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Chap 2 First Law of Thermodynamics
Table of Contents 目錄內容
2.11 Thermochemistry 熱化學
2.12 Enthalpy of Formation 生成熱焓
2.13 Calorimetry 卡計
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Chap 2 First Law of Thermodynamics
2.11 Thermochemistry :
Thermochemistry is a branch of thermodynamics which
deals with heats of reaction: energy evolved or
absorbed as heat when a chemical reaction or phase
change takes place. The study of the heat produced by
or required for a chemical reaction is called the
thermochemistry.
(熱化學是熱力學的一支,在化學反應或相變化中發生的能量吸
收或釋放,若以熱的形式表現.即為熱化學研究的對象.)
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The First Law: Thermochemistry
• 絕熱過程中溫度的改變用來量測反應熱的吸收或釋放
The heat released or required can be measured by the
temperature change in an adiabatic process.
• 若孤立系統內的反應發生造成系統昇溫. 產生的熱釋放到環境中,
q 為負值,就稱為放熱反應; 反之 q 為正值,就稱為吸熱反應.
If the temperature rises when a reaction occurs in an
isolated system, heat must flow to the surroundings,
such a reaction is called exothermic reaction, and the
heat q is negative. If the temperature falls when a
reaction occurs in an isolated system, heat must flow
from the surroundings to the system to restore the
system to its initial temperature, such a reaction is
called endothermic reaction, and the heat q is positive.
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The First Law: Thermochemistry
• 吸熱放熱反應中的反應熱，是系統與環境的熱交換，以卡計來
量測。定容或定壓過程中吸收或釋放的反應熱 q ，用內能U 或
熱焓 H 的改變表示。
We use calorimetry to measure the heat produced or
absorbed by a reaction, and can identify q with a change
in internal energy (constant volume) or a change in
enthalpy (constant pressure).
• 反之，若已知反應的內能ΔU 或熱焓ΔH ，則可預測反應熱。
Conversely, if we know the ΔU or ΔH for a reaction, we
can predict the heat the reaction can produce.
• 定壓下的放熱反應即表示反應熱焓的降低，屬於負的熱焓。
An exothermic process at constant pressure is for which
ΔH < 0.
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The First Law: Thermochemistry
• 熱焓的改變值是指標準狀態下發生的變化過程。標準熱焓變化
(ΔH ɵ )是在特定溫度下, 標準狀態起始與最終態的熱焓改變。
Standard enthalpy changes are the change in enthalpy for
a process (chemical or physical) where the initial and final
substances are in their standard states at the same
specific temperature.
• 物質的標準狀態是以純物質在1 bar 壓力下的狀態. 標準壓力為
1 bar, 以 p ɵ 表示。
The standard state of a substance at a specified
temperature is its pure form at a pressure of 1 bar. The
standard pressure of 1 bar is denoted p ɵ.
• 反向過程的熱焓變化是以正向過程的熱焓改變大小加上負號
The enthalpy change for a reverse process is of opposite
sign but equal magnitude to that for the forward process.
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The First Law: Thermochemistry
• d U = δq – P dV if PV-work only and Pex = P (isobaric
process).
• dH = dU + P dV + V dP = δq – P dV + P dV + V dP
= δq + V d P
= δq p
(constant P, T, dP =0)
• Now δqp is also a state function as dH.
• dH in terms of the partial molar enthalpy ( H i ):
N
dH = ∑H i dni = δqP
i =1
N : the number of species
H i : the molar enthalpy of species i
dni : the molar change of species i
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The First Law: Thermochemistry
• The stoichiometric number νi is a coefficient that occurs in
a balanced chemical equation. (計量數νi 是平衡反應式的係數)
• With the convention that:
νi > 0 for products and
νi < 0 for reactants,
• a chemical equation has the form: 0 = ∑ν i ⋅ Bi
i
• Example: the equation:
H2 (g) + ½ O2 (g) = H2O (l)
as written in the form 0 = -H2 (g) - ½ O2 (g) + H2O (l)
• Therefore the stoichiometric number are :
ν(H2)= -1, ν(O2)= -½ , ν(H2O)= +1
• The notation make it easy to write a general thermodynamic
equation for the expression of enthalpy of reaction, and the
equilibrium constant.
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The First Law: Thermochemistry
• The extent of reaction (ξ)用來指示反應進行的程度. 一般而言,
當反應進行程度ξ的變化以微分量 dξ表示時, 物種 i 的改變量為
dni =νi dξ, 其中νi 為反應式中 i 物種的計量數•
• 進行中的反應,在任一時間物種 i 的莫耳量以ξ的積分式表示為
ni = nio + νi ξ
在此 the extent of reaction ξ 以莫耳為單位
nio : 為species i 存在的反應的起始量
ni : 為species i 在反應中途的存在量
• 在任何反應開始前, extent of reaction 初始值為零,ξ=0. 若反應
式中, 限量反應物 (limiting reactant) L 的初始量為 nL, 則當限量反
應物被消耗完時, extent of reaction 為ξ=nL.
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The First Law: Thermochemistry
0 = ∑ν i ⋅ Bi
• 可得 dni = νi dξ 且
N
i
dH = δqP = ∑ν i H i dξ
i =1
 ∂H 
δqP N
∆
H
= 
= ∑ν i H i
 =
r
• 除以 dξ 可得
 ∂ξ T ,P dξ i =1
• 系統熱焓對反應程度ξ 的微分,即為反應熱焓值 ∆rH. 對系統大量的
改變, 可寫為:
∆H
= ∆r H
∆ξ
• 若以一莫耳的反應物發生反應, Δξ = 1 莫耳，且
ΔＨ =（ 1 莫耳）ｘΔｒＨ
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The First Law: Thermochemistry
• The enthalpy change for a
reverse process is of
opposite sign but equal
magnitude to that for the
forward process
 ∂H
∆r H = 
 ∂ξ
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
δqP N
= ∑ν i H i
 =
T ,P dξ i =1
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The First Law: Thermochemistry
• 標準反應熱焓是以特定溫度下,每莫耳標準狀態的純粹起始物
反應變成標準狀態的純粹最終物的熱焓改變。
The standard enthalpy of reaction (ΔrHo ) is the change
in enthalpy per mole that accompanies the conversion
of pure, separated reactants in their standard states
into pure, separated products in their standard states at
the temperature specified.
• 實際上,標準反應熱焓可表示為產物與反應物的標準熱焓差
In practical, the standard enthalpy of reaction, (ΔrHº )
is expressed as the difference of the standard
enthalpies of the products and reactions.
N
∆r H ° = ∑ν i ⋅ H i °
i =1
for reaction: 0 = ∑ν i ⋅ Bi
i
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The First Law: Thermochemistry
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The First Law: Thermochemistry
• For example, in the reaction:
H2 (g) + ½ O2 (g) = H2O
(l)
The standard enthalpy of reaction is
ΔrHº = Hm(H2O,l) – {Hm(H2,g) + ½ Hm(O2,g) }
Note that the use of molar standard enthalpies. So by
implication the enthalpy of reaction is also a molar quantity.
• 熱化學方程式是由化學式加上標準反應熱焓組成
A thermochemical equation is the combination of a
chemical equation and its standard enthalpy.
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The First Law: Thermochemistry
• 在化學熱力學中的標準狀態是由下列定義組成:
1. The standard state of a pure gases substance, denoted by
g, at a given temperature is the ideal gas at 1 bar pressure.
2. The standard state of a pure liquid substance, denoted by l,
at a given temperature is the pure liquid at 1 bar pressure.
3. The standard state of a pure crystalline substance at a
given temperature is the pure crystalline substance,
denoted by s, at 1 bar pressure.
4. The standard state of a substance in solution is the
hypothetical 1 of the substance ideal solution of standard
state molality (1 mol kg-1) at 1 bar pressure, at each given
temperature.
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The First Law: Thermochemistry
5. To indicate the standard state of an electrolyte, the NBS
Tables of Chemical Thermodynamic Properties (1982) use
two symbol. The thermodynamic property of completely
dissociated molecules in water are designated by ai.
6. The thermodynamic property of undissociated molecules in
water are designated by ao.
7. The thermodynamic property of ions in water are also
designated by ao to indicate that no further ionization
occurs.
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The First Law: Thermochemistry
• For example, it is impractical to measure the heat evolved
when carbon burns to carbon monoxide in a limited amount
of oxygen. However, one can measure the heat of reaction
for carbon to completely burn to carbon dioxide in an excess
of oxygen. And heat evolved can be readily measured when
carbon monoxide burns to carbon dioxide also.
• For graphite burns to CO at 25 ℃
C(graphite) + O2 (g) = CO2(g) ΔrHº = -393.509 kJ/mol
CO2(g) = CO(g) + ½ O2 (g) ΔrHº = 282.984 kJ/mol
--------------------------------------------C(graphite) + ½ O2 (g) = CO(g) ΔrHº = -110.525 kJ/mol
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The First Law: Thermochemistry
Fig.2.18 Enthalpy level
diagram for the
system C(s) + O2(g).
The differences in
level are standard
enthalpy changes at
25℃ and 1 bar.
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The First Law: Thermochemistry
• The diagram also shows the enthalpy changes that are
involved in vaporizing graphite to atoms and dissociating
oxygen into atoms at 25 ℃
C(graphite) = C(g)
ΔrHº = 716.682 kJ/mol
O2(g) = 2O(g)
ΔrHº = 498.340 kJ/mol
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The First Law: Thermochemistry
2.12 Enthalpy of Formation
• 標準形成熱焓是特定溫度下來自物質各元素在其參考狀態下的合成反
應產生的相對熱焓. The standard enthalpy of formation of a
substance J, ΔfH º (J), at a given temperature is the standard
enthalpy of reaction accompanying the formation of the
substance from its elements in their reference states.
• 依此定義,所有在參考狀態下的元素,在任何溫度下其標準形成熱焓皆
為零 By definition, the standard enthalpies of formation of
elements in their reference states are ZERO at all
temperatures.
• 參考狀態是元素在1 bar壓力下,特定溫度時, 的熱力學最穩定之形態
Reference states is the thermodynamically most stable state
of the element (its state of lowest energy) at the stated
temperature.
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The First Law: Thermochemistry
• Allotrope: (同素異性物)
O2 (g) = O3
ΔfHº = 143 kJ/mol
(g)
C (graphite) = C (diamond)
ΔfHº =
S8 (rhombic) = S8 (monoclinic) ΔfHº =
1.9 kJ/mol
2 kJ/mol
P (red) = P4 (white)
ΔfHº = 69.8 kJ/mol
Sn (white) = Sn (gray)
ΔfHº =
3 kJ/mol
• In practical, the standard enthalpy of reaction, (ΔrHº )
is expressed as the difference of the standard
enthalpies of formation of the products and reactions.
∆r H ° = ∑ν J ⋅ ∆f H °( J ) for reaction: 0 = ∑ν i ⋅ Bi
J
i
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The First Law: Thermochemistry
• Appendix C.2 for NBS Table of Chemical Thermodynamic
Properties (1982) at 298.15 K and 1 bar.
• Appendix C.3 gives enthalpies of formation under 1 bar
from 0 to 3000 K for substances from JANAF (Joint ArmyNavy-Air Force) Thermochemical Tables (1985)
• Come from 4 sources:
1. Calorimetrically measured enthalpies of reaction, fusion,
vaporization, sublimation, transition, solution, and
dilution;
2. Temperature variation of equilibrium constants.;
3. Spectroscopic ally determined dissociation energies;
4. Calculation from Gibbs energies and entropies.
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The First Law: Thermochemistry
• Example 2.8 Calculating the standard enthalpy of
reaction at constant temperature.
What are the standard enthalpy changes at 298.15 K
and 2000 K for the following reaction ?
CO2(g) + C(graphite) = 2CO(g)
• Answer: Using Appendix C.3 at 298.15 K,
∆r H ° = ∑ν J ⋅ ∆f H J ° = 2∆f H °(CO) − ∆f H °(CO2 )
J
= 2(-110.527 kJ mol-1) – (-393.522 kJ mol-1)
= 172.468 kJ mol-1
• at 2000 K,
ΔrH ̊ = 2(-118.896 kJ mol-1) – (-396.784 kJ mol-1)
= 158.992 kJ mol-1
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The First Law: Thermochemistry
Hess law: 黑斯定律
• 反應熱焓與反應發生的實際路徑無關. The standard
enthalpy of reaction, (ΔrHº ) is independent of the path
by which the reaction actually take place.
• 反應熱焓可表為各別(假想)步驟的標準熱焓之總和 Thus, we
may express the standard enthalpy of a reaction as the
sum of the standard enthalpies of the (possible
hypothetical) reactions into which the overall reactions
may be regarded as divided.
• 反應熱焓可表為各別步驟的標準熱焓的總和。
黑斯定律中各別步驟的總和必須用相同溫度的標準熱焓. All
the standard enthalpies of reaction used in a given
application of Hess’s law must refer to the same
temperature.
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The First Law of Thermodynamics
反應熱焓(ΔrHº )
• The standard enthalpy of reaction, (ΔrHº ) can be any
temperature other than at 298.15 K. Need both heat
capacities data of reactants and products to find the
standard enthalpy of reaction at some other
temperatures.
T
298
ΔrHTº = ∫T C °P,react dT + ∆r H 298 ° + ∫298 C °P,prod dT
Reactants -------→ Products
|
↑
| 298
| T
| ∫ C P,react dT
| ∫ C P ,prod dT
298
T
↓
|
Reactants -------→ Products
ΔrH298º
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The First Law: 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.
T
298
ΔrHTº = ∫ C °P,react dT + ∆r H 298 ° + ∫ C °P,prod dT
T
298
(
C °P,prod − C °P,react )dT
298
T
∆r HT ° = ∆r H 298 ° + ∫
T
= ∆r H 298 ° + ∫
298
∆C °P,prod dT
• Kirchhoff’s law simplifies to
ΔrHº (T2) = ΔrHº (T1) + (T2 - T1)ΔrCpº
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First Law of Thermodynamics
• Change in Reaction Enthalpy
When the temperature is
increased, the enthalpies of
the products and the reactants
both increase, but may do so
to different extents. In each
case, the change in enthalpy
depends on the heat capacities
of the substances. The change
in reaction enthalpy reflects
the difference in the changes
of the enthalpies. An
illustration of the content of
Kirchhoff's law.
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The First Law: Thermochemistry
The temperature dependence of reaction enthalpies
• The temperature variation of the ΔrHº is expressed
by Kirchhoff’s law:
T2
º
ΔrH (T2) = ΔrHº (T1) + ∫ ∆r C p °(T )dT
T1
∆r C p °(T ) = ∑ν J ⋅C P °( J ,T
J
)
• Where ∆r C P °( J ,T ) is the molar reaction heat capacity
at constant pressure of the substance J in its standard
 ∂H °( J ,T ) 
state at temperature T.

C p °( J ,T ) = 

T
∂
P =P °

• The law is a direct consequence of the heat capacity
at constant pressure.
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The First Law: Thermochemistry
• The temperature variation of the ΔrCPº is expressed by
power series in T:
ΔrCP º(T) = Δrα + (Δrβ) T + (Δrγ) T2
• Where ∆r α = ∑ν J ⋅α J and so on.
J
[∆ α + (∆ β )T + (∆ γ )T ]dT
∆β
∆γ
(
(
T − 298 ) +
° + ∆ α (T − 298 ) +
T
2
3
T
∆r HT ° = ∆r H 298 ° + ∫
298
= ∆r H 298
r
r
r
r
r
 ∆r β
∆r HT ° = ∆r H 0 ° + (∆r α )T + 
 2
2
2
2
r
3
− 298 3
)
 2  ∆rγ  3
T + 
T
 3 

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The First Law of Thermodynamics
Example Calculation of the amount of graphite to produce a
mole of hydrogen at constant temperature.
The reaction of heated coal with superheated steam absorbs
heat. This heat is usually provided by burning some of the
coal. Calculate ΔrH º(500 K) for both reactions?
a. C(graphite) + H2O(g) = CO(g) + H2(g)
b. C(graphite) + O2(g) = CO2(g)
How many of carbon are required to produce 1 mol of
hydrogen if all the heat is produced by reaction (b) ?
Answer: Using Appendix C.3 at 500 K,
(a) ΔrH º(500 K) = -110.003-(-243.826)=133.823 kJ mol-1
(b) ΔrH º(500 K) = -393.666 kJ mol-1 using Appendix C.3
Thus 133.823/393.666=0.340 mol of graphite is needed for
(b) and (1.0+0.340) mol of graphite for (a) + (b)
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The First Law: Thermochemistry
• Some tables give value of H T ° − H 298 ° to assist the in the
calculation of the ΔrHTº for chemical reaction or phase
transition :
T
H T ° − H 298 ° = ∫ C P °dT
298K
• Depending on the table, ΔtrHTrº for phase transition in the
intervening temperature range may be added to the righthand side of the equation:
Ttr
Ttr
298K
298K
H T ° − H 298 ° = ∫ C P °dT + ∆tr H Ttr ° + ∫ C P °dT
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Example 2.9 Calculation of the bond energy of molecular
hydrogen from the enthalpy of formation of hydrogen atoms
at 298 K.
What is the value of ΔrH º at 0 K for the following reaction?
H2(g) = 2H(g)
Answer: The calculation using ΔrH º (298 K) illustrates the use of
H 0 ° − H 298 °from Appendix C.3
ΔrH º(298 K) = 2(217.999 kJ/mol) = 435.998 kJ mol-1
H2(g) -------Æ 2H(g) ΔrH298º=435.998 kJ/mol
⇡
298 K
|
|
H 298 ° − H 0 ° =| 8.467 kJ / mol
H
0 ° − H 298 ° = −2x 6.197 kJ / mol
|
|
|
0K
⇣
H2(g) -------Æ 2H(g)
ΔrHº(0 K) = (8.467+435.998-12.394) kJ/mol=432.071 kJ/mol
From ΔHf of H(g) at 0 K
ΔrHº(0 K) = 2x(216.035) = 432.070 kJ/mol
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The First Law: Thermochemistry
• Mean Bond Enthalpy (平均鍵熱焓)
平均鍵熱焓是一系列相關化合物的鍵解離熱焓平均值. A mean
bond enthalpy, B(A-B) of an A-B bond is the mean value
of the bond dissociation enthalpy ΔHº (A-B ) average
over a series of related compounds.
Exercise Given the following averaged bond enthalpies:
H(C–Cl) = 339 kJ mol-1
H(C–H) = 414 kJ mol-1
H(Cl–Cl) = 243 kJ mol-1
H(H–Cl) = 432 kJ mol-1
How much reaction enthalpy would be released in the
reaction:
CH4 (g) + 2 Cl2 (g) ---> CH2Cl2 (g) + 2 HCl (g) ?
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2.15 Calorimetry
Heats of reaction using adiabatic calorimeter ΔHA=0
R(T2) + Cal(T2) ----------→ P(T2) + Cal(T2)
⇡
ΔrH (T2) ↗ ⇡
|
╱
|
|
╱
|
ΔHR | ΔHA=0 ╱
| ΔHP
|
╱
|
|
╱
|
|
╱
|
|╱
ΔrH (T1)
|
R(T1) + Cal(T1) ----------→ P(T1) + Cal(T1)
ΔHA = 0 = ΔrH (T1) + ΔHP = ΔrH (T2) + ΔHR
ΔrH (T2) + ΔHR = ΔrH (T1) +ΔHP
ΔrH (T2) = ΔrH (T1) + (ΔHP – ΔHR )
ΔrH (T2) = ΔrH (T1) + (T2 - T1) [Σ ɣJ Cp (BJ )]
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The First Law of Thermodynamics
Fig.2.19 Adiabatic
calorimeter operated
at constant pressure.
A reaction between
solutions A and B is
initiated by rotating
the reaction vessel
around the axis
indicated.
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• For adiabatic calorimeter, ΔHA=0 , since the heat capacities
of the reactants, products, and calorimeter may be assumed
constant over the range T1 to T2. and equations become
ΔrH (T1) = - ΔHP = - (T2 - T1) [Cp (P ) +Cp (Cal )]
ΔrH (T2) = - ΔHR = - (T2 - T1) [Cp (R ) +Cp (Cal )]
R(T2) + Cal(T2) ----------→ P(T2) + Cal(T2)
⇡
ΔrH (T2) ↗ ⇡
|
╱
|
|
╱
|
ΔHR | ΔHA=0 ╱
| ΔHP
|
╱
|
|
╱
|
|
╱
|
|╱
ΔrH (T1)
|
R(T1) + Cal(T1) ----------→ P(T1) + Cal(T1)
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The First Law of Thermodynamics
Fig.2.20 Adiabatic bomb
calorimeter for carrying
out combustions at
constant volume.
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The First Law of Thermodynamics
Bomb Calorimeter
A constant volume bomb
calorimeter. The bomb is the
central vessel, which is massive
enough to withstand high
pressures. The calorimeter (for
which the heat capacity must be
known) is the entire assembly
shown here. To ensure
adiabaticity, the calorimeter is
immersed in a water bath with a
temperature continuously
readjusted to that of the
calorimeter at each stage of the
combustion, The heater is for the
CP(R) measurement before firing.
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The First Law of Thermodynamics
Flame Calorimeter
A constant-pressure flame
calorimeter consists of this
element immersed in a stirred
water bath. Combustion
occurs as a known amount of
reactant is passed through to
fuel the flame, and the rise of
temperature is monitored.
The difference of heat
released is related by the
equation:
∆rH = ∆rU + RT ∑νg
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The First Law: Thermochemistry
• 標準蒸發熱焓是標準壓力下每莫耳純液體蒸發為標準壓力下氣體的熱
焓變化. The standard enthalpy of vaporation (ΔvapHº ) is the
enthalpy change per mole when a pure liquid at 1 bar
vaporized to a gas at 1 bar.
• 標準轉化熱焓是標準壓力下每莫耳物質發生物理變化的熱焓變
化.The standard enthalpy of transition (ΔtrsHº ) is the
enthalpy change that accompanies a change of physical state.
• 標準燃燒熱焓是1 bar標準壓力下每莫耳有機物質完全氧化反應的標
準反應熱焓變化.對某些物質,如 H2O(l), CO2(g), SO2(g), N2(g),人為
規定其標準燃燒熱為零. The standard enthalpy of combustion,
(ΔtrsHº ) is the standard reaction enthalpy for the complete
oxidation of an organic compound.
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The First Law: Thermochemistry
• The standard enthalpy of combustion ΔcHº of glucose
to CO2 gas and liquid H2O is –2808 kJ/mol
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The First Law of Thermodynamics
Example 2.10 Calculating the enthalpy of reactions from the
heat of combustion
The combustion of ethanol (C2H5OH(l)) in a constant-volume
calorimeter produces 1364.34 kJ mol-1 at 25 ºC. What is the
value of ΔrH º for the following combustion reaction ?
C2H5OH(l) + 3O2(g) = 2CO2(g) + 3H2O(l)
Answer: CalculateΔrH º (298 K) from ∆rH = ∆rU + RT ∑νg
= -1364.34 kJ mol-1+(8.3145x10-3 kJ K-1mol-1)x(298.15 K)x(-1)
= -1364.34 kJ mol-1 + (-2.4788 kJ mol-1)
= -1366.82 kJ mol-1
This is the quantity of heat that would be evolved at 25 ºC and
a constant pressure of 1 bar.
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The First Law of Thermodynamics
Example 2.11 Calculating the molar internal energy of
combustion
In an adiabatic bomb calorimeter, the combustion of
0.5173 g of ethanol causes the temperature to rise from
25.0 to 29.289 ºC. The heat capacity of the bomb, the
reactants and the other contents of the calorimeter is 3576
J K-1. What is the molar internal energy of combustion of
ethanol at 25.0 oC ?
Answer: The change of state can be written
{C2H5OH(l) + 3O2(g) + other contents, T = 25.0 oC. V } =
{2CO2(g) + 3H2O(l) + other contents, T = 29.289 oC, V }
for which q = 0 (adiabatic) and w = 0 (constant volume V)
so that ΔUA = 0.
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The First Law of Thermodynamics
Example 2.11 (continue) Calculating the molar internal
energy of combustion
Answer: This change in state can be written as the sum of
(1)
{C2H5OH(l) + 3O2(g) + other contents, T = 25.0 ºC. V}
|| ΔU1 = ΔrU
{2CO2(g) + 3H2O(l) + other contents, T = 25.0 ºC, V }
(2)
{2CO2(g) + 3H2O(l) + other contents, T = 25.0 ºC, V }
|| ΔU2 = ΔUP
{2CO2(g) + 3H2O(l) + other contents, T = 29.289 ºC, V }
where ΔU1 + ΔU2 = ΔUA =0 so that ΔU1 = -ΔU2
ΔrU (25.0 ºC) = - ΔUP = - ΔT [CV (P ) +CV (Cal )]
ΔrU = -(3576 J K-1) (4.289 K) (46.0 g mol-1) / 0.5173 g
= -1364 J mol-1
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The First Law of Thermodynamics
DTA:
Differential
Thermal
Analyzer
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The First Law of Thermodynamics
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The First Law of Thermodynamics
積分溶解熱 (Integral Heat of Solution)
• 定量物質溶於定量溶劑中所產生的熱效應稱為該物質的溶解熱。
• 一般而言,溶解熱與溶液的最終濃度有關。
• 積分溶解熱是一莫耳溶質溶於N莫耳溶劑中所產生的熱焓改變效應.
當溶劑增加時,積分溶解熱ΔsolHº趨近一個極限值. 在298 K時, 由C.2
表可得:
½ H2(g) + ½ Cl2(g) = HCl(g)
ΔfHº(HCl(g)) = -92.307 kJ mol-1
HCl(g)
5H2O ΔsolHº = -63.467 kJ mol-1
+ 5H2O
HCl(g) + 100H2O
HCl(g) + 200H2O
HCl(g)
+ ∞H2O
(l)
= HCl in
(l)
= HCl in 100H2O ΔsolHº = -73.618 kJ mol-1
(l)
= HCl in 200H2O ΔsolHº = -73.965 kJ mol-1
(l)
= HCl in ∞H2O
ΔsolHº = -74.852 kJ mol-1
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The First Law of Thermodynamics
積分溶解熱 (Integral Heat of Solution)
• 溶質在溶液中的標準狀態, 是在理論上,選定一個標準重量濃度為1 m
的理想溶液在定溫下,壓力為1 bar 的溶質.
• 電解質在溶液中的標準狀態, 是選定一個在水中完全解離的電解質為
ai, (means completely ionized) . 而選定一個在水中未解離的電解
質分子為 ao, (means not dissociated). 而在水中完全水合的離子
也是 (ao), 表示已不發生進一步的解離. 在NBS表中所依照的傳統約
定為: A(ao) + n H2O(l) = A·(H2O)n(ao) ΔrGº=ΔrHº=ΔrSº=0.
理由是在水中的稀薄溶液,無法決定水的濃度,故其平衡常數K值未定.
• 例如, extent of hydration of CO2(ao) + H2O(ℓ) = H2CO3(ao) 通
常是未知的,但在中性的pH 環境中,水解過程較慢, 可決定平衡常數
Kh = H2CO3(ao)/ CO2(ao) =2.6x10-3
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The First Law of Thermodynamics
• 醋酸液體溶於水中形成的水溶液，其中未姐離的醋酸分子維持在標準
狀態下:
CH3CO2H(l) = CH3CO2H(ao) ΔsolHº (298 K) = -1.3 kJ mol-1
• ao 在此表示為CH3CO2H(ao) 不再進一步解離為離子的狀態。
• 溶質溶於溶劑時,兩者的化學相似度高，且無離子化或水和的作用
時，溶解熱近乎溶質的溶化熱 (Fusion Energy)。
• 溶質溶於溶劑中，與溶劑分子的強作用力，所產生的熱效應稱為該物
質的水解熱 (Hydration Energy)。
• 一般而言，水解熱與溶劑的介電常數(εo)， 和溶劑分子電偶距有
關。
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The First Law of Thermodynamics
Heat of Neutralization
HCl in 100H2O + NaOH in 100H2O = NaCl in 200H2O + H2O(l)
ΔneutralHº (298 K) = -406.923 -285.830 +165.925 +469.646
= -57.182 kJ mol-1
• For dilute solutions, the heat of reaction of strong base with
strong acids is independent of the nature of acid or base,
due to the complete ionization of strong acids and base and
the salts formed by neutralization:
OH- (ao) + H+ (ao) = H2O(l) ΔrHº (298 K) = -55.835 kJ mol-1
• The heat of neutralization for a dilute solution of a weak
base or acid is some what less because of the absorption of
heat in the dissociation of the weak acid or base,
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The First Law of Thermodynamics
Heat of Neutralization
• For dilute solutions of strong electrolytes, the ions are
independent of the accompanying ions, it is convenient to
use the enthalpies of formation of individual ions.
• The sum of the enthalpies of formation of ions may be
calculated from
H2O(l) = OH- (ao) + H+ (ao) ΔrHº (298 K) = 55.835 kJ mol-1
H2(g) + ½ O2(g) = H2O (l) ΔrHº (298 K) = -285.830 kJ mol-1
________________________________________________________
H2(g) + ½ O2(g) = H+(ao) + OH-(ao) ΔrHº = -229.995 kJ mol-1
The separate enthalpies of formation of H+(ao)) or OH-(ao)),
∆fHº (H+(ao)) or ∆fHº (OH-(ao)), cannot be calculated,
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The First Law of Thermodynamics
• 定溫下由標準狀態的穩定單純物質1 mol 溶於極大量的水生成B的離
子時的焓變化稱該離子B的標準生成焓，符號為△fHº (B (aq))
• 對 H+(ao) 的標準生成焓△fHº (H+(ao)) 約定其為零。因符合△H=qp
的條件，故又叫離子的標準生成熱.
• From equation ½ H2(g) = H+(ao) + e- ΔrHº = 0, where e- is a
formal electron not dissolved in water. So △fHº (e-) =0
• The enthalpy of formation of OH- is given by:
½H2(g) + ½O2(g) + e- = OH- (ao) ΔrHº = -229.995 kJ mol-1
• Based on these values for enthalpies of formation of H+ and
OH-, other ions of strong electrolytes may be calculated.
• From ΔrHº (HCl(ai)) it is possible to calculate the ΔrHº (Cl-(ao))
½H2(g) + ½Cl2(g) = H+ (ao) + Cl- (ao) ΔrHº = -167.159 kJ mol-1
½Cl2(g) + e- = Cl- (ao) ΔrHº = -167.159 kJ mol-1
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