CHAPTER 4
Thermochemistry
(熱化學是熱力學的一支，在化學反應或相變化
過程中發生的能量吸收或釋出，若以吸放熱的形
式表現，即為熱化學研究的對象)
Chap. 4 Thermochemistry
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Chap. 4 Thermochemistry
2
4.1 Energy Stored in Chemical Bonds Is Released
or Taken Up in Chemical Reactions
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I 2.2 Food and Energy reserves
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4.2
Internal Energy and Enthalpy Changes Associated with
Chemical Reactions
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0
0
∆H reaction
=
ν
∆
H
∑ i f ,i
i
Figure 4.2
Equation (4.12) follows from
the fact that the enthalpy of
both paths is the same
because they connect the
same initial and final states.
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(c) Hess Law: 黑斯定律
• Hess law: 黑斯定律
“The standard enthalpy of reaction, (ΔrH ɵ ) is
independent of the path by which the reaction actually
take place.” (標準反應熱焓,與發生反應的實際路徑無關)
• Thus, the standard enthalpy of an overall reaction is the
sum of the standard enthalpies of the (possible
hypothetical) individual reactions into which the overall
reaction may be regarded as divided. (任一假想或實際的平
衡總反應標準熱焓,是分開串連反應步驟的各別標準熱焓總和)
• All the standard enthalpies of reaction used in a given
application of Hess’s law must refer to the same
temperature. (黑斯定律中,反應標準熱焓的各別步驟,必須取相
同溫度的標準熱焓)
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P4.5) Several reactions and their standard reaction enthalpies at 25ºC
are given here.
(kJ mol–1)
CaC2(s) + 2H2O(l) → Ca(OH) 2(s) + C2H2 (g)
–127.9
Ca(s) + ½ O2 (g) → CaO(s)
–635.1
CaO(s) + H2O(l) → Ca(OH) 2 (s)
–65.2
The standard enthalpies of combustion of graphite and C2H2(g) are
–393.51 and –1299.58 kJ mol–1, respectively.
Calculate the standard enthalpy of formation of CaC2(s) at 25ºC.
Chap. 4 Thermochemistry
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(kJ mol–1)
Ca(OH)2(s) + C2H2(g) → CaC2(s) + 2H2O(l)
+127.9
CaO(s) + H2O(l) → Ca(OH)2(s)
–65.2
2CO2(g) + H2O(l) → C2H2(s) + 5/2 O2(s)
1299.58
2C(s) + 2O2(g) → 2CO2(g)
2 × (–393.51)
Ca(s) + 1/2O2(g) → CaO(s)
–635.1
2C(s) + Ca(s) → CaC2(s)
= –59.8 kJ mol–1
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21
P4.10) Calculate the average C–H bond enthalpy in methane
using the data tables. Calculate the percent error in
equating the average C–H bond energy in Table 4.3
with the bond enthalpy.
CH4(g) → C(g) + 4H(g)

∆H reaction
= 4∆H f ( H, g ) + ∆H f ( C, g ) − ∆H f ( CH 4 , g )
4 × 218.0 kJ mol−1 + 716.7 kJ mol−1 + 74.6 kJ mol−1 =
1663 kJ mol−1
=
1663 kJ mol−1
Average Bond
Enthalpy = 415.8 kJ mol−1
=
4
415.8 kJ mol−1 − 411 kJ mol−1
Relative Error
= 100 ×
= 1.2%
−1
415.8 kJ mol
Chap. 4 Thermochemistry
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P4.11) Use the average bond energies in Table 4.3 to estimate
for the reaction C2H4(g) + H2(g) → C2H6(g). Also
calculate from the tabulated values of for reactant and
products (Appendix B, Data Tables). Calculate the percent
error in estimating from the average bond energies for this
reaction.
∆Ureaction = –(C–C bond energy + 6 C–H bond energy – H–H bond energy –
C=C bond energy – 4 C–H bond energy)
∆Ureaction = –(346 kJ mol–1 + 6 411 kJ mol–1 – 432 kJ mol–1 – 602 kJ mol–1 – 4
× 411 kJ mol–1) = –134 kJ mol–1.
Chap. 4 Thermochemistry
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



∆H reaction
298.15
K
=
∆
H
C
H
,
g
−
∆
H
C
H
,
g
−
∆
H
(
)
)
)
f ( 2 6
f ( 2 4
f ( H2 , g )

∆H reaction
−84.0 kJ mol−1 − 52.4 kJ mol−1 =
−136.4 kJ mol−1
( 298.15 K ) =


∆U reaction
298.15
K
=
∆
H
(
)
reaction ( 298.15 K ) − ∆nRT
=
−136.4 kJ mol−1 + 8.314 J mol−1 K −1 × 298.15 K = − 133.9 kJ mol−1
+134 kJ mol−1 − 133.9 kJ mol−1
Relative Error =
100 ×
≈ 0%
−1
−133.9 kJ mol
Chap. 4 Thermochemistry
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Thermochemistry
Kirchhoff’s law:
• It is often sufficient to assume that the heat capacity
difference (not the individual heat capacities) is
independent of temperature over the range of
temperature of interest.
Δr HT ɵ = 298C θ P,reactdT + Δ H θ + T C θ P,proddT
∫T
r
T
ΔrHT = ΔrH 298 + ∫
θ
θ
298
T
= ΔrH 298 + ∫
θ
298
(C
298
θ
P,prod
∫
298
)
− C θ P,react dT
ΔC θ P,proddT
• Kirchhoff’s law simplifies to
ΔrH ɵ (T2) = ΔrH ɵ (T1) + (T2 - T1)ΔrCpɵ
Chap. 4 Thermochemistry
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Chap. 4 Thermochemistry
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Thermochemistry
The temperature dependence of reaction enthalpies
• The temperature variation of the Δ rH ɵ is expressed by
Kirchhoff’s law:
T2
θ
ɵ
ɵ
Δ
C
Δ rH (T2) = ΔrH (T1) + ∫T
r p (T ) dT
∆rC p (T ) = ∑ν J ⋅C P
θ
J
θ
1
( J ,T )
• Where ∆rC P ( J ,T ) is the molar reaction heat capacity at
constant pressure of the substance J in its standard state at
θ
temperature T.

(
H
J ,T ) 
∂
θ

C p ( J ,T ) =


∂T
 θ

θ
P =P
• The law is a direct consequence of the heat capacity at
constant pressure. Although the individual heat capacities
may vary, their difference varies less significantly.
Chap. 4 Thermochemistry
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Thermochemistry
• The temperature variation of the ΔrCP ɵ is expressed by
power series in T:
ΔrCP ɵ (T) = Δra + (Δrb) T + (Δrc) /T 2
• Where Δr a = ∑ν J ⋅ a J
J
T
ΔrHT = ΔrH 298 + ∫
θ
θ
298
and so on
[Δ a + (Δ b)T + (Δ c)T ] dT
r
r
= ΔrH 298 + Δr a(T − 298) +
θ
r
-2
Δrb 2
1 
1
T − 2982 - Δr c −

T
2
298


(
)
• From absolute zero temperature:
ΔrHT
θ
 1
 Δrb  2
= ΔrH 0 ° + (Δr a )T + 
 T - Δr c 
T 
 2 
Chap. 4 Thermochemistry
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Using Kirchhoff’s law
Example
The standard enthalpy of formation of gases H2O at 298 K is
-241.82 kJ mol-1. Estimate its value at 100 ℃ given the
following values of the molar heat capacities at constant
pressure:
H2O(g): 33.58 J K-1 mol-1; H2(g): 28.84 J K-1 mol-1;
O2(g): 29.37 J K-1 mol-1; Assume that the heat capacities are
independent of temperature.
Answer:
The reaction is H2(g) + ½ O2(g) → H2O(g)
ΔrC = C
θ
P
θ
P,m
1 θ
 θ

(H2O, g ) - C P,m (H2, g ) + C P,m (O2, g )
2


= -9.94 J K-1 mol-1
ΔfH ɵ (373 K) = -241.82 kJ mol-1+ (75 K)x(-9.94 J K-1 mol-1)
= -242.6 kJ mol-1
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• P4.4) Calculate for N2(g) at 650 K using the temperature dependence
of the heat capacities from the data tables. How large is the relative
error if the molar heat capacity is assumed to be constant at its value
of 298.15 K over the temperature interval?
∆H f ( N 2 , g , 650 K ) =
∆H f ( N 2 , g , 298.15 K ) +
T  T
C
P ,m 
d
∫
K K
298.15
650
2
3
 650 

T
−5 T
−8 T  T
−1
−1
d
=  ∫  30.81 − 0.01187 + 2.3968 ×10
−
×
1.0176
10
J
K
mol

2
3 
K
K
K

 K
 298.15
= (10841 − 1980 + 1982 − 434.0 ) J mol−1= 10.41 kJ mol−1
If it is assumed that the heat capacity is constant at its value at 298 K,
 650
T
10.25 kJ mol−1
∆H ≈  ∫ ( 29.13) d  J K −1mol−1 =
K
 298.15
10.25 kJ mol−1 − 10.41 kJ mol−1
Error =
100 ×
=
−1.54%
−1
10.41 kJ mol

f
Chap. 4 Thermochemistry
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P4.12) Calculate the standard enthalpy of formation of FeS2(s) at
300ºC from the following data at 25ºC. Assume that the heat
capacities are independent of temperature.
You are also given that for the reaction
2FeS2(s) + 11/2O2(g) → Fe2O3(s) + 4 SO2(g), = –1655 kJ mol–1.
Substance
∆H
Fe(s)
FeS2(s)
Fe2O3(s)
S (rhombic)
–824.2

f
SO2(g)
–296.81
(kJ mol–1)
3.02
7.48
2.72
CP ,m /R
Chap. 4 Thermochemistry
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
∆H reaction
2FeS2(s) + 11/2O2(g) → Fe2O3(s) + 4SO2(g) = –1655 kJ mol–1
−1655 kJ mol−1 =∆H f ( Fe 2 O3 ,s ) + 4∆H f ( SO 2 , g ) − 2∆H f ( Fe 2S2 , s )
−1


1655
kJ
mol
Fe
O
,
4
H
s
H
+
∆
+
∆
(
)
2 3
f
f ( SO 2 , g )

∆H f ( Fe 2S2 , s,298 K ) =
2
1655 − 824.2 − 4 × 296.81 kJ mol−1
=
2
= −178.2 kJ mol−1
Chap. 4 Thermochemistry
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The enthalpy of formation at 300°C is given by
∆H f ( FeS2 ( s ) ,573 K ) − ∆H f ( FeS2 ( s ) , 298 K ) +
573 K
∫
∆C p (T ) d
298 K
Because the heat capacities are assumed to be independent of T,
∆H f ( FeS2 ( s ) ,573 K ) =
∆H f ( FeS2 ( s ) , 298 K )
+ CP ,m ( FeS2 , s ) − CP ,m ( Fe, s ) − 2CP ,mS ( s )  [573 K − 298 K ]
=−178.2 kJ mol−1 + 8.314 JK −1 mol−1 × ( 7.48 − 3.02 − 2 × 2.70 ) × [573 K − 298 K ]
= −180.0 kJ mol−1
Chap. 4 Thermochemistry
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4.5 The Experimental Determination of ∆U and
∆H for Chemical Reactions
Chap. 4 Thermochemistry
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Chap. 4 Thermochemistry
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4.5 The Experimental Determination of ∆U and
∆H for Chemical Reactions
mH 2O
ms
°
∆U °=
∆U reaction ,m +
CH 2O ,m ∆T + Ccalorimeter ∆T= 0
Ms
M H 2O
°
°
∆H reaction
= ∆U reaction
+ ∆nRT
∆H
mH 2O
ms
°
=
∆H reaction ,m +
CH 2O ,m ∆T + Ccalorimeter ∆T= 0
Ms
M H 2O
°
reaction
∆T change in temperature of the inner water bath
mH
O
2
mass of water in the inner bath
MH O molecular weight of water in the inner bath
2
CH O,m heat capacity of water in the inner bath
2
mH
O
mass of the sample
O
molecular weight of the sample
2
MH
2
Chap. 4 Thermochemistry
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∆H
mH 2O
ms
°
=
∆H reaction ,m +
CH 2O ,m ∆T + Ccalorimeter ∆T= 0
Ms
M H 2O
°
reaction
°
∆H reaction
=
qP
Figure 4.4
Schematic diagram of a constant pressure
calorimeter suitable for measuring the
enthalpy of solution of a salt in a solution.
Chap. 4 Thermochemistry
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4.6 Differential Scanning Calorimetry
Chap. 4 Thermochemistry
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Figure 4.5
A differential scanning calorimeter
consists of a massive enclosure and lid
that are heated to the temperature T
using a resistive heater. A support disk
in good thermal contact with the
enclosure supports multiple samples
and a reference material. The
temperature of each of the samples and
reference are measured with a
Chap. 4 Thermochemistry
thermocouple.
Differential Scanning
Calorimetry
49
Chap. 4 Thermochemistry
50
 dqP
∫  dt

 dt = qP = ∆H

Figure 4.6
At temperature different from the
melting temperature, the rate of heat
uptake of the sample and the
reference are nearly identical.
However, at the time corresponding
to T=Tmelting, the uptake rate of heat
for the sample differs significantly
from that of the reference. Kinetic
delays due to finite rates of heat
conduction broaden the sharp
melting transition to finite range.
Chap. 4 Thermochemistry
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The following table is in Appendix B in the text book.
Table 4.1
Thermodynamic Data for Inorganic Compounds
Chap. 4 Thermochemistry
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Table 4.1a
see Appendix B
Chap. 4 Thermochemistry
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