mol water vapour one

would assume S to be zero (no change in number of gas molecules).

However, S is large and positive, why?

Due to formation of a more complexmolecule.

Remember? The distribution of energy, not of molecules is the important

The H2O molecule can vibrate and rotate in several ways.


factor when considering entropy!!

‘Rules’ for Entro Chan es



Liquid H2O has a relatively lowentropy. Why?

Remember intermolecular forces?

Just another manifestation of the.

High BP, Low entropy – H-bonds


• The entropy for gas is larger than for liquid which is larger thanfor a solid

• The more com lex the molecule the hi her the entro

• Forming a solution from a molecular solid yields an increase in

• Forming a solution from a salt can cause either an increase or

• Pressure and Temperature affect entropy


Pressure and EntroWhich has the higher entropy?

N as at 1 atm or N as at 1x10-2 atm?

pV = nRT

One mole of N2 gas at 1x10-2 atm has a volume 100 timesthat of one mole of N2 gas at 1 atm. Thus, N2 gas at 1x10-2

atm has the higher entropy (particle in a box)

1x10-2 atm >>

1 atm

Slow pressure > Shigh pressure


Tem erature and Entro

Why is temperature so important?

l e v e l s

e e n e r g

c c e s s i b l

Absolute zero

(BEC)

Temp 1 Temp 2

A change in temperature can make a huge difference

to accessible energy levels and thus the distribution


o energy n ose eve s.

What ou should be able to do at thisstage:

Inter la of S and S determines

the sign of Suniv



g

‘ ’ . e ne w a s mean y a spon aneous reac on .

2. Define the Second Law of Thermodynamics.

3. Explain what is meant by ‘entropy’.

4. Describe how entropy changes for a spontaneous reaction.

5. Predict entro chan es based on anal sis of athermodynamically balanced equation (e.g. formation of gases,higher complexity molecules).

6. Use thermochemical data to predict the entropy change for asystem.


7. Describe how temperature affects entropy.

the sign of Suniv



Zumdahl, 2006

We can determine the system entropy8.4, 309

change, but what about entropy changes inthe surroundings?

To determine spontaneity, we need to know Ssys

+ Ssurr . How does the entropy of the

surroundings change during a process?

The only way we can exchange energy with the.

The thermodymanic definition of entropy is:

q / T


.

Heat flow to the surroundin s - S ?8.4, 309

T e entropy c ange in t e surroun ings is ue toheat transfer from the system (conservation of

.

Ssurr = q /T

Therefore spontaneity is often dependent ontemperature. Eg an en ot ermic reaction may espontaneous at high temperature, but not at low.


Remember Enthal ?

Recall:

. surr

flow.



Heat flow = H at constant

+ sign: endothermic (into the system)- sign: exothermic (out of the system)

So: At constant temperature andpressure.

Ssurr =-H sys

T

+ sign: exothermic

- sign: endothermic

S = S + S > 0


2. The magnitude of Ssurr depends on

temperature

Driving forceprovided by =

Magnitude ofthe entro =

Quantity of heat (J)

energy flow(heat)

change of thesurroundings

Temperature (K)

=Quantity of heat (J)+SsurrExothermic

=Quantity of heat (J)-SEndothermic


Temperature (K)

=

=H

- -ve si n because enthal with res ect to s stemT

surr

1. The sign of Ssurr depends on direction of heat flow.

2. The magnitude of Ssurr depends on temperature and

magnitude of enthalpy

As T increases, Ssurr decreases

“The impact of the transfer of a given quantity of energy as heat to

or from the surroundings will be greater at lower temperatures”


The roblem of rust

4Fe(s) + 3O2( g) 2Fe2O3(s) S0 < 0 (prediction)

4 mol Fe

But experimentally, rusting is spontaneous

-H sys

+ surrT

3 mol O2

- . ,

Ssurr = -(-1648.4 x 103J)/298K = +5529 J/K

Ssys = -549.4 J/K

2 mol Fe2O3

Suniv = -549.4 + 5529= 4980 J/K


The Gibbs Free Energy

8.2, 283

S = S + S

Converting the ‘general’ equation to one involving only

system variables.



Suniv = Ssys + Ssurr.

But it’s inconvenient as a measure of spontaneity, since requiredto know about system and surroundings.

We want a simple thermodynamic function related tospontaneity which only requires knowledge of system.

Thanks to J. Williard Gibbs we do:

The Gibbs fun c t i on ( G)

In 1876 Gibbs published the first part of the work for which he is


,

publishing the second part of this work in 1878.

S = S + S

HsysS

universe

= S s stem

-

TS universe = Hsystem - TS system

G = H - TThis is the 2nd equation

G is the Gibbs Function of a process at const ant pressure and

o remem er

temperature

Just a more convenient wa of determinin if a rocess is


spontaneous……but it is much more useful than that.

Wh is this useful?8.2, 284

If the process is spontaneous:

S universe = S system + S surr > 0

Hsys

TS universe = S system - > 0

TS universe = Hsystem – TS system < 0

G = H – TS < 0

G w ill be negative for a spontaneous process!


H - T S < 0

i.e. if G < 0 process is spontaneous

if G > 0 reverse process is spontaneous

if G = 0 system is at equilibrium

Two functions that can predict spontaneity:

+,

ΔSuniv: all processes


Exam le: Wh does ice melt at 10°C but not

at -10°C?

kJ03.6)(OH)(OH0

22 H ls



H2O(s) H2O(l) H° = 6.01 kJ

º

t < 0ºC


T T H° S° S S G°surr un v

C K kJ J K-1 J K-1 J K-1 J mol-1

-10 263 6.03 22.1 -22.9 -0.8 +220

-. . .

+10 283 6.03 22.1 -21.3 +0.8 -220

us a , ce me s spon aneous y, u a - ,water freezes spontaneously.


8.5, 314

ummary o e e ec s o e s gns o

and S on spontaneity

Only when the negative

va ue o s arger an

the negative value of TS

H2O(l)→ H2O(s)

This is an exothermic change

accompan e y a ecrease

in entropy so is spontaneouso


8.5, 314

Various combinations are possible:

.

H S G Example

____________________________________________________________

+ + Spont. at high temp. CaCO3(s) everse spon . a ow emp. a s 2 g

Spont. at low temp, reverse CaO(s) + CO2(g) spont. at high temp. CaCO3(s)

+ G always +. 3O2(g) 2O3 (g)

Reverse rxn spont. at all temps.

+ G alwa s . 2O 3O

Spont. at all temps.______________________________________________________________________________________________________________

= –


At what temperature would processes with the

Solution:

a)G = 10 000 J + 220 J K-1 x T = 0



p p

a) H = +10 kJ; S = -220 J K-1

b) H = +10 kJ; S = +184 J K-1

At equilibrium we know G = 0, so given the data we

G = H - T S

for the temperature T.


- s canno e zero a any empera ure

This process has G +ve at all temperatures.

- -

- this equals zero at T = 54.3 K.

Below 54.3 K, G is > 0 (not spontaneous)

. ,

i.e. we can determine the temperature at which a


non-spontaneous reaction becomes spontaneous.

Module 1: Lecture 9


Standard Gibbs Free Ener Chan es8.5, 314

As for H °, we define G° (standard Gibbs Free Energy

change) as the change for a reaction carried out under

standard state conditions, with all reactants and products in

.

Also use standard free energies of formation (G° f ).

For aA + bB cC + dD,

G°rxn = [cG°

f (C) + dG°f (D)] - [aG°

f (A) + bG°f (B)]

Exactly analogous to use of H


The Standard Free Energy of Formation

Gf °

Gf

°



Gf ° is the change in the Gibbs Free Energy when one mole of a

substance is created from its constituent elements, with all species

.

Like Hf °, Gf

° has units of kJ mol-1.

Like Hf ° for an element in its standard state, Gf

°

is defined to be zero.


• G° cannot be measured directly bymeans of any instrumentation

• G° can be calculated from othermeasure quantities

• H can e etermine y measuringheat flow in a calorimeter

• S° can be determined experimentally


8.5, 314 - 315

Three ways to determine G°:

1. G°rxn = npG°

f,prod – nrG°f,react

2. G°

= H°

– TS°

3. Reaction Manipulation (calculating from other rns.)

earn ese ways


Methods for Calculatin G° 1

2CH3OH(g) + 3O2(g) 2CO2(g) + 4H2O(g)

G f º -163 kJ/mol 0 kJ/mol -394 kJ/mol -229 kJ/mol

G° = 2Gf °(CO2(g)) + 4Gf

°(H2O(g)) – (2Gf °(CH3OH(g)) - 3Gf

°(O2(g))

= 2mol(-394 kJ/mol) + 4mol(-229 kJ/mol)


– 2mol(-163 kJ/mol) – 3mol(0 kJ/mol) = -1378 kJ


G° = H° - T S° 22))((Of))((SOf(g))(SOf

ggH H H H



Consider the reaction

2SO2(g ) + O2(g ) 2SO3(g )

carried out at 25C and 1 atm. Calculate H , S , and G using the

following data:

Substance H (kJ/mol) S (J/K mol)

f

SO2(g) -297 248

SO3(g) -396 257

O2(g) 0 205


))((Of ))((SOf (g))3

(SOf 22 gg

kJ594kJ792

0kJ/mol)297(mol2kJ/mol)396(mol2

kJ198

22 )(O)(SO)(SO 223 ggg SSSS

J/K205-J/K496-J/K514

mol)J/Kmol(2051mol)(248J/Kmol2mol)J/K(257mol2

J/K187-


ST H G

K187)K298(J000,198 J700,55J000,198



The Free Energy is a State Function so we can manipulate any’

Cdiamond(s) Cgraphite(s)

Cdiamond(s) + O2(g) CO2(g) Gº = -397kJ

CO2(g) Cgraphite(s) + O2(g) Gº = 394kJ

Cdiamond(s) Cgraphite(s) Gº = -3kJ

This reaction is spontaneous but VERY slow

º

° ’

.


.

Why is G so useful?The Gibbs Free Ener and Chemical

8.5, 318

Equilibrium



1.Its relationship to equilibrium

. ts re at ons p to wor .


The real power of G lies in its relationship to equilibrium.

We can predict in which direction a system under standard

conditions will a roach e uilibrium G = 0 .

if G < 0, process is spontaneous (products favoured)

if G > 0, reverse process is spontaneous (reactants)

if G = 0, system is at equilibrium


The Gibbs Free Energy and ChemicalEquilibrium

G < 0

In terms of energy: atequilibrium G = 0 andthe system no longer has

the capacity to do work.

productsreactants

G = 0

G ° = - RT lnK This is the 3rd equationto remember!!!

where K is the equilibrium constant


(This concept will be covered in more depth in module 2)

2. The Gibbs Free Ener and Work.8.5, 316

A reversible rocess is done in a s ecific wa so thatwhen undone there is no observable change in thesystem or surroundings.

A Breversible

Remember the First Law?

U = q + w

For the perfect reversible process we would be able toextract the maximum work from the internal energy –


.relate to?

This is the entropy.

The Gibbs Free Ener is w

U

,



T=

U

q r w r,max

For a perfectly reversible process, qr , thus the gain inentropy, is minimized.


r,max

q r w r,max

G = H - T S

G = wr,max

The thermodynamic limit to efficiency.

But: perfectly reversible processes are unattainable,

yet we can determine G for any reaction usingthermodynamic tables??


G, H and S are State Functions.

G = w

When G is negative, the magnitude of G is equal to

,

.

When G is ositive the ma nitude of G is e ual to

the minimum work that must be expended to make the

process spontaneous.


Wh the ‘Gibbs Free Ener ’ ?

‘ ’ –work, not pV work done on/by the system, butelectrical, mechanical etc work.

’ n s as s you a so see en ropy e ne asthe ‘energy that can’t do useful work’ or ‘low

’ .


The other extreme – total irreversibilit

8.3, 287

U

–

8.5, 317

U



q

For a perfectly irreversible process, q is maximizedand work is minimized.

e.g. irreversible battery discharge.


q r w

In a real process there will always be losses of heat to.

Extra losses will occur due to friction etc.

‘As long as a body retains the ability to transfer heat to another body – a s, a any empera ure a ove a so u e zero – row n an mo on s no

only possible but also i nev i tab le .’ Einstein, 1905.


Another statement of the second law:

In an real c clic rocess work is chan ed toheat in the surroundings, and the entropy of the

universe increases.


Example: What is the maximum work obtainable from

2CH OH l + 3O 2CO + 4H O l

the combustion of methanol

G f º -166.4 kJ/mol 0 kJ/mol -394.4 kJ/mol -237.2 kJ/mol

Gº = (2 x -394.4 + 4 x -237.2) – (2 x -166.3) kJ

= -1404.8 kJ

Ho = (2 x -393.5 + 4 x -285.8) – (2 x -238.7) kJ

= - .

We can expect no better than 96.7 % conversion of the

energy the reaction can provide.

i.e. even done reversibly we still lose 3.3% as heat to the


surroundings – entropy has increased.

Example:

How much work can be gained from a reactionwith the following parameters?

Example: The combustion of glucose provides energy for

nervous an muscu ar ac v y. a s e max mum wor

available from combustion of 1 mole of glucose at 37oC (310K)o = -1



g p

H = +10 kJ; S = +184 J K-1; T = 50K

G = H – TS

= 10 000 – 184 50

=

.

in order for reaction to be at equilibrium (e.g. raise thetem .


. .

C6H12O6 s + 6O2 6CO2 + 6H2O l

Know: wmax = G, and G = H-TS

So: Hrxn = [6xH(H2O(l)) + 6xH(CO2(g))] – [6xH(O2(g)) + H(C6H12O6)]

= -2802.04 kJ/mol

G0 = H0 - TS0

= -2802.04 x103 J - (310 K x 182.4 J/K)

- .

Thus: Wmax = 2858.6 kJ/mol glucose burnt.


MW glucose = 180g/mol, then 2858.6/180 = 15.9 kJ/gram glucose.

. ,

need: (70 x 9.8) x 3 = 2.06kJ of energy.

We have 15.9 kJ/gram of glucose, so 70kg person would

need to burn: 2.06/15.9 = 0.13 grams glucose.

In reality, much more than this would be required due to.

Other examples: See Page 317, Blackman.


4Fe(s) + 3O2( g) 2Fe2O3(s) Suniv = 4980J/K

4 mol Fe

+

G = -TSuniv = H – TS

3 mol O2 = -298 K 4980 J/K

= - 1484.0 kJ

2 mol Fe2O3






Affected by:

tem erature Links to

Equilibriumpressure,

concentrations

Thermodynamics Note: In the next few lectures we use specific examples to teach key ideas

(concepts) – you are expected to be able to apply these ideas to any

equilibrium reaction.



Homogeneous

Module 1

Heterogeneous

Module 2

ChemicalReactions

Acid/Base:Module 4

PhaseChanges

SolubilityElectro-

chemistry:Module 2


Underpinning CHEM2002, CHEM2056, BIOC2000

Examples of equilibrium processes in real life tend to be complex systems so we will be in with sim le exam les. .


http://www.rsc.org/education/teachers/learnnet/cfb/transport.htm

Explore your existing understanding of equilibrium …

Write the balanced equation for the decomposition of dinitrogen tetroxide

to nitro en dioxide:

2NO

Write the balanced equation for the dimerisation reaction of nitrogen

ox e o orm n rogen e rox e:

42 g2 g


42 2 g422

Blackman Chapt er 9

Consider . . .

dinitrogen

tetroxidenitrogen

dioxide

•N2O4(g ) is in a sealed tube – closed system - begins to dissociate

o orm 2 g , w c n urn s mu aneous y eg ns o recom ne o

form N2O4(g )


•Equilibrium is a dynamic process (it can respond to change!)p 334

We can observe the progress of a reaction by monitoring concentration as

a function of time.)(ON 42 g )(2NO2 g



• The initial concentration of N2O4 decreases with time• Simultaneously the concentration of NO2 increases with time


• qu r um s reac e w en ere s no c ange

The same equilibrium will be reached by starting with either N2O4(g) or

with NO2(g)

This is an example of a Homogeneous equilibrium


The position of chemical equilibrium can be expressed in terms of

the equilibrium constant:

)(ON 42 g )(2NO2 g

After concentrations have stopped changing, find that:

C25at1061.4

NO o3

2

2 K ON 42

c

regar ess o s ar ng concen ra ons. s ra o s nown as e

equilibrium expression for the reaction and results in the

equilibrium const ant , K c .


Generic re resentation of the ex ression for the

Equilibrium constant:

p 335

Note:

1. [A] refers to the equilibrium concentration of A in mol L-1 (orM).

2. K c is dimensionless in the sub-discipline of physical chemistry.


The value of Kc is dependent on temperature

)(ON 42 g )(2NO2 g

K ... What does the ma nitude tell us?

Compare:

2H (g) + O (g) 2H O(g)



]NO[ 2

2

.]ON[ 42c

C25at1061.4]ON[

]NO[ 3

42

2 cK

Kc ... What does the magnitude tell us?

At equilibrium, concentrations of reactants and products will be of thesame order of magnitude at 100 oC.


http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/lechv17.swf

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

298Kat101.9][O]H[

80

2

2

2

2 cK

2 2

298Kat108.4]NO[ 31

2K

][O]N[ 22


equilibrium

K >> 1 : products favoured

K << 1 : reactants avoure

=

will be comparable

When is an equilibrium positioned in the middle?

Rule of thumb …

When K is in the ran e of 0.01-100 10-2 – 102


The equilibrium constant for the following reaction is

= 5c

CO(g) + Cl2(g) COCl2(g)

What is the position of this equilibrium

at this temperature?

1. Far to the left

.

3. Close to the middle

e uivalent concentrations

of reactants and products)


An equi l ibr i um expression can be wr i t ten in more than one way … What happens t o value of c f or a r ever se reaction?

If th it ti i id d (i th ti i itt i

Using the Equilibrium Constant to predict composition

The equilibrium constant is useful because it enables us to calculate

e uilibrium concentrations of reactants and roducts:



If the opposite reaction is considered (i.e the equation is written inreverse) then K c is inverted.

e.g. )(ON 42 g )(2NO2 g

C25at1061.4]ON[

]NO[ 3

42

2 cK

inverted

)(2NO2 g )(ON 42 g

C25at217]ON[2

42 cK


2

)(ON 42 g )(2NO2 g

The equilibrium expression is:

2

C25at10x4.61]ON[

-

42

2 cK

If, at equilibrium M01.0NO2 , then

M0.0217NOON

2

242


c

Using the Equilibrium Constant to predict composition

The equilibrium constant also enables us to calculate equilibrium

concentrations, more complex:

C25at10x4.61]ON[

3-

42

2 cK )(ON42

g )(2NO2

g

If the initial concentration of N2O4 is 0.50 M, what are the equilibrium

To solve this t e of roblem we can use an ‘ICE Table’ where werepresent the concentrations of reactant and product species as a

variable and use algebra!!


Strate :

The concentration of the reactant will have decreased by the an amount

,

increased to be (2 x ).

Initial conc: 0.50 0.00

)(ON 42 g )(2NO2 g

Change: - x +2 x

Equilibrium conc: (0.50 – x ) 2 x

Use the stoichiometry of the equation to determine the relative amounts

at equilibrium


The equilibrium constant is given by:

2

C25at10x4 61 3-2 K

o e equ r um concen ra ons are:

= – =. . .

[NO ] = 2 x 0 023 M = 0 046 M



C25at10x4.61]ON[ 42

cK

2

x

x

50.01061.4 3

substituting

Which becomes a quadratic equation and when solved for x:

x = 0.023


Homework: Write out the steps to solve this equation.

[NO2] = 2 x 0.023 M = 0.046 M

)(ON 42 g )(2NO2 g

. .


A l in understandin of e uilibrium:

The amount of nitrogen fixed naturally by plantsis too low to meet our demands for food supply -

so the Haber process is one of the most important

commercial rocesses.

Liquid ammonia produced by the Haber process is either added directly to

soil as a fertiliser or is converted into ammonium salts (phosphates &.

)(H)(N gg )(NH3 g

Homework: Balance this equation:


At 127 ºC the following equilibrium concentrations were found:

[H2] = 3.1x10-3 M

-12 .

[NH3] = 3.1x10-2 M

• Write the equilibrium expression for this reaction

º• c .


Module 1: Lecture 11

Where we left off …

)(H3)(N 22 gg )(NH2 3 g




or

)(H)(N 2222 gg )(NH3 g

At 127 ºC the following equilibrium concentrations were

found:

2 = . x -

[N2] = 8.5x10-1 M

NH3

= 3.1x10- M

• Write the equilibrium expression for this reaction


• Calculate the value of K c at 127 ºC for this reaction.

Writing and substituting into the equilibrium expression –

there is more than one correct answer!

222

22 3

4

3313

22

3 108.3

101.3105.8

.

HN

cK

What is the value of K* for the reverse reaction?

)(H3)(N 22 gg )(NH2 3 g

5

3

22* 106.211HN

K


3108.3NH cK

How will the equilibrium constant change if we alter the

coefficients that we use to balance the equation for the

Haber process?

)(H)(N 223

221 gg )(NH3 g

1

K cWhen would we use

this stoichiometry?

2

3

2

33

23

21

NH NH K

24

2222

109.1108.3 21

21

cK K


Note that (xa) = xa

For the following reaction, Kc = 7.9 x 1011 at 500 K

(2) g + r 2 g r g

Gas Phase Equilibrium (Homogeneous)

concentration … when we consider gases we can also express K

in terms of the artial ressures of the ases.



( )

c 4HBr 2H2 + 2Br 2

. .

2. (7.9 x 1011)2

3. 1/[(7.9 x 1011

)2

]


)(ON 42 g )(2NO2 g

)(2

2 NOP

)(42

O N p

P

sua y use s an ar s a e o a m, so pressures expresse n

units of atmospheres, and equilibrium constants quoted with no

.

Alternatively: pressure in Pa divided by standard pressure of 5 ‘ ’


.

How are K and K related?

Remember that pressure is related to concentration

Since concentration is Pn

t roug t e ea gas equat on.

RT V

K RT

P

2NO

2

2NO n

RT RT

Pc

4O2N42ON

cP

where n is change in moles of gas as reactants are


converted to products

Exam le:)(ON 42 g )(2NO2 g

C25at10x4.61]ON[

]NO[ 3-2

2 cK

What is the value of K ?


p337 (9.3)

n

cP RT K K )(

Kc = 4.61 x 10-3

One possible way of removing NO

to react it with CO in the presence of



K c 4.61 x 10

R = 0.08206 atm L mol-1 K-1

T = 298 K (273 + 25 oC)

n = +1

113 1013.129808206.01061.4 PK


to react it with CO in the presence of

suitable catalyst.

2 2

At 300 oC this reaction has K c = 2.2 x 1059.o p



Caution!!

It is very important that correct units are used in this

equation. This relates to units of the gas constant, R.

- c p

be in atm, respectively, we must use R = 0.08206 atm L mol-1

K-1

Otherwise a conversion must be used.

Note:

‘ ’

very complex due to unit conversions. Refer to p 338-9.




Usin Q to redict the res onse of e uilibrium to

change in concentration:

Qualitatively: If we

- 2

2]NO[



Qualitatively: If we

,

will the system

respond?

)(ON 42 g )(2NO2 g

It will move to restore equilibrium . . .

We can use the reaction quotient, Q, to predict in whichdirection it will move


2 ]NO[

42ON

By adding more NO2 gas we will increase the concentration

.

So Q > Kc and the reaction will proceed in the reverse

This means that more reactants form and we end up with


.

c


…

2SO (g) 2SO (g) + O (g)

= -3 , c .

r te t e express on or c

If we mix together the following reactants:

[SO3] = 2 x 10-3 M

2 = x -

[O2] = 3 x 10-2 M


In which direction will the reaction proceed towards equilibrium?

Module 1: Lecture 12

Effect of Chan e in Pressure (or Volume) on Equi l ibrium

Very important if gases are involved . . . pressure can be

c ange n t ree separate ways:




1. Add or remove a gaseous reactant or product

2. Add an inert gas (eg helium)

3. Change the volume


p 352

1. Add or remove a gaseous reactant or product

Changes the concentration, since concentration = n/V , so

treat as e ore.

.

No change in concentrations, so equilibrium position

unaffected.

Concentration of aseous s ecies chan es but ma or

3. Change the volume

may not change equilibrium. This depends on the

stoichiometry of the reaction.


Experimental Observation:

What is the effect of decreasing the volume of this system

at equilibrium at constant temperature?

)(ON 42 g )(2NO2 g

Observe: The concentration of both species increases.

S stem res onds to chan e b shiftin balance to the side

with fewer moles of gases, ie left.


Vol & equilLe Chatelier!

,

2

2NOQ (any conditions)

Effect of decreasin volume b increasin ressure

Volume change:Initial

equilibriumno onger a

equilibriumFinal

equilibrium



Q (any conditions)

= >

42

, , , .

System adjusts by decreasing [NO2] and increasing [N2O4]

until Q = K once more.

The value of K remains constant.


q


Compare the number of molecules at equilibrium in (a) and (c)

H3N NH2


reducing the number of molecules.

vessel in which the following equilibrium existed:

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


Eff ect of Temperat ure on Equil ibri um

If temperature is changed, the value of K

c anges. rect on o c ange epen s on

the sign of H

OH6CoCl2 Cl4OHCo2

2 + kJ



g

2OH6CoCl 24 o

62

blue pink

Forward reaction is exothermic (ΔH < 0).


Cobalt temp final

OH6CoCl 24 Cl4OHCo62 + kJ

blue pink

Heating forces equilibrium to left, cooling to right, ie


Le Chatelier!

o = + .

What effect would raising the temperature have?

1. No effect2. An increase in the amount of NO2(g)

. 2 4


N2O4 effect of T

An increase in the number of NO2 molecules. The

reaction is endothermic and so the input of heatenergy as a reactant shifts the equilibrium to the

right.


How do we ex lain effect of tem erature uantitativel

The 1st winner of the Nobel

’

p 355

2 RT dT



’

For his work in thermodynamics

But A so etermine t at car on

forms tetrahedral rather than


.

H is positive H is negativeExothermic

Endothermic


ncreas ng ecreases soreactants are favoured

ncreas ng ncreases soproducts are favoured

2

ln H K d

Integrate

212

1ln

T T R K

When we know the standard enthalpy and K at one

temperature, we are able to calculate the K at a different

empera ure.


Temperature Dependence of K

We can deduce the temperature dependence of the

equi i rium constant using t e o owing:

000

S H 00

S H

R RT 00

1

RT R

,

slope = -H ° / R and y -intercept S° / R


.p 350

00

RT RK

ln

c xm y

If at temp. T 1 we have K 1 and at temp. T 2, K 2, then

S H K

00

11

ln



y

l n K


1/T

1

S H 00

1 RT R 2

u rac equa on rom :

0

1

2ln T T R


Subtract SECOND equation from FIRST:

R

S

T R

H K

00

1

1ln

S H

K 00

2

1ln

2

00 11 H H

21

21T R T R

0

1 11H K

212 T T R K o:


2 2 3

• By using the equation below, explain the observed values

R

S

RT

H K ln

R

S

T R

H 00

1


empera ure epen ence o

N2(g) + 3H2(g) 2NH3(g) - 92.38 kJ

• Exothermic process: H 0 < 0

F th i th l i

The a licabilit of the van’t Hoff e uation at two

0

specified temperatures:

1ln

TTR

T1 = 500 K

K1 = 90



• For an exothermic rocess the slo e is

positive and therefore as the temperature

increases the equilibrium constant

decreases

00 K

R RT K ln

l

R

S

T R

H 1


Slope!1/T

T T R

2

K2 = ?

Ho = - 92 kJ 0 11H

21

21T T R

ln(90)- ln K2 = – – 92 x 103 1 – 1

8.314 500 800

4.50 – lnK2 = 8.30ln K = -3.80


.

K2 = 0.02

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CHEM1020 Module 1 2011 Notes