1

Second Year Chemistry• 1st semester: Organic• 1st semester: Physical (2005-

2006)• December exams

• 2nd: Analytical & Environmental• 2nd: Inorganic

• Summer exams• Physical: 3 lecturers 8 topics• Dónal Leech: four topics

• Thermodynamics• Gases, Laws & Phases, Equilibrium

2

Introduction Energetics and Equilibria

What makes reactions “go”!

This area of science is called THERMODYNAMICS

Thermodynamics is expressed in a mathematical language

BUT

Don’t, initially anyway, get bogged down in the detail of the equations: try to picture the physical principle expressed in the equations

We will develop ideas leading to one important Law, and explore practical applications along the way

The Second Law of Thermodynamics000

0 ln

STHG

KRTG

rrr

r

3

Lecture Resources12 lectures leading to four exam questions (section A, you must answer two from this section)

• Main Text: “Elements of Physical Chemistry”

Atkins & de Paula, 4th Edition (Desk reserve)http://www.oup.com/uk/booksites/content/0199271836/OTHERS. “Physical Chemistry” Atkins & de Paula, 7th Edition or any other

PChem textbook

These notes available on NUI Galway web pages athttp://www.nuigalway.ie/chem/degrees.htm

See also excellent lecture notes from James Keeler, Cambridge, although topics are treated in a different running order than here.

4

Thermodynamics: the 1st law The internal energy of an isolated system is

constant

Energy can neither be created nor destroyed only inter-converted Energy: capacity to do work

Work: motion against an opposing force

System: part of the universe in which we are interested

Surroundings: where we make our observations (the

universe)Boundary: separates above

two

5

System and Surroundings

Systems

• Open: energy and matter

exchanged

• Closed: energy exchanged

• Isolated: no exchange

• Diathermic wall: heat

transfer permitted

• Adiabatic wall: no heat

transfer

6

Work and Heat• Work (w): transfer of

energy that changes motions of atoms in the surroundings in a uniform manner

• Heat (q): transfer of energy that changes motions of atoms in the surroundings in a chaotic manner

• Endothermic: absorbs heat• Exothermic: releases heat

7

Work• Mechanical work can generally be described by dw = -F.dz

• Gravitational work (mg.dh) • Electrical work (.dq) • Extension work (f.dl)• Surface expansion work (.d)

As chemists we will concentrate on EXPANSION WORK

(many chemical reactions produce gases)

w = -F.z but pex = F/A

therefore w = -pex.V

Expansion against constant external pressure

8

Expansion Work

• In thermodynamics “reversible” means a process that can be reversed by an infinitesimal change of a variable. • A system does maximum expansion work when the external pressure is

equal to that of the system at every stage of the expansion

Expansion against zero external pressure (free expansion)

w = -pex.V = 0 (external pressure = 0)Reversible isothermal expansion

9

Isothermal reversible expansion

i

f

if

if

V

V

V

V

V

V

ex

V

VnRTw

VVnRTw

VVnRTw

VnRTw

dVV

nRTpdVw

dVpw

f

i

f

i

f

i

ln

)ln-(ln

)constantln(-constant)ln(

)constant(ln

1

Exercise: Calculate the work done when 1.0 mol Ar(g) confined in a cylinder of volume 1.0 dm3 at 25°C expands isothermally and reversibly to 2.0 dm3.

10

11

1st Law of ThermodynamicsThe internal energy of an isolated system is

constant

Energy can neither be created nor destroyed only inter-converted U =

q+wExercise: A car battery is charged by supplying 250 kJ of energy to it as electrical work, but in the process it loses 25kJ of energy as heat to the

surroundings. What is the change in internal energy of the battery?

How do we measure

heat?

Use calorimetry. If we enclose our system in a constant volume container

(no expansion), provided no other kind of work can be done, then w = 0.

U = qV

INTERNAL ENERGY is a State Function

12

Bomb calorimetry• By measuring the change in Temperature

of the water surrounding the bomb, and knowing the calorimeter heat capacity, C, we can determine the heat, and hence U.

Heat CapacityAmount of energy required to raise the temperature of a substance by 1°C (extensive property)

For 1 mol of substance: molar heat capacity (intensive property)

For 1g of substance: specific heat capacity (intensive property)

VV

VV

qTCU

T

UC

13

Calorimeter calibrationCan calibrate the calorimeter, if its heat capacity is unknown, by passing a known electrical current for a given time to give rise to a measured temperature change.

IVtq Amperes.Volts.Sec = Coulombs.Volts = Joules

Exercise: In an experiment to measure the heat released by the combustion of a fuel, the compound was burned in an oxygen atmosphere inside a calorimeter and the temperature rose by 2.78°C. When a current of 1.12 A from an 11.5 V source was passed through a heater in the same calorimeter for 162 s, the temperature rose by 5.11°C. What is the heat released by the combustion reaction?

14

EnthalpyMost reactions we investigate occur under

conditions of constant PRESSURE (not Volume)ENTHALPY: Heat of reaction at constant pressure!

PqH

Vpbut

VpUH

pVUH

- w

Use a “coffee-cup” calorimeter to measure it

PP

PP

qTCH

T

HC

Heat capacity

Exercise: When 50mL of 1M HCl is mixed with 50mL of 1M NaOH in a coffee-cup calorimeter, the temperature increases from 21°C to 27.5°C. What is the enthalpy change, if the density is 1g/mL and specific heat 4.18 J/g.K?

15

Perfect gas enthalpy

RTUH

pVUH

mm

mmm

• Use intensive property of molar enthalpy and internal energy

• At 25°C, RT = 2.5 kJ/mol

Thermicity-RevisionEndothermic reaction (q>0) results in an increase in

enthalpy (H>0)Exothermic reaction (q<0) results in an increase in

enthalpy (H<0)

NB: Internal energy and Enthalpy are STATE FUNCTIONS

16

Temperature variation of enthalpy

2, T

cbTaC mp

Convenient empirical expression to use for heat

capacity is:

Exercise: What is the change in molar enthalpy of N2 when it is heated from 25°C to 100 °C, given that:

2,

5000000377.058.28

TTC mp

17

Relation between heat capacities

RCC

RT

U

T

H

TRUH

RTUH

RTUH

mVmp

mm

mm

mm

mm

,,

18

ThermochemistryChemists report data for a set of standard conditions:

The standard state of a substance (°) is the pure substance at exactly 1 bar

It is conventional (though not obligatory) to report data for a T of 298.15K

Standard enthalpies of phase transition

Energy that must be supplied (or is evolved) as heat, at constant pressure, per mole of molecules that undergo the phase transition under standard conditions (pure phases),

denoted H°

Note: the enthalpy change of a reverse transition is the

negative of the enthalpy change of the forward

transition

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H°

Substance Freezing point, Tf/K fusHo/(kJ mol 1) Boiling point, Tb/K vapHo/(kJ mol 1)

Ammonia, NH3 195.3 5.65 239.7 23.4

Argon, Ar 83.8 1.2 87.3 6.5

Benzene, C6H6 278.7 9.87 353.3 30.8

Ethanol, C2H5OH 158.7 4.60 351.5 43.5

Helium, He 3.5 0.02 4.22 0.08

Mercury, Hg 234.3 2.292 629.7 59.30

Methane, CH4 90.7 0.94 111.7 8.2

Methanol, CH3OH 175.5 3.16 337.2 35.3

Propanone, CH3COCH3 177.8 5.72 329.4 29.1

Water, H2O 273.15 6.01 373.2 40.7

* For values at 298.15 K, use the information in the Data section.

20

Sublimation Direct conversion of a solid to a vapour

The enthalpy change of an overall process is the sum of the enthalpy changes for the steps

into which it may be divided

21

Enthalpies of ionisation (kJ/mol)1 2 13 14 15 16 17 18

H He

1312 2370

5250

Li Be B C N O F Ne

519 900 799 1 090 1400 1310 1680 2080

7300 1760 2 420 2 350 2860 3390 3370 3950

14 800 3 660

25 000

Na Mg Al Si P S Cl Ar

494 738 577 786 1060 1000 1260 1520

4560 1451 1 820

7740 2 740

11 600

ionH°(T)= Ionisation energy(0) + (5/2)RT (see Atkins & de Paula, Table 3.2)

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Problems Ethanol is brought to the boil at 1 atm. When the electric

current of 0.682 A from a 12.0 V supply is passed for 500 s through a heating coil immersed in the boiling liquid, it is found that the temperature remains constant but 4.33 g of ethanol is vapourised. What is the enthalpy of vapourisation of ethanol at its boiling point at 1 atm?

Calculate the standard enthalpy of sublimation of ice at 0°C given that fusH° is 6.01 kJ/mol and vapH° is 45.07 kJ/mol, both at 0°C.

subH° for Mg at 25°C is 148 kJ/mol. How much energy as heat must be supplied to 1.00 g of solid magnesium metal to produce a gas composed of Mg2+ ions and electrons?

23

Bond enthalpies (kJ/mol) H C N O F Cl Br I S P Si

H 436

C 412 348 (1)

612 (2)

838 (3)

518 (a)

N 388 305 (1) 163 (1)

613 (2) 409 (2)

890 (3) 945 (3)

O 463 360 (1) 157 146 (1)

743 (2) 497 (2)

F 565 484 270 185 155

Cl 431 338 200 203 254 242

Br 366 276 219 193

I 299 238 210 178 151

S 338 259 496 250 212 264

P 322 200

Si 318 374 466 226

Values are for single bonds except where otherwise stated (in parentheses). (a) Denotes aromatic.

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Problem

Estimate the standard reaction enthalpy for the formation of liquid methanol from its elements as 25°C

25

Enthalpies of combustionEnthalpies (heats) of combustion: complete reaction of compounds with oxygen.

Measure using a bomb calorimeter.

Most chemical reactions used for the production of heat are combustion reactions. The energy released when 1g of material is combusted is its Fuel Value. Since all heats of combustion are exothermic, fuel values are reported as positive.

Most of the energy our body needs comes from fats and carbohydrates. Carbohydrates are broken down in the intestines to glucose. Glucose is

transported in the blood to cells where it is oxidized to produce CO2, H2O and energy:

C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l) cH°=-2816 kJ The breakdown of fats also produces CO2 and H2O Any excess energy in the body is stored as fats

RTnUH

P

RTn

P

nRTV

VPUH

g

g

26

Heats of formationIf one mole of the compound is formed under standard conditions from its elements in their reference state then the resulting enthalpy change is said to be the standard molar enthalpy (Heat) of formation, fH° where the subscript indicates this.

The reference state is the most stable form under the prevailing conditions.

27

Hess’s LawTo evaluate unknown heats of reaction

The standard enthalpy of a reaction is the sum of the standard enthalpies for the reactions into which the overall reaction may be divided

rxnHo = fHom(products) - fHo

m(reactants)

28

Variation of rH° with T

rH°(T2) = rH°(T1) + rCp°(T2-T1)

If heat capacity is temperature dependent, we need to integrate

over the temperature range

2

1

d)()( 12

T

T

opr

or

or TCTHTH

rCp° = Cp,m°(products) - Cp,m°(reactants)

Kirchoff’s Law

29

Thermodynamics: the 2nd lawDeals with the direction of spontaneous change

(no work required to bring it about)

Kelvin Statement

No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work

Impossible!

30

EntropyThe apparent driving force for spontaneous change is the

dispersal of energy

A thermodynamic state function, Entropy, S, is a

measure of the dispersal of energy (molecular disorder)

of a system2nd Law: The Entropy

of an isolated system increases in

the course of spontaneous change

Stot>0

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Thermodynamic definition of S Concentrates on the change in entropy:

S = qrev/T

Can use this equation to quantify entropy changes.

We will see later (3rd & 4th year) a statistical description of entropy

S = k lnW (Boltzmann formula)

32

Heat EnginesAll heat engines have a hot region “source” and a cold region “sink”: some energy must be discarded into the

cold sink as heat and not used to do work

h

c

h

c

h

ch

hh

cc

c

c

h

htot

T

T

q

q

q

qq

heat

work

qT

Tq

T

q

T

qS

11

33

Expansion entropy Intuitively can guess that entropy increases with gas

expansion. Thermodynamic definition allows us to quantify this

increase

Recall that: w = -nRT ln (Vf/Vi)

BUT qrev = -w (U = 0 for isothermal processes)

S = nR ln (Vf/Vi)

Note: independent of TAlso: Because S is a state function, get the same

value for an irreversible expansion

34

Heating Entropy

capacityheat constant for

lndCd

Cdd

ddd

d T in change malinfinitesifor or

T in change malinfinitesifor d

d

i

fT

T

T

T

rev

T

TC

T

TC

T

TS

T

TS

TCqT

qC

T

qC

T

qS

f

i

f

i

35

Entropy of phase transition

Entropy of fusion

Entropy of vapourisation

f

ffusfus T

THS

)(

b

bvapvap T

THS

)(

Trouton’s rule

The entropy of vapourisation is approximately the same (85

J/K.mol) for all non-polar liquids

36

Phase transitions

To evaluate entropies of transition at T other than the transition temperature

Entropy of vapourisation of water at 25°C? Sum of S for heating from 25°C to 100°C,

S for vapourisation at 100°C, and S for cooling vapour from 100°C to 25°C. Try it! (+118 J/K.mol).

37

Entropy changes in the surroundings

T

qS

T

qS

T

qS

sur

sursur

revsursur

,

Stot = Ssys + Ssur

Stot = Ssys – q/T

Example: Water freezing to ice.

Entropy change of system is -22 J/K.mol, and heat evolved is -6.01

kJ/mol.

Entropy change in surroundings must be positive for this process to occur

spontaneously.

Check this for different temperatures.

Note that Stot = 0 at equilibrium

38

Spontaneity of water freezing

At 5°C: Stot = -22 JK-1mol-1 – (6,010Jmol-1/278K) = -0.38 JK-1mol-1

At -5°C: Stot = -22 JK-1mol-1 – (6,010Jmol-1/268K) = +0.43 JK-1mol-1

At 0°C: Stot = -22 JK-1mol-1 – (6,010Jmol-1/273K) = 0.01 JK-1mol-1

To find transition temperature, set Stot = 0 and solve for T.

273.18 K (slight error because of rounding of entropy and heats).

Stot = Ssys - qsys/T

39

Problem

Typical person heats the surroundings at a rate of 100W (=J/s). Estimate entropy change in one day at 20°C.

qsur = 86,400 s × 100 J/s

Ssur = qsur/T = (86,400 × 100 J)/293 K

= 2.95 × 104 J/K

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3rd Law Entropy of sulfur phase transition

is 1.09 J/K.mol. Consider plot at left. Subtract

entropy for phase transition (to give plot at right)

T=0 intercept is the same.

Entropies of all perfectly crystalline substances are the same at T=0.

41

Absolute and standard molar entropies (S and S0

m)Absolute entropies can be determined by integration

of areas under heat capacity/T as a function of

T, and including entropies of phase transitions.

Standard molar entropies are the molar entropies of

substances at 1bar pressure (and usually 298 K)

42

Standard molar entropies

43

Standard reaction entropies

Difference in molar entropy between products and reactants in their standard states is called the standard reaction entropy and can be expressed (like enthalpy) as:

Note: absolute entropies, S, and standard molar entropies, S0

m, are discussed in section 4.7 of the textbook

rxnSo = Som(products) - So

m(reactants)

44

Spontaneity of reactionsConsider the reaction:

2H2(g) + O2(g) 2H2O (l)rS0 = 2(70 J/K.mol) –[2(131 J/K.mol) + (205 J/K.mol)]

= -327 J/K.molBut this reaction is spontaneous (explosive even!)

When considering the implications of entropy, we must always consider the total change of the system and its surroundings

rH0 = -572 kJ/mol. Therefore rSsur = +1920 J/K.molrStot is positive +1590 J/K.mol (spontaneous reaction!).

45

Gibbs Energy Introduced by J.W. Gibbs to combine the

calculations of 2 entropies, into one. Because Stot = S – H/T (constant T and P) Introduce G = H – TS (Gibbs “free” energy) Then G = H – TS (constant T) So that G = – TStot (constant T and P)

G = H – TSIn a spontaneous change at constant temperature and pressure, the Gibbs

energy decreases

46

Maximum non-expansion work

Can derive (see box 4.5 in textbook) that G = w’max

Example: formation of water: enthalpy -286kJ, free energy -237kJ

Example: suppose a small bird has a mass of 30 g. What is the minimum mass of glucose that it must consume to fly to a branch 10 m above the ground?

(G for oxidation of glucose to carbon dioxide and water is -2828 kJ at 25°C)

Exercise: A human brain operates at about 25 W (J/s). What mass of glucose must be consumed to sustain that power for 1 hour?

47

Problems solved

w’ = (30 × 10-3 kg) × (9.81 m s-2) × 10 mNote 1J = 1kg m2 s-2

n = 2.943 J/(2828 × 103 J/mol)

m = nM = (1.04 × 10-6 mol) × 180 g/mol

= 1.9 × 10-4 g

Answer 2: 5.7g

48

Phase Equilibria Phase transitions

Changes in phase without a change in chemical composition

Gibbs Energy is at the centre of the discussion of transitions Molar Gibbs energy

Gm = G/n

Depends on the phase of the substance

A substance has a spontaneous tendency to change into a phase with the lowest molar Gibbs

energy

49

Variation of G with pressure We can derive (see

derivation 5.1 in textbook) that Gm = Vmp

Therefore Gm>0 when p>0

Can usually ignore pressure dependence of G for condensed states

Can derive that, for a gas:

Gm = RT ln(pf/pi)

50

Proof-go back to fundamental definitions

G = H – TS; H = U + pV; dU = dq + dwFor an infinitesimal change in G:

G + dG = H + dH – (T + dT)(S + dS)= H + dH – TS – TdS – SdT – dTdS

dG = dH – TdS – SdTAlso can write: dH = dU + pdV + Vdp

dU = TdS – pdV (dS = dqrev/T and dw = -pdV)

dG = Vdp – SdTMaster Equations

51

Variation of G with temperature

Gm = -SmTCan help us to understand why transitions occur

The transition temperature is the temperature when the molar Gibbs energy of the two phases are equal.

The two phases are in EQUILIBIRIUM at this temperature

52

Phase diagrams

Map showing conditions of T and p at which various phases are thermodynamically stable

At any point on the phase boundaries, the phases are in dynamic equilibrium

53

Location of phase boundaries Clapeyron equation (see derivation 5.4)

Clausius-Clapeyron equation (derivation 5.5)

TVT

Hp

trs

trs

constant11

lnln

ln

1212

2

TTR

Hpp

TRT

Hp

vap

vap

Constant is

vapS/R

54

Derivations

dGm = Vmdp – SmdTdGm(1) = dGm(2)

Vm(1)dp – Sm(1)dT = Vm(2)dp – Sm(2)dT

{Vm(2) – Vm(1)}dp = {Sm(2) – Sm(1)}dT

trsV dp = trsS dT

T trsV dp = trsH dT

dp/dT = trsH/(T trsV)

55

Derivations: liquid-vapour transitions

dp/dT = vapH/(T vapV)≈ vapH/{T Vm(g)} = vapH/{T (RT/p)}

(dp/p)/dT = vapH/(RT2)

(d lnp)/dT = vapH/(RT2)

122

1

2

2

ln

ln

2

111ln

ln

ln

2

1

2

1

2

1

TTR

HdT

TR

H

p

p

dTRT

Hpd

dTRT

Hpd

vapT

T

vap

T

T

vapp

p

vap

+ constant

56

Using the equation The vapour pressure of mercury is 160 mPa at 20°C.

What is its vapour pressure at 50°C given that its enthalpy of vapourisation is 59.3 kJ/mol?

The vapour pressure of pyridine is 50.0 kPa at 365.7 K and the normal boiling point is 388.4 K. What is the enthalpy of vapourisation of pyridine?

Estimate the normal and standard boiling point of benzene given that its vapour pressure is 20.0kPa at 35°C and 50.0kPa at 58.8°C.

Remember: BP: temperature at which the vapour pressure of the

liquid is equal to the prevailing atmospheric pressure. At 1atm pressure: Normal Boiling Point (100°C for water) At 1bar pressure: Standard Boiling Point (99.6°C for

water; 1bar=0.987atm, 1atm = 1.01325bar)

57

Summary Thermodynamics tells which way a process

will go

• Internal energy of an isolated system is constant

(work and heat). We looked at expansion work

(reversible and irreversible).

• Thermochemistry usually deals with heat at constant

pressure, which is the enthalpy.

• Spontaneous processes are accompanied by an

increase in the entropy (disorder?) of the universe

• Gibbs free energy decreases in a spontaneous process