CHAPTER 6 THERMODYNAMICS: THE FIRST...

2009년도 제1학기 화 학 1 담당교수: 신국조 thTextbook: P. Atkins / L. Jones, Chemical Principles, 4 ed., Freeman (2008) Chapter 6

CHAPTER 6 THERMODYNAMICS: THE FIRST LAW

Steam Locomotive : Thermal energy Mechanical Energy

Diesel Locomotive : Chemical energy Electrical Energy


3 Laws of Thermodynamics

Phenomenological (Macroscopic)

Cannot be derived or proved

Universal

Operational

First law of thermodynamics : Energy conservation

~ Black, Davy, Rumford, Mayer(1842), Joule, Helmholtz

Second law of thermodynamics : Irreversibility

~ Carnot, Clausius, Thomson (Lord Kelvin), Boltzmann

Third law of thermodynamics : Unavailability of 0 K

~ Nernst, Planck

• Zeroth law of thermodynamics : Concept of temperature

~ Thermal equilibrium at contact (A,B,C)

The First Law of Thermodynamics

Julius Robert von Mayer (獨,1814–1878)

~ German physician ~ law of energy conservation (1842) ~ crude experiment on

mechanical equivalent of heat

James Prescott Joule (英,1818–1889 ) ~ accurate experiment on

mechanical equivalent of heat

~ Joule-Thomson effect

~ 1 cal = 4.184 J

2009년도 제1학기 화 학 1 담당교수: 신국조 Textbook: P. Atkins / L. Jones, Chemical Principles, 4th ed., Freeman (2008) Chapter 6


Perpetual motion machines of 1st and 2nd kinds

“Waterfall”(1961) by Maurice C. Escher (1898-1972), Dutch artist


SYSTEMS, STATES, AND ENERGY Heat: thought to be a fluid “caloric” flowing from a hot substance to cooler one Wrong !

Work: resulted from the flow of caloric (Carnot) Wrong !

Both of them are forms of energy (Joule) True !

Fig. 6.1 Sadi Carnot Fig. 6.2 James Joule

(佛,1796-1832) (英,1818-1889)

6.1 Systems

▶ System : Anything of our interest

▶ Surroundings: Everything else

▶ Universe = system + boundary + surroundings

Fig. 6.3 System & Surroundings

▷ Open system : Exchange of both matter and heat with the surroundings

▷ Closed system: Exchange only heat

▷ Isolated system: Exchange nothing

Fig. 6.4 Open, closed, isolated systems

6.2 Work and Energy

◈ Work : motion against an opposing force Work = opposing force x distance moved

Lifting a weight against the gravity

Chemical reaction in a battery against the flow of current through a circuit

Expansion of a gas against the piston

◈ Internal energy, U : Capacity of a system to do work

Change in internal energy: ᇫU = Uf – Ui

▶ Unit of energy and work: joule, J 1J = 1 kg·m2·s–2

☺ Sign convention:

Work done on the system, , Work done by the system, 0w> 0w<


6.3 Expansion Work (or pV-work), w

(1) Expansion against an external constant pressure irreversible process

Fig. 6.5 Expansion work work force distance exP A d= × = − × V A d∆ = ×,

∴ exP V= − ∆w

▷ Free expansion: Expansion against vacuum ( ex 0P = ) No work is done. 0=w

▷ Units of pV-work:

1 Pa·m3 =1 kg·m–1·s–2 x 1 m3 = 1 kg·m2·s–2 = 1 J

1 L·atm = 10–3 m3 x 101 325 Pa = 101.325 Pa·m3 = 101.325 J (exactly)

(2) Expansion against a changing external pressure.

◈ Reversible isothermal expansion of an ideal gas

External pressure, Pex, matches to the Internal pressure of a gas, P,

at every stage

( / )exd P dV PdV nRT V dV= − = − = −w


lnf

i

V f

Vi

VdVdw nRT nRTV V

= = − = −∫ ∫w ∴ ln f

i

VnRT

V= −w

Expansion: f iV V> 0<w Work done by the system

Compression: f iV V< Work done on the system 0w>

Fig. 6.6 Reversible expansion of a gas Fig. 6.7 Work done by the system

6.4 Heat, q

◈ Heat : energy transferred as a result of temperature difference

1 cal = 4.184 J, 1 Cal = 1 kcal

▶ Exothermic process : most chemical reactions, all combustions

▶ Endothermic process : vaporization, dissolution of ammonium nitrate in water (cold pack for sports injuries)

Fig. 6.8 Thermite reaction. Fig. 6.9 Endothermic reaction between

Al reacts with Fe2O3 molten iron sparks NH4SCN and Ba(OH)2·8H2O

6.5 The Measurement of Heat

◈ Heat capacity (열용량,熱容量) : qCT

=∆

extensive property

▶ Specific heat capacity (비열,比熱) : heat capacity per gram intensive property /sC C m=▶ Molar heat capacity : intensive property m /C C= n

sC m C= ×

s mq C T mC T nC T= ∆ = ∆ = ∆


Fig. 6.10 A large object has a larger heat capacity than a small object of the same material. An object with a larger heat capacity needs more heat to bring about a given rise in temperature.

▶ liquid water, Cs = 4.184 J·(oC) –1·g–1 , Cm =75 J·K–1·mol–1 at 25oC

◈ Calorimeter (열량계,熱量計) : measurement of heat

▶ Constant pressure calorimeter

▶ Constant volume calorimeter : Bomb calorimeter [통(筒)열량계] or combustion calorimeter

Fig. 6.11 Energy released by a reaction Fig. 6.12 A bomb calorimeter at constant pressure. at constant volume.

: heat capacity of the calorimeter calq C T= ∆ calC


6.6 The First Law

◆ Energy conservation:

U q∆ = +w for a closed system

▶ Work: causes a concerted motion of molecules in a definite direction

▶ Heat: causes a random motion of molecules with no definite direction

Fig. 7.7 (Oxtoby) Falling weight turns a paddle Fig. 6.13 Transfer of heat by more rigorously that does work on the system. moving particles.

☺ Internal energy of an isolated system is constant. 0U∆ =

▷ Perpetual motion machine of 1st kind exception to the first law !

No one has been successful making this machine.

★ State function : U, P, V, T, d, m, …

A property that depends only on the current state of the system and is independent of

how that state was prepared.

Fig. 6.14 (a)Altitude of a mountain spot. (b) Internal energy as a thermodynamic state function.

★ Path function: w , q, …

A property that depends on the paths leading to the current state.


Fig. 6.15 Two different paths between the same initial and final states. (a) Isothermal expansion against an external pressure. (b) No work is done during isothermal expansion against the vacuum.

▶ Isothermal expansion of an ideal gas: 3

2 0kU E nR T∆ = ∆ = ∆ =

Ex. 6.5 Calculating the work, heat, and change in internal energy for the expansion of an ideal gas via two paths.

(1 mol of an ideal gas at 292 K and 3.00 atm expands from 8.00 L to 20.00 L. The final pressure is 1.20 atm.)

[Path A] Isothermal reversible expansion.

0U∆ =

1 1 20.00ln (1.00 mol) (8.3145 J K mol ) (292 K) ln 2.22 kJ8.00

f

i

VnRT

V− − ⎛ ⎞= − = − × ⋅ ⋅ × × = −⎜ ⎟

⎝ ⎠w

0 2q U= ∆ − = − = − = +w w w .22 kJ

[Path B] Irreversible expansion in two steps.

Fig. 6.16 (a) Isothermal reversible work done. (b) Irreversible work done. Step 1: 1 0 0V= ∆∵w =

w

Step 2: 2 ext (1.20 atm) (20.00 8.00)L 14.4 L atm 1.46 kJP V= − ∆ = − × − = − ⋅ = −w

Total: w w 1 2 0 ( 1.46 kJ) 1.46 kJ= + = + − = − From , 0U q∆ = + =w 1.46 kJq = − = +w


6.7 A Molecular Interlude: The Origin of Internal Energy

Fig. 6.17 Average energies of each mode at a temperature T. (a) translation (b) rotation of a linear molecule (c) rotation of a nonlinear molecule.

◈ Equipartition theorem

Average value of each quadratic contribution to the energy of a molecules in a sample

at a temperature T is equal to 12 k TB .

▶ 23 1B 1.381 10 J Kk − −= × ⋅ : Boltzmann constant A BR N k=

▶ Quadratic contributions: 212 mv , 21

2 Iω , 212 kx

▶ Classical theory: Applies only to translation and rotation at room temperature

3m A Bk T2(translation)U N= m(rotation, linear), U RT= , 3

m 2(rotation, nonlinear) T=U R

ENTHALPY

At constant volume, ex 0P V= − ∆ =w . U q∆ =

At constant pressure, . 0≠w ? q∆ =

6.8 Heat Transfers at Constant Pressure

Define a state function, enthalpy H : H U PV= +

At constant pressure, H U P V q P V∆ = ∆ + ∆ = + + ∆w

No work other than expansion: exH q P V P V∆ = − ∆ + ∆

System is open to the atmosphere: exP P= H q∆ =

.

Fig. 6.18 An exothermic reaction, . Fig. 6.19 An endothermic reaction, . 0H∆ < 0H∆ >


6.9 Heat Capacities at Constant Volume and Constant Pressure

▶ Heat capacity at constant volume: VV

U UCT T

∂ ∆⎛ ⎞= ≈⎜ ⎟∂ ∆⎝ ⎠

▶ Heat capacity at constant pressure: PP

H HCT T

∂ ∆⎛ ⎞= ≈⎜ ⎟∂ ∆⎝ ⎠

◈ For an ideal gas, H U PV U nRT= + = + H U nR T∆ = ∆ + ∆

PH U nR T UC nRT T T

∆ ∆ + ∆ ∆= = = +∆ ∆ ∆ P VC C nR = +

∴ C C P,m V,m R= +

6.10 A Molecular Interlude: The Origin of the Heat Capacities of Gases

▶ Ideal monatomic gas:

3m 2 3

V,m 2U R TC RT T

∆ ∆= = =

∆ ∆ 5

P,m 2C R=

▶ Ideal gas of linear molecules:

3m 2 5

V,m 2( )U R R TC R

T T∆ + ∆

= = =∆ ∆

7P,m 2C R=

▶ Ideal gas of nonlinear molecules:

3 3m 2 2

V,m( ) 3U R R TC R

T T∆ + ∆

= = =∆ ∆

P,m 4C R=

◇ Variation of with temperature for iodine vapor, V,mC 2I (g)

At very low temperature, 3V,m 2C R=

translation only

At slightly increased temperature, 5V,m 2C R=

rotational motion comes in

Further increased temperature, 7V,m 2C R=

vibrational motion comes in

Fig. 6.20 Variation of molar heat capacity with temperature for iodine vapor at constant volume.

Ex. 6.6 Calculating the energy change when heating an ideal gas.

Calculate the final temperature and the change in internal energy when 500 J of heat is transferred to

0.900 mol O2(g) at 298 K and 1.00 atm at (a) constant volume; (b) constant pressure.


1 15V,m 2 20.79 J K molC R − −= = ⋅ ⋅

3 translational + 2 rotaional modes

1 17P,m 2 29.10 J K molC R − −= = ⋅ ⋅

(a) 1 1V,m

500 J(0.900 mol) (20.79 J K mol )

qTnC − −∆ = =

× ⋅ ⋅

26.7 K= +

Final temperature: (298 26.7) K 325 KT = + =

From , U q∆ = 500 JU∆ = +

(b) 1 1P,m

500 J(0.900 mol) (29.10 J K mol )

qTnC − −∆ = =

× ⋅ ⋅

19.1 K= +

Final temperature: (298 19.1) K 317 KT = + =

Heat: Transfer enough energy at constant volume to raise the

temperature to its final value (317 K), and use U q∆ = .

1 1(0.900 mol) (20.79 J K mol ) (19.1 K)U q − −∆ = = × ⋅ ⋅ ×

+357 J=

Expand: Allow the sample to expand isothermally to its final volume.

Because U is independent of volume for an ideal gas,

step2 0U∆ =

Add two changes in internal energy:

+357 JU∆ =


6.11 The Enthalpy of Physical Change

▶ Enthalpy of vaporization, vap m m(vapor) (liquid)H H H∆ = −

▶ Enthalpy of fusion, fus m m(liquid) (solid)H H H∆ = −

▷ Enthalpy of freezing,

freez m m fus(solid) (liquid)H H H H∆ = − = −∆

▶ Enthalpy of sublimation,

sub m m fus vap(vapor) (solid)H H H H H∆ = − = ∆ + ∆

Fig. 6.21 Potential energy curve of a liquid. The read bar is vapH∆ .

Fig. 6.22 Endothermic melting process.


Fig. 6.23 Enthalpy of a reversible process. Fig. 6.24 The polar ice caps on Mars. Fig. 6.25 Enthalpy of sublimation. Sublimation of solid CO2.

6.12 Heating Curves Steep slopes (ice, water vapor) low heat capacities

Fig. 6.26 The heating curve of water.

Box 6.1 HOW DO WE KNOW … THE SHAPE OF A HEATING CURVE? ◈ Differential thermal analysis (DTA) [시차열분석법,示差熱分析法]

Reference material : undergoes no phase change ex. Al2O3

Thermocouples (열전기쌍,熱電氣雙)

Heating both sample and reference

in the same large metal heat sink

Constant temperatures during phase transitions for the sample

Continuous increase in temperature for the reference

Generation of electric signals

Output of DTA Thermogram (열분석도,熱分析圖)

◈ Differential scanning calorimetry (DSC), [시차주사열량계법,示差走査熱量計法]

Heating the sample and reference in separate, but identical, metal heat sinks

Temperatures of the sample and reference are kept the same by varying the power supplied to two heaters.

The output is the difference in power as a function of heat added.


A thermogram of DSC The peak indicates a phase transition. The step in the baseline is due to different heat capacities.

THE ENTHALPY OF CHEMICAL CHANGE

6.13 Reaction Enthalpies

Thermochemical equations (열화학반응식)

Combustion of methane gas:

CH4(g) + 2 O2(g) CO⎯⎯→ 2(g) + 2 H2O(l) 890 kJH∆ = −

2 CH4(g) + 4 O2(g) 2 CO⎯⎯→ 2(g) + 4 H2O(l) 1780 kJH∆ = −

CO2(g) + 2 H2O(l) CH⎯⎯→ 4(g) + 2 O2(g) 890 kJH∆ = +

Ex. 6.7 Determining a reaction enthalpy from experimental data.

Combustion of 0.113 g benzene in a constant pressure calorimeter with the heat capacity of 551 J·(oC)–1.

The temperature of the calorimeter increased by 8.60 oC. exothermic, 0H∆ <

2 C6H6(l) + 15 O2(g) 12 CO⎯⎯→ 2(g) + 6 H2O(l)

Find the heat transferred to the calorimeter:

o 1 ocal cal [551 J ( C) ] (8.60 C)

8.60 551 J 4.74 kJq C T −= ∆ = ⋅ ×

= × =

Calculate the amount of benzene that reacts: /n m M=

1

0.113 g 0.113 mol78.12 g mol 78.12

n −= =⋅


Calculate for 2 mol of benzene: 0H∆ <

3(2 mol) (2 mol) (4.74 kJ) 6.55 10 kJ(0.113/78.12) mol

H qn

×∆ = × = − = − ×

6.14 The Relation Between △H and △U

For an ideal gas,

initial initial initial initial initialH U PV U n R= + = + T

T

T

final final final final finalH U PV U n R= + = +

final initial final initial( )H H H U n n RT∆ = − = ∆ + −

∴ gasH U n R∆ = ∆ + ∆

Sef-Test 6.12B Aluminum powder as a rocket fuel.

4 Al(s) + 3 O2(g) 2 Al⎯⎯→ 2O3(s)

Fig. 6.27 Preparation of rocket fuel.

Al powder is mixed with oxidizing agent in a liquid polymer base that hardens inside the booster rocket shell.

1.00 mol of Al produced 3378 kJ of heat in a constant-pressure vessel at 1000oC. =? U∆

gas

1 1(4 mol) (3378 kJ mol ) ( 3 mol) (8.3145 J K mol ) (1273 K)=13512 kJ 31.753 kJ 13544 kJ

U H n RT− − −

∆ = ∆ − ∆

= × ⋅ − − × ⋅ ⋅ ×+

1

6.15 Standard Reaction Enthalpies

CH4(g) + 2 O2(g) CO⎯⎯→ 2(g) + 2 H2O(g)

1 802 kJH∆ = −

CH4(g) + 2 O2(g) CO⎯⎯→ 2(g) + 2 H2O(l)

2 890 kJH∆ = −

Water: at 25water 1vap 44.0 kJ molH −∆ = ⋅ oC

12 1 2watervap

802 kJ ( 890 kJ) 88 kJ 2

H H HH

∆ = ∆ − ∆ = − − − =

= ×∆

Fig. 6.28 △H’s of reactions of combustion of CH4 to give H2O(g) (left) and H2O(l) (right).


▶ Standard state: pure form at 1 bar (and usually at 25oC)

Standard state of water is pure water at 1 bar.

Standard state of ice is pure ice at 1 bar.

Standard state of a solute in a liquid solution : When its concentration is 1 mol·L–1

▶ Standard reaction enthalpy, oH∆ Reaction enthalpy when reactants in their standard states change into products in their standard states

Fig. 6.29 Standard enthalpy of reaction: Combustion of CH4 to give CO2 and liquid water.

6.16 Combining Reaction Enthalpies: Hess’s Law

Enthalpy : a state function.

◈ Hess’s law

The overall reaction enthalpy is the sum of the reaction enthalpies of the steps

into which the reaction can be divided. ▶ Oxidation of carbon, C(gr), to carbon dioxide:

(1) C(gr) + O2(g) CO⎯⎯→ 2(g) o1 ?H∆ =

(2) C(gr) + ½ O2(g) CO(g) ⎯⎯→ o2 110.5 kJH∆ = −

(3) CO(g) + ½ O2(g) CO⎯⎯→ 2(g) o3 283.0 kJH∆ = −

(1) = (2) + (3) o o o

1 2 3

110.5 kJ ( 283.0) kJ 393.5 kJ

H H H∆ = ∆ + ∆= − + −= −

Fig. 6.30 Hess’s law

6.17 The Heat Output of Reactions

◈ Combustion reactions:

▷ Oxidation of glucose in human body

▷ Burning of fossil fuels (coal, petroleum, natural gas) for heating, for industries, for cars


Box 6.2 WHAT HAS THIS TO DO WITH… THE ENVIRONMENT? ◈ Alternatives of fossil fuels

▶ Sustainable alternatives: Hydroelectric power, Windmill power, Solar energy cells

▶ Non-sustainable alternatives:

◆ Hydrogen: Electrolysis of ocean water

◆ Ethanol: Fermenting biomass (starches in corn, cellulose in straw and cornstalks)

◆ Methane: Anaerobic bacterial digestion of sewage and agricultural wastes

☺ Enthalpy density : enthalpy of combustion per liter

Digestion of cellulose-based biomass by special enzymes. Water treatment tanks generate methane by anaerobic digestion of sewage.


◈ Standard enthalpy of combustion, ocH∆

Change in enthalpy per mole of a substance that is burned in a combustion reaction

under standard conditions.

Ex. 6.10 Calculating the heat output of a fuel

Calculate the mass of butane to produce 350 kJ of heat to heat 1 L of water from 17oC to boiling at sea level.

2 C4H10(g) + 13 O2(g) 8 CO⎯⎯→ 2(g) + 10 H2O(l) o 5756 kJH∆ = −

Relation of the enthalpy change and amount of fuel:

5756 kJ 2 mol C4H10

Convert heat output into moles of fuel:

butane2 mol butane(350 kJ)

5756 kJn ⎛ ⎞= × ⎜ ⎟

⎝ ⎠

Find the mass of fuel:

1butane

2 mol butane(350 kJ) (58.12 g mol )5756 kJ

7.07 g butane

m −⎛ ⎞= × × ⋅⎜ ⎟⎝ ⎠

=

6.18 Standard Enthalpies of Formation

◈ Standard enthalpy of formation, , of a substance is the standard reaction enthalpy per mole of

formula units for the formation of a substance from its elements in their most stable form.

ofH∆

Ex: 2 C(gr) + 3 H2(g) + ½ O2(g) C⎯⎯→ 2H5OH(l) o 1f 2 5(C H OH) 277.69 kJ molH −∆ = − ⋅

⋅

C(gr) C(diamond) ⎯⎯→ o 1f [C(diamond)] 1.9 kJ molH −∆ = +


◈ Standard reaction enthalpy, oH∆

o oof f(products) (reactants)H n H n H∆ = ∆ − ∆∑ ∑

Fig. 6.31 Construction of reaction enthalpy from enthalpy of formations.

Ex. 6.11 Using standard enthalpies of formation to calculate a standard enthalpy of reaction.

2 NH2CH2COOH(s) + 3 O2(g) H⎯⎯→ 2NCONH2(s) + 3 CO2(g) + 3 H2O(l) o ?H∆ = Glycine Urea

o o of f 2 2 f 2(products) (1 mol) (H NCONH ,s) (3 mol) (CO ,g) (3 mol) (H O,l)n H H H H∆ = ∆ + ∆ + ∆∑ o

f 2

1−

of 2

of 2

1 1(1 mol)( 333.51 kJ mol ) (3 mol)( 393.51 kJ mol ) (3 mol)( 285.83 kJ mol )− −= − ⋅ + − ⋅ + − ⋅

2371.53 kJ= −

o of f 2 2(reactants) (2 mol) (NH CH COOH,s) (3 mol) (O ,g)n H H H∆ = ∆ + ∆∑

1 1(2 mol)( 532.9 kJ mol ) (3 mol)( kJ mol )0 1065.8 kJ− −= − ⋅ + ⋅ = −

∴ o oof f(products) (reactants) 2371.53 ( 1065.8) kJ 1305.7 kJH n H n H∆ = ∆ − ∆ = − − − = −∑ ∑

Ex. 6.12 Using the enthalpy of combustion to calculate an enthalpy of formation.

C3H8(g) + 5 O2(g) 3 CO⎯⎯→ 2(g) + 4 H2O(l) enthalpy of combustion o 2220 kJH∆ = −

o of f 2(products) (3 mol) (CO ,g) (4 mol) (H O,l)n H H H∆ = ∆ + ∆∑

1 1(3 mol)( 393.51 kJ mol ) (4 mol)( 285.83 kJ mol ) 2323.85 kJ− −= − ⋅ + − ⋅ = −

o of f

o23 8 f(reactants) (1 mol) (C H ,g) (5 m (O ,g)ol) Hn H H= ∆ ∆∆ +∑ o

f 3 8(1 mol) (C H ,g)H= ∆

o o oof f f 3 8(products) (reactants) 2323.85 kJ (1 mol) (C H ,g) 2220 kJH n H n H H∆ = ∆ − ∆ = − − ∆ = −∑ ∑

∴ o 1 1f 3 8(C H ,g) 2323.85 ( 2220) kJ mol 104 kJ molH − −∆ = − − − ⋅ = − ⋅


6.19 The Born-Haber Cycle

▶ Lattice enthalpy, : heat required to vaporize the solid LH∆

L m m(ions,g) (solid)H H H∆ = −

◈ Born-Haber cycle :

A thermodynamic cycle that allows the experimental determination of lattice energy of ionic crystals

Fig. 6.32 A Born-Haber cycle Fig. 6.33 The Born-Haber cycle to determine the lattice enthalpy of KCl.

(1) Ionic solid is converted to the elements in their standard states. – (Enthalpy of formation)

(2) Elements are transformed into gas-phase atoms. Enthalpy of atomization

(3) Electron transfer between atoms generating gaseous ions Enthalpy of ionization

(4) (3) (1) – (Lattice enthalpy)

– (Enthalpy of formation) + Enthalpy of atomization + Enthalpy of ionization – (Lattice enthalpy) = 0


Ex. 6.13 Using a Born-Haber cycle to calculate the lattice enthalpy of KCl crystal.

(1) K(s) + Cl2(g) KCl(s) ⎯⎯→ o 1f (KCl,s) 437 kJ molH −∆ = − ⋅

KCl(s) K(s) + Cl⎯⎯→ 2(g) o 1f 2(K,Cl ;elements) ( 437 kJ mol )H −∆ = − − ⋅

(2) K(s) K(g) ⎯⎯→ o 1f (K,g) 89.2 kJ molH −∆ = + ⋅

½ Cl2(g) Cl(g) ⎯⎯→ o 1f (Cl,g) 122 kJ molH −∆ = + ⋅

(3) K(g) K⎯⎯→ +(g) + e– o + 1f (K ,g) 418 kJ molH −∆ = + ⋅ ionization energy

Cl(g) + e– Cl⎯⎯→ –(g) o 1f (Cl ,g) 349 kJ molH − −∆ = − ⋅

L

– (electron affinity)

(4) KCl(s) K⎯⎯→ +(g) + Cl–(g) (KCl,s)H∆

K+(g) + Cl–(g) KCl(s) ⎯⎯→ L(KCl,s)H−∆

Over a Born-Haber cycle:

(437 + 89.2 + 122 + 418 – 349) kJ·mol–1 – L (KCl,s)H∆ = 0

∴ = + 717 kJ·molL (KCl,s)H∆ –1

6.20 Bond Enthalpies

◈ Bond enthalpy, > 0 B(X Y)H∆ −

X–Y (g) X(g) + Y(g) ⎯⎯→ { }o o oB m m m(X Y) (X,g) (Y,g) (XY,g)H H H H∆ − = + −

H2(g) 2 H(g) ⎯⎯→ 1(H-H) 436 kJ molHB−∆ = ⋅

Fig. 6.34 An electrostatic potential diagram of ethanol.

More electronegative O atom draws electrons from more

distant parts of the molecule.


▶ Mean bond enthalpy

Average enthalpy of a specific bond in different molecules

Refers to the dissociation of a molecule in the gas phase

For a liquid phase, X2(l) 2 X(g) ⎯⎯→ o oovap 2 B(X ,l) (X X)H H H∆ = ∆ + ∆ −

Ex. 6.14 Using mean bond enthalpies to estimate the enthalpy of a reaction

CH3CH2I(g) + H2O(g) CH⎯⎯→ 3CH2OH(g) + HI(g) o ?H∆ =

Reactants:

Break 1.00 mol C–I bonds in CH3CH2I : 1B(C-I) 238 kJ molH −∆ = ⋅

Break 1.00 mol O–H bonds in H2O : 1B(O-H) 463 kJ molH −∆ = ⋅

Products:

Form 1.00 mol C–O bonds : 1B(C-O) 360 kJ molH −−∆ = − ⋅

Form 1.00 mol H–I bonds : 1B(H-I) 299 kJ molH −−∆ = − ⋅

Overall enthalpy change:

oB B B B(C-I) (O-H) (C-O) (H-I)

238 463 ( 360 299) kJ 701 ( 659) kJ 42 kJH H H H H∆ = ∆ + ∆ − ∆ − ∆

= + + − − = + − = +


6.21 The Variation of Reaction Enthalpy with Temperature

P PH q C T∆ = = ∆ >0 as △T >0

Enthalpy increases with temperature

Amount of increase depends on the value of heat capacities for a given △T

◈ Kirchoff’ law

o o2 1 2 1( ) ( ) ( ) PH T H T T T C∆ = ∆ + − ∆

where ,m ,m(products) (reactants)P P PC nC nC∆ = −∑ ∑

Fig. 6.35 Temperature dependence of enthalpies Fig. 6.36 Enthalpy of reaction, N2+3H2 2NH3 of reactants and products.

Ex. 6.15 Predicting the reaction enthalpy at a different temperature.

2 2N (g) 3 H (g) 2 NH (g)+ ⎯⎯→ 3 o 192.22(298 kJ o K) m lH −− ⋅∆ = o o(450 C) ?H∆ =

P,m 3 P,m 2 P,m 2(2 mol) (NH ,g) (1 mol) (N ,g) (3 mol) (H ,g)PC C C C∆ = − −

1 1 1 1 1 1

1

(2 mol)(35.06 J K mol ) (1 mol)(29.12 J K mol ) (3 mol)(28.82 J K mol )45.46 J K

− − − − − −

−

= ⋅ ⋅ − ⋅ ⋅ ⋅ ⋅

= − ⋅

−

2 1 (450 273) K 2 425 K98 KT T T∆ = − = + − =o o

2 1 2 1( ) ( ) ( ) PH T H T T T C∆ = ∆ + − ∆

192.22 kJ (425 K) ( 45.46 J K ) 111.54 kJ−−= + × − ⋅ =

o 1 12 ,m(products, ) (products) (425 K) (2 mol)(35.06 J K mol ) 29.80 kJPH T T nC − − +∆ = ∆ × = × ⋅ ⋅ =∑

o 1 12 ,m

1 1

(reactants, ) (reactants) (425 K) {(1 mol)(29.12 J K mol )

(3 mol)(28.82 J K mol ) 4 12 } 9. kJPH T T nC − −

− −

∆ = ∆ × = × ⋅

+ ⋅ +⋅ =

∑ ⋅

CHAPTER 6 THERMODYNAMICS: THE FIRST...

Documents

Transcript of CHAPTER 6 THERMODYNAMICS: THE FIRST...