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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/s 2)
(1 calorie = 4.184 J)
•
Potential Energy: is stored energy.
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Chapter 08
Thermodynamics
•
Slide 2
02
Thermal Energy is the kinetic energy of molecular
motion (translational, rotational, and vibrational) .
•
Thermal energy is proportional to the temperature
in degrees Kelvin. Ethermal α T(K)
•
Heat is the amount of thermal energy transferred
between two objects at different temperatures.
Prentice Hall ©2004
Chapter 08
Slide 3
1
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).
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Chapter 08
Thermodynamics
Slide 4
04
•
Closed System: Only energy can be lost or gained.
•
Isolated System: No matter or energy is exchanged.
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Chapter 08
Thermodynamics
Slide 5
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.
∆E = Efinal – Einitial = E2 – E1 = q + w
q = heat, w = work
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Chapter 08
Slide 6
2
Thermodynamics
•
Pressure is the force per unit area.
Pressure =
•
06
Force F
=
Area A
(1 N/m2 = 1 Pa)
(1 atm = 101,325 Pa)
Work is a force (F) that produces an object’s
movement, times the distance moved (d):
Work = Force x Distance
Prentice Hall ©2004
Chapter 08
Slide 7
Thermodynamics
07
w = –P∆V
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Prentice Hall ©2004
Chapter 08
Thermodynamics
•
Slide 8
08
How much work is done (in kilojoules), and in which
direction, as a result of the following reaction?
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Chapter 08
Slide 9
3
Thermodynamics
•
09
The amount of heat exchanged between the
system and the surroundings is given the symbol q.
q = ∆E + P∆V
At constant volume (∆V = 0): qv = ∆E
At constant pressure: qp = ∆E + P∆V = ∆H
Enthalpy change: ∆H = Hproducts – Hreactants
Prentice Hall ©2004
Chapter 08
Thermodynamics
•
Slide 10
10
The following reaction has ∆ E = –186 kJ/mol.
•
•
Is the sign of P∆V positive or negative?
What is the sign and approximate magnitude of ∆H?
Prentice Hall ©2004
Chapter 08
Thermodynamics
Slide 11
11
•
The reaction between hydrogen and oxygen to yield water
vapor has ∆H° = –484 kJ. How much PV work is done, and
what is the value of ∆E (in kilojoules) for the reaction of 0.50
mol of H2 with 0.25 mol of O 2 at atmospheric pressure if the
volume change is –5.6 L?
•
The explosion of 2.00 mol of solid TNT with a volume of
approximately 0.274 L produces gases with a volume of 489
L at room temperature. How much PV (in kilojoules) work is
done during the explosion? Assume P = 1 atm, T = 25°C.
2 C7H5N3O6(s) → 12 CO(g) + 5 H2(g) + 3 N2(g) + 2 C(s)
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Chapter 08
Slide 12
4
State Functions
•
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.
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Chapter 08
State Functions
•
01
Slide 13
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.
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Chapter 08
Enthalpy Changes
•
Slide 14
01
Enthalpies of Physical Change:
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Chapter 08
Slide 15
5
Enthalpy Changes
•
02
Enthalpies of Chemical Change: Often called
heats of reaction (∆Hreaction) .
Endothermic: Heat flows into the system from the
surroundings and ∆H has a positive sign.
Exothermic: Heat flows out of the system into the
surroundings and ∆H has a negative sign.
Prentice Hall ©2004
Chapter 08
Enthalpy Changes
•
Slide 16
03
Reversing a reaction changes the sign of ∆H for a
reaction.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) ∆H = –2219 kJ
3 CO2(g) + 4 H2O(l) → C3H8(g) + 5 O 2(g) ∆H = +2219 kJ
•
Multiplying a reaction increases ∆ H by the same factor.
3 C3H8(g) + 15 O 2(g) → 9 CO2(g) + 12 H2O(l) ∆H = –6657 kJ
Prentice Hall ©2004
Chapter 08
Enthalpy Changes
•
Slide 17
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 O 2( g) → 3 CO2( g) + 4 H 2O(l)
∆H = –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)
∆H = +80.3 kJ
Prentice Hall ©2004
Chapter 08
Slide 18
6
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 ∆ H°.
Prentice Hall ©2004
Chapter 08
Slide 19
Hess’s Law
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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) ∆H° = –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)
∆H°1 = ?
(b)
N2H4(g) + H2(g)
2 NH3(g)
∆H°2 = –187.6 kJ
(c)
3 H2(g) + N2(g)
2 NH3(g)
∆H°3 = –92.2 kJ
∆H°1 = ∆H°3 – ∆H°2 = (–92.2 kJ) – (–187.6 kJ) = +95.4 kJ
Prentice Hall ©2004
Chapter 08
Slide 21
7
Hess’s Law
03
The industrial degreasing solvent methylene
chloride (CH2 Cl2, dichloromethane) is prepared
from methane by reaction with chlorine:
•
CH4(g) + 2 Cl 2(g)
CH 2Cl2(g) + 2 HCl( g)
Use the following data to calculate ∆H° (in kilojoules)
for the above reaction:
•
CH4(g) + Cl 2(g)
∆H° = –98.3 kJ
CH3Cl(g) + Cl 2(g)
∆H° = –104 kJ
CH3Cl(g) + HCl( g)
CH2Cl2(g) + HCl(g)
Prentice Hall ©2004
Chapter 08
Slide 22
Standard Heats of Formation
01
•
Standard Heats of Formation (∆ H°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.
•
∆ H°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
02
∆H°f = –286 kJ/mol
H2(g) + 1/2 N2(g) → NH3(g)
2 C(s) + H2(g) → C2H2(g)
∆H°f = –46 kJ/mol
∆H°f = +227 kJ/mol
2 C(s) + 3 H2(g) + 1/2 O2(g) → C2H5OH(g) ∆H°f = –235 kJ/mol
Prentice Hall ©2004
Chapter 08
Slide 24
8
Standard Heats of Formation
•
03
Calculating ∆ H° for a reaction:
∆H° = ∆H°f (Products) – ∆H°f (Reactants)
•
For a balanced equation, each heat of formation must
be multiplied by the stoichiometric coefficient.
aA + bB
cC + dD
∆H° = [c∆H°f (C) + d∆H°f (D)] – [a∆H°f (A) + b∆H°f (B)]
Prentice Hall ©2004
Chapter 08
Slide 25
Standard Heats of Formation
04
Some Heats of Formation, ∆ Hf ° (kJ/mol)
CO(g)
-111
C 2 H2 (g)
227
Ag+(aq)
106
-240
CO2 (g)
-394
C 2 H4 (g)
52
Na+(aq)
H2 O(l)
-286
C 2 H6 (g)
-85
NO3 -(aq)
-207
Cl-(aq)
-167
NH3 (g)
-46
CH3 OH(g)
-201
N2 H4 (g)
95.4
C 2 H5 OH(g)
-235
AgCl(s)
-127
HCl(g)
-92
C 6 H6 (l)
49
Na2 CO3 (s)
-1131
Prentice Hall ©2004
Chapter 08
Standard Heats of Formation
•
Slide 26
05
Calculate ∆ H° (in kilojoules) for the reaction of
ammonia with O2 to yield nitric oxide (NO) and
H2 O(g), a step in the Ostwald process for the
commercial production of nitric acid.
•
Calculate ∆ H° (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
9
Bond Dissociation Energy
•
01
Bond Dissociation Energy: Can be used to
determine an approximate value for ∆ H° f .
∆H = D (Bonds Broken) – D (Bonds Formed)
•
For the reaction between H2 and Cl2 to form HCl:
∆ H = D(H–Cl) – ? {D(H–H) + D(O=O)}
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Chapter 08
Bond Dissociation Energy
Prentice Hall ©2004
Chapter 08
Bond Dissociation Energy
•
Slide 28
02
Slide 29
03
Calculate an approximate ∆H° (in kilojoules) for the
synthesis of ethyl alcohol from ethylene:
C2 H4(g) + H2O(g) → C2 H5OH(g)
•
Calculate an approximate ∆H° (in kilojoules) for the
synthesis of hydrazine from ammonia:
2 NH3(g) + Cl2(g) → N2 H4(g) + 2 HCl(g)
Prentice Hall ©2004
Chapter 08
Slide 30
10
Calorimetry and Heat Capacity
•
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 = ∆E.
•
Constant Pressure Calorimetry: A constant pressure
calorimeter measures the heat change at constant
pressure such that q = ∆H.
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
•
01
03
Heat capacity (C) is the amount of heat required to
raise the temperature of an object or substance a
given amount.
C =
q
∆T
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
11
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 H2SO 4 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 ∆H for the
reaction.
Prentice Hall ©2004
Chapter 08
Calorimetry and Heat Capacity
Prentice Hall ©2004
Chapter 08
Introduction to Entropy
Slide 34
05
Slide 35
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
12
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).
∆S = Sfinal – Sinitial
Positive value of ∆S indicates increased disorder.
Negative value of ∆S indicates decreased disorder.
Prentice Hall ©2004
Chapter 08
Introduction to Entropy
Prentice Hall ©2004
Slide 37
03
Chapter 08
Introduction to Entropy
Slide 38
04
•
To decide whether a process is spontaneous, both
enthalpy and entropy changes must be considered:
•
Spontaneous process:
Decrease in enthalpy (–∆H).
Increase in entropy (+∆S).
•
Nonspontaneous process:
Increase in enthalpy (+∆H).
Decrease in entropy (–∆S).
Prentice Hall ©2004
Chapter 08
Slide 39
13
Introduction to Entropy
•
05
Predict whether ∆S° is likely to be positive or
negative for each of the following reactions. Using
tabulated values, calculate ∆ S° for each:
a. 2 CO(g) + O2(g) → 2 CO2(g)
b. 2 NaHCO3(s) → Na 2CO3(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
Introduction to Free Energy
•
Slide 40
01
Gibbs Free Energy Change (∆ G): Weighs the
relative contributions of enthalpy and entropy to the
overall spontaneity of a process.
∆G = ∆H – T∆S
∆G < 0
Process is spontaneous
∆G = 0
Process is at equilibrium
∆G > 0
Process is nonspontaneous
Prentice Hall ©2004
Chapter 08
Introduction to Free Energy
•
Slide 41
02
Situations leading to ∆G < 0:
∆H is negative and T∆S is positive
∆H is very negative and T∆S is slightly negative
∆H is slightly positive and T∆S is very positive
•
Situations leading to ∆G = 0:
∆H and T∆S are equally negative
∆H and T∆S are equally positive
•
Situations leading to ∆G > 0:
∆H is positive and T∆S is negative
∆H is slightly negative and T∆S is very negative
∆H is very positive and T∆S is slightly positive
Prentice Hall ©2004
Chapter 08
Slide 42
14
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)
∆G° = –55.7 kJ
b. 2 C(s) + 2 H2(g) → C2H4(g)
∆G° = 68.1 kJ
c. N2(g) + 3 H2(g) → 2 NH3(g)
∆H° = –92 kJ; ∆S° = –199 J/K
Prentice Hall ©2004
Chapter 08
Introduction to Free Energy
•
Slide 43
04
Equilibrium (∆ G° = 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)
∆H° = –92.0 kJ
∆S° = –199 J/K
Equilibrium is the point where ∆G° = ∆H° – T∆S° = 0
Prentice Hall ©2004
Chapter 08
Introduction to Free Energy
•
Slide 44
05
Benzene, C6 H6 , has an enthalpy of vaporization,
∆H vap , equal to 30.8 kJ/mol and boils at 80.1°C.
What is the entropy of vaporization, ∆ S vap, for
benzene?
Prentice Hall ©2004
Chapter 08
Slide 45
15