thermochemistry energy flow and chemical change

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Chapter 6
Thermochemistry:
Energy Flow and Chemical Change

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Thermochemistry: Energy Flow and Chemical Change
6.1 Forms of Energy and Their Interconversion
6.2 Enthalpy: Heats of Reaction and Chemical Change
6.3 Calorimetry: Laboratory Measurement of Heats of Reaction
6.4 Stoichiometry of Thermochemical Equations
6.5 Hess’s Law of Heat Summation
6.6 Standard Heats of Reaction (ΔH0
rxn)

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Figure 6.1 A chemical system and its surroundings.
the system
the surroundings

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Thermochemistry is a branch of thermodynamics that deals with
the heat involved with chemical and physical changes.
Thermodynamics is the study of heat and its transformations.
Fundamental premise
When energy is transferred from one object to another,
it appears as work and/or as heat.
For our work we must define a system to study; everything else
then becomes the surroundings.
The system is composed of particles with their own internal energies (E or U).
Therefore the system has an internal energy. When a change occurs, the
internal energy changes.

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ΔE = Efinal - Einitial = Eproducts - Ereactants
Figure 6.2 Energy diagrams for the transfer of internal energy (E)
between a system and its surroundings.
1. Heat, q – is the energy transferred as a result of difference in temperature
between the system and surroundings.
2. Work, q – the energy transferred when an object is moved by
force.

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Figure 6.3 A system transferring energy as heat only.

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Figure 6.4 A system losing energy as work only.
Energy, E
Zn(s) + 2H+(aq) + 2Cl-(aq)
H2(g) + Zn2+(aq) + 2Cl-(aq)
ΔE<0
work done on
surroundings

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Table 6.1 The Sign Conventions* for q, w and ΔE
q w
+ = ΔE
+
+
-
-
-
-
+
+
+
-
depends on sizes of q and w
depends on sizes of q and w
* For q: + means system gains heat; - means system loses heat.
* For w: + means word done on system; - means work done by system.

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ΔEuniverse = ΔEsystem + ΔEsurroundings
Units of Energy
Joule (J)
Calorie (cal)
British Thermal Unit
1 cal = 4.18J
1 J = 1 kg*m2/s2
1 Btu = 1055 J

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Figure 6.5
Some interesting
quantities of energy.

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Sample Problem 6.1 Determining the Change in Internal Energy of a
System
PROBLEM: When gasoline burns in a car engine, the heat released causes
the products CO2 and H2O to expand, which pushes the pistons
outward. Excess heat is removed by the car’s cooling system.
If the expanding gases do 451 J of work on the pistons and the
system loses 325 J to the surroundings as heat, calculate the
change in energy (ΔE) in J, kJ, and kcal.
SOLUTION:
PLAN: Define system and surroundings, assign signs to q and w and calculate
ΔE. The answer should be converted from J to kJ and then to kcal.
q = - 325 J w = - 451 J
ΔE = q + w = -325 J + (-451 J) =-776 J
-776J
103J
kJ
= -0.776kJ -0.776kJ
4.18kJ
kcal
= -0.185 kcal

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Figure 6.6 Two different paths for the energy change of a system.

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Figure 6.7
Pressure-volume
work.
The two most important types of chemical work
1.Electrical work – done by moving particles
2.Mechanical work – done when the volume of the system changes in
the presence of external pressure. (Pressure volume work).

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The Meaning of Enthalpy
w = - PΔV
ΔH = ΔE +
PΔV
qp = ΔE + PΔV = ΔH
ΔH ≈ ΔE in
1. Reactions that do not involve gases.
2. Reactions in which the number of
moles of gas does not change.
3. Reactions in which the number of
moles of gas does change but q is >>>
PΔV.
H = E + PV
where H is enthalpy
Change in enthalpy (ΔH) is the change in internal energy plus the
product of the pressure (which is constant)

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Figure 6.8
Enthalpy diagrams for exothermic and endothermic processes.
Enthalpy,
H
Enthalpy,
H
CH4 + 2O2
CO2 + 2H2O
Hinitial
Hinitial
Hfinal
Hfinal
H2O(l)
H2O(g)
heat out heat in
ΔH < 0 ΔH > 0
A Exothermic process B Endothermic process
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g)

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Sample Problem 6.2 Drawing Enthalpy Diagrams and Determining
the Sign of ΔH
PROBLEM: In each of the following cases, determine the sign of ΔH, state
whether the reaction is exothermic or endothermic, and draw
and enthalpy diagram.
SOLUTION:
PLAN: Determine whether heat is a reactant or a product. As a reactant,
the products are at a higher energy and the reaction is
endothermic. The opposite is true for an exothermic reaction
(a) H2(g) + 1/2O2(g) H2O(l) + 285.8kJ
(b) 40.7kJ + H2O(l) H2O(g)
(a) The reaction is exothermic.
H2(g) + 1/2O2(g) (reactants)
H2O(l) (products)
EXOTHERMIC
(products)
H2O(g)
(reactants)
H2O(l)
ΔH = -285.8kJ ΔH = +40.7kJ
ENDOTHERMIC
(b) The reaction is endothermic.

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Some Important Types of Enthalpy Change
heat of combustion (ΔHcomb)
heat of formation (ΔHf)
heat of fusion (ΔHfus)
heat of vaporization (ΔHvap)
C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g)
K(s) + 1/2Br2(l) KBr(s)
NaCl(s) NaCl(l)
C6H6(l) C6H6(g)

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Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and
Materials
Specific Heat
Capacity (J/g*K)
Substance
Specific Heat
Capacity (J/g*K)
Substance
Compounds
water, H2O(l)
ethyl alcohol, C2H5OH(l)
ethylene glycol, (CH2OH)2(l)
carbon tetrachloride, CCl4(l)
4.184
2.46
2.42
0.864
Elements
aluminum, Al
graphite,C
iron, Fe
copper, Cu
gold, Au
0.900
0.711
0.450
0.387
0.129
wood
cement
glass
granite
steel
Materials
1.76
0.88
0.84
0.79
0.45

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Measuring the heat of a chemical or physical
change
• Heat Capacity – the quantity of heat required to
change its temperature by 1 Kelvin.
Heat capacity = q / ΔT
• Specific heat capacity ( c ) - the quantity of heat
required to change the temperature of 1 gram of
the object by 1 Kelvin
c = q / (mass x ΔT)

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Measuring the heat of a chemical or physical
change
• Molar heat capacity ( C ) = quantity of heat
required to change the temperature of 1 mole of
a substance by 1 Kelvin
C = q / (amount (mol) x ΔT)

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Sample Problem 6.3 Finding the Quantity of Heat from Specific Heat
Capacity
PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g.
How much heat is needed to raise the temperature of the
copper layer from 250C to 300.0C? The specific heat capacity
(c) of Cu is 0.387 J/g*K.
SOLUTION:
PLAN: Given the mass, specific heat capacity and change in temperature, we
can use q = c x mass x ΔT to find the answer. ΔT in 0C is the same as for
K.
q =
0.387 J
g*K
125 g (300-25)0C
x x = 1.33x104 J

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Finding the Quantity of Heat from Specific Heat Capacity
Find the heat released ( in kJ) when 5.50 L of ethylene glycol (d =
1.11 g/mL) in a car radiator cools from 37.0 ⁰C to 25.0 ⁰C.
Ans: -177 kJ

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Finding the Quantity of Heat from Specific Heat Capacity
Find the heat released ( in kJ) when 5.50 L of ethylene glycol (d =
1.11 g/mL) in a car radiator cools from 37.0 ⁰C to 25.0 ⁰C.
Ans: -177 kJ

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Figure 6.9
Coffee-cup calorimeter.
Calorimetry is used to measure the heat released by a physical or chemical
process.
This device is used to find the
specific heat capacity of solid that
does not react with or dissolve in
water.

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This device is used to measure the
heat of combustion reaction, such
as for fuels and foods.

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Figure 6.10
A bomb calorimeter

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Sample Problem 6.5 Calculating the Heat of Combustion
PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer
than 10 Calories per serving.” To test the claim, a chemist at
the Department of Consumer Affairs places one serving in a
bomb calorimeter and burns it in O2(the heat capacity of the
calorimeter = 8.15 kJ/K). The temperature increases 4.9370C.
Is the manufacturer’s claim correct?
SOLUTION:
PLAN: - q sample = qcalorimeter
qcalorimeter = heat capacity x ΔT
= 8.151 kJ/K x 4.937 K
= 40.24 kJ
40.24 kJ kcal
4.18 kJ
= 9.63 kcal or Calories
The manufacturer’s claim is true.

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Ans: 10.85 kJ/K

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Ans: 10.85 kJ/K

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STOICHIOMETRY OF THERMOCHEMICAL EQUATIONS
A thermochemical equation is a balanced equation that includes the
enthalpy change of the reaction (ΔH), it refers only to the amounts
(mol) of substances and their states of matter in that equation.
Two aspects:
1.Sign. The sign of ΔH depends on whether the reaction is
exothermic or endothermic.
2.Magnitude. The magnitude of ΔH is proportional to the amount
of substance.

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Two key points to understand about thermochemical
equations are as follows:
1. Balancing coefficients
2. Thermochemical equivalence

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AMOUNT (mol)
of compound A
AMOUNT (mol)
of compound B
HEAT (kJ)
gained or lost
molar ratio from
balanced equation
ΔHrxn (kJ/mol)
Figure 6.11
Summary of the relationship between
amount (mol) of substance and the heat
(kJ) transferred during a reaction.

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Sample Problem 6.6 Using the Heat of Reaction (ΔHrxn) to Find
Amounts
SOLUTION:
PLAN:
PROBLEM: The major source of aluminum in the world is bauxite (mostly
aluminum oxide). Its thermal decomposition can be represented by
If aluminum is produced this way, how many grams of aluminum can
form when 1.000x103 kJ of heat is transferred?
Al2O3(s) 2Al(s) + 3/2O2(g) ΔHrxn = 1676 kJ
heat(kJ)
mol of Al
g of Al
1676kJ=2mol Al
x M
1.000x103 kJ x 2 mol Al
1676 kJ
26.98 g Al
1 mol Al
= 32.20 g Al

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Hydrogenation reactions, in which H2 and an
“unsaturated” organic compound combine are used in
the food, fuel and polymer industries. In the simplest
case, ethene and H2 form ethane. If 137 kJ is given off
per mole of ethene reaction, how much heat is released
when 15.0 kg of ethane forms?
Ans: - 6.83 x 10^(4) kJ

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Hess’s Law
the enthalpy change of an overall process is the sum of the enthalpy
changes of its individual steps:
ΔH(overall) = ΔH1 + ΔH2+ ………. ΔHn
To summarize, calculating an unknown ΔH involves three steps
1.Identify the target equation
2.Manipulate each equation
– Change the sign of ΔH when you reverse an equation
– Multiply amount (mol) and ΔH by the same factor
3.Add the manipulated equation and their resulting ΔH values to get the
target equation and ΔH

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Sample Problem 6.7 Using Hess’s Law to Calculate an Unknown ΔH
SOLUTION:
PLAN:
PROBLEM: Two gaseous pollutants that form in auto exhaust are CO and
NO. An environmental chemist is studying ways to convert
them to less harmful gases through the following equation:
CO(g) + NO(g) CO2(g) + 1/2N2(g) ΔH = ?
Given the following information, calculate the unknown ΔH:
Equation A: CO(g) + 1/2O2(g) CO2(g) ΔHA = -283.0 kJ
Equation B: N2(g) + O2(g) 2NO(g) ΔHB = 180.6 kJ
Equations A and B have to be manipulated by reversal and/or
multiplication by factors in order to sum to the first, or target, equation.
Multiply Equation B by 1/2 and reverse it.
ΔHB = -90.3 kJ
CO(g) + 1/2O2(g) CO2(g) ΔHA = -283.0
kJ
NO(g) 1/2N2(g) + 1/2O2(g)
ΔHrxn = -373.3 kJ
CO(g) + NO(g) CO2(g) + 1/2N2(g)

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STANDARD ENTHALPIES OF REACTION (ΔHrxn)
Set of conditions called standard states:
1.For a gas, the standard state is 1 atm and ideal behaviour
2.For a substance in aqueous solution, the standard state is 1 M
concentration
3.For a pure substance (element or compound), the standard state is
usually the most stable form of the substance at 1 atm and the
temperature is usually 25 ⁰C (298 K)
Standard states – indicates that the variable has been measured with all
the substance in their standard states.
Standard enthalpy of reaction, ΔHrxn
- is measured at the standard state

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Standard enthalpy of formation
- is the enthalpy change for the formation equation
when all the substances are their in their standard states.
Need to remember:
1.For an element in its standard state, ΔH⁰f = 0
2.Most compounds have a negative ΔH⁰f.

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Table 6.5 Selected Standard Heats of Formation at 250C(298K)
Formula ΔH0
f(kJ/mol)
calcium
Ca(s)
CaO(s)
CaCO3(s)
carbon
C(graphite)
C(diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(l)
HCN(g)
CSs(l)
chlorine
Cl(g)
0
-635.1
-1206.9
0
1.9
-110.5
-393.5
-74.9
-238.6
135
87.9
121.0
hydrogen
nitrogen
oxygen
Formula ΔH0
f(kJ/mol)
H(g)
H2(g)
N2(g)
NH3(g)
NO(g)
O2(g)
O3(g)
H2O(g)
H2O(l)
Cl2(g)
HCl(g)
0
0
0
-92.3
0
218
-45.9
90.3
143
-241.8
-285.8
107.8
Formula ΔH0
f(kJ/mol)
silver
Ag(s)
AgCl(s)
sodium
Na(s)
Na(g)
NaCl(s)
sulfur
S8(rhombic)
S8(monoclinic)
SO2(g)
SO3(g)
0
0
0
-127.0
-411.1
2
-296.8
-396.0

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Figure 6.13
The general process for determining ΔH0
rxn from ΔH0
f values.
Enthalpy,
H
Elements
Reactants
Products
ΔH0
rxn = Σ mΔH0
f(products) - Σ nΔH0
f(reactants)
decomposition
-ΔH0
f ΔH0
f
formation
ΔH0
rxn
Hinitial
Hfinal

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Sample Problem 6.9 Calculating the Heat of Reaction from Heats of
Formation
SOLUTION:
PLAN:
PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion
kilograms, is used to make many products, including fertilizer, dyes,
and explosives. The first step in the industrial production process is
the oxidation of ammonia:
Calculate ΔH0
rxn from ΔH0
f values.
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Look up the ΔH0
f values and use Hess’s Law to find ΔHrxn.
ΔHrxn = Σ mΔH0
f (products) - Σ nΔH0
f (reactants)
ΔHrxn = [4(ΔH0
f NO(g) + 6(ΔH0
f H2O(g)] - [4(ΔH0
f NH3(g) + 5(ΔH0
f O2(g)]
= (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) -
[(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)]
ΔHrxn = -906.0 kJ

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