Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 10
Energy Changes in
Chemical Reactions
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Energy is the capacity to do work
Radiant energy comes from the sun and is earth’s
primary energy source
Thermal energy is the energy associated with the
random motion of atoms and molecules
Chemical energy is the energy stored within the
bonds of chemical substances
Nuclear energy is the energy stored within the
collection of neutrons and protons in the atom
Potential energy is the energy available
virtue of an object’s position
by
6.1
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that
are at different temperatures.
Temperature is a measure of the thermal energy.
Temperature = Thermal Energy
400C
greater thermal energy
900C
Energy and Energy Changes
Thermochemistry is the study of heat (the transfer of thermal
energy) in chemical reactions.
Heat is the transfer of thermal energy.
Heat is either absorbed or released during a process.
heat
Surroundings
10.1
Energy and Energy Changes
The system is a part of the universe that is of specific interest.
The surroundings constitute the rest of the universe outside the
system.
System
Surroundings
Universe = System + Surroundings
The system is usually defined as the substances involved in
chemical and physical changes.
Energy and Energy Changes
An exothermic process occurs
when heat is transferred from the
system to the surroundings.
“Feels hot!”
2H2(g) + O2(g)
System
2H2O(l) + energy
heat
Surroundings
Universe = System + Surroundings
Energy and Energy Changes
An endothermic process occurs
when heat is transferred from the
surroundings to the system.
“Feels cold”
energy + 2HgO(s)
System
2Hg(l) + O2(g)
heat
Surroundings
Universe = System + Surroundings
10.2
Introduction to Thermodynamics
Thermodynamics is the study of the interconversion of heat and
other kinds of energy.
In thermodynamics, there are three types
of systems:
An open system can exchange mass and
energy with the surroundings.
A closed system allows the transfer
of energy but not mass.
An isolated system does not exchange
either mass or energy with its
surroundings.
Introduction to Thermodynamics
State functions are properties that are determined by the state of the
system, regardless of how that condition was achieved.
The magnitude of change depends only on the initial and final states
of the system.
 Energy
 Pressure
 Volume
 Temperature
The First Law of Thermodynamics
The first law of thermodynamics states that energy can be converted
from one form to another, but cannot be created or destroyed.
ΔUsys + ΔUsurr = 0
ΔU is the change in the internal energy.
“sys” and “surr” denote system and surroundings, respectively.
ΔU = Uf – Ui; the difference in the energies of the initial and final
states.
ΔUsys = –ΔUsurr
Work and Heat
The overall change in the system’s internal energy is given by:
ΔU = q + w
q is heat
 q is positive for an endothermic process (heat absorbed by the
system)
 q is negative for an exothermic process (heat released by the
system)
w is work
 w is positive for work done on the system
 w is negative for work done by the system
Work and Heat
ΔU = q + w
Worked Example 10.1
Calculate the overall change in internal energy, ΔU, (in joules) for a system that
absorbs 188 J of heat and does 141 J of work on its surroundings.
Strategy Combine the two contributions to internal energy using ΔU = q + w
and the sign conventions for q and w.
Solution The system absorbs heat, so q is positive. The system does work on
the surroundings, so w is negative.
ΔU = q + w = 188 J + (-141 J) = 47 J
Think About It Consult Table 10.1 to make sure you have used the proper sign
conventions for q and w.
10.4 Calorimetry
Calorimetry is the measurement of heat changes.
Heat changes are measured in a device called a calorimeter.
The specific heat (s) of a substance is the amount of heat required to
raise the temperature of 1 g of the substance by 1°C.
Specific Heat and Heat Capacity
The heat capacity (C) is the amount of heat required to raise the
temperature of an object by 1°C.
The “object” may be a given quantity of a particular substance.
heat capacity of 1 kg of water =
4.184 J
 1000 g = 4184 J/C
1 g  C
Specific heat capacity of water
Specific heat capacity has units of J/(g • °C)
Heat capacity has units of J/°C
heat capacity of
1 kg water
Specific Heat and Heat Capacity
The heat associated with a temperature change may be calculated:
q = msΔT
q = CΔT
m is the mass.
S (c) is the specific heat.
ΔT is the change in temperature (ΔT = Tfinal – Tinitial).
C is the heat capacity.
Worked Example 10.4
Calculate the amount of heat (in kJ) required to heat 255 g of water from 25.2°C
to 90.5°C.
Strategy Use q = msΔT to calculate q. s = 4.184 J/g∙°C, m = 255 g,
ΔT = 90.5°C – 25.2°C = 65.3°C.
Solution
4.184 J
4 J or 69.7 kJ
q=
×
255
g
×
65.3°C
=
6.97×10
g∙°C
Think About It Look carefully at the cancellation of units and make sure that
the number of kilojoules is smaller than the number of joules. It is a common
error to multiply by 1000 instead of dividing in conversions of this kind.
Calorimetry
Calculate the amount of heat required to heat 1.01 kg of water from
0.05°C to 35.81°C.
Solution:
Step 1:Use the equation q = msΔT to calculate q.
q  1.01 kg 
1000 g 4.184 J

 [35.81C  0.05C] = 151000 J = 151 kJ
1 kg
g  C
Calorimetry
A coffee-cup calorimeter may be used to measure the heat
exchange for a variety of reactions at constant pressure:
Heat of neutralization:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Heat of ionization:
H2O(l) → H+(aq) + OH‒(aq)
Heat of fusion:
H2O(s) → H2O(l)
Heat of vaporization:
H2O(l) → H2O(g)
Calorimetry
Concepts to consider for coffee-cup calorimetry:
qP = nΔH
System:
reactants and products (the reaction)
Surroundings: water in the calorimeter
For an exothermic reaction:

the system loses heat

the surroundings gain (absorb) heat
qsys = −msΔT
qsurr = msΔT
qsys = −qsurr
The minus sign is used to keep
sign conventions consistent.
Worked Example 10.5
A metal pellet with a mass of 100.0 g, originally at 88.4°C, is dropped into 125
g of water originally at 25.1°C. The final temperature of both pellet and the
water is 31.3°C. Calculate the heat capacity C (in J/°C) of the pellet.
Strategy Water constitutes the surroundings; the pellet is the system. Use qsurr =
msΔT to determine the heat absorbed by the water; then use q = CΔT to determine
the heat capacity of the metal pellet.
mwater = 125 g, swater = 4.184 J/g∙°C, and ΔTwater = 31.3°C – 25.1°C = 6.2°C.
The heat absorbed by the water must be released by the pellet: qwater = -qpellet,
mpellet = 100.0 g, and ΔTpellet = 31.3°C – 88.4°C = -57.1°C.
Worked Example 10.5 (cont.)
Solution
4.184 J
qwater =
g∙°C × 125 g × 6.2°C = 3242.6 J
Thus,
qpellet = -3242.6 J
From q = CΔT we have
-3242.6 J = Cpellet × (-57.1°C)
Thus,
Cpellet = 57 J/°C
Think About It The units cancel properly to give appropriate units for heat
capacity. Moreover, ΔTpellet is a negative number because the temperature of the
pellet decreases.
Constant-Volume Calorimetry
Constant volume calorimetry is carried out in a device known as a
constant-volume bomb.
A constant-volume
calorimeter is an isolated
system.
Bomb calorimeters are
typically used to determine
heats of combustion.
qcal = −qrxn
Constant-Volume Calorimetry
To calculate qcal, the heat capacity of the calorimeter must be
known.
qcal = CcalΔT
qrxn = −qcal
qrxn = −CcalΔT
Qsystem = - Qsurroundings
If the system (reaction) gives off heat it is said to be exothermic
this means the surroundings get warmer
If the system (reaction) takes in heat it is said to be endothermic
this means the surroundings get colder
Bomb Calorimetry
A device that enables us to measure the heat transfer
Between the system and the surroundings
When finding the specific heat of a metal
Q of the surroundings is equal to the specific heat of the
the water X the mass of the water X the change
in temperature of the water
Q of the metal is equal to –Q of the surroundings
Then, the specific heat (c) of the metal is equal to
–Q of the surroundings
(mass of the metal) X (the change in temp of the metal)
A piece of stainless steel of mass 25.0 g at 88.0 °C
was placed in a calorimeter that contained 150.0 g
of water at 20.0 °C. If the temperature of the water
rose to 21.4 °C, what is the specific heat capacity
of the stainless steel? The specific heat capacity of
water is 4.184 J/ g·°C
Ans:
0.53 J/ g ·°C
A piece of graphite of mass 75.0 g at 92.0 °C was
placed in a calorimeter that contained 200.0 g of
water at 12.0 °C. If the temperature of the water rose
to 16.8 °C, what is the specific heat capacity of
graphite? The specific heat capacity of water is
4.184 J/ g·°C
Ans: 0.71 J/ g·°C
Calculate the heat required to raise the temperature of
50.0 g of water from 25.0 °C to 39.0 °C. The specific
heat capacity of water is 4.184 J/ g·°C
Ans: 2.93 kJ
How much heat is absorbed by a 25.0-g sample of gold
at 25.0 °C when it is immersed in boiling water? The
specific heat capacity of gold is 0.128 J/ g·°C and the
specific heat capacity of water is 4.184 J/ g·°C.
Ans: 0.240 kJ
A reaction occurs in a calorimeter that contains
120.0 g of water at 25.00 °C, and 3.60 kJ of heat
is evolved. If the specific heat capacity of water is
4.184 J/ g·°C, what is the final temperature of the
water?
Ans:
32.17 °C
Bomb Calorimetry and DH
Bomb calorimetry can be used to determine the Heat of Reaction (DHrxn)
the Heat of formation (DHf) and the Heat of solution (DHsoln)
Q of the surroundings is equal to the total mass of the solution X
specific heat of the solution X the change in temp of the solution
Q of the reaction is equal to – Q of the surroundings
Q of the reaction is equal to n DH. Where n is equal to the moles of reaction
Therefore DH equals - (mass of soln X D temp of soln X spec. heat of soln)
n of reaction
When 5.00 g of ammonium nitrate are dissolved in
1.00 L of water in a calorimeter with a heat
capacity of 2.30 kJ/ C, the temperature drops
0.760 C. What is the heat of solution of
ammonium nitrate?
Ans:
+28.0 kJ/ mol
If 100 mL of 1.00 M barium chloride are added to 100
mL of 1.00 M potassium sulfate in a calorimeter with a
heat capacity of 1.52 kJ/ C, the temperature rises by
1.73 C. Calculate the heat of reaction for
Ba2+(aq) + SO42–(aq)  BaSO4(s)
Ans:
–26.3 kJ
The standard enthalpy of formation (heat of
formation) of H2O2(l) is –187.8 kJ/ mol and of H2O(l) is
–285.8 kJ/ mol. (Oxygen is in it’s standard state)
Calculate the reaction enthalpy for
2H2O2(l)  2H2O(l) + O2(g)
Ans:
–196.0 kJ
Chemistry in Action:
Fuel Values of Foods and Other Substances
C6H12O6 (s) + 6O2 (g)
6CO2 (g) + 6H2O (l) DH = -2801 kJ/mol
1 cal = 4.184 J
1 Cal = 1000 cal = 4184 J
Worked Example 10.6
A Famous Amos bite-sized chocolate chip cookie weighing 7.25 g is burned in a
bomb calorimeter to determine its energy content. The heat capacity of the
calorimeter is 39.97 kJ/°C. During the combustion, the temperature of the water
in the calorimeter increases by 3.90°C. Calculate the energy content (in kJ/g) of
the cookie.
Think About It According to the label on the cookie package, a
service
size
or 29 g, the
andheat
eachreleased
servingby
contains
150
Strategy
Use
qrxnis=four
-Ccalcookies,
ΔT to calculate
the combustion
of
Cal. Convert
the heat
energy
per gram
to Calories
per cookie
servingtotodetermine
verify its
the cookie.
Divide the
released
by the
mass of the
result.per gram C = 39.97 kJ/°C and ΔT = 3.90°C.
energythe
content
cal
21.5 kJ
1 Cal
29 g
= 1.5×102 Cal/serving
×
×
g
4.184 kJ
serving
Solution
qrxn = -CcalΔT = -(39.97 kJ/°C)(3.90°C) = -1.559×102 kJ
Because energy content is a positive quantity, we write
1.559×102 kJ
energy content per gram =
= 21.5 kJ/g
7.25 g
Enthalpy and Enthalpy Changes
The enthalpy of reaction (ΔH) is the difference between the
enthalpies of the products and the enthalpies of the reactants:
ΔH = H(products) – H(reactants)
Assumes reactions in the lab occur at constant pressure
ΔH > 0 (positive) endothermic process
ΔH < 0 (negative) exothermic process
Without looking up the values, What is the
standard heat of reaction for
2HN3(g)  H2(g) + 3N2(g)
Ans
–2 DHf°[HN3(g)]
Thermochemical Equations
H2O(s)
H2O(l)
ΔH = +6.01 kJ/mol
Concepts to consider:
Is this a constant pressure process?
What is the system?
What are the surroundings?
ΔH > 0 endothermic
Thermochemical Equations
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
ΔH = −890.4 kJ/mol
Concepts to consider:
Is this a constant pressure process?
What is the system?
What are the surroundings?
ΔH < 0 exothermic
Thermochemical Equations
Enthalpy is an extensive property.
Extensive properties are dependent on the amount of matter involved.
H2O(l) → H2O(g)
Double the amount of matter
2H2O(l) → 2H2O(g)
ΔH = +44 kJ/mol
Double the enthalpy
ΔH = +88 kJ/mol
Units refer to
mole of reaction as written
Thermochemical Equations
The following guidelines are useful when considering
thermochemical equations:
1) Always specify the physical states of reactants and products
because they help determine the actual enthapy changes.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
different states
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
ΔH = −802.4 kJ/mol
different enthalpies
ΔH = +890.4 kJ/mol
Thermochemical Equations
The following guidelines are useful when considering
thermochemical equations:
2) When multiplying an equation by a factor (n), multiply the ΔH
value by same factor.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
ΔH = − 802.4 kJ/mol
2CH4(g) + 4O2(g)
2CO2(g) + 4H2O(g)
ΔH = − 1604.8 kJ/mol
3) Reversing an equation changes the sign but not the magnitude of
ΔH.
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
ΔH = − 802.4 kJ/mol
CO2(g) + 2H2O(g)
CH4(g) + 2O2(g)
ΔH = +802.4 kJ/mol
Worked Example 10.3
Given the thermochemical equation for photosynthesis,
6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g)
ΔH = +2803 kJ/mol
calculate the solar energy required to produce 75.0 g of C6H12O6.
Strategy The thermochemical equation shows that for every mole of C6H12O6
produced, 2803 kJ is absorbed. We need to find out how much energy is absorbed
for the production of 75.0 g of C6H12O6. We must first find out how many moles
there are in 75.0 g of C6H12O6.
The molar mass of C6H12O6 is 180.2 g/mol, so 75.0 g of C6H12O6 is
75.0 g C6H12O6 ×
1 mol C6H12O6
180.2 g C6H12O6 = 0.416 mol C6H12O6
We will multiply the thermochemical equation, including the enthalpy change, by
0.416, in order to write the equation in terms of the appropriate amount of
C6H12O6.
Worked Example 10.3 (cont.)
Solution
(0.416 mol)[6H2O(l) + 6CO2(g) → C6H12O6(s) + 6O2(g)]
and
(0.416 mol)(ΔH) = (0.416 mol)(2803 kJ/mol) gives
2.50H2O(l) + 2.50CO2(g) → 0.416C6H12O6(s) + 2.50O2(g)
ΔH = +1.17×103 kJ
Therefore, 1.17×103 kJ of energy in the form of sunlight is consumed in the
production of 75.0 g of C6H12O6. Note that the “per mole” units in ΔH are
canceled when we multiply the thermochemical equation by the number of moles
of C6H12O6.
Think About It The specified amount of C6H12O6 is less than half a mole.
Therefore, we should expect the associated enthalpy change to be less than half
that specified in the thermochemical equation for the production of 1 mole of
C6H12O6.
10.5 Hess’s Law
Hess’s law states that the change in enthalpy for a stepwise process is
the sum of the enthalpy changes for each of the steps.
CH4(g) + 2O2(g)
2H2O(l)
CH4(g) + 2O2(g)
CO2(g) + 2H2O(l)
ΔH = −890.4 kJ/mol
2H2O(g)
ΔH = +88.0 kJ/mol
CO2(g) + 2H2O(g)
ΔH = −802.4 kJ/mol
Enthalpy
CH4(g) + 2O2(g)
ΔH = −802.4 kJ
ΔH = −890.4 kJ
CO2(g) + 2H2O(g)
CO2(g) + 2H2O(l)
ΔH = +88.0 kJ
Hess’s Law
When applying Hess’s Law:
1) Manipulate thermochemical equations in a manner that gives the
overall desired equation.
2) Remember the rules for manipulating thermochemical equations:
Always specify the physical states of reactants and products because
they help determine the actual enthalpy changes.
When multiplying an equation by a factor (n), multiply the ΔH value by
same factor.
Reversing an equation changes the sign but not the magnitude of ΔH.
3) Add the ΔH for each step after proper manipulation.
4) Process is useful for calculating enthalpies that cannot be found
directly.
Worked Example 10.7
Given the following thermochemical equations,
NO(g) + O3(g) → NO2(g) + O2(g)
3
O3(g) → O2(g)
2
O2(g) → 2O(g)
ΔH = –198.9 kJ/mol
ΔH = –142.3 kJ/mol
ΔH = +495 kJ/mol
determine the enthalpy change for the reaction
NO(g) + O(g) → NO2(g)
Strategy Arrange the given thermochemical equations so that they sum to the
desired equation. Make the corresponding changes to the enthalpy changes, and
add them to get the desired enthalpy change.
Worked Example 10.7 (cont.)
Solution The first equation has NO as a reactant with the correct coefficients, so
we will use it as is.
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = –198.9 kJ/mol
The second equation must be reversed so that the O3 introduced by the first
equation will cancel (O3 is not part of the overall chemical equation). We also
must change the sign on the corresponding ΔH value.
3
ΔH = +142.3 kJ/mol
O (g) → O3(g)
2 2
These two steps sum to give:
NO(g) + O3(g) → NO2(g) + O2(g)
1
3
+ O2(g) O2(g) → O3(g)
2
2
1
NO(g) + O2(g) → NO2(g)
2
ΔH = –198.9 kJ/mol
ΔH = +142.3 kJ/mol
ΔH = –56.6 kJ/mol
Worked Example 10.7 (cont.)
1
Solution We then replace 2 O2 on the left with O by incorporating the last
equation. To do so, we divide the third equation by 2 and reverse its direction. As
a result, we must also divide ΔH value by 2 and change its sign.
1
O(g) → O2(g)
ΔH = –247.5 kJ/mol
2
Finally, we sum all the steps and add their enthalpy changes.
NO(g) + O3(g) → NO2(g) + O2(g)
3
O (g) → O3(g)
2 2
1
+ O(g) → O2(g)
2
NO(g) + O(g) → NO2(g)
ΔH = –198.9 kJ/mol
ΔH = +142.3 kJ/mol
ΔH = –247.5 kJ/mol
ΔH = –304 kJ/mol
Think About It Double-check the cancellation of identical items–especially
where fractions are involved.
The standard enthalpy of formation of NO(g) is +90.3 kJ/
mol and of NO2(g) is +33.2 kJ/ mol. (Oxygen is in
It’s standard state)
Calculate the reaction enthalpy (heat of reaction) for
2NO(g) + O2(g)  2NO2(g)
Ans:
–114.2 kJ
The heat of reaction for
3PbO2(s)  Pb3O4(s) + O2(g) is +22.8 kJ/ mol.
The standard heat of formation of Pb3O4(s) is –175.6 kJ/ mol.
What is the standard heat of formation of PbO2(s)?
Ans:
–66.1 kJ/mol
The heat of reaction for 2SO3(g)  2SO2(g) + O2(g ) is
+198.4 kJ
The standard heat of formation of SO2(g) is –296.8 kJ/ mol.
What is the standard heat of formation of SO3(g)?
Ans:
–396.0 kJ/ mol
Calculate the heat of reaction for
H2O(l)  H2(g) + ½ O2(g)
from the data
2H2(g) + O2(g)  2H2O(l)
Ans:
DH° = –571.6 kJ
+285.8 kJ
For the combustion of methane, DHr = -890 kJ/mol.
What is the heat of reaction for the following reaction?
4H2O(l) + 2CO2(g)  2CH4(g) + 4O2(g)
Ans:
+1780 kJ
Calculate the heat of reaction for
CH4(g) + 4S(s)  CS2(l) + 2H2S(g)
from the data
C(s) + 2H2(g)  CH4(g)DH° = –74.8 kJ
C(s) + 2S(s)  CS2(l) DH° = +87.9 kJ
S(s) + H2(g)  H2S(g) DH° = –20.6 kJ
Ans:
+121.5 kJ
Calculate the heat of reaction for
2PbO(s) + O2(g)  2PbO2(s)
from the data
2Pb(s) + O2(g)  2PbO(s)
DH° = –435.8 kJ
Pb(s) + O2(g)  PbO2(s)
DH° = –276.6 kJ
Ans:
–117.4 kJ
Calculate the heat of reaction for
C6H6(l)  3C2H2(g)
from the data
2C2H2(g) + 5O2(g)  4CO2(g) + 2H2O(l) DH° = –2600 kJ
2C6H6(l) + 15O2(g)  12CO2(g) + 6H2O(l)
Ans:
+632 kJ
DH° = –6536 kJ
Calculate the heat of reaction for the combustion of
methane, CH4(g),
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
from the data
C(s) + O2(g)  CO2(g) DH° = –393.5 kJ
H2(g) + ½ O2(g)  H2O(l) DH° = –285.8 kJ
C(s) + 2H2(g)  CH4(g)DH° = –74.8 kJ
Ans:
–890.3 kJ
10.6
Standard Enthalpies of Formation
The standard enthalpy of formation (ΔH °)
f is defined as the heat
change that results when 1 mole of a compound is formed from its
constituent elements in their standard states.
C(graphite) + O2(g)
CO2(g)
Elements in standard
states
1 mole of product
ΔH f° = −393.5 kJ/mol
Standard Enthalpies of Formation
The standard enthalpy of formation (ΔH f°) is defined as the heat
change that results when 1 mole of a compound is formed from its
constituent elements in their standard states.
The superscripted degree sign denotes standard conditions.
1 atm pressure for gases
1 M concentration for solutions
“f” stands for formation.
ΔH f° for an element in its most stable form is zero.
ΔH f° for many substances are tabulated in Appendix 2 of the
textbook.
Standard Enthalpies of Formation
The standard enthalpy of reaction (ΔH°rxn) is defined as the
enthalpy of a reaction carried out under standard conditions.
aA + bB → cC + dD
ΔH°rxn = [cΔH °(C)
+ dΔH °(D)
] – [aΔH °(A)
+ bΔH °(B)]
f
f
f
f
ΔH°rxn = ΣnΔH °(products)
– ΣmΔH °(reactants)
f
f
n and m are the stoichiometric coefficients for the reactants and
products.
Worked Example 10.8
Using data from Appendix 2, calculate ΔH°rxn for Ag+(aq) + Cl-(aq) → AgCl(s).
Strategy Use ΔH°rxn = ΣnΔH °(products)
– ΣmΔH °(reactants)
and ΔH °f values
f
f
from Appendix 2 to calculate ΔH°rxn. The ΔH °f values for Ag+(aq), Cl-(aq), and
AgCl(s) are +105.9, –167.2, and –127.0 kJ/mol, respectively.
Solution
ΔH°rxn = ΔH f°(AgCl) – [ΔH f°(Ag+) + ΔH f°(Cl-)]
= –127.0 kJ/mol – [(+105.9 kJ/mol) + (–167.2 kJ/mol)]
= –127.0 kJ/mol – (–61.3 kJ/mol) = –65.7 kJ/mol
Think About It Watch out for misplaced or missing minus signs. This is an easy
place to lose track of them.
Worked Example 10.9
Given the following information, calculate the standard enthalpy of formation of
acetylene (C2H2) from its constituent elements:
C(graphite) + O2(g) → CO2(g)
1
H2(g) + O2(g) → H2O(l)
2
2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(l)
ΔH°rxn = –393.5 kJ/mol (1)
ΔH°rxn = –285.5 kJ/mol (2)
ΔH°rxn = –2598.8 kJ/mol (3)
Strategy Arrange the equations that are provided so that they will sum to the
desired equation. This may require reversing or multiplying one or more of the
equations. For any such change, the corresponding change must also be made to
the ΔH°rxn value. The desired equation, corresponding to the standard enthalpy of
formation of acetylene, is
2C(graphite) + H2(g) → C2H2(g)
Worked Example 10.9 (cont.)
Solution We multiply Equation (1) and its ΔH°rxn value by 2:
2C(graphite) + 2O2(g) → 2CO2(g)
ΔH°rxn = –787.0 kJ/mol
We include Equation (2) and its ΔH°rxn value as is:
1
ΔH°rxn = –285.5 kJ/mol
H2(g) + 2 O2(g) → H2O(l)
We reverse Equation (3) and divide it by 2 (i.e., multiply through by 1/2):
5
Think
About
It Remember
only a=ΔH°
there is
rxn isΔH°
f whenkJ/mol
+1299.4
2CO2(g)
+ H2O(l)
→ C2H2(g)that
+ 2aOΔH°
(g)
rxn
2
just one product, just one mole produced,
and all the reactants are
elements
in their equations
standard states.
Summing
the resulting
and the corresponding ΔH° values:
rxn
2C(graphite) + 2O2(g) → 2CO2(g)
ΔH°rxn = –787.0 kJ/mol
ΔH°rxn = –285.5 kJ/mol
1
H2(g) + 2 O2(g) → H2O(l)
5
2CO2(g) + H2O(l) → C2H2(g) + 2 O2(g)
2C(graphite) + H2(g) → C2H2(g)
ΔH°rxn = +1299.4 kJ/mol
ΔH°f = +226.6 kJ/mol
10.7
Bond Enthalpy and the Stability of Covalent
Molecules
The bond enthalpy is the enthalpy change associated with breaking
a bond in 1 mole of gaseous molecule.
H2(g) → H(g) + H(g) ΔH° = 436.4 kJ/mol
The enthalpy for a gas phase reaction is given by:
ΔH° = ΣBE(reactants) – ΣBE(products)
ΔH° = total energy input – total energy released
bonds broken
bonds formed
Bond Enthalpy and the Stability of Covalent Molecules
Bond enthalpy change in an exothermic reaction.:
Bond Enthalpy and the Stability of Covalent Molecules
Bond enthalpy change in an endothermic reaction:
Worked Example 10.10
Use bond enthalpies from Table 10.4 to estimate the enthalpy of reaction for the
combustion of methane:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)
Think
About
It structures
Use equation
ΔH°rxn = ΣnΔH
Strategy
Draw
Lewis
to determine
what °(products)
fbonds are to–be broken and
ΣmΔHare
°(reactants)
and data from Appendix 2 to calculate this
what bonds
f to be formed.
enthalpy of reaction again; then compare your results using the two
approaches. The difference in this case is due to two things: most
tabulated bond enthalpies are averages and, by convention, we show
the product of combustion as liquid water–but average bond
species
Bonds enthalpies
to break: 4apply
C–H to
and
2 O=Oin the gas phase, where there is little or
influence
exerted
neighboring molecules.
Bonds no
to form:
2 C=O
and 4byH–O
Bond enthalpies from Table 10.4: 414 kJ/mol (C–H), 498.7 kJ/mol (O=O),
799 kJ/mol (C=O in CO2), and 460 kJ/mol (H–O).
Solution
[4(414 kJ/mol) + 2(498.7 kJ/mol)] – [2(799 kJ/mol) + 4(460 kJ/mol)] = –785 kJ/mol