Chapter 9Energy, Enthalpy and Thermochemistry

The study of energy and its interconversions is called thermodynamics.

Kinetic Energy: energy due to the motion of the object (1/2 mv2)

Potential Energy: energy due to position or composition

Heat: the transfer of energy between two objects due to a temperature difference

Work: a force acting over a distance

Figure 9.1 (a): Initial position of balls

Figure 9.1 (b): Final position of balls

Energy of Matters

E=T+V+U

E:total energy

T:kinetic energy

V:potential energy

U:internal energy

State Function

A property of the system depends only on its present state. A state function does not depend in any way on the system’s past.

Energy is a state function, but work and heat are not state function

Combustion of methane

Nitrogen/oxygen

First Law of Thermodynamics

The energy of the universe is constant

In closed system∆E= ∆U=q+w ( 內能 =熱能 +功 )q: the heat added to the system during the processw: the work done on the system during the processq>0 heat flows into the system from the surroundingsq<0 an outflow of heat from the system to the surroundingsw>0 work is done on the system by the surroundingsw<0 the system does work on the surroundings

P-V WorkP-V Work

P-V Work

2

1

0 0

rev ext

rev ext

w F X

FP w P A x P V

Adw P dV

w P dV

expansion

dV dW negative work

Enthalpy ( 焓 )

2 2

1 12 1 2 1

2 2 1 1 2 1

( )

( ) ( )

constant P, closed system, P-V work only

V V

p pV V

p

p

H U PV

U U q w q PdV q P dV q P V V

q U PV U PV H H

H q

The heat qp absorbed in a constant-pressure

process equals the system’s enthalpy change.

For a chemical reaction ΔH=ΔHproducts-ΔHreactants

If ΔHreactants<ΔHproducts (endothermic)If ΔHreactants>ΔHproducts (exothermic)

Consider a constant-volume processdw=-PdV=0 (體積固定不做功)∆U=q+w=qv

∆U=qv

Heat Capacity ( 熱容量 )

)./( KmolkJdT

dqC

Thermodynamics of Ideal Gases

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pressureconstant at capacity heat

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Heat Capacity of Heating an Ideal Monatomic Gas Under constant volume, the energy flowing into the

gas is used to increase the translational energy of the gas molecules.

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Heat Capacity of Ideal Monatomic Gases

Gas Cv

(J/K mol)

Cp

(J/K mol)

He 12.47 20.8

Ne 12.47 20.8

Ar 12.47 20.8

Heat Capacity of Diatomic Gases

Gas Cv

(J/K mol)

Cp

(J/K mol)

H220.54 28.86

N220.71 29.03

Heat Capacity of Polyatomic Gases

Gas Cv

(J/K mol)

Cp

(J/K mol)

N2O 30.38 38.70

CO228.95 37.27

C2H644.60 52.92

Heat Capacity of Heating a Polyatomic Gas

Polyatomic gases have observed values for Cv that are significantly greater than 3/2 R.

This larger value for Cv results because polyatomic molecules absorb energy to excite rotational and vibrational motions in addition to translational motions.

Cv and Cp of molecules

3translational energy

2rotational energy

3

2

tr rot vib el

tr

rot

vib

U U U U U

U RT

U RT for linear molecules

RT for nonlinear molecules

vibrational energy U electronic energy

0 ( )elU at low T

Cv and Cp of Monatomic Gas

,

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3

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V rot nonlin

V V tr V rot nonlin

P V

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C

C C C R

C C R R

Cv and Cp of H2O at 373K

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V tr

V rot nonlin

V V tr V rot nonlin

P V

C R

C RT

C C C R

C C R

R cal mol K cal mol K

2 mol of monatomic ideal gasCalculate q, w, ∆U and ∆H for both pathway

TA=122K, TC=366K, TB=183K, TD=61K

Cv=3/2R, Cp=5/2R

Path 1(A→C)

w1=-P∆V=-2atm×(30-10)L×101.3J/Latm=-4.05×103J

q1=qp=nCp∆T=2×5/2(R)×(366-122)=1.01×104J=∆H1

∆U1=nCv∆T= 2×3/2(R)×(366-122)=6.08×103J

Path 2(C→B)

q2=qv=nCv∆T=2×3/2(R)×(183-366)=-4.56×103J=∆U2

∆H2=nCp∆T= 2×5/2(R)×(183-366)=-7.6×103J

∆V=0 w2 =-P∆V =0

Path 3(A→D)q3=qv=nCv∆T=2×3/2(R)×(61-122)=-1.52×103J=∆U3

∆H3=nCp∆T= 2×5/2(R)×(61-122)=-2.53×103J

∆V=0 w2 =-P∆V =0

Path 4(D→B)w1=-P∆V=-1atm×(30-10)L×101.3J/Latm=-2.03×103J

q4=qp=nCp∆T=2×5/2(R)×(183-61)=5.08×104J=∆H4

∆U4=nCv∆T= 2×3/2(R)×(183-61)=3.05×103J

SummaryPath 1

qpath1=q1+q2=5.5 ×103J

wpath1=w1+w2= -4.05×103J

∆Hpath1= ∆H1 +∆H2= 2.55×103J

∆Upath1= ∆U1 +∆U2= 1.52×103J

Path 2

qpath2=q3+q4=3.56×103J

wpath2=w3+w4= -2.03×103J

∆Hpath2= ∆H3 +∆H4= 2.55×103J

∆Upath2= ∆U3 +∆U4= 1.52×103J

Calorimetry

Specific heat capacity with unit JK-1g-1

Molar heat capacity with unit JK-1mol-1

turein tempera increase

absorbedheat (C)capacity heat

Coffee Cup Calorimeter

A constant-pressure calorimetry is used in determining the change in enthalpy equals the heat.

∆H=qp=nCp∆T

∆V=0∆U=qv+0=qv

∆U=qv=nCv ∆ T

Bomb Calorimeter

2SO2(g)+O2(g)→2SO3(g) ∆H=-198 KJ2 mol. 1 mol. 2 mol.Calculate ∆H and ∆U

P is constant, ∆H=qp=-198 KJ (energy flow out of system)

∆U= qp +w

w=-P∆V and ∆V=∆n(RT/P)

T and P are constant, ∆n=nfinal-ninitial=-1 mol

So w=-P∆V=-P×∆n× (RT/P)

=- ∆nRT=-(-1)(8.314)(298)=2.48 kJ∆U= qp +w=-198 kJ+2.48 kJ=-196 kJ

Hess’s Law

If a reaction is carried out in a series of steps, ∆H for the reaction will be equal to the sum of the enthalpy changes for the individual steps

The overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out.

Consider the combustion reaction of methane to

form CO2 and liquid H2O

CH4(g) + 2O2(g)→ CO2(g) + 2H2O(l)

∆H1 =-890KJ/mol

This reaction can be thought of as occurring in

two steps:

CH4(g) + 2O2(g)→ CO2(g) + 2H2O(g)

∆H2 = -802 kJ/mol

2H2O(g)→2H2O(l)

∆H3 =-88KJ/mol

∆H1 = ∆H2 + ∆H3

Standard Enthalpies of Formation

The change in enthalpy that accompanies the formation of 1 mole of a compound from its elements with all substances in their standard states.

The superscript zero indicates that the corresponding process has been carried out under standard conditions.

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Petroleum and Natural Gas Coal

Present Sources of EnergyPresent Sources of Energy

New Energy SourcesNew Energy Sources

Coal ConversionHydrogen as a fuel

CO2 capture