George Mason University
General Chemistry 211
Chapter 6
Thermochemistry: Energy Flow and Chemical Change
Acknowledgements
Course Text: Chemistry: the Molecular Nature of Matter and
Change, 7th edition, 2011, McGraw-Hill
Martin S. Silberberg & Patricia Amateis
The Chemistry 211/212 General Chemistry courses taught at
George Mason are intended for those students enrolled in a science
/engineering oriented curricula, with particular emphasis on
chemistry, biochemistry, and biology The material on these slides is
taken primarily from the course text but the instructor has modified,
condensed, or otherwise reorganized selected material.
Additional material from other sources may also be included.
Interpretation of course material to clarify concepts and solutions to
problems is the sole responsibility of this instructor.
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1
Thermochemistry

Whenever matter changes composition, such as in a
chemical reaction, the energy content of the matter
changes also

In some reactions the energy that the reactants contain is
greater than the energy contained by the products
This excess energy is released as heat

In other reactions it is necessary to add energy (heat)
before the reaction can proceed
The energy contained by the products in these reactions is
greater than the energy of the original reactants

Physical changes can also involve a change in energy such
as when ice melts
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2
Thermochemistry

Thermodynamics is the science of the relationship
between heat and other forms of energy

Thermochemistry is the study of the quantity of heat
absorbed or evolved by chemical reactions

Energy is the potential or capacity to move matter (do
work); energy is a property of matter

Energy can be in many forms:
 Radiant Energy - Electromagnetic radiation
 Thermal Energy - Associated with random motion of a
molecule or atom
 Chemical Energy - Energy stored within the structural
limits of a molecule or atom
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3
Energy

There are three broad concepts of energy:
 Kinetic Energy (Ek) is the energy associated with an
object by virtue of its motion,
Ek = ½mv2
 Potential Energy (Ep) is the energy an object has by
virtue of its position in a field of force,
Ep = mgh
 Internal Energy (Ei or Ui) is the sum of the kinetic
and potential energies of the particles making up a
substance
E(i) = Ek + Ep
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4
Energy

SI (metric) unit of energy is the Joule:
(J) = kgm2/s2

1 watt
= 1 J/s
1 cal
= amount of energy needed to raise 1 g
of water 1 oC (common energy unit)
1 cal
= 4.181 J
1 Btu
= 1055 J
(Btu - British Thermal Unit)
The Law of Conservation of Energy: Energy may be
converted from one form to another, but the total
quantities of energy remain constant
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5
Practice Problem
Thermal decomposition of 5.0 metric tons of limestone
(CaCO3) to Lime (CaO) and Carbon Dioxide (CO2) requires 9.0
x 106 kJ of heat (E)
Convert this energy to:
a. Joules
b. calories
c. British Thermal Units (Btu)
(5 tons) CaCO 3 (s)  9.0 x 106 kJ  CaO(s)  CO 2 (g)
3
10
J
6
ΔE(J)   9.0 x 10 kJ) 
 9.0 x 109 J
1kJ
3

10
J   1 cal 
6
9
9
ΔE(cal)   9.0 x 10 kJ  

2.151105
x
10

2.2
x
10
cal


 1 kJ   4.181 J 
3

10
J   1 Btu 
6
6
6
ΔE(Btu)  9.0 x 10 kJ 
  8.5308 x 10  8.5x 10 Btu

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 1 kJ   1055 J 


6
Energy

When a chemical system changes from reactants to
products and the products are allowed to return to the
starting temperature, the Internal Energy (E) has changed
(E)

The difference between the system internal energy after
the change (Efinal) and before the change (Einitial) is:
E = Efinal - Einitial = Eproducts - Ereactants

If energy is lost to the surroundings, then
Efinal < Einitial

If energy is gained from the surroundings, then
Efinal > Einitial
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E < 0
E > 0
7
Heat of Reaction

In chemical reactions, heat is often transferred from the
“system” to its “surroundings,” or vice versa

The substance or mixture of substances under study in
which a change occurs is called the thermodynamic
system (or simply system)

The surroundings are everything in the vicinity of the
thermodynamic system

Heat is defined as the energy that flows into or out of a
system because of a difference in temperature between
the system and its surroundings
 Heat flows from a region of higher temperature to one
of lower temperature
 Once the temperatures become equal, heat flow stops
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8
Heat of Reaction
When heat (q) is released from the system (heat out) to the
surroundings (q < 0 negative), the reaction is defined as an
Exothermic reaction
When the surroundings deliver heat (heat in) to the system
(q > 0 positive), the reaction is defined as an Endothermic
reaction
Exothermic
Endothermic
q<0
q>0
Energy
System
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Surroundings
Energy
Surroundings
System
9
Heat of Reaction

Heat of a system is denoted by the symbol “q”
 The sign of “q” is positive (q > o) if heat is absorbed by
the system, i.e., system temperature increases
 The sign of “q” is negative (q < o) if heat is evolved by
the system, i.e., system temperature decreases

Heat of Reaction (at a given temperature) is the value
of “q” required to return a system to the given
temperature when the reaction stops at the completion of
the reaction
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10
Heat of Reaction
 Consider a chemical reaction that begins with the system
and surroundings temperature at 25oC
 If the temperature of a system decreases during the
reaction, heat flows from the surroundings into the
system
 When the reaction stops, heat continues to flow until
the system temperature returns to the temperature of
the surroundings at 25oC
 Heat has been absorbed by or added to the system
from the surroundings
 The value of “q” is positive, that is: q > 0
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11
Heat of Reaction
 If the temperature of the system rises, heat flows from
the system to the surroundings
 When the reaction stops, heat continues to flow to the
surroundings until the system returns to the
temperature of its surroundings (at 25oC)
 Heat has flowed out of the system; it has evolved (lost)
heat; thus, “’q” is negative, that is q < 0
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12
Heat Flow and Phase Changes
Predict the sign of q for each of the processes below
1. H2O(g)  H2O(l)
Condensation - Energy (heat) is lost by water vapor
Exothermic reaction - q is negative
2. CO2(s)  CO2(g)
Evaporation – Energy (heat) is absorbed (added)
by the system from its surroundings
Endothermic reaction - (q is positive)
3. CH4(g) + O2(g)  CO2(g) + H2O(g)
Combustion – Burning (oxidation) of Methane
releases (evolves) heat to surroundings
Exothermic reaction - (q is negative)
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Work & Internal Energy

Internal Energy
 The Internal Energy of a system, E, is precisely
defined as the heat at constant pressure (qp) plus any
work (w) done by the system
 Work is the energy transferred when an object is
moved by a force
w = - P Δ V = - P (Vfinal - Vinitial )
ΔE = qp + w
ΔE = qp + (-P ΔV)
Internal Energy used
to expand volume by
increasing pressure is
lost to the
surroundings, thus
the negative sign
qp = ΔE + P ΔV
 Adiabatic Process – Thermodynamic Process Without
the Gain or Loss of Heat (∆q = 0)
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Practice Problem
A system delivers 225 J of heat to the surroundings
while delivering 645 J of work.
Calculate the change in the internal energy, ∆E, of the
system
qp = heat delivered to surroundings from system = - 225 J
w = work delivered to surroundings from system = - 645 J
ΔE = change in internal energy
ΔE = qp + w
ΔE = - 225 J + (-645 J) = - 870 J
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Heat is lost
to the surroundings
15
Pressure-Volume Work
Sign conventions for q, w, and ∆E:
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q + w (-P∆V)
+
+
+
—
—
+
—
—
=
q:
q:
w:
w:
+
—
+
—
system gains heat
system loses heat
work done on system
work done by system
∆E
+
Depends on sizes of q and w
Depends on sizes of q and w
—
16
Practice Problem

A system expands in volume from 2.00 L to
24.5 L at constant temperature.
Calculate the work (w), in Joules (J), if the expansion
occurs against a constant pressure of 5.00 atm
w = - pΔV
kg
2
1.01325×105 Pa
10-3 m 3
m
•
s
w = - 5.00 atm ×
×
× (24.5 L - 2.0 L)
atm
Pa
L
1


2
 kg • m 2 

  1J 
kg
•
m
4
w = - 1,1399.1 

 = - 1.14×10 
 ×
2
2
2
 s

 s
  kg • m 


2


s
w = - 1.14×104 J
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17
Practice Problem
A system that does no work but which transfers heat to the
surrounding has:
a. q < 0, ∆E > 0
d. q > 0, ∆E < 0
b. q < 0, ∆E < 0
e. q < 0, ∆E = 0
c. q > 0, ∆E > 0
A system that does no work but receives heat from the surroundings
has:
a. q < 0, ∆E > 0
d. q = - ∆E
b. q > 0, ∆E < 0
e. w = ∆E
c. q = ∆E
A system which undergoes an adiabatic change (i.e., ∆q = 0) and
does work on the surroundings has:
a. w < 0, ∆E = 0,
d. w < 0, ∆E > 0
b. w > 0, ∆E > 0
e. w < 0, ∆E < 0
c. w > 0, ∆E < 0
A system which undergoes an adiabatic change (i.e., ∆q = 0) and has
work done on it by the surroundings has:
a. w = ∆E
d. w < 0, ∆E > 0
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b. w = -∆E
e. w > ∆E
c. w > 0, ∆E < 0
18
Enthalpy and Enthalpy Change

Enthalpy, denoted H, is an extensive property of a
substance that can be used to obtain the heat absorbed or
evolved in a chemical reaction
 An extensive property is one that depends on the
quantity of substance
 Enthalpy is a state function, a property of a system
that depends only on its present state and is
independent of any previous history of the system

Enthalpy represents the heat energy tied up in chemical
bonds
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19
Enthalpy and Enthalpy Change

The change in Enthalpy for a reaction at a given
temperature and pressure, called the Enthalpy of
Reaction, is obtained by subtracting the Enthalpy of the
reactants from the Enthalpy of the products.
ΔHrxn = H(products) - H(reactants)
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20
Enthalpy and Enthalpy Change

Enthalpy is defined as the internal energy plus the product
of the pressure and volume (work)
H  E  PV

The change in Enthalpy is the change in internal energy
plus the product of constant pressure and the change in
Volume
ΔH = ΔE + P ΔV
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
Enthalpy and Enthalpy Change
Recall:
qp = ΔE + P ΔV
ΔH = ΔE + P ΔV
Thus :
ΔH = q p
(At Constant Pressure)
 The change in Enthalpy equals the heat gained or lost
(heat of reaction, Hrxn) at constant pressure
 This represents the entire change in internal energy
(E) minus any expansion “work” done by the system
(PV would have negative sign)
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22
Practice Problem
An ideal gas (the system) is contained in a flexible balloon at a
pressure of 1 atm and is initially at a temperature of 20.0oC.
The surrounding air is at the same pressure, but its temperature is
25oC. When the system is equilibrated with its surroundings, both
systems and surroundings are at 25oC and 1 atm.
In changing from the initial to the final state, which of the
following relationships regarding the system is correct?
a.
∆E = 0
Heat is added, internal energy increases
 ∆E > 0
b.
∆E < 0
Heat is added, internal energy increases
 ∆E > 0
c.
∆H = 0
∆E increases and P∆V work is done by system
 ∆H > 0
d.
w > 0
P∆V work is done by system (volume increase)  W < 0
e.
q > 0
Temperature (heat) in system increases
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23
Practice Problem
In which of the following processes is ∆H = ∆E, i.e. P∆V = 0?
a. 2HI(g)  H2(g) + I2(g) at atmospheric pressure
(P∆V = 0 no change in moles, volume)
b. Two moles of Ammonia gas are cooled from 325oC to 300oC at 1.2 atm
(P∆V ≠ 0 Vol decreases)
c. H2O(l)  H2O(g) at 100oC at atmospheric pressure
(P∆V ≠ 0 Vol increases)
d. CaCO3(s)  CaO(s) + CO2 (g) at 800oC at atmospheric pressure
(P∆V ≠ 0 Vol increases)
e. CO2(s)  CO2(g) at atmospheric pressure
(P∆V ≠ 0 Vol increases)
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24
Comparing E & H



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Reactions that do not involve gases
 Reactions such as precipitation, acid-base, many
redox, etc., do not produce gases
 Since the change in volumes of liquids and solids are
quite small:
V  0
P V  0
H  E
Reactions in which the amount (mol) of gas does not
change
(Vol of Gaseous Reactants = Vol Gaseous Products
V = 0
P V = 0
H = E
Reactions in which the amount (mol) of gas does change
PV  0
However, qp is usually much greater than PV
Therefore: H  E
25
Comparing E & H
Example:
2H2(g) + O2(g)  2H2O(g)
Change in moles: 3 mol  2 mol
H = -483.6 kJ
 PV  0
and PV = -2.5kJ
E = H - PV = -483.6 kJ - (-2.5 kJ) = -481.1 kJ
Most of E occurs as Heat (H = qp)
 H  E
For many reactions, even when PV  0, H is close to
E
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Comparing E & H
For which one of the following reactions will ∆H be approximately (or
exactly) equal to ∆E?

a. H2(g) + Br2(g)
2HBr(g)
(No change in volume; no change in work, PV = 0)
b. H2 O(l)

H2O(g)
(Change in volume; change in work due to gas expansion, PV  0)
c. CaCO3(s)

CaO(s) + CO2(g)
(Change in volume; change in work due to gas expansion, PV  0
d. 2H(g) + O(g)

H2O(l)
(Change in volume; condensation, heat (q) released, PV  0)
e. CH4(g) + 2O2(g)

CO2(g) + 2H2O(l)
(Change in volume; condensation, heat (q) released, PV  0)
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27
Exothermic and Endothermic Processes

Energy (E), Pressure (P), and Volume (V) are “state”
functions

Enthalpy (H) is also a state function, which means that
H depends only on the difference between Hfinal &
Hinitial

The Enthalpy change of a reaction, also called the
Heat of Reaction (Hrxn), always refers to
Hrxn = Hfinal - Hinitial = Hproducts - Hreactants

Hproducts can be either more or less than Hreactants

The resulting sign of H indicates whether heat is
absorbed from the surroundings (heat in) or released
to the surroundings (heat out) in the process
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Exothermic and Endothermic Processes

An Exothermic reaction releases heat (heat out) to
surroundings with a decrease in system Enthalpy
CH4(g) + 2O2  CO2(g) + 2H2O(g) + heat
Exothermic: Hfinal < Hinitial

H < 0 (negative)
An Endothermic reaction absorbs heat (heat in) from
the surroundings resulting in an increase in system
Enthalpy
Heat + H2O(s)  H2O(l)
Endothermic
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Hfinal > Hinitial
H > 0 (positive)
29
Types of Enthalpy Changes

When a compound is produced from its elements, the
Enthalpy change (Heat of Reaction) is called:
Heat of Formation (∆Hf)
K(s) + ½Br2()l)  KBr(s) ∆H = ∆Hf

When a substance melts, the Enthalpy change is called:
Heat of Fusion (∆Hfus)
NaCl(s)  NaCl(l)

∆H = ∆H(fus)
When a substance vaporizes, the Enthalpy change is
called:
Heat of Vaporization
C6H6(l)  C6H6(g)
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∆H = ∆H(vap)
30
Thermochemical Equations

A Thermochemical Equation is the chemical equation
for a reaction (including phase labels) in which the
equation is given a molar interpretation, and the Enthalpy
of Reaction (∆Hrxn) for these molar amounts is written
directly after the equation.
N 2 (g) + 3 H 2 (g)  2 NH 3 (g)
ΔH rxn = - 91.8 kJ
H is negative; heat is lost to surroundings
1 mol N2 + 3 mol H2 yields 91.8 kJ of heat
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31
Practice Problem
Sulfur, S8, burns in air to produce Sulfur Dioxide. The
reaction evolves (releases) 9.31 kJ of heat per gram of Sulfur
at constant pressure. Write the thermochemical equation for
this reaction.
S8  O 2  SO 2 + Heat
Exothermic Reaction
Balance the Reaction
S 8  8 O 2  8 SO 2
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ΔH  - 9.31 kJ
32
Practice Problem
In a phase change of water between the liquid and the gas
phases, 770.1 kJ of energy was released by the system.
What was the product, and how much of it was formed in
the phase change.
(Data: H2O(l) 
H2O(g)
∆H = 44.01 kJ/mol)
a. 315 g of water vapor was produced
b. 17.5 g of water vapor was produced
c. 17.5 mol of water vapor was produced
d. 17.5 mol of liquid water was produced
∆H is positive (endothermic reaction)
Since energy was released, the gas condensed to liquid
e. 17.5 g of liquid water was produced
770.1 kJ / 44.01 kJ / mol = 17.5 mols
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33
Thermochemical Equations

The following are two important rules for manipulating
Thermochemical equations:
 When a thermochemical equation is multiplied by
any factor, the value of H for the new equation is
obtained by multiplying the H in the original equation
by that same factor
 When a chemical equation is reversed, the value of
H is reversed in sign
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34
Practice Problem
When White Phosphorus burns in air, it produces
Phosphorus (V) Oxide (Change in Oxidation state)
P4(s) + 5O2(g)  P4O10(s)
H = -3010 kJ
What is H for the following equation?
P4O10(s)  P4(s) + 5O2(g)
H = ?
Ans:
The original reaction is reversed
 Change the Sign !!
H = + 3010 kJ
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35
Practice Problem
Carbon Disulfide (CS2(l)) burns in air, producing Carbon
Dioxide and Sulfur Dioxide
CS2(l) + 3 O2(g)  CO2(g) + 2 SO2(g)
H = -1077 kJ
What is H for the following equation?
1/2 CS2(l) + 3/2 O2(g)  1/2 CO2(g) + SO2(g)
Ans: The new reaction uses ½ the original amounts
 Divide H by 2
H = (-1077 / 2) = - 538.5 kJ
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36
Applying Stoichiometry and
Heats of Reactions
Consider the reaction of Methane, CH4, burning in the
presence of Oxygen at constant pressure. Given the following
equation, how much heat could be obtained by the
combustion of
10.0 grams CH4?
CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(l) ΔH o = - 890.3 kJ
1 mol CH
kJ
10.0 g CH 4 × 16.0 g CH4 × 1-890.3
mol CH = - 556 kJ
4
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4
37
Measuring Heats of Reaction

Heats of Reactions (Enthalpy change of reaction – Hrxn)
are determined from the heat required to raise the
temperature of a substance
A thermochemical measurement is based on the
relationship between heat and temperature change; that is
the quantity of heat (q) absorbed or released by an object
is proportional to its temperature change
q
Heat (q)  ΔT or q = constant × ΔT or
= constant
ΔT
 Each object has its own heat capacity, which is the
quantity of heat required to change its temperature by 1
Kelvin (K)

Thus, proportionality constant above is the Heat Capacity:
q
= Constant = Heat Capacity
[in units of J / K]
ΔT
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38
Measuring Heats of Reaction

The specific heat capacity, S, (or “specific heat”) is the
heat required to raise the temperature of
one gram of a substance by one degree Celsius
S is in units of J/g  oK
q = m S ΔT

m = grams of sample
T = Tfinal - Tinitial
The molar heat capacity, C, of a sample of substance is
the quantity of heat required to raise the temperature of
one mole of substance one degree Celsius
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C is in units of J/mol  oK,
n = moles of substance
q = n C ΔT
T = Tfinal - Tinitial
39
Measuring Heats of Reaction

Bomb Calorimeter used to measure heats of combustion
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40
Measuring Heats of Reaction
Specific Heats and Molar Heat Capacities of some substances
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41
Practice Problem
Suppose you mix 20.5 g of water at 66.2 oC with 45.4 g
of water at 35.7 oC in an insulated cup. What is the
maximum temperature of the solution after mixing?
Ans: The heat lost by the water at 66.2 oC is balanced
by the heat gained by the water at 35.7 oC
q = m S ΔT
ΔT1 = Tf - 66.2o C ΔT2 = Tf - 35.7o C
m1 = 20.5 g
m 2 = 45.4 g
-q lost = q gained = - m1SΔT1 = m 2SΔT2
-m1 *  Tf - 66.2o C  = m 2 *  Tf - 35.7o C 
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 m2 
 Tf - 66.2  =    Tf - 35.7 
 - m1 
 45.4 
T
66.2
=
 f
 
  Tf - 35.7 
 -20.5 
Tf = 45.2 o C
42
Practice Problem
A piece of copper with a temperature of 100oC is dropped
into a beaker containing 50.0 grams of water at 20oC. When
the reaction is completed the temperature of the solution is
25oC. Assuming no loss of heat and the heat capacity of the
solution is the same as water, 4.184 J/(g * oK), what is the
heat capacity of Copper in J/K?
ΔT1 = (25 – 100) = -75oC
ΔT2 = (25 – 20) = 5oC
Heat change by increasing water temperature by 5oK
q2 = 50g * 4.184 J/(g * oK) * 5oK = 1,046 J
Heat lost by copper (q1) is equal to heat gained by water
-q1 = +q2
Therefore, The Heat Capacity of copper is:
-q2
-1,045 J
q1 =
=
= 13.9 J / K
o
ΔT
-75 C
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43
Practice Problem
How much heat is gained by Nickel when 500 g of Nickel is
warmed from 22.4oC to 58.4°C?
[The specific heat of Nickel is 0.444 J/(g • °C)]
a. 2000 J
b. 4000 J
d. 8000 J
e. 10000 J
Ans: d
c. 6000 J
q = m s ΔT
ΔT = Tf (58.4o C) - Ti (22.4o C) = 36.0 oC
s = Specific Heat Nickel = 0.444 J / g •o C
m = 500 g
o
1/13/2015
o
q = 500 g * 0.444 J / g • C * 36.0 C
q = 7992 = 8000 J
44
Practice Problem
When 25.0 mL of 0.5 M H2SO4 is added to 25.0 mL of 1.00 M KOH in a
calorimeter at 23.5 oC, the temperature rises to 30.17oC
Calculate Hrxn for each reactant. Assume density (d) and specific heat
of the solution (s) are the same as water
2 KOH(aq) + H 2 SO 4 (aq)  K 2 SO 4 (aq) + 2 H 2 O(l)
qsoln = m s ΔT (Δ T = Tf - Ti )
d = 1 g / mL
s = 4.184 J / g • oC
 1.00 g   4.184 J 
o  1 kJ 
qsoln =  25.0 + 25.0  mL 
30.17
23.5
C
 o 

 = 1.395354 kJ
mL
1000
J

 g• C 




Calculate moles
 25.0 mL  
0.500 mol H 2SO 4  1 L
= 0.0125 mol H 2SO 4

L
1000
mL


 25.0 mL  
1.00 mol KOH  1 L

L

 1000 mL
= 0.0250 mol KOH
Both KOH or H 2SO 4 are limiting (2 moles KOH / 1 mol H 2SO4 )
1/13/2015
Con’t
45
Practice Problem (Con’t)
When 25.0 mL of 0.5 M H2SO4 is added to 25.0 mL of 1.00 M KOH in a
calorimeter at 23.5 oC, the temperature rises to 30.17oC.
Calculate Hrxn for each reactant. Assume density (d) and specific heat
of the solution are the same as water.
Temperature of water increased (23.5oC  30.17oC)
The Reaction is Exothermic (heat released to surroundings (water))
Thus, qrxn is negative
qsoln = qrxn = - 1.395634 kJ
-1.395364 kJ
H rxn (H 2SO 4 ) 
 -111.62912  -112 kJ / mol H 2SO4
0.0125 mol H 2SO 4
-1.395364 kJ
H rxn (KOH) 
0.0250 mol KOH
1/13/2015
 - 55.81456  - 55.8 kJ / mol KOH
46
Hess’s Law

Hess’s law of Heat Summation
For a chemical equation that can be written
as the sum of two or more steps, the
Enthalpy change for the overall equation
is the sum of the Enthalpy changes for the individual
steps

In coupled reactions, the Enthalpy change for the overall
reaction is the sum of the Enthalpy changes for the
coupled reactions
Note: It is often necessary to reverse chemical equations
to couple them so chemical species are on the correct side
of yield sign, or multiply through by a coefficient to cancel
common chemical species
1/13/2015
47
Hess’s Law
For example, suppose you are given the following data:
S(s) + O 2 (g)  SO 2 (g)
2 SO 3 (g)  2 SO 2 (g) + O 2 (g)
ΔH = -297 kJ
o
ΔH o = 198 kJ
Could you use these data to obtain the Enthalpy change
for the following reaction?
2 S(s) + 3 O 2 (g)  2 SO 3 (g)
ΔH = ?
o
Con’t
1/13/2015
48
Hess’s Law
If we multiply the first equation by 2 and reverse the second
equation, they will sum together to become the third
2 S(s) + 2 O 2 (g)  2 SO 2 (g)
ΔHo = (-297 kJ)×(2)
2 SO 2 (g) + O 2 (g)  2 SO 3 (g) ΔH = (198 kJ)×(-1)
o
2 S(s) + 3 O 2 (g)
 2 SO 3 (g)
o
ΔH = (-792 kJ)
Note : ( - 297 × 2) + (198 × - 1) = - 594 kJ - 198 kJ = - 792 kJ
Note the change in H values with the changes in the molar
coefficients to balance equation 1 and the reversal of
equation 2
1/13/2015
49
Practice Problem
Given the following data,
A(s) + O2(g)  AO2(g)
H° = – 105 kJ/mol
A(g) + O2(g)  AO2(g)
H° = – 1200 kJ/mol
Find the heat required for the reaction converting:
A(s) to A(g) at 298 K and 1 atm pressure.
A(s) + O 2 (g)  AO 2 (g)
AO 2 (g)

A(s)
 A(g)
1/13/2015
A(g) + O 2 (g)
- 105 kJ
+ 1200 kJ
+ 1095 kJ
50
Standard Enthalpies of Formation

The term standard state refers to the standard
thermodynamic conditions chosen for substances when
listing or comparing thermodynamic data:
 Pressure
- 1 atmosphere (760 mm Hg)
 Temperature - (usually 25oC).

The Enthalpy change for a reaction in which reactants are
in their standard states is denoted as the
“Standard Heat of Reaction”
o
ΔH rxn
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51
Standard Enthalpies of Formation

Standard Enthalpy of Formation of Substance
ΔH
o
f
The Enthalpy change for the formation of
one mole of a substance in its standard state from its
component elements in their standard states
Note: The standard Enthalpy of Formation for a
“Pure Element”
(C, Fe, Au, N, etc.)
in its standard state is zero
1/13/2015
52
Standard Enthalpies of Formation
Law of Summation of Heats of Formation
The Enthalpy of a reaction i.e., the “Standard Heat of
Reaction:
(∆Horxn)”
is equal to the total formation energy of the products
minus that of the reactants
ΔH orxn =
o
 nΔH f (products)
-
o
 mΔH f (reactants)
Where  is the mathematical symbol meaning
“the sum of”
and m and n are the coefficients of the substances in the
chemical equation, i.e., the relative number of moles of each
substance
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53
Standard Enthalpies of Formation
Formula
Hof
Formula
(kJ/mol)
Calcium
Ca(s)
CaO(s)
CaCO3(s)
Selected Standard
Heats of Formation
(Enthalpies)
At 25oC (298oK)
Carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(l)
HCN(g)
CS2
0
1.9
-110.5
-393.5
-74.9
-238.6
135
87.9
Chlorine
Cl2(g)
Cl(g)
Cl-(aq)
Cl-(g)
HCl(g)
0
121.0
167.2
-234.0
-92.31
Bromine
Br2(l)
Br(g)
Br2(g)
Br-(ag)
Br-(g)
HBr(g)
1/13/2015
0
-635.1
-1206.9
0
111.9
30.91
-121.5
-219.0
-36.44
Hof
Formula
(kJ/mol)
Hydrogen
H2(g)
H(g)
0
218.0
Oxygen
O2(g)
O3(g)
H2O(g)
H2O(l)
0
143
-241.8
-285.8
Nitrogen
N2(g)
NH3(g)
NO(g)
0
-45.9
90.3
Sulfur
S8(rhombic)
S8(monoclinic)
SO2(g)
SO3(g)
Hof
(kJ/mol)
Silver
Ag(s)
AgCl(s)
0
-127.0
Sodium
Na(s)
Na(g)
NaCL(s)
0
107.8
-411.1
0
0.3
-296.8
-396.0
54
Practice Problem
Calculate the Heat of Reaction, H°rxn, for the combustion of
C3H6(g):
C3H6(g) + 9/2 O2(g)  3 CO2(g) + 3 H2O(l)
Hof values in kilojoules per mole are as follows:
C3H6(g) = 21
CO2(g) = –394
a. –2061 kJ
b. –2019 kJ
d. 2019 kJ
e. 2061 kJ
Ans: a
o
ΔHrxn
ΔHorxn
=
1/13/2015
c. –701 kJ
o
 n ΔH f (products)
-
o
 m ΔH f (reactants)
=  3×  -394   +  3×  -286    -  21 + 0
ΔHorxn =
o
ΔHrxn
H2O(l) = –286
 -1182 - 858
- 21
= - 2061 kJ
55
Practice Problem
Acetylene burns in air according to the equation below.
Given: Hof CO2(g) = -395.5 kJ/mol
Hof H2O(g) = -241.8 kJ/mol
C2 H 2 (g) + 5 / 2 O 2 (g)  2 CO 2 (g) + H 2O(g)
o
ΔHrxn
= -1255.8 kJ
Calculate Hof of C2H2(g)
ΔH orxn =  2 mol (Δ H fo , CO 2 (g)) + 1 mol (Δ H fo , H 2O(g))  1 mol (Δ H of , C2 H 2 (g)) + 5 / 2 mol (Δ H fo , O 2 (g)) 
-1255.8 kJ =  2 mol (-393.5 kJ / mol) + 1 mol (-241.826 kJ / mol) 1 mol ( (Δ H of , C2 H 2 (g) + 5 / 2 mol (0.0) 
- 1255.8kJ = - 787.0kJ - 241.8kJ - 1 mol (Δ H of , C2 H 2 (g))
ΔH of , C2 H 2 (g) =
1/13/2015
-227.0 kJ
= 227.0 kJ / mol
-mol
56
Summary – Equations & Relationships
Ek = 1 / 2 m v 2
ΔE = Efinal - Einitial
= Eproducts - Ereactants
w = - P Δ V = - P (Vfinal - Vinitial )
ΔE = qp + w
ΔE = qp + (-P ΔV)
qp = ΔE + P ΔV
ΔH = ΔE + P ΔV
ΔH = qp
(at Constant P)
q = sm ΔT
-q lost (exothermic)
1/13/2015
ΔH orxn =
o
=
q gained (endothermic)
 nΔ H f (products)
-
o
 mΔ H f (reactants)
57