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5.4
Figure 1
Some reactions are too slow to be
studied experimentally.
Hess’s Law of Additivity of
Reaction Enthalpies
Calorimetry is an accurate technique for determining enthalpy
changes, but how do chemists deal with chemical systems that
cannot be analyzed using this technique? For example, the rusting
of iron (Figure 1) is extremely slow and, therefore, the resulting
temperature change would be too small to be measured using a
conventional calorimeter. Similarly, the reaction to produce
carbon monoxide from its elements is impossible to measure
with a calorimeter because the combustion of carbon produces
both carbon dioxide and carbon monoxide simultaneously.
Chemists have devised a number of methods to deal with this
problem. These methods are based on the principle that net (or
overall) changes in some properties of a system are independent
of the way the system changes from the initial state to the final
state. An analogy for this concept is shown in Figure 2. The net
vertical distance that the bricks rise is the same whether they go
up in one stage or in two stages. The same principle applies to
enthalpy changes: If a set of reactions occurs in different steps but the initial reactants
and final products are the same, the overall enthalpy change is the same (Figure 3).
Figure 2
In this illustration, bricks on a construction site are being moved from
the ground up to the second floor,
but there are two different paths
that the bricks could follow. In one
path, they could move from the
ground up to the third floor and
then be carried down to the second
floor. In the other path, they would
be carried up to the second floor in
a single step. In both cases, the
overall change in position — one
floor up — is the same.
322 Chapter 5
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Section 5.4
Potential Energy Diagram Showing Additive Enthalpy Changes
NO
–56 kJ
Ep
+90 kJ
NO2
+34 kJ
N2 and O2
Reaction Progress
Figure 3
In this potential energy diagram,
nitrogen gas and oxygen gas combine to form nitrogen dioxide, but
there are two different paths to
reach the products. In one path,
nitrogen (N2) and oxygen (O2) gases
react to form nitrogen monoxide
(NO), a reaction for which
H +90 kJ. Then, nitrogen
monoxide and more oxygen react to
form nitrogen dioxide (NO2) gas, a
reaction for which H 56 kJ. In
the other path, nitrogen (N2) and
oxygen (O2) gases react directly to
form nitrogen dioxide (NO2) gas. In
both cases, the overall enthalpy
change, H 34 kJ, is the same.
Predicting H Using Hess’s Law
Based on experimental measurements of enthalpy changes, the Swiss chemist G. H. Hess
suggested that there is a mathematical relationship among a series of reactions leading
from a set of reactants to a set of products. This generalization has been tested in many
experiments and is now accepted as the law of additivity of reaction enthalpies, also
known as
Hess’s Law
The value of the H for any reaction that can be written in
steps equals the sum of the values of H for each of the
individual steps.
Another way to state Hess’s law is: If two or more equations with known enthalpy
changes can be added together to form a new “target” equation, then their enthalpy
changes may be similarly added together to yield the enthalpy change of the target equation.
Hess’s law can also be written as an equation using the uppercase Greek letter
(pronounced “sigma”) to mean “the sum of.”
Htarget H1 H2 H3 …
or
Htarget Hknown
Hess’s discovery allowed chemists to determine the enthalpy change of a reaction
without direct calorimetry, using two familiar rules for chemical equations and enthalpy
changes:
• If a chemical equation is reversed, then the sign of ∆H changes.
• If the coefficients of a chemical equation are altered by multiplying or dividing
by a constant factor, then the ∆H is altered in the same way.
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Thermochemistry 323
SAMPLE problem
Using Hess’s Law to Find H
1.
What is the enthalpy change for the formation of two moles of nitrogen
monoxide from its elements?
N2(g) O2(g) → 2 NO(g)
H° ?
This reaction, which may be called the target equation to distinguish it clearly from the
other equations, is difficult to study calorimetrically since the combustion of nitrogen produces nitrogen dioxide as well as nitrogen monoxide. However, we can measure the
enthalpy of complete combustion in excess oxygen (to nitrogen dioxide) for both nitrogen
and nitrogen monoxide by calorimetry. Consider the following two known reference equations:
LEARNING
TIP
Generally, a subscript on a H
value indicates a molar enthalpy
value, expressed in kJ/mol. The
“known” equations in Hess’s law
problems are exceptions. The subscript is used as a convenience in
distinguishing equations from
each other and the units of the
Hn values are kilojoules.
1
2
(1) N2(g) O2(g) → NO2(g)
(2) NO(g) 1
2
O2(g) → NO2(g)
H°1 +34 kJ
H°2 –56 kJ
If we work with these two equations, which may be called known equations, and then
add them together, we obtain the chemical equation for the formation of nitrogen
monoxide.
The first term in the target equation for the formation of nitrogen monoxide is one mole
of nitrogen gas. We therefore need to double equation (1) so that N2(g) will appear on the
reactant side when we add the equations. However, from equation (2) we want 2 mol of
NO(g) to appear as a product, so we must reverse equation (2) and double each of its
terms (including the enthalpy change). Effectively, we have multiplied known equation (1)
by +2, and multiplied known equation (2) by –2.
2 (1): N2(g) 2 O2(g) → 2 NO2(g)
H° 2(+34) kJ
–2 (2): 2 NO2(g) → 2 NO(g) O2(g)
H° –2(–56) kJ
Note that the sign of the enthalpy change in equation (2) will change, since the equation
has been reversed.
Now add the reactants, products, and enthalpy changes to obtain a net reaction equation. Note that 2 NO2(g) can be cancelled because it appears on both sides of the net
equation. Similarly, O2(g) can be cancelled from each side of the equation, yielding the
target equation:
N2(g) 2 O2(g) + 2 NO
2(g) → 2 NO2(g) 2 NO(g) + O2(g)
or
N2(g) O2(g) → 2 NO(g)
Now we can apply Hess’s law: If the known equations can be added together to form the
target equation, then their enthalpy changes can be added together.
H ° (2 34) kJ + (–2 (–56)) kJ
68 kJ 112 kJ
H ° +180 kJ
The enthalpy change for the formation of two moles of nitrogen monoxide from its
elements is 180 kJ.
When manipulating the known equations, you should check the target equation and
plan ahead to ensure that the substances end up on the correct sides and in the correct
amounts.
324 Chapter 5
NEL
Section 5.4
2.
What is the enthalpy change for the formation of one mole of butane ( C4H10 )
gas from its elements? The reaction is:
4 C(s) + 5 H2(g) → C4H10(g)
H° ?
The following known equations, determined by calorimetry, are provided:
(1) C4H10(g) +
13
2
O2(g) → 4 CO2(g) + 5 H2O(g)
H°1 2657.4 kJ
(2) C(s) + O2(g)
→ CO2(g)
H°2 393.5 kJ
(3) 2 H2(g) + O2(g)
→ 2 H2O(g)
H°3 483.6 kJ
Reversing known equation (1), which will require multiplying its H by 1, will make
C4H10(g) a product; multiplying known equation (2) by 4 will provide the required amount
of C(s) reactant; and multiplying known equation (3) by 5/2 will provide the required
amount of H2(g) reactant. Cancellation when the equations are added will determine
whether the required amount of O2(g) remains.
→ C4H10(g) 1 (1): 4 CO2(g) + 5 H2O(g)
13
2
O2(g)
H° 1(657.4) kJ
4 (2): 4 C(s) 4 O2(g)
→ 4 CO2(g)
H° 4(393.5) kJ
5
2
→ 5 H2O(g)
H° (3): 5 H2(g) 5
2
O2(g)
4
CO2(g) 5 H2O(g) + 4 C(s) 13
2
2(g) 5 H2(g) → C4H10(g) O
13
2
5
(483.6)
2
kJ
2(g)
O
+4
CO2(g) 5 H2O(g)
or
4 C(s) 5 H2(g)
→ C4H10(g)
If the known equations can be added together to form the target equation, then their
enthalpy changes can be added together. In this case,
H °total (+2657.4) + ( 1574.0) + ( 1209.0) kJ
H °total 125.6 kJ
The enthalpy change for the formation of one mole of butane is 125.6 kJ.
Example
Determine the enthalpy change involved in the formation of two moles of liquid propanol.
6 C(s) 8 H2(g) O2(g) → 2 C3H7OH(l)
The standard enthalpies of combustion of propanol, carbon, and hydrogen gas at SATP
are 2008, 394, and 286 kJ/mol, respectively.
Solution
The known equations are
(1) C3H7OH(l) 9
2
(2) C(s) O2(g)
(3) H2(g) +
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1
2
O2(g)
O2(g)
→ 3 CO2(g) 4 H2O(l)
H°1 –2008 kJ
→ CO2(g)
H°2 –394 kJ
→ H2O(l)
H°3 –286 kJ
Thermochemistry 325
2 (1): 6 CO2(g) 8 H2O(l)
→ 2 C3H7OH(l) 9 O2(g) H° 2(2008 kJ)
6 (2):
6 C(s) 6 O2(g)
→ 6 CO2(g)
H° 6(394 kJ)
8 (3):
8 H2(g) 4 O2(g)
→ 8 H2O(l)
H° 8(286 kJ)
→ 2 C3H7OH(l)
H° –636 kJ
6 C(s) 8 H2(g) O2(g)
The enthalpy change involved in the formation of propanol is 636 kJ.
Practice
Understanding Concepts
Answers
1. The enthalpy changes for the formation of aluminum oxide and iron(III) oxide from
1. 851.5 kJ
their elements are:
2. 131.3 kJ
3
2
H°1
–1675.7 kJ
3
2
H°2
–824.2 kJ
(1) 2 Al(s) O2(g) → Al2O3(s)
3. 524.8 kJ
(2) 2 Fe(s) O2(g) → Fe2O3(s)
Calculate the enthalpy change for the following target reaction.
Fe2O3(s) 2 Al(s)
Figure 4
Electric power generating stations
that use coal as a fuel are only
30% to 40% efficient. Coal gasification and combustion of the coal
gas provide one alternative to
burning coal. Efficiency is
improved by using both a combustion turbine and a steam turbine to produce electricity.
→ Al2O3(s) 2 Fe(s)
H2O(g) C(s) → CO(g) H2(g)
H° ?
Calculate the standard enthalpy change for this reaction from the following chemical equations and enthalpy changes.
(1) 2 C(s) O2(g)
→ 2 CO(g)
H°1 221.0 kJ
(2) 2 H2(g) O2(g)
→ 2 H2O(g)
H°2 483.6 kJ
coal
gas
heat recovery
steam generator
gasifier
coal gas
combustion
turbine
coal
slurry
pulverized coal
326 Chapter 5
?
2. Coal gasification converts coal into a combustible mixture of carbon monoxide and
hydrogen, called coal gas (Figure 4), in a gasifier.
oxygen
mixed with
water
H°
mineral
slag
ash and
sulfur
removal
to exhaust
stack
steam
generator
generator
power
out
power
out
boiler water
NEL
Section 5.4
3. The coal gas described in the previous question can be used as a fuel, for example,
in a combustion turbine.
CO(g) H2(g) O2(g) → CO2(g) H2O(g)
H° ?
Predict the change in enthalpy for this combustion reaction from the following
information.
(1) 2 C(s) O2(g)
→ 2 CO(g)
H°1 –221.0 kJ
(2) C(s) O2(g)
→ CO2(g)
H°2 –393.5 kJ
(3) 2 H2(g) O2(g)
→ 2 H2O(g)
H°3 –483.6 kJ
INVESTIGATION 5.4.1
Hess’s Law (p. 351)
Use calorimetry to determine your
own “known” equations and use
them to calculate the molar
enthalpy of combustion of
magnesium.
Multistep Energy Calculations
In practice, energy calculations rarely involve only a single-step calculation of heat or
enthalpy change. Several energy calculations might be required, involving a combination
of energy change definitions such as
• heat flows,
q mc∆T
• enthalpy changes,
∆H n∆Hr
• Hess’s law,
∆Htarget ∆Hknown
In these multi-step problems, ∆H is often found by using standard molar enthalpies or
Hess’s law and then equated to the transfer of heat, q. As shown in the following sample
problem, if we know the enthalpy change of a reaction and the quantity of reactant or
product, we can predict how much energy will be absorbed or released.
Solving Multistep Enthalpy Problems
SAMPLE problem
In the Solvay process for the production of sodium carbonate (or washing soda),
one step is the endothermic decomposition of sodium hydrogen carbonate:
2 NaHCO3(s) 129.2 kJ → Na2CO3(s) CO2(g) H2O(g)
What quantity of chemical energy, H, is required to decompose 100.0 kg of
sodium hydrogen carbonate?
First, calculate the energy absorbed per mole of NaHCO3, that is, the molar enthalpy of
reaction with respect to sodium hydrogen carbonate.
H nHr
H
Hr n
129.2 kJ
2 mol
Hr 64.6 kJ/mol
This means that 64.6 kJ of energy is required for every mole of NaHCO3 decomposed.
Converting 100.0 kg to an amount in moles and multiplying by the molar enthalpy will give
us the required enthalpy change, H, for the equation.
NEL
Thermochemistry 327
1 mol
nNaHCO3 100.0 kg
84.01 g
1.190 kmol
H nHr
64 .6 kJ
1.190 kmol
1
mol
H 76.9 MJ
The decomposition of 100 kg of sodium hydrogen carbonate requires 76.9 MJ of energy.
Example
How much energy can be obtained from the roasting of 50.0 kg of zinc sulfide ore?
ZnS(s) 3
2
O2(g) → ZnO(s) SO2(g)
You are given the following thermochemical equations.
1
2
(1) ZnO(s)
→ Zn(s) (2) S(s) O2(g)
→ SO2(g)
H°2 296.8 kJ
(3) ZnS(s)
→ Zn(s) S(s)
H°3 206.0 kJ
O2(g)
H°1 350.5 kJ
Solution
MZnS 97.44 g/mol
1 (1): Zn(s) 1
2
O2(g) → ZnO(s)
H° 1(350.5 kJ)
1 (2):
S(s) O2(g)
→ SO2(g)
H° 1(296.8 kJ)
1 (3):
ZnS(s)
→ Zn(s) S(s)
H° 1(206.0 kJ)
_____________________________________________________________________
ZnS(s) 3
2
O2(g)
→ ZnO(s) SO2(g)
H° 441.3 kJ
According to Hess’s law, 441.3 kJ of energy can be obtained from the roasting of 1 mol of
ZnS for which reaction H°r 441.3 kJ/mol.
1 mol
nZnS 50.0 kg
97.44 g
513 mol
H ° nZnSH°r
(441.3 kJ)
513 mol 1
mol
H ° 2.26 105 kJ or 226 MJ
According to Hess’s law, 226 MJ of energy can be obtained from the roasting of 50.0 kg
of zinc sulfide ore.
328 Chapter 5
NEL
Section 5.4
Practice
Understanding Concepts
4. Ethyne gas may be reduced by reaction with hydrogen gas to form ethane gas in
Answers
the following reduction reaction:
4. 2.39 MJ
C2H2(g) 2 H2(g) → C2H6(g)
5. 2.20 MJ
Predict the enthalpy change for the reduction of 200 g of ethyne, using the following information.
5
(1) C2H2(g) O2(g)
2
1
(2) H2 O2(g)
2
7
(3) C2H6(g) + O2(g)
2
→ 2 CO2(g) H2O(l)
H°1 1299 kJ
→ H2O(l)
H°2 286 kJ
→ 2 CO2(g) 3 H2O(l)
H°3 1560 kJ
5. As an alternative to combustion of coal gas described earlier in this section, coal
gas can undergo a process called methanation.
3 H2(g) CO(g) → CH4(g) H2O(g)
H ?
Determine the enthalpy change involved in the reaction of 300 g of carbon
monoxide in this methanation reaction, using the following reference equations
and enthalpy changes.
(1) 2 H2(g) O2(g)
→ 2 H2O(g)
H°1 483.6 kJ
(2) 2 C(s) + O2(g)
→ 2 CO(g)
H°2 221.0 kJ
(3) CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
(4) C(s) + O2(g)
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→ CO2(g)
H°3 802.7 kJ
H°4 393.5 kJ
Thermochemistry 329
Section 5.4 Questions
Understanding Concepts
Applying Inquiry Skills
1. (a) Write three balanced thermochemical equations to
represent the combustions of one mole each of octane,
hydrogen, and carbon, given that their molar enthalpies
of combustion are, respectively, –5.47 MJ, –285.8 kJ,
and –393.5 kJ/mol.
(b) Use Hess’s law to predict the enthalpy change for the
formation of octane from its elements.
8 C(s) 9 H2(g) → C8H18(l)
using the following information:
(2) NO NO2 + Na2O → 2 NaNO2
Question
Can Hess’s law be verified experimentally by combining
enthalpies of reaction?
Three calorimetry experiments are performed, with the
choice of chemical systems such that the enthalpy changes
of two of the reactions should equal the enthalpy change of
the third. The three thermochemical reactions are:
HCl(g) + NaNO2(s) → HNO2 (g) + NaCl(s)
→ 2 HCl Na2O
Hess’s law. Complete the Experimental Design, Prediction,
and Analysis sections of the investigation report.
Experimental Design
H ?
2. Predict the enthalpy change for the reaction
(1) 2 NaCl H2O
4. A series of calorimetric experiments is perfomed to test
H °1 507 kJ
H °2 –427 kJ
(3) NO NO2
→ N2O + O2
H °3 –43 kJ
(4) 2 HNO2
→ N2O + O2 + H2O
H °4 34 kJ
3. Bacteria sour wines and beers by converting ethanol
(C2H5OH) into acetic acid (CH3COOH). The reaction is
C2H5OH O2 → CH3COOH H2O
(1) HBr(aq) KOH(aq) → H2O(l) + KBr(aq)
(2) KOH(s)
→ KOH(aq)
(3) KOH(s) HBr(aq) → H2O(l) KBr(aq)
H1 ? kJ
H2 ? kJ
H3 ? kJ
(a) Use Hess’s law to show how two of the thermochemical equations can be added together to yield the third
thermochemical equation.
Evidence
See Table 1.
The molar enthalpies of combustion of ethanol and acetic
acid are, respectively, 1367 kJ/mol and 875 kJ/mol.
Write thermochemical equations for the combustions, and
use Hess’s law to determine the enthalpy change for the
conversion of ethanol to acetic acid.
Analysis
(b) Use the experimental values to calculate the enthalpy
change in each system.
Evaluation
(c) Calculate a percentage error in the experiment, given
that the H for one equation should equal exactly the
sum of the other two.
Table 1 Observations for Hess’s Law Investigation
330 Chapter 5
Observation
Experiment 1
Experiment 2
Experiment 3
quantity of
reactant 1
100.0 mL of
1.00 mol/L KOH(aq)
5.61 g KOH(s)
5.61 g KOH(s)
quantity of
reactant 2
100.0 mL of
1.00 mol/L HBr(aq)
N/A
200.0 mL of
0.50 mol/L HBr(aq)
initial
temperature
20.0°C
20.0°C
20.0°C
final
temperature
22.5°C
24.1°C
26.7°C
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