OFB Chap. 10 16/4/2003

OFB Chapter 10Thermochemistry

• 10-1 Heat• 10-2 Calorimetry• 10-3 Enthalpy• 10-4 Standard-State Enthalpies• 10-5 Bond Enthalpies• 10-6 The First Law of

Thermodynamics (skip)

• 6-3 Born-Haber Cycle (Rayner text)

OFB Chap. 10 26/4/2003

OFB Chap. 10 36/4/2003

Thermite Reaction2 Al (s) + Fe2O3 (s) → Al2O3 (s) + 2Fe (s)

OFB Chap. 10 46/4/2003

Chapter 10Thermochemistry

•Heat Capacity, Cp , is the amount of heat required to raise the temperature of a substance by one degree at constant pressure.

•Cp is always a positive number•Q and ∆T must be both negative or positive•If q is negative, then heat is evolved or given off and the temperature decreases•If q is positive, then heat is absorbed and the temperature increases

OFB Chap. 10 56/4/2003

Heat Capacity (continued)• Cp units are energy per temp. change• Cp units are Joule/oK or JK-1

• Units are J/K/mole or JK-1mol-1

• Molar heat capacity

OFB Chap. 10 66/4/2003

OFB Chap. 10 76/4/2003

Specific Heat Capacity• The word Specific before the name

of a physical quantity very often means divided by the mass

4.18H2O0.719O2

0.739SiO2

0.140Hg (l)cs

• Specific Heat Capacity is the amount of heat required to raise the temperature of one gram of material by one degree Kelvin (at constant pressure)

OFB Chap. 10 86/4/2003

OFB Chap. 10 96/4/2003

∆Tmcq∆Tm

qm

q/∆/mCc

ipsrealtionsh someEquating

s

ps

=

===

sp

pps

ppsp

MccmC

Mcc

cMmncmcC

=

==

===

ipsrelationsh some Equating

OFB Chap. 10 146/4/2003

Equilibration Temperature

• When 2 bodies are placed in contact, heat is exchanged until they reach a common final temperature

• Heat gained by the cooler body equals the heat lost by the warmer body

• If q is positive, heat is gained• If q is negative, heat is lost

• Can also write similar equations for cs

2ifp221ifp11

2p221p11

21

TTcnTTcn∆Tcn∆Tcn

qq

−−=−

−=−=

OFB Chap. 10 176/4/2003

• For processes carried out at constant pressure, the heat absorbed equals a change in Enthalpy, H.

• Enthalpy is a state property, which means it depends on its initial and final state, not the path to get there.

• Enthalpy is the heat content of a substance

OFB Chap. 10 186/4/2003

OFB Chap. 10 196/4/2003

Enthalpy of Reaction• If a chemical reaction occurs at

constant pressure then• (Heat absorbed) = qp = ∆H• ∆H may be either positive or

negative– If ∆H positiveMeans heat is absorbedAnd is called If ∆H negativeMeans heat is given offAnd is called

OFB Chap. 10 206/4/2003

CO (g) + ½ O2 (g) → CO2 (g) ∆H= -283kJ

An exothermic reaction

If doubled (x2), double ∆H2CO (g) + 1O2 (g) → 2CO2 (g) ∆H= -566kJ

If the original reaction is reversedCO2 (g) → CO (g) + ½ O2 (g) ∆H= +283kJ

An endothermic reaction

OFB Chap. 10 216/4/2003

Phase Changes• Phase changes are not chemical reactions

but involve enthalpy change

H2O(s)→ H2O(l)

– if reversed

H2O(l)→ H2O(s)

• Note that there are no T units, these occur at constant temperatures for freezing, fusion, vaporization or condensation

OFB Chap. 10 226/4/2003

OFB Chap. 10 236/4/2003

How much heat energy is needed to decompose 9.74g of HBr (g) (M =80.9 g/mol) into its elements?

H2 + Br2 → 2HBr (g)

∆H = -72.8 kJmol-1

Strategy:• Decomposition is actually

the reverse reaction• ∆H to decompose 2 moles of

HBr is +72.8kJ/mol

OFB Chap. 10 246/4/2003

How much heat energy is needed to decompose 9.74g of HBr (g) (M =80.9 g/mol) into its elements?

H2 + Br2 → 2HBr (g)

∆Hr = -72.8 kJmol-1

Solution:2HBr (g) → H2 + Br2

∆H = +72.8 kJmol-1

OFB Chap. 10 256/4/2003

Hess’s Law• Sometimes its difficult or even

impossible to conduct some experiments in the lab.

• Because Enthalpy (H) is a State Property, it doesn’t matter what path is taken as long as the initial reactants lead to the final product.

• Hess’s Law is about adding or subtracting equations and enthalpies.

• Hess’s Law: If two or more chemical equations are added to give a new equation, then adding the enthalpies of the reactions that they represent gives the enthalpy of the new reaction.

OFB Chap. 10 266/4/2003

Hess’s Law?H (g) CO (g) O 1/2 (graphite) CUnknown 2 =→+ ∆

kJ 283H (g) 1/2O (g) CO (g) CO (2.)kJ (-393.5) H (g) CO (g) O (graphite) C (1.)

Known

22

22

+=+→

=→+

∆

∆

→

(g) CO (g) 1/2O (graphite) C

Simplify

(g) 1/2O (g) CO (g) CO(g) CO (g) O (graphite) C

(2) (1) Add

2

22

22

→+

++++

+

This is Hess’s Law. It always works no matter how complicated the path.

OFB Chap. 10 276/4/2003

Standard State Enthalpies• Designated using a superscript °

(pronounced naught)• This is needed because Enthalpy varies

with conditions• Is written as ∆H°

• If no temperature is indicated in the sub-script, assume 25°C.

• OFB text appendix D lists standard enthalpies of formation for a variety of chemical species at 1 atm and 25°C.

• Standard State conditions– All concentrations are 1M or 1 mol/L– All partial pressures are 1 atm

οοK298.15

οC25 ∆Hor ∆Hor ∆H oo

OFB Chap. 10 286/4/2003

Enthalpy of Formation

• Enthalpy of Formation is defined as the change in heat content when 1 mol of a compound is formed from its elements’ most stable forms in their standard phases at 25°C and 1 atm (or 101.325 kPa)

• By definition , the enthalpy of formation of an element in its most stable form is Zero.

οοK298.15

οC25 ∆Hor ∆Hor ∆H oo f

OFB Chap. 10 296/4/2003

Standard State Enthalpies

• ∆Hr° is the sum of products minus the sum of the reactants

• For a general reactiona A + b B → c C + d D

• ∆Hr° = ΣH°products- ΣH°reactants

• from Hess’s Law

OFB Chap. 10 306/4/2003

Chapter 10Thermochemistry

• Exercise 10-7 Suppose hydrazine and oxygen react to give dinitrogen pentaoxide and water vapor:

2N2H4 (l) + 7O2 (g) →2N2O5 (s) + 4H2O (g)

Calculate the ∆ H of this reaction, given that the reaction:N2 (g) + 5/2O2 (g) ⇒N2O5 (s)has a ∆ H of -43 kJ.

OFB Chap. 10 316/4/2003

10-7 Exercise2N2H4 (l) + 7O2 (g) ⇒

2N2O5 (s) + 4H2O (g)• ∆Hr° =

• What values to use?– ∆Hf°N2O5 (s) given ∆H°=-43.1 kJmol-1

– ∆Hf°H2O (g) look up in App. D– ∆Hf° N2H4 (l) look up in App. D– ∆Hf°O2 (g) look up in App. D

•

• ∆Hr°

OFB Chap. 10 326/4/2003

Given the following thermo chemical equations, calculate the standard enthalpy of formation for propane, C3H8 (g).

1-r

22283

1-r

222

1-r

22

kJmol (-2452)∆H

O(l)4H(g)3CO(g)5O(g)HC 3.kJmol (-286)∆H

O(l)H(g)1/2O(g)H 2.kJmol (-393)∆H

(g)CO(g)OC(s) 1.

=

+→+

=

→+

=

→+

o

o

o

OFB Chap. 10 336/4/2003

BHbAHaDHdCHcHdDcCbBaA

General

ooooo ∆∆∆∆∆ −−+=

+→+

1-r

22283

1-2f

222

1-2f

22

kJmol (-2452)∆H

O(l)4H(g)3CO(g)5O(g)HC 3.kJmol (-286)O)(H∆H

O(l)H(g)1/2O(g)H 2.kJmol (-393))(CO∆H

(g)CO(g)OC(s) 1.

=

+→+=

→+

=

→+

o

o

o

OFB Chap. 10 346/4/2003

Bond Enthalpies• The breaking of chemical bonds in

stable substances often generates highly reactive products (or intermediates)CH4 → ·CH3 + ·H ∆H°= + 439 kJmol-1

• Called Bond Enthalpy• OFB Table 10-3 p 462 gives Average

Bond Enthalpies

OFB Chap. 10 356/4/2003

Bond Enthalpy• Bond Enthalpy (∆H Dissociation)

is the enthalpy change in a reaction in which a chemical bond is broken in the gas phase.

• Bond Energy is the energy needed to break 1 mol of the particular bond.–

–

OFB Chap. 10 366/4/2003

• Average Bond Enthalpies

C2H6 → ·C2H5 + ·H ∆H°= + 410 kJmol-1

CHF3 → ·CF3 + ·H ∆H°= + 429 kJmol-1

CHCl3 → ·CCl3 + ·H ∆H°= + 380 kJmol-1

CHBr3 → ·CBr3 + ·H ∆H°= + 377 kJmol-1

average ∆H°C-H = + 412 kJmol-1

• Applications of Bond Enthalpy– Given a reaction– 1st Step is break all bonds to give free atoms in

the gas phase (Endothermic)– 2nd Step is form new bonds for the products.

(Exothermic)

OFB Chap. 10 376/4/2003

• Applications of Bond Enthalpy– Given a reaction– 1st Step is break all bonds to give free atoms

in the gas phase (Endothermic)– 2nd Step is form new bonds for the products.

(Exothermic)

OFB Chap. 10 386/4/2003

Exercise 10-10• Estimate the Standard Enthalpy of

reaction for the gas-phase reaction that forms methanol from methane and water and compare it with the ∆H°r obtained from the data in Appendix D.

CH4(g) + H2O(g)→CH3OH(g) + H2(g)

OFB Chap. 10 396/4/2003

H

C

H H

H

+O

H

H

H

HH

+ H HC

O H

3 C-H 1 O-H 1 C-O 1 H-H

4 C-H 2 O-H

FormedBroken

CH4(g) + H2O(g)→CH3OH(g) + H2(g)

OFB Chap. 10 406/4/2003

Lattice Energy• Lattice Energy is the energy change

for the formation of 1 mol of an ionic solid from its constituent gaseous ions.

Na+ (g) + Cl- (g) → Na+ Cl- (s)(Note that the formation is exothermic)

• A measure of the electrostatic attractions and repulsions of the ions in the crystal lattice

OFB Chap. 10 416/4/2003

Enthalpy of Atomization

• Energy of atomization is the energy needed to produce 1 mol of gaseous atoms of that element form the element in its normal phase at room temperature.Cu (s) → Cu (g) (sublimation)I2 (s) → 2 I (g) (bond energy)

(Note that the formation is endothermic)

OFB Chap. 10 436/4/2003

The Born-Haber Cycle• A graphical display of the thermodynamic

processes representing the formation of a species or compound

• Enthalpy Diagram– To gain a visual image of the key enthalpy terms

in the formation of the compound– To determine any one unknown enthalpy value in

the thermodynamic cycle– The sum of the component processes should equal

the overall enthalpy change for the formation process

• Individual processes include– Heat of atomization– Heat of dissociation– Ionization Energy– Electron Affinity– Lattice Energy– Energy of Hydration (solutions)

OFB Chap. 10 446/4/2003

The Born-Haber Cycle– Rayner-Canham Overton text

groups these individual processes as either

• Endothermic (left) or increasing energy

• Exothermic (right) or decreasing energy

– In this course, we prefer to base the Born Haber cycle on logical transformations (or individual processes) rather than collecting the endothermic and exothermic reactions

– Thus if heat is added for a transformation a + sign is used for the endothermic process and if heat is given off a – sign is used for the exothermic process. The individual processes are summed

OFB Chap. 10 456/4/2003

Born-Haber Cycle Formation of NaCl (s)Na (s) + ½ Cl2 (g) → Na+Cl- (s)

1. Energy of Atomization of Na to change solid to a gas

∆H °= +108 kJ/mol

2. Energy of dissociation of gaseous Cl2 requires ½the bond energy of Cl2 (or 242 kJ/mol)

∆H °= +121 kJ/mol

3. Ionization of the sodium atom, the first ionization energy

∆H °= +502 kJ/mol

4. Chlorine atoms gain an electron, electron affinity∆H °= -349 kJ/mol

5. Free gaseous ions then associate to form an ionic compound. This is the lattice energy.

∆H °= -793 kJ/mol

OFB Chap. 10 466/4/2003

Born-Haber Cycle Formation of NaCl (s)Na (s) + ½ Cl2 (g) → Na+Cl- (s)

1. Energy of Atomization Na (s) → Na (g) ∆H °= +108 kJ/mol

2. Energy of dissociation ½ Cl2 (g) → Cl (g) ∆H °= +121 kJ/mol

3. Ionization Energy Na (g) → Na+(g) + e- (g) ∆H °= +502 kJ/mol

4. Electron affinityCl (g) + e- (g) → Cl- (g) ∆H °= -349 kJ/mol

5. Lattice energy.Na+ (g) + Cl- (g) → Na+Cl- (s) ∆H °= -793 kJ/mol

Σ (individual processes) = ∆Hf OFB Chap. 10 476/4/2003

Na (s) + ½ Cl2 (g) → Na+Cl- (s)

Na (g)

Na+(g)

Na+ (g) + Cl- (g)Cl (g)

Σ (individual processes) = ∆Hf

OFB Chap. 10 486/4/2003

Enthalpy Factors (kJmol-1) NaCl NaCl2

Na atomization 108.00 108.00Cl-Cl bond dissociation 121.00 242.00Na ionization (total) 502.00 5002.00Cl electron affinity (349.00) (700.00)Lattice energy (793.00) (2400.00)

calculated

∆H°f (estimated) (411.00) 2252.00favored not favored

Na (s) + ½ Cl2 (g) → Na+Cl- (s)

Why not NaCl2?

Na (s) + Cl2 (g) → NaCl2 (s)

OFB Chap. 10 496/4/2003

Why not NaCl2?Na (s) + Cl2 (g) → NaCl2 (s)

∆H°Atomization

= +108

Na (g)

Na+(g)

∆H°1st ionization energy = +502

Na++ (g)+ 2 Cl- (g)2Cl (g)

Σ (individual processes) = ∆Hf

OFB Chap. 10 616/4/2003

Chapter 10Thermochemistry Summary

if

pressureconstant at CapacityHeat denotes Cp where

p

T- T ∆T

∆Tabsorbedheat

∆TqCCapacityHeat

=

===

11-pp molJK are units

nCc

CapacityHeat Molar

−= 1-1-ps gJK are units

mCc

CapacityHeat Specific

=

( ) ( )2if p221if p11

2 p221 p11

21

TTcnTTcn∆Tcn∆Tcn

qqeTemperatur ionEquilibrat

−−=−−=−=

OFB Chap. 10 626/4/2003

– If ∆H positive = ∆H > 0 = q > 0 Means heat is absorbedAnd is called Endothermic– If ∆H negative = ∆H < 0 = q < 0 Means heat is given offAnd is called Exothermic

• Enthalpy, H.

qp= ∆H Enthalpy Change (note p denotes constant pressure)

• Enthalpy is a state property, which means it depends on its initial and final state, not the path to get there.

•Hess’s Law: If two or more chemical equations are added to give a new equation, then adding the enthalpies of the reactions that they represent gives the enthalpy of the new reaction.

OFB Chap. 10 636/4/2003

Standard State Enthalpies• ∆H° is the sum of products minus the

sum of the reactants• For a general reaction

a A + b B → c C + d D

(B)H(A)H

(D)H(C)H∆Hof

of

of

of

o

∆∆

∆∆

ba

dc

−−

+=

Bond Enthalpies∆H= ∆H bonds broken + ∆H bonds formed

Born-Haber Cycle• A graphical display of the thermodynamic

processes representing the formation of a species or compound

• Enthalpy DiagramΣ (individual processes) = ∆Hf