THERMODYNAMICSTHERMODYNAMICS

Chapter 19 Chapter 19

A process that occurs without ongoing outside intervention.

Examples

• Nails rusting outdoors

SPONTANEOUS PROCESSSPONTANEOUS PROCESS

• Ice melting at room temperature

• Formation of water from O2(g) and H2(g):

• Expansion of gas into an evacuated space

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

Why are some processes spontaneous and others not?

We know that temperature has an effect on the spontaneity of a process.

e.g. T>0oC ice melts spontaneous at this temp.

H2O (s) H2O (l)

T<0oC water freezes spontaneous at this temp.

H2O (l) H2O (s)

T=0oC water and ice in equilibriumH2O (l) H2O (s)

Exothermic processes tend to be spontaneous.

Rusting of nail - SPONTANEOUS!4Fe(s) + 3O2(g) Fe2O3(s)

H = - 822.2 kJ.mol-1

Example

Formation of water - SPONTANEOUS!2H2(g) + O2(g) 2H2O(l)

H = - 285.8 kJ.mol-1

Pt cat

However, the dissolution of ammonium nitrate is also spontaneous, but it is also endothermic.

NH4NO3(s) NH4+(aq) + NO3

-(aq)H = +25.7 kJ.mol-1

So is:2N2O5(s) 4NO2(g) + 2O2(g)

H = +109.5 kJ.mol-1

a process does not have to be exothermic to be spontaneous.

something else besides sign of H must contribute to determining whether a process is spontaneous or not.

That something else is:

ENTROPY (S)ENTROPY (S)

extent of disorder!

More disordered larger entropy

S = Sfinal - Sinitial

Entropy is a state function

Units: J K-1mol-1

Entropy’s effects on the mindEntropy’s effects on the mind

NH4NO3(s) NH4+(aq) + NO3

-(aq)H = +25.7 kJ.mol-1

2N2O5(s) 4NO2(g) + 2O2(g)H = +109.5 kJ.mol-1

Examples of spontaneous processes where entropy increases:

Dissolution of ammonium nitrate:

Decomposition of dinitrogen pentoxide:

Spontaneous

Non-spontaneous

However, entropy does not always increase for a spontaneous process

At room temperature:

SECOND LAW OF THERMODYNAMICSSECOND LAW OF THERMODYNAMICS

The entropy of the universe increases in any spontaneous process.

Suniverse = Ssystem + Ssurroundings

Spontaneous process: Suniverse > 0

Process at equilibrium: Suniverse = 0

Thus Suniv is continually increasing!

Suniv must increase during a spontaneous process, even if Ssyst decreases.

Q: What is the connection between sausages and Q: What is the connection between sausages and the second law of thermo? the second law of thermo?

A: Because of the 2nd law, you can A: Because of the 2nd law, you can put a pig into a machine and get put a pig into a machine and get sausage, but you can't put sausage sausage, but you can't put sausage into the machine and get the pig into the machine and get the pig back. back.

For example:

Rusting nail = spontaneous process4Fe(s) + 3O2(g) 2Fe2O3(s)Ssyst<0

BUT reaction is exothermic,

entropy of surroundings increases as heat is evolved by the system thereby increasing motion of molecules in the surroundings.

Suniv>0

Ssurr>0

Thus for Suniv = Ssyst + Ssurr >0

Ssurr > Ssyst

-ve +ve+ve

Special circumstance = Isolated system:

Does not exchange energy nor matter with surroundings

Ssurr = 0

Spontaneous process: Ssyst > 0

Process at equilibrium: Ssyst = 0

Spontaneous Thermodynamically favourable(Not necessarily occur at observable rate.)

Thermodynamics direction and extent of reaction, not speed.

EXAMPLE

State whether the processes below are spontaneous, non-spontaneous or in equilibrium:

•CO2 decomposes to form diamond and O2(g)

•Water boiling at 100oC to produce steam in a closed container

•Sodium chloride dissolves in water

NON-SPONTANEOUS

SPONTANEOUS

EQUILIBRIUM

MOLECULAR INTEPRETATION OF SMOLECULAR INTEPRETATION OF S

Decrease in number of gaseous molecules decrease in S

e.g. 2NO(g) + O2(g) 2NO2(g)

3 moles gas 2 moles gas

Molecules have 3 types of motion:

Translational motion - Entire molecule moves in a direction (gas > liquid > solid)

Vibrational motion – within a molecule

Rotational motion – “spinning”

Greater the number of degrees of freedom greater entropy

Decrease in temperature decrease in thermal energy decrease in translational, vibrational

and rotational motion decrease in entropy

As the temperature keeps decreasing, these motions “shut down” reaches a point of perfect order.

EXAMPLE

Which substance has the great entropy in each pair? Explain.

• C2H5OH(l) or C2H5OH(g)

• 2 moles of NO(g) or 1.5 moles of NO(g)

• 1 mole O2(g) at STP or 1 mole NO2(g) at STP

THIRD LAW OF THERMODYNAMICSTHIRD LAW OF THERMODYNAMICS

The entropy of a pure crystalline substance at absolute zero is zero.

S(0 K) = 0 perfect order

Ginsberg's TheoremGinsberg's Theorem

(The modern statement of the three laws of (The modern statement of the three laws of thermodynamics)thermodynamics)

1. You can't win. 1. You can't win. 2. You can't even break even. 2. You can't even break even. 3. You can't get out of the game. 3. You can't get out of the game.

Entropy increases for s l g

EXAMPLE

Predict whether the entropy change of the system in each reaction is positive or negative.

• CaCO3(s) CaO(s) + CO2(g)

• 2SO2(g) + O2(g) 2SO3(g)

• N2(g) + O2(g) 2NO(g)

•H2O(l) at 25oC H2O(l) at 55oC

+ve

-ve

?

+ve

3 mol gas 2 mol gas

2 mol gas 2 mol gas Can’t predict, but it is close to zero

Increase thermal energy

Standard molar entropy (SStandard molar entropy (Soo) )

= molar entropy for substances in their standard state

NOTE• So 0 for elements in their standard state• So(gas) > So(liquid) > So(solid)• So generally increases with increasing molar

mass• So generally increases with increasing number of atoms in the formula of the substance

Calculation of Calculation of S for a reaction S for a reaction

(So from tabulated data)

)(reactantsmS(products)nSS ooo

Stoichiometric coefficients

EXAMPLE

Calculate So for the synthesis of ammonia from N2(g) and H2(g):

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

So/J.K-1.mol-1

N2(g) 191.5

H2(g) 130.6

NH3(g) 192.5

Calculation of Calculation of S for the surroundingsS for the surroundings

For a process that occurs at constant temperature and pressure, the entropy change of the surroundings is:

(T &P constant) T

H-S sys

surr

GIBB’S FREE ENERGY (G)GIBB’S FREE ENERGY (G)

Defined as: G = H – TS

- state function- extensive property

Suniv = Ssys + Ssurr

At constant T and P:

- TSuniv = - TSsys + Hsys

(at constant T & P)

Suniv = Ssys -Hsys

T

G = H – TS

Spontaneous process: Suniv > 0

Process at equilibrium: Suniv = 0

Gsyst = -TSuniv

We know:

Therefore:

Spontaneous process:

Process at equilibrium:

< 0

= 0

-TSuniv

-TSuniv

Spontaneity involves

S H

G = H – TS

Spontaneity is favoured by increasing S and H is large and negative.

T

G allows us to predict whether a process is spontaneous or not (under constant temperature and pressure conditions):

G < 0 spontaneous in forward direction

G > 0 non-spontaneous in forward direction/spontaneous in reverse direction

G = 0 at equilibrium

But nothing about rateBut nothing about rate

Standard free energy (Standard free energy (GGoo))

Go = Ho – TSo

Standard states:Gas - 1 atmSolid - pure substanceLiquid - pure liquidSolution - Concentration = 1M

Gfo = 0 kJ/mol for elements in their standard

states

Tabulated data of Gfo can be used to calculate

standard free energy change for a reaction as follows:

)(reactantsGm(products)GnG of

of

o

Stoichiometric coefficients

Substance H (kJ mol-1)

S (J K-1 mol-1)

G (kJ mol-1)

Substance H (kJ mol-1)

S (J K-1 mol-1)

G (kJ mol-1)

AgS 0 +42.6 0 I2(s) 0 +116.1 0

AgCl(s) 127.1 +96.2 -109.8 I2(g) +62.4 +260.6 +19.4

Al(s) 0 28.32 0 MgO(s) -601.5 +27.0 -569.2

AlCl3(s) -704.2 +110.7 -628.8 MnO2(s) -520.0 +53.1 -465.2

Al2O3(s) -1669.8 +51.0 -1576.5 N2(g) 0 +191.5 0

Br2() 0 +152.2 0 N2O4(g) +9.3 +304.2 +97.8

BrF3(g) -255.6 +292.4 -229.5 Na(s) 0 +51.3 0

C(g) +716.7 +158.0 +671.3 NaF(s) -569.0 +51.3 -546.3

C(graphite) 0 +5.8 0 NaCl(s) -411.1 +72.4 -384.3

C(diamond) +1.9 +2.4 +2.9 NaBr(s) -361.1 +87.2 -349.1

CO(g) -110.5 +197.6 -137.2 NaI(s) -287.8 +98.5 -282.4

CO2(g) -393.5 +213.7 -394.4 NaOH(s) -425.6 +64.5 -379.5

CH4(g) -74.5 +186.1 -50.8 NH3(g) -46.2 +192.7 -16.4

C3H8(g) -103.8 +269.9 -23.4 N2H4() +50.6 +121.2 +149.2

Ca(s) 0 41.4 0 NO(g) +90.3 +210.6 +86.6

CaO(s) -635.1 +38.1 -603.5 NO2(g) +33.2 +240.0 +51.3

CaCO3(s)(calcite) -1206.9 +92.9 -1128.8 HNO3() -174.1 +155.6 -80.8

Cl2(g) 0 +223.0 0 O2(g) 0 +205.0 0

Cu(s) 0 +33.2 0 O3(g) +142.7 +238.8 +163.2

F2(g) 0 +202.7 0 P(s)(white) 0 +41.1 0

Fe(s) 0 +27.3 0 P4O10(s) -3010.0 +231.0 -2724.0

Fe2O3(s)(hematite) -824.2 +87.4 -742.2 PCl3(g) -287.0 +311.7 -267.8

H(g) +218.0 +114.6 +203.3 PCl5(g) -374.9 +364.5 -305.0

H2(g) 0 +130.6 0 PbO2(s) -277.4 +68.6 -217.4

HCl(g) -92.3 +186.8 -95.3 S(s)(orthorhombic) 0 +32.0 0

HF(g) -271.1 +173.8 -273.2 H2S(g) -20.6 +205.6 -33.4

HI(g) +26.4 +206.5 +1.6 SiO2(s)(quartz) -910.7 +41.5 -856.3

HBr(g) -36.4 +198.6 -53.5 SiCl4() -687.0 +239.7 -619.9

HCN(g) +135.1 +201.7 +124.7 SO2(g) -296.8 +248.1 -300.2

H2O(g) -241.8 +188.7 -228.6 SO3(g) -395.7 +256.6 -371.1

H2O() -285.8 +70.0 -237.2 Zn(s) 0 +41.6 0

H2O2() -187.8 +109.6 -120.4 ZnO(s) -350.5 +43.6 -320.5

Hg() 0 +75.9 0

EXAMPLE

The combustion of propane gas occurs as follows:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Using thermodynamic data for Go, calculate the standard free energy change for the reaction at 298 K.

Gfo/kJ.mol-1

C3H8(g) -23.47

CO2(g) -394.4

H2O(g) -228.57

H2O(l) -237.13

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Gfo/kJ.mol-1

C3H8(g) -23.47

CO2(g) -394.4

H2O(g) -228.57

H2O(l) -237.13

)(reactantsGm(products)GnG of

of

o

Go = [3(-394.4) + 4(-237.13)] – [(-23.47) – 5(0)]

Go = -2108 kJ

Free Energy and TemperatureFree Energy and Temperature

How is change in free energy affected by change in temperature?

G = H – TS

H S -TS G = H - TS

+

+ +

+

+

+---

-

-

-

-

+ at all temp

- at all temp

- at high temp + at low temp

+ at high temp - at low temp

Note:

For a spontaneous process the maximum useful work that can be done by the system:

wmax = G

“free energy” = energy available to do work

EXAMPLE

The combustion of propane gas occurs as follows at 298K:

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Ho = -2220 kJ.mol-1

a) Without using thermodynamic data tables, predict whether Go, for this reaction is more or less negative than Ho.

b) Given that So = -374.46 J.K-1.mol-1 at 298 K for the above reaction, calculate Go. Wasyour prediction correct?

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Ho = -2220 kJ.mol-1

a) Without using thermodynamic data tables, predict whether Go, for this reaction is more or less negative than Ho.

Go = Ho – TSo

-ve -ve6 moles gas 3 moles gas

– TSo > 0

Ho – TSo will be less negative than Ho

i.e. Go will be less negative than Ho

C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)

Ho = -2220 kJ.mol-1

b) Given that So = -374.46 J.K-1.mol-1 at 298 K for the above reaction, calculate Go. Was your prediction correct?

Go = Ho – TSo

Go = (-2220 kJ) – (298 K)(-374.46x10-3 kJ.K-1.mol-1)

Go = -2108 kJ.mol-1

Prediction was correct.

EXAMPLE (TUT no. 5a)

At what temperature is the reaction below spontaneous?

AI2O3(s) + 2Fe(s) 2AI(s) + Fe2O3(s)

Ho = 851.5 kJ; So = 38.5 J K-1

At what temperature is the reaction below spontaneous? AI2O3(s) + 2Fe(s) 2AI(s) + Fe2O3(s)

Ho = 851.5 kJ; So = 38.5 J K-1

Go = Ho – TSo

Assume H and S do not vary that much with temperature.

0 = (851.5 kJ) – T(38.5x10-3 kJ.K-1)

T = 22117 K

equilibriumSet G = 0

at equilibrium

For spontaneous reaction: G < 0

T > 22117 K

Free Energy and the equilibrium constantFree Energy and the equilibrium constant

Recall: G = Change in Gibb’s free energy under standard conditions.

G can be calculated from tabulated values.

BUT most reactions do not occur under standard conditions.

Calculate G under non-standard conditions:

QlnRTGG

Q = reaction quotient

R = gas constant = 8.314 J.K-1.mol-1

Under standard conditions: (1 M, 1 atm)Q = 1 ln Q = 0 G = Go

At equilibrium: G = 0 and Q = Keq

KlnRTG0 KlnRTG

If Go < 0 ln Keq > 0 Keq > 1 i.e. the more negative Go, the larger K etc.

Go < 0 Keq > 1 Go > 0 Keq < 1 Go = 0 Keq = 1

RT/GeK Also

QlnRTGG

EXAMPLECalculate K for the following reaction at 25oC:

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

Gfo/kJ.mol-1

H2O(g) -228.57

H2O(l) -237.13

)(reactantsGm(products)GnG of

of

o

Go = [2(0) + (0)] – [2(-237.13)]

Go = 474.26 kJ.mol-1

KlnRTG 474.26x103 J.mol-1 = -(8.314 J.K-1.mol-1)(298 K) lnK

]O[]H[K 22

2 lnK = -191.4 K = 7.36x10-84