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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
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
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
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
Thermodynamics of Ideal Gases
R C-C
dTCqΔU )dT
dU(
dT
dqC
C
dTCqΔH)dT
dH(
dT
dqC
C
TnCUE T nCH
vp
T
T vvvv
v
v
T
T pppp
p
p
VP
2
1
2
1
pressureconstant at capacity heat
pressureconstant at capacity heat
ΔΔΔΔΔ
),(:n
CC
n
CCnote v
vp
p
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.
RCTCTRTRPVUH
RCTCPVTREU
nFor
nRTPVnRTE
pp
vvk
k
2
5
2
32
3
2
3
mole 1
,2
3
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 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
,
, ,
, , ,
3
20
3
25
2
V tr
V rot nonlin
V V tr V rot nonlin
P V
C R
C
C C C R
C C R R
Cv and Cp of H2O at 373K
,
, ,
, , ,
3
23
23
4 4 1.987( / ) 7.95( / )
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
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
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