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Chap 6: Thermochemistry
Bushra Javed
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Contents
1. Introduction to Thermochemistry
2. Energy and it’s units
3. Understanding Heats of Reaction
2. Enthalpy and Enthalpy Changes
3. Thermochemical Equations
4. Applying Stoichiometry to Heats of Reaction
5. Measuring Heats of Reaction
6. Using Heats of Reaction; Hess’s Law
7. Standard Enthalpies of Formation
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Thermodynamics
The science of the relationship between heat
and other forms of energy.
Thermochemistry
An area of thermodynamics that concerns
the study of the heat absorbed or evolved by
a chemical reaction.
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Energy
Energy
The potential or capacity to move matter.
One form of energy can be converted to another
form of energy: electromagnetic, mechanical,
electrical, or chemical.
Next, we’ll study kinetic energy, potential
energy, and internal energy
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Kinetic Energy, EK
The energy associated with an object by virtue
of its motion.
EK 
1
mv
2
2
m = mass (kg)
v = velocity (m/s)
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Energy & it’s units
The SI unit of energy is the joule, J, pronounced
“jewel.”
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J 
kg  m
s
2
The calorie is a non-SI unit of energy commonly
used by chemists. It was originally defined as the
amount of energy required to raise the temperature
of one gram of water by one degree Celsius.
The exact definition is given by the equation:
1 cal  4.184 J (exact)
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Kinetic Energy, EK
Example 1
A person weighing 75.0 kg (165 lbs) runs a
course at 1.78 m/s (4.00 mph). What is the
person’s kinetic energy?
m = 75.0 kg
v = 1.78 m/s
EK
m

 (75.0 kg)  1.78

2
s 

1
2
EK = ½ mv2 =
119kg.m2 /s2
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Kinetic Energy KE
Example 2
What is the kinetic energy of a 2100-lb car
traveling at 48 miles per hour? (1 lb = 0.4536
kg, 1 mi = 1.609 km)
a) 3.3 × 10–8 J
b)2.2 × 105 J
c) 3.7 × 1019 J
d)1.1 × 106 J
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Potential Energy, EP
The energy an object has by virtue of its position in a field
of force, such as gravitational, electric or magnetic field.
Gravitational potential energy is given by the equation
E P  mgh
m = mass (kg)
g = gravitational constant (9.80 m/s2)
h = height (m)
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Internal Energy
Internal Energy, U
The sum of the kinetic and potential energies of the
particles making up a substance.
Total Energy
Etot = EK + EP + U
In the laboratory, assuming flasks and test tubes are
at rest, both Ek and Ep are equal to zero.
Etotal = U
(internal energy due to molecular motion)
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Law of Conservation of Energy
Energy may be converted from one form to
another, but the total quantity of energy
remains constant.
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Thermodynamic System and Surroundings
Thermodynamic system:
the substance or mixture of substances under study
in which a change occurs surroundings.
Thermodynamic surroundings:
Everything outside the system is surroundings
A system is separated from its surroundings by a boundary
across which matter and/or energy is transferred.
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Heat, q
The energy that flows into or out of a system
because of a difference in temperature between
the thermodynamic system and its surroundings.
Heat flows spontaneously from a region of higher
temperature to a region of lower temperature.
• q is defined as positive if heat is absorbed by the
system (heat is added to the system)
• q is defined as negative if heat is evolved by a
system (heat is subtracted from the system)
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Heat of Reaction
The value of q required to return a system to the
given temperature at the completion of the
reaction (at a given temperature).
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Heat of Reaction
Endothermic Process
A chemical reaction or process in which heat is
absorbed by the system (q is positive). The reaction
vessel will feel cool.
Exothermic Process
A chemical reaction or process in which heat is
evolved by the system (q is negative). The reaction
vessel will feel warm.
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Heat of Reaction
In an endothermic reaction:
The reaction vessel cools.
Heat is absorbed.
Energy is added to the system.
q is positive.
In an exothermic reaction:
The reaction vessel warms.
Heat is evolved.
Energy is subtracted from the system.
q is negative.
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Enthalpy of Reaction
The change in enthalpy for a reaction at a given
temperature and pressure:
DH = H(products) – H(reactants)
Note: D means “change in.”
Enthalpy change is equal to the heat of reaction at
constant pressure:
DH = qP
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Enthalpy of Reaction
Enthalpy, H
An extensive property of a substance that can be
used to obtain the heat absorbed or evolved in a
chemical reaction.
Extensive Property
A property that depends on the amount of
substance. Mass and volume are extensive
properties.
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Enthalpy change ∆H
Example 3
The phrase “the heat absorbed or released by a system
undergoing a physical or chemical change at constant
pressure” is
a) the definition of a state function.
b) the change in enthalpy of the system.
c) a statement of Hess’s law.
d) the change in internal energy of the system.
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Enthalpy change ∆H
Example 4
If ∆ H = –31 kJ for a certain process, that process
a) occurs rapidly.
b) is exothermic.
c) is endothermic.
d) cannot occur.
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The diagram illustrates the enthalpy change for
the reaction
2Na(s) + 2H2O(l)  2NaOH(aq) + H2(g)
The reactants are at
the top. The products
are at the bottom.
The products have less
enthalpy than the
reactants, so enthalpy
is evolved as heat.
The signs of both q and
DH are negative.
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Thermochemical Equations
The 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 for these molar
amounts is written directly after the equation.
For the reaction of sodium metal with water, the
thermochemical equation is:
2Na(s) + 2H2O(l) 
2NaOH(aq) + H2(g); DH = –368.6 kJ
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Thermochemical Equations
A thermochemical equation expresses the enthalpy of the
reaction next to a balanced chemical equation.
1. N2(g) + 3H2(g)
→
2. C6H12O6(s) + 6O2(g) →
3. H2(g) + O2(g)
→
2NH3(g); ΔH = –91.8kJ
CO2(g) + H2O(l); ΔH = –2803kJ
H2O (g); ΔH = –242kJ
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Thermochemical Equation
Manipulating a Thermochemical Equation
• When the equation is multiplied by a factor,
the value of DH must be multiplied by the
same factor.
• When a chemical equation is reversed, the
sign of DH is reversed.
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Manipulating Thermochemical
Equations
Example 5
When sulfur burns in air, the following reaction
occurs:
S8(s) + 8O2(g)  8SO2(g);
DH = – 2.39 x 103 kJ
Write the thermochemical equation for the
dissociation of one mole of sulfur dioxide into
its elements.
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S8(s) + 8O2(g)  8SO2(g); DH = –2.39 × 103 kJ
We want SO2 as a reactant, so we reverse the
given reaction, changing the sign of DH:
8SO2(g)  S8(g) + 8O2(g) ; DH = +2.39 × 103 kJ
We want only one mole SO2, so now we divide
every coefficient and DH by 8:
SO2(g)  1/8S8(g) + O2(g) ; DH = +299 kJ
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Manipulating Thermochemical
Equations
Example 6
You burn 15.0 g sulfur in air. How much heat
evolves from this amount of sulfur? The
thermochemical equation is
S8(s) + 8O2(g)  8SO2(g); DH = -2.39 × 103 kJ
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Manipulating Thermochemical
Equations
S8(s) + 8O2(g)  8SO2(g); DH = -2.39  103 kJ
Molar mass of S8 = 256.52 g
q  1 5 .0 g S 8 
1 m ol S8
 2 .3 9  1 0 k J
3

2 5 6 .5 g S 8
1 m ol S8
q = –1.40  102 kJ
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Manipulating Thermochemical
Equations
Example 7
Given:
4AlCl3(s) + 3O2(g) → 2Al2O3(s) + 6Cl2(g); ∆H = –529.0 kJ
determine ∆H for the following thermochemical
equation.
Cl2(g) + ⅓Al2O3(s) → ⅔AlCl3(s) + ½O2(g)
a) +529.0 kJ
b) +88.2 kJ
c) +176.3 kJ
d) +264.5 kJ
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Applying Stoichiometry to Heats of
Reaction
Example 8
How much heat is evolved upon the complete
oxidation of 9.41 g of aluminum at 25°C and 1
atm pressure? (∆H for Al2O3 is –1676 kJ/mol.)
4Al(s) + 3O2(g) → 2Al2O3(s)
a) 146 kJ
b) 1169 kJ
c) 292 kJ
d) 585 kJ
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Measuring Heats of Reaction
We will first look at the heat needed to raise the
temperature of a substance because this is the
basis of our measurements of heats of reaction.
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Measuring Heats of Reaction
Heat Capacity, C, of a Sample of Substance
The quantity of heat needed to raise the
temperature of the sample of substance by one
degree Celsius (or one Kelvin).
Molar Heat Capacity
The heat capacity for one mole of substance.
Specific Heat Capacity, s (or specific heat)
The quantity of heat needed to raise the
temperature of one gram of substance by one
degree Celsius (or one Kelvin) at constant pressure.
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Measuring Heats of Reaction
The heat required can be found by using the
following equations.
Using specific heat capacity:
q = s x m x Dt
s = specific heat of the substance
m = mass in grams of substance
∆ t = tfinal – tinitial (change in temperature)
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Specific Heat
Example 9
The units for specific heat are
a)J/(g · °C).
b)(J · °C).
c) (J · g).
d)J/°C.
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Example10
A piece of zinc weighing 35.8 g was heated from
20.00°C to 28.00°C. How much heat was
required? The specific heat of zinc is 0.388 J/(g°C).
m = 35.8 g
s = 0.388 J/(g°C)
Dt = 28.00°
q = m  s  Dt
 0.388 J 
 8.00  C 
q  35.8 g  
 g C 
q = 111 J
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Specific Heat
Example 11
How much heat is gained by nickel when 29.2 g of
nickel is warmed from 18.3°C to 69.6°C? The
specific heat of nickel is 0.443 J/(g · °C).
a) 2.37 × 102 J
b)9.00 × 102 J
c) 22.7 J
d)6.64 × 102 J
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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 equals the
sum of the enthalpy changes for the individual
steps
• ∆Hreaction =
∆H1 + ∆H2 + ∆H3
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Hess’s Law of Heat Summation
Suppose we want DH for the reaction
2C(graphite) + O2(g)  2CO(g)
It is difficult to measure directly.
However, two other reactions are known:
C(graphite) + O2(g)  CO2(g); DH = -393.5 kJ
2CO2(g)  2CO(g) + O2(g); DH = – 566.0 kJ
• In order for these to add to give the reaction
we want, we must multiply the first reaction
by 2.
• Note that we also multiply DH by 2.
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Hess’s Law of Heat Summation
we can add the reactions and the DH values.
Cancel the species that appear on both sides.
2C(graphite) + 2O2(g)  2CO2(g); DH = -787.0 kJ
2CO2(g)  2CO(g) + O2(g); DH = – 566.0 kJ
2 C(graphite) + O2(g)  2 CO(g); DH = –1353.0 kJ
2C(graphite) + O2(g)  2CO(g)
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Hess’s Law of Heat Summation
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Hess’s law
Example 12
Consider the following changes:
H2O(s) → H2O(l); ∆H1
H2O(l) → H2O(g); ∆H2
H2O(g) → H2O(s); ∆H3
Using Hess’s law, the sum ∆H1 + ∆H2 + ∆H3 is
a) greater than zero.
b) equal to zero.
c) less than zero.
d) sometimes greater than zero and sometimes less
than zero.
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Hess’s law
Example 13 Given:
Pb(s) + PbO2(s) + 2H2SO4(l) → 2PbSO4(s) + 2H2O(l); ∆H° = –509.2 kJ
SO3(g) + H2O(l) → H2SO4(l); ∆H° = –130. kJ
determine ∆H° for the following equation.
Pb(s) + PbO2(s) + 2SO3(g) → 2PbSO4(s)
a) –3.77 × 103 kJ
b) 3.77 × 103 kJ
c) –639 kJ
d) –521 kJ
e) –769 kJ
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Hess’s law
Example 14 Given:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g); ∆H = –26.8 kJ
FeO(s) + CO(g) → Fe(s) + CO2(g); ∆H = –16.5 kJ
determine ∆H for the following thermochemical
equation.
Fe2O3(s) + CO(g) → 2FeO(s) + CO2(g)
a) –43.3 kJ
b) –10.3 kJ
c) 6.2 kJ
d) 10.3 kJ
e) 22.7 kJ
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Standard Enthalpies of Formation
The term standard state refers to the standard
thermodynamic conditions chosen for substances
when listing or comparing thermodynamic data: 1
atm pressure and the specified temperature
(usually 25°C). These standard conditions are
indicated with a degree sign (°).
When reactants in their standard states yield
products in their standard states, the enthalpy of
reaction is called the standard enthalpy of reaction,
DH°. (DH° is read “delta H zero.”)
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Allotropes
Elements can exist in more than one physical state
Some elements exist in more than one distinct
form in the same physical state. For example,
carbon can exist as graphite or as diamond; oxygen
can exist as O2 or as O3 (ozone).
These different forms of an element in the same
physical state are called allotropes.
The reference form is the most stable form of the
element (both physical state and allotrope).
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Standard enthalpy of formation, DHf°
is the enthalpy change for the formation of one
mole of the substance from its elements in their
reference forms and in their standard states. DHf°
for an element in its reference and standard state is
zero.
For example, the standard enthalpy of formation for
liquid water is the enthalpy change for the reaction
H2(g) + 1/2O2(g)  H2O(l)
DHf° = –285.8 kJ
Other DHf° values are given in Table 6.2 and
Appendix C.
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Finding heat of reaction from Standard
enthalpy of formation, DHf°
Example 15
What is the heat of vaporization of methanol,
CH3OH, at 25°C and 1 atm?
Use standard enthalpies of formation (Appendix
C).
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We want DH° for the reaction:
CH3OH(l)  CH3OH(g)


Δ H reaction 
products
For liquid methanol
For gaseous
Δ H v ap
: ΔH
methanol

f


nΔH f 

nΔH f
reactants
  238.7
: ΔH

f
kJ
mol
  200.7
kJ
mol

kJ   
kJ  


 1 mol   200.7
   1 mol   238.7

mol   
mol  



DHvap= +38.0 kJ
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Finding ΔH from Standard
Heats of Formation
Example 16
All of the following have a standard enthalpy of
formation value of zero at 25°C except
a) CO(g).
b) Fe(s).
c) C(s).
d) F2(g).
e) Ne(g).
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Finding ΔH from Standard
Heats of Formation
Example 17
What is ∆H° for the following phase change?
NaI(s) → NaI(l)
∆H°f (kJ/mol)
NaI(s)
–287.86
NaI(l)
–266.51
a) 554.37 kJ
b) –554.37 kJ
c) 0 kJ
d) –21.35 kJ
e) 21.35 kJ
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Finding ΔH from Standard Heats of Formation
Example 18
What is ∆H° for the following reaction?
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)
C2H2(g)
CO2(g)
H2O(l)
∆H°f (kJ/mol)
+226.7
–393
–285.8
a)
b)
c)
d)
e)
+1692.2 kJ
–2599.0 kJ
+2599.0 kJ
–1692.2 kJ
–452.6 kJ
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