Standard Enthalpies of Formation
Calorimetry and Hess’s law are two ways of determining enthalpies of reaction. A third
method uses tabulated enthalpy changes (standard enthalpies of formation) for a special set of reactions called formation reactions, in which compounds are formed from
their elements. For example, the formation reaction and standard enthalpy of formation
for carbon dioxide are:
C(s) + O2(g) → CO2(g)
H f̊ 393.5 kJ/mol
Both the elements on the left side of the equation are in their standard states — their
most stable form at SATP (25°C and 100 kPa). Note also that the units of standard
enthalpies of formation are kJ/mol because they are always stated for that quantity of substance.
5.5
standard enthalpy of formation
the quantity of energy associated
with the formation of one mole of a
substance from its elements in their
standard states
LEARNING
TIP
Although the name “standard
enthalpies of formation” does not
include the word molar, it is always
a quantity of energy per mole.
Writing Formation Equations
Formation equations are always written for one mole of a particular product, which
may be in any state or form, but the reactant elements must be in their standard states.
For example, the standard states of most metals are monatomic solids (Mg(s), Ca(s), Fe(s),
Au(s), Na(s)), some nonmetals are diatomic gases (N2(g), O2(g), H2(g)), and the halogen
family shows a variety of states (F2(g), Cl2(g), Br2(l), I2(s)). The periodic table at the back
of this text identifies the states of elements.
Writing Formation Equations
SUMMARY
Step 1: Write one mole of product in the state that has been specified.
Step 2: Write the reactant elements in their standard states.
Step 3: Choose equation coefficients for the reactants to give a balanced equation yielding one mole of product.
Writing Formation Equations
SAMPLE problem
Write the formation equation for liquid ethanol.
Start with one mole of product.
C2H5OH(l)
Write the reactant elements in their standard states.
C(s) H2(g) O2(g) → C2H5OH(l)
Balance the equation to yield one mole of product.
2 C(s) 3 H2(g) NEL
1
2
O2(g) → C2H5OH(l)
LEARNING
TIP
The standard state of most elements in the periodic table is solid.
There are five common gaseous
elements at SATP that form compounds readily: H2, O2, N2, F2, and
Cl2. There are only two liquid elements at SATP: Hg and Br2.
Thermochemistry 331
Example
What is the formation equation for liquid carbonic acid?
Solution
H2(g) C(s) 3
2
O2(g) → H2CO3(l)
Practice
Understanding Concepts
1. Write formation equations for the compounds:
(a)
(b)
(c)
(d)
benzene (C6H6), used as a solvent
potassium bromate, used in commercial bread dough
glucose (C6H12O6), found in soft drinks
magnesium hydroxide, found in antacids
Using Standard Enthalpies of Formation
Consider the equation for the formation of hydrogen gas:
H2(g) → H2(g)
The product and reactant are the same, so there is no change in the enthalpy of the
system. This observation can be generalized to all elements in their standard states:
H f̊ for Elements
The standard enthalpy of formation of an element already in
its standard state is zero.
Relative Potential Energies of
Graphite and Diamond
diamond
Ep
∆H = 1.9 kJ/mol
graphite
Reaction Progress
Figure 1
Graphite is the more stable form of
carbon. The formation of diamond
requires an increase in potential
energy.
332 Chapter 5
Thus, the standard enthalpies of formation of, for example, Fe(s), O2(g), and Br2(l) are
all zero.
Standard molar enthalpies of formation give us a means of comparing the stabilities
of substances. For example, the element carbon exists in two solid forms at SATP: diamond, used in jewellery and mining drill bits; and graphite, the black substance used in
pencil “leads” and composite plastics. Graphite is the more stable form of carbon at
SATP, so
º
H f(graphite)
0 kJ/mol
Diamond is slightly less stable at SATP and has a greater potential energy than graphite
(Figure 1). For the formation of diamond,
C(graphite) → C(diamond)
º
H f(diamond)
+1.9 kJ/mol
You have seen that Hess’s law may be applied to a set of known equations to find an
unknown enthalpy change (Section 5.4). We can apply this problem-solving method to
predict the energy changes for many reactions.
NEL
Section 5.5
Using Enthalpies of Formation to Find H
SAMPLE problem
What is the thermochemical equation for the reaction of lime (calcium oxide) and
water?
We can express the target equation as
CaO(s) H2O(l) → Ca(OH)2(s)
H ?
Consider the following set of formation reactions for each compound in the equation, each
with its corresponding standard enthalpy of formation. (Tables of standard enthalpies of
formation are readily available, including a sample reference in Appendix C6 in this text.)
Use these with Hess’s law to find the enthalpy change H for the target equation.
(1) Ca(s) 1
2
(2) H2(g) 1
2
O2(g)
O2(g)
(3) Ca(s) H2(g) O2(g)
→ CaO(s)
H1 634.9 kJ/mol
→ H2O(l)
H2 285.8 kJ/mol
→ Ca(OH)2(s)
H3 986.1 kJ/mol
Manipulating and adding the equations according to Hess’s law results in a sum for the
thermochemical equation
1
2
→ CaO(s)
–1 H1
1
2
→ H2O(l)
–1 H2
1 (1): Ca(s) O2(g)
1 (2): H2(g) O2(g)
1 (3): Ca(s) H2(g) + O2(g)
→ Ca(OH)2(s)
1 H3
___________________________________________________________________________
→ Ca(OH)2(s)
CaO(s) H2O(l)
H H3 (–H2) (–H1)
Notice that the enthalpy change for the target equation equals the enthalpy of formation
for the products (calcium hydroxide) minus the enthalpies of formation of the reactants
(calcium oxide and water). This observation can be generalized to any chemical equation:
The enthalpy change for any given equation equals the sum of the enthalpies of formation
of the products minus the sum of the enthalpies of formation of the reactants, or, symbolically
H nH°
f(products)
nH°
f(reactants)
where n represents the amount (in moles) of each particular product or reactant.
Substitute our known molar enthalpies of formation into this equation:
H nH °f(Ca(OH) (s)) (nH °f(CaO(s)) n H °f(H2O) (l))
KEY EQUATION
H nH °
f(products)
nH °
–
f(reactants)
986 .1 kJ
63 4.9 kJ
28 5.8 kJ
mol ) (1 mol )
(1 mol ) ((1 1m
ol
1
mol
1
mol
H 65.4 kJ
The thermochemical equation for the slaking of lime with water is
CaO(s) H2O(l) → Ca(OH)2(s)
NEL
H 65.4 kJ
Thermochemistry 333
Example 1
The main component in natural gas used in home heating or laboratory burners is
methane. What is the molar enthalpy of combustion of methane fuel?
Solution
CH4(g) 2 O2(g) → CO2(g) 2 H2O(l)
LEARNING
TIP
Enthalpies of combustion of
hydrocarbons generally assume
production of CO2(g) and H2O(l),
the states of these compounds
at SATP.
H nH °f(products)
nH °f(reactants)
393.5 kJ
285.8kJ
74.4 kJ
0 kJ
(1 mol ) 2 mol ) (1 mol 2 mol )
1
mol
1
mol
1
mol
1
mol
965.1 kJ (74.4 kJ)
H 890.7 kJ
890.7 kJ
H
Hc 1 mol CH4
n
890.7 kJ/mol CH4
The molar enthalpy of combustion of methane fuel is –890.7 kJ/mol.
A variation on the application of standard enthalpies of formation to thermochemical
equations is a problem in which the enthalpy change of reaction is provided, and you are
asked to find one of the H °f values.
LEARNING
TIP
Note the importance of using
the standard enthalpy of formation appropriate to the state of a
substance. The standard
enthalpy of formation of H2O(g)
(241.8 kJ/mol) is different from
that of H2O(i) (285.8 kJ/mol).
Example 2
The standard enthalpy of combustion of benzene (C6H6(l)) to carbon dioxide and liquid
water is –3273 kJ/mol. What is the standard enthalpy of formation of benzene, given the
tabulated values for carbon dioxide and liquid water (Appendix C6)?
Solution
15
C6H6(g) O2(g)
2
H → 6 CO2(g) 3 H2O(l)
nH°f(products) nH°f(reactants)
393.5 kJ
285.8 kJ
3273 kJ (6 mol 3 mol 1 mol
1 )
mol
15
0 kJ
1 mol H °f(benzene) + mol
2
1
mol
3273 kJ 3217.5 kJ 1 mol H °f(benzene)
3217.5 kJ 3273 kJ
H°f(benzene) 1 mol
H°f(benzene) 56 kJ/mol
The standard enthalpy of formation of benzene is +56 kJ/mol.
334 Chapter 5
NEL
Section 5.5
Practice
Understanding Concepts
2. Use standard enthalpies of formation to calculate:
(a) the molar enthalpy of combustion for pentane to produce carbon dioxide gas
and liquid water;
(b) the enthalpy change that accompanies the reaction between solid iron(III)
oxide and carbon monoxide gas to produce solid iron metal and carbon
dioxide gas.
3. The standard enthalpy of combustion of liquid cyclohexane to carbon dioxide and
liquid water is –3824 kJ/mol. What is the standard enthalpy of formation of cyclohexane (C6H12(l))?
4. Methane, the major component of natural gas, is used as a source of hydrogen gas
to produce ammonia. Ammonia is used as a fertilizer and a refrigerant, and is used
to manufacture fertilizers, plastics, cleaning agents, and prescription drugs. The
following questions refer to some of the chemical reactions of these processes. For
each of these equations, use standard enthalpies of formation to calculate H:
(a) The first step in the production of ammonia is the reaction of methane with
steam, using a nickel catalyst.
Answers
2. (a) 3509 kJ
(b) 24.8 kJ
3. 252 kJ/mol
4. (a) 249.7 kJ
(b) 2.8 kJ
(c) 91.8 kJ
5. (a) 906.4 kJ
(b) 114.2 kJ
(c) 71.8 kJ
Ni
CH4(g) H2O(g) → CO(g) 3 H2(g)
(b) The second step of this process is the further reaction of water with carbon
monoxide to produce more hydrogen. Both iron and zinc–copper catalysts are
used.
Fe, Zn Cu
CO(g) H2O(g) → CO2(g) H2(g)
(c) After the carbon dioxide gas is removed by dissolving it in water, the hydrogen
reacts with nitrogen in the air to form ammonia.
N2(g) + 3 H2(g) → 2 NH3(g)
5. Nitric acid, required in the production of nitrate fertilizers, is produced from
ammonia by the Ostwald process. Use standard enthalpies of formation to calculate the enthalpy changes in each of the following systems.
(a) 4 NH3(g) 5 O2(g) → 4 NO(g) 6 H2O(g)
(b) 2 NO(g) + O2(g) → 2 NO2(g)
(c) 3 NO2(g) H2O(l) → 2 HNO3(l) + NO(g)
Making Connections
6. Energy is used in the manufacture of fertilizers, to grow crops. We then extract
food energy from these crops.
(a) Trace the energy path through the various steps in this process.
(b) If more thermal energy is put into the process than food energy is gained from
the process, should we abandon the practice of manufacturing fertilizers?
Discuss.
NEL
Thermochemistry 335
Multistep Energy Calculations
Using Standard Enthalpies of Formation
Chemical engineers frequently need to do calculations in which they determine the heats
produced by an internal-combustion engine. Such calculations involve bringing together
many of the problem-solving skills that you are developing. In Section 3.4 you learned
how to solve multistep problems where enthalpy changes, heats transferred, and masses
of reactants were interrelated. Two key relationships that you applied were
1. enthalpy change in the system heat transferred to/from the surroundings
H q
and
2. H nHr
The following sample problem illustrates a situation involving both a multistep
problem approach and the technique of referring to standard enthalpies of formation
introduced earlier in this section.
SAMPLE problem
Solving Multistep Problems Using Standard
Enthalpies of Formation
When octane burns in an automobile engine, heat is released to the air and to the
metal in the car engine, but a significant portion is absorbed by the liquid in the
cooling system—an aqueous solution of ethylene glycol. What mass of octane is
completely burned to cause the heating of 20.0 kg of aqueous ethylene glycol
automobile coolant from 10°C to 70°C? The specific heat capacity of the aqueous
ethylene glycol is 3.5 J/(g •°C). Assume water is produced as a gas and that all the
heat flows into the coolant.
First, write and balance the combustion equation for octane. To simplify later calculations,
write the equation so there is 1 mol of octane:
25
C8H18(g) O2(g) → 8 CO2(g) 9 H2O(g)
2
Use a reference (such as Appendix C6) to find the standard enthalpies of formation for the
products and reactants:
H°f (CO2(g)) 393.5 kJ/mol
H °f (C H
H°f (H O(g)) 241.8 kJ/mol
H °f (O
2
8
18(g)
2(g)
)
250.1 kJ/mol
)
0.0 kJ/mol
Next, use the coefficients from the combustion equation and the standard molar
enthalpies to calculate the molar enthalpy of combustion of octane:
H nH °f(products) nH °f(reactants)
(8 mol H °f (CO2(g)) 9 mol H °f (H2O(g)))
25
2 mol H °f (C8H18(g)) mol H °f (O2(g))
2
39 3.5 kJ
241.8 kJ
H 8 mol 9 mol 1
mol
1
mol
250.1 kJ
25
0 kJ
1
mol mol
1
mo l
2
1
mol
H 5074.1 kJ
Note that if you check tabulated values, you will find a listing of 5471 kJ/mol for
octane, because the standard reaction is for H2O(l), not H2O(g).
336 Chapter 5
NEL
Section 5.5
Next, we assume that the enthalpy change in the reaction in the engine equals the heat
flow into the aqueous ethylene glycol coolant, or
Hoctane qcoolant
nHc mcT
Rearrange to solve for the required amount of octane, noctane:
mcT
noctane Hc
Substitute the givens from the problem and the calculated value of H c:
mcoolant 20.0 kg
T 70°C (10°C) 80°C
ccoolant 3.5 J/(g•°C)
noctane 3.5 J
20.0 kg 80°C
•°C
g
5074.1 k
J
1 mol
1.1 mol
To find the required mass of octane, convert the amount into mass, using the molar mass
of octane (114.26 g/mol):
moctane
114 .26 g
1.1 mol 1m
ol
moctane
0.13 kg
If you are well practised in this technique, you may want to do the final steps in one calculation. Since
moctane
noctane Moctane
we can rewrite
noctane
mcT
Hc
moctane
mcoolant ccoolant TMoctane
Hc(octane)
as
moctane
3.5 J
114 .26 g
20.0 kg 80°C g
•°C
1m
ol
5074.1 kJ
1
mol
0.13 kg
According to the molar enthalpy of formation method and the law of conservation of
energy, the mass of octane required is 0.13 kg.
Example
One way to heat water in a home or cottage is to burn propane. If 3.20 g of propane burns,
what temperature change will be observed if all of the heat from combustion transfers
into 4.0 kg of water?
NEL
Thermochemistry 337
Solution
C3H8(g) 5 O2(g) → 3 CO2(g) 4 H2O(l)
H °f (CO2(g) 393.5 kJ/mol
H °f (H2O(l)) 285.8 kJ/mol
H °f (C3H8(g)) 104.7 kJ/mol
H °f (O2(g) 0.0 kJ/mol
mpropane 3.20 g
mH2O 4.0 kg
cH2O 4.18 J/(g•°C)
The enthalpy of change for the reaction is:
H nH°f(products) nH °f(reactants)
393.5 kJ
285.8 kJ
H 3 mol 4 mol 1 mol
1 mol
104.7 kJ
0.0 kJ
1
mol 5 mol
1 mol
1 mol
H 2219 kJ
LEARNING
TIP
Specific heat capacities may be
expressed in various units for
convenience. For example, the
specific heat capacity of water is
4.18 J/(g•°C) (convenient for a
student lab investigation) or
4.18 kJ/(kg•°C) (e.g., in an automobile engine) or
4.18 MJ/(Mg•°C) (e.g., in power
plants).
Therefore, the molar enthalpy of combustion of propane is
Hc(propane) 2219 kJ/mol
Hc(propane) qwater
nHc mcT
nHc
T mc
mpropaneHc
Mpropanemwaterc
2219 kJ
3.20 g 1m
ol
44 .11 g
4.18 J
4.0 kg 1m
ol
g•° C
T 9.6°C
The temperature change of the water is 9.6°C.
Practice
Making Connections
7. Ammonium nitrate fertilizer is produced by the reaction of ammonia with nitric acid:
NH3(g) HNO3(l) → NH4NO3(s)
Figure 2
The production of canola crops
are dependent on the use of fertilizers such as ammonium nitrate.
338 Chapter 5
Ammonium nitrate is one of the most important fertilizers for increasing crop yields
(Figure 2).
(a) Use standard enthalpies of formation to calculate the standard enthalpy
change of the reaction used to produce ammonium nitrate.
NEL
Section 5.5
(b) Sketch a potential energy diagram for the reaction of ammonia and nitric acid.
(c) Calculate the heat that would be produced or absorbed in the production of
50 T of ammonium nitrate.
8. Coal is a major energy source for electricity, of which industry is the largest user
(Figure 3). Anthracite coal is a high-molar-mass carbon compound with a compo-
sition of about 95% carbon by mass. A typical simplest-ratio formula for anthracite
coal is C52H16O(s). The standard enthalpy of formation of anthracite can be estimated at 396.4 kJ/mol. What is the quantity of energy available from burning
100.0 kg of anthracite coal in a thermal electric power plant, according to the following equation?
2 C52H16O(s) 111 O2(g) → 104 CO2(g) 16 H2O(g)
9. Alternative transportation fuels include methanol and hydrogen.
(a) Use standard enthalpies of formation to calculate the energy produced per
mole of methanol burned.
(b) Use standard enthalpies of formation to calculate the energy produced per
mole of hydrogen burned.
(c) In terms of energy content, how do these two alternative fuels compare with
gasoline, which is mostly octane? The molar enthalpy of combustion of octane
is 5.07 MJ/mol.
(d) What factors other than energy content are important when comparing
different automobile fuels? Include several perspectives.
Figure 3
Coal is an important current
source of electrical energy in
Ontario but its use has serious
environmental consequences.
Answers
7. (a) 145.6 kJ
(c) 9.10 104 MJ
Making Connections
10. In a typical household, about one-quarter of the energy consumed is used to heat
water.
(a) What mass of methane undergoing complete combustion is required to heat
100.0 kg of water from 5.0°C to 70.0°C in a gas water heater?
(b) How might we heat water more efficiently?
(c) What alternative energy resources are available for heating water?
8. 3.34 103 MJ
9. (a) 22.7 MJ/mol
(b) 142 MJ/mol
10. (a) .48 kg
Section 5.5 Questions
Understanding Concepts
1. Write balanced equations for the formation of the following:
(a)
(b)
(c)
(d)
acetylene gas (C2H2), used in welding
creatine (C4H9N3O2), used as a food supplement
potassium iodide (KI), used as a salt substitute
iron(II) sulfate (FeSO4), used as a diet supplement
(a) Write the balanced equation for this reaction.
(b) Use standard enthalpies of formation to calculate the
enthalpy change associated with the cracking of one
mole of octane.
2. Use standard enthalpies of formation to calculate the
enthalpy changes in each of the following equations:
(a) Magnesium carbonate decomposes when strongly
heated.
(b) Ethene burns in air.
(c) Sucrose (C12H22O11(s)) decomposes to carbon and
water vapour when concentrated sulfuric acid is
poured onto it (Figure 4).
3. When octane is strongly heated with hydrogen gas in the
presence of a suitable catalyst, it “cracks,” forming a mixture of hydrocarbons. A typical cracking reaction yields
1 mol of methane, 2 mol of ethane, and 1 mol of propane
from each mole of octane.
NEL
Figure 4
The decomposition of sucrose produces black carbon and water.
Thermochemistry 339
Applying Inquiry Skills
4. A sample of acetone (C3H6O) is burned in an insulated
calorimeter to produce carbon dioxide gas and liquid water.
Question
What is the molar enthalpy of combustion of acetone?
Experimental Design
(d) Does the evidence support the law of conservation of
energy?
(e) If heat is lost to the surroundings instead of being
transferred only into the aluminum can and water, will
the experimental molar enthalpy be higher or lower?
Explain.
Extension
A sample of liquid acetone is burned such that the heat
from the burning is transferred into an aluminum
calorimeter and its water contents.
Prediction
(a) Use tabulated standard enthalpies of formation to calculate a theoretical value for the molar enthalpy of
combustion of acetone.
Evidence
Table 1 Observations on the Combustion of Acetone
Quantity
Measurement
mass of water
100.0 g
specific heat capacity of aluminum
0.91 J/(g • °C)
mass of aluminum can
50.0 g
initial temperature of calorimeter
20.0°C
final temperature of calorimeter
25.0°C
mass of acetone burned
0.092 g
5. Remote controlled model boats are powered by burning
methanol or a racing mixture of 80% methanol and 20%
nitromethane by mass (Figure 5). The standard enthalpies of
formation of liquid methanol and nitromethane are, respectively, –238.6 kJ/mol and –74.78 kJ/mol.
(a) Draw structural diagrams for all reactants and products.
(b) Calculate the percentage change in energy output
achieved by using a racing mixture.
(c) Considering the various desirable characteristics of a
fuel, why do you suppose such a racing mixture is
used?
Analysis
(b) Calculate the molar enthalpy of combustion of acetone,
using the experimental evidence.
Evaluation
(c) Calculate the percentage error by comparing the predicted and experimental values.
340 Chapter 5
Figure 5
NEL