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Chapter
8
Thermochemistry:
Chemical Energy
Chemistry 4th Edition
McMurry/Fay
Dr. Paul Charlesworth
Michigan Technological University
Thermodynamics
•
01
Energy: is the capacity to do work, or supply heat.
Energy = Work + Heat
•
Kinetic Energy: is the energy of motion.
EK = 1/2 mv2
(1 Joule = 1 kgm2/s2)
(1 calorie = 4.184 J)
•
Potential Energy: is stored energy.
Prentice Hall ©2004
Chapter 08
Slide 2
Thermodynamics
•
03
In an experiment: Reactants and products are the
system; everything else is the surroundings.
•
Energy flow from the system to the surroundings
has a negative sign (loss of energy).
•
Energy flow from the surroundings to the system
has a positive sign (gain of energy).
Prentice Hall ©2004
Chapter 08
Slide 4
Thermodynamics
04
Closed System: Only energy can be lost or gained.
• Isolated System: No matter or energy is exchanged.
•
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Chapter 08
Slide 5
Thermodynamics
05
•
The law of the conservation of energy: Energy
cannot be created or destroyed.
•
The energy of an isolated system must be
constant.
•
The energy change in a system equals the work
done on the system + the heat added.
DE = Efinal – Einitial = E2 – E1 = q + w
q = heat, w = work
Prentice Hall ©2004
Chapter 08
Slide 6
State Functions
•
01
State Function: A function or property whose
value depends only on the present state (condition)
of the system.
•
The change in a state function is zero when the
system returns to its original condition.
•
For nonstate functions, the change is not zero if
the path returns to the original condition.
Prentice Hall ©2004
Chapter 08
Slide 13
State Functions
•
02
State and Nonstate Properties: The two paths
below give the same final state:
N2H4(g) + H2(g)  2 NH3(g) + heat (188 kJ)
N2(g) + 3 H2(g)  2 NH3(g) + heat (92 kJ)
•
State properties include temperature, total energy,
pressure, density, and [NH3].
•
Nonstate properties include the heat.
Prentice Hall ©2004
Chapter 08
Slide 14
Enthalpy Changes
•
01
Enthalpies of Physical Change:
Prentice Hall ©2004
Chapter 08
Slide 15
Enthalpy Changes
•
02
Enthalpies of Chemical Change: Often called
heats of reaction (DHreaction).
Endothermic: Heat flows into the system from the
surroundings and DH has a positive sign.
Exothermic: Heat flows out of the system into the
surroundings and DH has a negative sign.
Prentice Hall ©2004
Chapter 08
Slide 16
Enthalpy Changes
•
03
Reversing a reaction changes the sign of DH for a
reaction.
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l) DH = –2219 kJ
3 CO2(g) + 4 H2O(l)  C3H8(g) + 5 O2(g) DH = +2219 kJ
•
Multiplying a reaction increases DH by the same factor.
3 C3H8(g) + 15 O2(g)  9 CO2(g) + 12 H2O(l) DH = –6657 kJ
Prentice Hall ©2004
Chapter 08
Slide 17
Enthalpy Changes
•
04
How much heat (in kilojoules) is evolved or absorbed in
each of the following reactions?
•
Burning of 15.5 g of propane:
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
DH = –2219 kJ
•
Reaction of 4.88 g of barium hydroxide octahydrate with
ammonium chloride:
Ba(OH)2·8 H2O(s) + 2 NH4Cl(s)  BaCl2(aq) + 2 NH3(aq) + 10 H2O(l)
DH = +80.3 kJ
Prentice Hall ©2004
Chapter 08
Slide 18
Enthalpy Changes
05
•
Thermodynamic Standard State: Most stable
form of a substance at 1 atm pressure and 25°C;
1 M concentration for all substances in solution.
•
These are indicated by a superscript ° to the
symbol of the quantity reported.
•
Standard enthalpy change is indicated by the
symbol DH°.
Prentice Hall ©2004
Chapter 08
Slide 19
Hess’s Law
•
01
Hess’s Law: The overall enthalpy change for a
reaction is equal to the sum of the enthalpy
changes for the individual steps in the reaction.
3 H2(g) + N2(g)  2 NH3(g) DH° = –92.2 kJ
Prentice Hall ©2004
Chapter 08
Slide 20
Hess’s Law
•
02
Reactants and products in individual steps can be
added and subtracted to determine the overall
equation.
(a)
2 H2(g) + N2(g)
N2H4(g)
DH°1 = ?
(b)
N2H4(g) + H2(g)
2 NH3(g)
DH°2 = –187.6 kJ
(c)
3 H2(g) + N2(g)
2 NH3(g)
DH°3 = –92.2 kJ
DH°1 = DH°3 – DH°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ
Prentice Hall ©2004
Chapter 08
Slide 21
Hess’s Law
•
The industrial degreasing solvent methylene
chloride (CH2Cl2, dichloromethane) is prepared
from methane by reaction with chlorine:
CH4(g) + 2 Cl2(g)
•
03
CH2Cl2(g) + 2 HCl(g)
Use the following data to calculate DH° (in kilojoules)
for the above reaction:
CH4(g) + Cl2(g)
DH° = –98.3 kJ
CH3Cl(g) + Cl2(g)
DH° = –104 kJ
Prentice Hall ©2004
CH3Cl(g) + HCl(g)
CH2Cl2(g) + HCl(g)
Chapter 08
Slide 22
Standard Heats of Formation
01
•
Standard Heats of Formation (DH°f): The
enthalpy change for the formation of 1 mole of
substance in its standard state from its constituent
elements in their standard states.
•
The standard heat of formation for any element in
its standard state is defined as being ZERO.
•
DH°f = 0 for an element in its standard state
Prentice Hall ©2004
Chapter 08
Slide 23
Standard Heats of Formation
H2(g) + 1/2 O2(g)  H2O(l)
3/
2 H2(g)
02
DH°f = –286 kJ/mol
+ 1/2 N2(g)  NH3(g)
2 C(s) + H2(g)  C2H2(g)
DH°f = –46 kJ/mol
DH°f = +227 kJ/mol
2 C(s) + 3 H2(g) + 1/2 O2(g)  C2H5OH(g) DH°f = –235 kJ/mol
Prentice Hall ©2004
Chapter 08
Slide 24
Standard Heats of Formation
•
03
Calculating DH° for a reaction:
DH° = DH°f (Products) – DH°f (Reactants)
•
For a balanced equation, each heat of formation must
be multiplied by the stoichiometric coefficient.
aA + bB
cC + dD
DH° = [cDH°f (C) + dDH°f (D)] – [aDH°f (A) + bDH°f (B)]
Prentice Hall ©2004
Chapter 08
Slide 25
Standard Heats of Formation
04
Some Heats of Formation, DHf° (kJ/mol)
CO(g)
-111
C2H2(g)
227
Ag+(aq)
106
CO2(g)
-394
C2H4(g)
52
Na+(aq)
-240
H2O(l)
-286
C2H6(g)
-85
NO3-(aq)
-207
NH3(g)
-46
CH3OH(g)
-201
Cl-(aq)
-167
N2H4(g)
95.4
C2H5OH(g)
-235
AgCl(s)
-127
HCl(g)
-92
C6H6(l)
49
Na2CO3(s)
-1131
Prentice Hall ©2004
Chapter 08
Slide 26
Standard Heats of Formation
05
•
Calculate DH° (in kilojoules) for the reaction of
ammonia with O2 to yield nitric oxide (NO) and
H2O(g), a step in the Ostwald process for the
commercial production of nitric acid.
•
Calculate DH° (in kilojoules) for the photosynthesis
of glucose from CO2 and liquid water, a reaction
carried out by all green plants.
Prentice Hall ©2004
Chapter 08
Slide 27
Bond Dissociation Energy
•
01
Bond Dissociation Energy: Can be used to
determine an approximate value for DH°f .
DH = D (Bonds Broken) – D (Bonds Formed)
•
For the reaction between H2 and Cl2 to form HCl:
DH = D(H–Cl) – ∑ {D(H–H) + D(O=O)}
Prentice Hall ©2004
Chapter 08
Slide 28
Bond Dissociation Energy
Prentice Hall ©2004
Chapter 08
02
Slide 29
Bond Dissociation Energy
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03
Calculate an approximate DH° (in kilojoules) for the
synthesis of ethyl alcohol from ethylene:
C2H4(g) + H2O(g)  C2H5OH(g)
•
Calculate an approximate DH° (in kilojoules) for the
synthesis of hydrazine from ammonia:
2 NH3(g) + Cl2(g)  N2H4(g) + 2 HCl(g)
Prentice Hall ©2004
Chapter 08
Slide 30
Calorimetry and Heat Capacity
•
01
Calorimetry is the science of measuring heat
changes (q) for chemical reactions. There are two
types of calorimeters:
•
Bomb Calorimetry: A bomb calorimeter measures the
heat change at constant volume such that q = DE.
•
Constant Pressure Calorimetry: A constant pressure
calorimeter measures the heat change at constant
pressure such that q = DH.
Prentice Hall ©2004
Chapter 08
Slide 31
Calorimetry and Heat Capacity
Constant Pressure
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02
Bomb
Chapter 08
Slide 32
Calorimetry and Heat Capacity
•
03
Heat capacity (C) is the amount of heat required to
raise the temperature of an object or substance a
given amount.
q
C =
DT
Specific Heat: The amount of heat required to raise the
temperature of 1.00 g of substance by 1.00°C.
Molar Heat: The amount of heat required to raise the
temperature of 1.00 mole of substance by 1.00°C.
Prentice Hall ©2004
Chapter 08
Slide 33
Calorimetry and Heat Capacity
04
•
What is the specific heat of lead if it takes 96 J to
raise the temperature of a 75 g block by 10.0°C?
•
When 25.0 mL of 1.0 M H2SO4 is added to 50.0 mL
of 1.0 M NaOH at 25.0°C in a calorimeter, the
temperature of the solution increases to 33.9°C.
Assume specific heat of solution is 4.184 J/(g–1·°C–1),
and the density is 1.00 g/mL–1, calculate DH for the
reaction.
Prentice Hall ©2004
Chapter 08
Slide 34
Calorimetry and Heat Capacity
Prentice Hall ©2004
Chapter 08
05
Slide 35
Introduction to Entropy
01
•
Second Law of Thermodynamics: Reactions
proceed in the direction that increases the entropy
of the system plus surroundings.
•
A spontaneous process is one that proceeds on its
own without any continuous external influence.
•
A nonspontaneous process takes place only in the
presence of a continuous external influence.
Prentice Hall ©2004
Chapter 08
Slide 36
Introduction to Entropy
02
•
The measure of molecular disorder in a system is
called the system’s entropy; this is denoted S.
•
Entropy has units of J/K (Joules per Kelvin).
DS = Sfinal – Sinitial
Positive value of DS indicates increased disorder.
Negative value of DS indicates decreased disorder.
Prentice Hall ©2004
Chapter 08
Slide 37
Introduction to Entropy
Prentice Hall ©2004
Chapter 08
03
Slide 38
Introduction to Entropy
04
•
To decide whether a process is spontaneous, both
enthalpy and entropy changes must be considered:
•
Spontaneous process:
Decrease in enthalpy (–DH).
Increase in entropy (+DS).
•
Nonspontaneous process:
Increase in enthalpy (+DH).
Decrease in entropy (–DS).
Prentice Hall ©2004
Chapter 08
Slide 39
Introduction to Entropy
•
05
Predict whether DS° is likely to be positive or
negative for each of the following reactions. Using
tabulated values, calculate DS° for each:
a. 2 CO(g) + O2(g)  2 CO2(g)
b. 2 NaHCO3(s)  Na2CO3(s) + H2O(l) + CO2(g)
c. C2H4(g) + Br2(g)  CH2BrCH2Br(l)
d. 2 C2H6(g) + 7 O2(g)  4 CO2(g) + 6 H2O(g)
Prentice Hall ©2004
Chapter 08
Slide 40
Introduction to Free Energy
•
01
Gibbs Free Energy Change (DG): Weighs the
relative contributions of enthalpy and entropy to the
overall spontaneity of a process.
DG = DH – TDS
DG < 0
Process is spontaneous
DG = 0
Process is at equilibrium
DG > 0
Process is nonspontaneous
Prentice Hall ©2004
Chapter 08
Slide 41
Introduction to Free Energy
•
02
Situations leading to DG < 0:
DH is negative and TDS is positive
DH is very negative and TDS is slightly negative
DH is slightly positive and TDS is very positive
•
Situations leading to DG = 0:
DH and TDS are equally negative
DH and TDS are equally positive
•
Situations leading to DG > 0:
DH is positive and TDS is negative
DH is slightly negative and TDS is very negative
DH is very positive and TDS is slightly positive
Prentice Hall ©2004
Chapter 08
Slide 42
Introduction to Free Energy
•
03
Which of the following reactions are spontaneous
under standard conditions at 25°C?
a. AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
DG° = –55.7 kJ
b. 2 C(s) + 2 H2(g)  C2H4(g)
DG° = 68.1 kJ
c. N2(g) + 3 H2(g)  2 NH3(g)
DH° = –92 kJ; DS° = –199 J/K
Prentice Hall ©2004
Chapter 08
Slide 43
Introduction to Free Energy
•
04
Equilibrium (DG° = 0): Estimate the temperature at
which the following reaction will be at equilibrium.
Is the reaction spontaneous at room temperature?
N2(g) + 3 H2(g)  2 NH3(g)
DH° = –92.0 kJ
DS° = –199 J/K
Equilibrium is the point where DG° = DH° – TDS° = 0
Prentice Hall ©2004
Chapter 08
Slide 44
Introduction to Free Energy
•
05
Benzene, C6H6, has an enthalpy of vaporization,
DHvap, equal to 30.8 kJ/mol and boils at 80.1°C.
What is the entropy of vaporization, DSvap, for
benzene?
Prentice Hall ©2004
Chapter 08
Slide 45