Chapter 17.1 : Thermochemistry

Chapters 1. Reaction Energy and Reaction Kinetics
2. Chemical Equilibrium
3. Oxidation-Reduction Reactions

1. Define temperature and state the units in which it
is measured.
2. Define heat and state its units.
3. Perform specific-heat calculations.
4. Explain enthalpy change, enthalpy of reaction,
enthalpy of formation, and enthalpy of combustion.
5. Solve problems involving enthalpies of reaction,
enthalpies of formation, and enthalpies of
combustion.

Thermochemistry
• Virtually every chemical
reaction is accompanied by a
change in energy.
• Chemical reactions usually
either absorb or release energy
as heat.
• Thermochemistry is the study of
the transfers of energy as heat
that accompany chemical
reactions and physical changes.

Heat and Temperature
• The energy absorbed or released as heat in a chemical or
physical change is measured in a calorimeter.
• In one kind of calorimeter, known quantities of reactants
are sealed in a reaction chamber that is immersed in a
known quantity of water.
• Energy given off by the reaction is absorbed by the
water, and the temperature change of the water is
measured.
• From the temperature change of the water, it is
possible to calculate the energy as heat given off
by the reaction.

• Temperature is a measure of the average kinetic energy
of the particles in a sample of matter.
• The greater the kinetic energy of the particles in a
sample, the hotter it feels.
• For calculations in thermochemistry, the Celsius
and Kelvin temperature scales are used.
K = 273.15 + °C

• The amount of energy transferred as heat is usually
measured in joules.
• A joule is the SI unit of heat as well as all other forms
of energy.
• Heat can be thought of as the energy transferred
between samples of matter because of a difference in
their temperatures.
• Energy transferred as heat always moves
spontaneously from matter at a higher temperature
to matter at a lower temperature.

Specific Heat
• The amount of energy transferred as heat during a
temperature change depends
1. Nature of the material changing temperature,
2. Mass.
• The specific heat of a substance is the amount of
energy required to raise the temperature of one gram
by one Celsius degree (1°C) or one kelvin (1 K).
• Values of specific heat are usually given in units of
joules per gram per Celsius degree, J/(g•°C), or
joules per gram per kelvin, J/(g•K).

Chapter 17.1 : Thermochemistry

• Specific heat is calculated according to the
equation given below.
p
q
c
m T

 
pq c m T   
• cp is the specific heat at a given pressure, q is the
energy lost or gained, m is the mass of the sample,
and ∆T is the difference between the initial and
final temperatures.
• The above equation can be rearranged to given an
equation that can be used to find the quantity of
energy gained or lost with a change of temperature.

Sample Problem A
A 4.0 g sample of glass was heated from 274 K to 314 K, a
temperature increase of 40. K, and was found to have absorbed 32
J of energy as heat.
a. What is the specific heat of this type of glass?
b. How much energy will the same glass sample
gain when it is heated from 314 K to 344 K?
 Given: m = 4.0 g
 ∆T = 40. K
 q = 32 J
 Unknown: a. cp in J/(g•K)
 b. q for ∆T of 314 K → 344 K
 Solution:
 a. 32 J
(4.0 g)(40. K)
0.20 J/(g K)p
q
c
m T
  




 Sample Problem A Solution, continued
 Solution:
 b.
0.20 J
(4.0 g)(30 K)
(g K)
24 Jq  

pq c m T   
0.20 J
(4.0 g)(344 K 314 K)
(g K)
q  


• The energy absorbed as heat during a chemical
reaction at constant pressure is represented by ∆H. H
is the symbol for a quantity called enthalpy.
• Only changes in enthalpy can be measured. ∆H is read
as “change in enthalpy.”
• An enthalpy change is the amount of energy absorbed
by a system as heat during a process at constant
pressure.

• Enthalpy change is always the difference between
the enthalpies of products and reactants.
∆H = Hproducts – Hreactants
• A chemical reaction that releases energy is
exothermic, and the energy of the products is less
than the energy of the reactants.
• example:
2H2(g) + O2(g) → 2H2O(g) + 483.6 kJ

2H2(g) + O2(g) → 2H2O(g) + 483.6 kJ
• The expression above is an example of a thermochemical
equation, an equation that includes the quantity of energy
released or absorbed as heat during the reaction as written.
• Chemical coefficients in a thermochemical equation
should be interpreted as numbers of moles and never as
numbers of molecules.
• The quantity of energy released is proportional to the
quantity of the reactions formed.
 Producing twice as much water in the equation shown on
the previous slide would require twice as many moles of
reactants and would release 2 × 483.6 kJ of energy as heat.

• In an endothermic reaction, the products have a
higher energy than the reactants, and the reaction
absorbs energy.
• example:
2H2O(g) + 483.6 kJ → 2H2(g) + O2(g)
• The physical states of reactants and products must
always be included in thermochemical equations,
because the states of reactants and products
influence the overall amount of energy exchanged.

• Thermochemical equations are usually written
by designating a ∆H value rather than writing
the energy as a reactant or product.
• For an exothermic reaction, ∆H is negative
because the system loses energy.
• The thermochemical equation for the
exothermic reaction previously discussed
will look like the following:
2H2(g) + O2(g) → 2H2O(g) ∆H = –483.6 kJ

• In an exothermic reaction, energy is evolved, or given
off, during the reaction; ∆H is negative.

• In an endothermic reaction, energy is absorbed; in
this case, ∆H is designated as positive.

• The molar enthalpy of formation is the enthalpy
change that occurs when one mole of a compound is
formed from its elements in their standard state at
25°C and 1 atm.
• Enthalpies of formation are given for a standard
temperature and pressure so that comparisons between
compounds are meaningful.
• To signify standard states, a 0 sign is added to the
enthalpy symbol, and the subscript f indicates a
standard enthalpy of formation:

• Some standard enthalpies of formation are given in
the appendix of your book (pg. 902).
• Each entry in the table is the enthalpy of formation
for the synthesis of one mole of the compound from
its elements in their standard states.
• The thermochemical equation to accompany an
enthalpy of formation shows the formation of one
mole of the compound from its elements in their
standard states.

• Compounds with a large negative enthalpy of
formation are very stable.
• example: the of carbon dioxide is –393.5 kJ
per mol of gas produced.
• Elements in their standard states are defined as
having = 0.
• This indicates that carbon dioxide is more stable
than the elements from which it was formed.
0
fH
0
fH

• Compounds with positive values of enthalpies of
formation are typically unstable.
• example: hydrogen iodide, HI, has a of +26.5
kJ/mol.
• It decomposes at room temperature into violet
iodine vapor, I2, and hydrogen, H2.
0
fH

• The enthalpy change that occurs during the complete
combustion of one mole of a substance is called the
enthalpy of combustion of the substance.
• Enthalpy of combustion is defined in terms of one
mole of reactant, whereas the enthalpy of formation
is defined in terms of one mole of product.
• ∆H with a subscripted c, ∆Hc, refers specifically to
enthalpy of combustion.

• A combustion calorimeter, shown below, is a
common instrument used to determine enthalpies
of combustion.

• The basis for calculating enthalpies of reaction is
known as Hess’s law: the overall enthalpy change
in a reaction is equal to the sum of enthalpy
changes for the individual steps in the process.
• This means that the energy difference between
reactants and products is independent of the route
taken to get from one to the other.

• If you know the reaction enthalpies of individual steps in an
overall reaction, you can calculate the overall enthalpy
without having to measure it experimentally.
• To demonstrate how to apply Hess’s law, we will work
through the calculation of the enthalpy of formation for the
formation of methane gas, CH4, from its elements, hydrogen
gas and solid carbon:
C(s) + 2H2(g) → CH4(g) fH0
? 

• The component reactions in this case are the
combustion reactions of carbon, hydrogen, and
methane:
0
393.5 kJcH  
0
285.8 kJcH  
0
890.8 kJcH  
H2(g) + ½O2(g) → H2O(l)
C(s) + O2(g) → CO2(g)
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
• The overall reaction involves the formation rather than the
combustion of methane, so the combustion equation for
methane is reversed, and its enthalpy changed from negative
to positive:
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ∆H0 = +890.8 kJ

• Because 2 moles of water are used as a reactant in the above
reaction, 2 moles of water will be needed as a product.
• Therefore, the coefficients for the formation of water reaction, as
well as its enthalpy, need to be multiplied by 2:
2H2(g) + O2(g) → 2H2O(l) cH0
2( 285.8 kJ)  
• We are now ready to add the three equations together using
Hess’s law to give the enthalpy of formation for methane and
the balanced equation.
0
393.5 kJcH  
0
2( 285.8 kJ)cH  
0
890.8 kJH  
0
74.3 kJfH  
2H2(g) + O2(g) → 2H2O(l)
C(s) + O2(g) → CO2(g)
C(s) + 2H2(g) → CH4(g)
CO2(g) + 2H2O(l) → CH4(g) + 2O2(g)

• Using Hess’s law, any enthalpy of reaction may be
calculated using enthalpies of formation for all the
substances in the reaction of interest, without knowing
anything else about how the reaction occurs.
• Mathematically, the overall equation for enthalpy
change will be in the form of the following equation:
∆H0 = sum of [( of products) × (mol of products)]
– sum of [( of reactants) × (mol of reactants)]
fH0

fH0


Calculate the enthalpy of reaction for the combustion of
nitrogen monoxide gas, NO, to form nitrogen dioxide gas, NO2,
as given in the following equation.
NO(g) + ½O2(g) → NO2(g)
Use the enthalpy-of-formation data in the appendix. Solve by
combining the known thermochemical equations.
Given:
Unknown:
g g g k01
2 2 2 f2
N ( ) + O ( ) NO ( ) ΔH =+33.2 J
 g + g g =+ k1 1
2 22 2
0
fN ( ) O ( ) NO( H 90.29) J
0
H g + g g1
2 22
for NO( ) O ( ) NO ( ) 
Solution:
Using Hess’s law, combine the given thermochemical equations in
such a way as to obtain the unknown equation, and its ∆H0 value.

The desired equation is:
g + g g1
2 22
NO( ) O ( ) NO ( )
 g g + g = k1 1
22 2 f2
0
NO( ) N ( ) O ( H – 90.29) J
g g g k01
2 2 2 f2
N ( ) + O ( ) NO ( ) ΔH =+33.2 J
The other equation should have NO2 as a product, so we
can use the second given equation as is:
Reversing the first given reaction and its sign yields the
following thermochemical equation:

We can now add the equations and their ∆H0 values to
obtain the unknown ∆H0 value.
g g g k01
2 2 2 f2
N ( ) + O ( ) NO ( ) H =+33.2 J 
 g g + g = k1 1
22 2 f2
0
NO( ) N ( ) O ( H – 90.29) J
0
57.1 kJH  g + g g1
2 22
NO( ) O ( ) NO ( )

• When carbon is burned in a limited supply of
oxygen, carbon monoxide is produced:
s + g g1
22
C( ) O ( ) CO( )
• The above overall reaction consists of two reactions:
1) carbon is oxidized to carbon dioxide
2) carbon dioxide is reduced to give carbon
monoxide.

• Because these two reactions occur simultaneously, it is not
possible to directly measure the enthalpy of formation of CO(g)
from C(s) and O2(g).
• We do know the enthalpy of formation of carbon dioxide and
the enthalpy of combustion of carbon monoxide:
fH0
2 2C(s) + O (g) CO (g) 393.5 kJ/mol   
cg g g H01
2 22
CO( ) + O ( ) CO ( ) 283.0 kJ/mol   
g g g H01
2 22
CO ( ) CO( ) + O ( ) 283.0 kJ/mol   
H0
2 2C(s) + O (g) CO (g) 393.5 kJ/mol   
• We reverse the second equation because we need CO as a
product. Adding gives the desired enthalpy of formation of
carbon monoxide.
0
110.5 kJH  s + g g1
22
C( ) O ( ) CO( )

• The graph below models the process just described. It
shows the enthalpies of reaction for CO2 and CO.

Calculate the enthalpy of formation of pentane, C5H12, using
the information on enthalpies of formation and the
information on enthalpies of combustion in the appendix.
Solve by combining the known thermochemical equations.
0
393.5 kJfH  
0
3535.6 kJcH  
0
285.8 kJfH  
Given: C(s) + O2(g) → CO2(g)
Unknown: for 5C(s) + 6H2(g) → C5H12(g)
Solution:
Combine the given equations according to Hess’s law.
H2(g) + ½O2(g) → H2O(l)
C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(l)
0
fH

0
5( 393.5 kJ)H  
0
6( 285.8 kJ)H  
0
145.7 kJfH  
0
3535.6 kJH  
5C(s) + 5O2(g) → 5CO2(g)
6H2(g) + 3O2(g) → 6H2O(l)
5CO2(g) + 6H2O(l) → C5H12(g) + 8O2(g)
5C(s) + 6H2(g) → C5H12(g)

Chapter 17.1 : Thermochemistry

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Chapter 17.1 : Thermochemistry