Transcript 17.4 Calculating Heats of Reaction

17.4 Calculating Heats of Reaction >
Chapter 17
Thermochemistry
17.1 The Flow of Energy
17.2 Measuring and Expressing
Enthalpy Changes
17.3 Heat in Changes of State
17.4 Calculating Heats of
Reaction
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17.4 Calculating Heats of Reaction >
CHEMISTRY
& YOU
How much heat is released when a
diamond changes into graphite?
Diamonds are gemstones
composed of carbon.
Over a time period of
millions and millions of
years, diamond will break
down into graphite, which
is another form of carbon.
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17.4 Calculating Heats of Reaction > Hess’s Law
Hess’s Law
How can you calculate the heat of
reaction when it cannot be directly
measured?
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17.4 Calculating Heats of Reaction > Hess’s Law
Hess’s law of heat summation states that
if you add two or more thermochemical
equations to give a final equation, then you
can also add the heats of reaction to give
the final heat of reaction.
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17.4 Calculating Heats of Reaction > Hess’s Law
Hess’s law allows you to determine
the heat of reaction indirectly by using
the known heats of reaction of two or
more thermochemical equations.
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17.4 Calculating Heats of Reaction > Hess’s Law
C(s, diamond) → C(s, graphite)
Although the enthalpy change for this reaction
cannot be measured directly, you can use Hess’s
law to find the enthalpy change for the conversion
of diamond to graphite by using the following
combustion reactions.
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a. C(s, graphite) + O2(g) → CO2(g)
ΔH = –393.5 kJ
b. C(s, diamond) + O2(g) → CO2(g)
ΔH = –395.4 kJ
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17.4 Calculating Heats of Reaction > Hess’s Law
C(s, diamond) → C(s, graphite)
a. C(s, graphite) + O2(g) → CO2(g)
ΔH = –393.5 kJ
b. C(s, diamond) + O2(g) → CO2(g)
ΔH = –395.4 kJ
Write equation a in reverse to give:
c. CO2(g) → C(s, graphite) + O2(g)
ΔH = 393.5 kJ
When you reverse a reaction, you must also
change the sign of ΔH.
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17.4 Calculating Heats of Reaction > Hess’s Law
C(s, diamond) → C(s, graphite)
b. C(s, diamond) + O2(g) → CO2(g)
ΔH = –395.4 kJ
c. CO2(g) → C(s, graphite) + O2(g)
ΔH =
393.5 kJ
If you add equations b and c, you get the equation
for the conversion of diamond to graphite.
C(s, diamond) + O2(g) → CO2(g)
ΔH = –395.4 kJ
CO2(g) → C(s, graphite) + O2(g)
ΔH =
393.5 kJ
C(s, diamond) → C(s, graphite)
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17.4 Calculating Heats of Reaction > Hess’s Law
C(s, diamond) → C(s, graphite)
If you also add the values of ΔH for equations b
and c, you get the heat of reaction for this
conversion.
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C(s, diamond) + O2(g) → CO2(g)
ΔH = –395.4 kJ
CO2(g) → C(s, graphite) + O2(g)
ΔH =
393.5 kJ
C(s, diamond) → C(s, graphite)
ΔH =
–1.9 kJ
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17.4 Calculating Heats of Reaction > Hess’s Law
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C(s, diamond) + O2(g) → CO2(g)
ΔH = –395.4 kJ
CO2(g) → C(s, graphite) + O2(g)
ΔH =
393.5 kJ
C(s, diamond) → C(s, graphite)
ΔH =
–1.9 kJ
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17.4 Calculating Heats of Reaction >
CHEMISTRY
& YOU
How can you determine ΔH for the
conversion of diamond to graphite
without performing the reaction?
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17.4 Calculating Heats of Reaction >
CHEMISTRY
& YOU
How can you determine ΔH for the
conversion of diamond to graphite
without performing the reaction?
You can use Hess’s law by adding
thermochemical equations in which the
enthalpy changes are known and whose sum
will result in an equation for the conversion of
diamond to graphite.
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17.4 Calculating Heats of Reaction > Hess’s Law
Another case where Hess’s law is useful is
when reactions yield products in addition to
the product of interest.
• Suppose you want to determine the enthalpy
change for the formation of carbon monoxide
from its elements.
C(s, graphite)+ 12 O2(g) → CO(g) ΔH = ?
• Carrying out the reaction in the laboratory as
written is virtually impossible.
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17.4 Calculating Heats of Reaction > Hess’s Law
You can calculate the desired enthalpy
change by using Hess’s law and the
following two reactions that can be carried
out in the laboratory:
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C(s, graphite) + O2(g) → CO2(g)
ΔH = –393.5 kJ
CO2(g) → CO(g) + 12 O2(g)
ΔH =
C(s, graphite)+ 12 O2(g) → CO(g)
ΔH = –110.5 kJ
283.0 kJ
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17.4 Calculating Heats of Reaction > Hess’s Law
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C(s, graphite) + O2(g) → CO2(g)
ΔH = –393.5 kJ
CO2(g) → CO(g) + 12 O2(g)
ΔH =
C(s, graphite)+ 12 O2(g) → CO(g)
ΔH = –110.5 kJ
283.0 kJ
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17.4 Calculating Heats of Reaction >
According to Hess’s law, it is possible to
calculate an unknown heat of reaction
by using which of the following?
A. heats of fusion for each of the compounds in
the reaction
B. two other reactions with known heats of
reaction
C. specific heat capacities for each compound in
the reaction
D. density for each compound in the reaction
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17.4 Calculating Heats of Reaction >
According to Hess’s law, it is possible to
calculate an unknown heat of reaction
by using which of the following?
A. heats of fusion for each of the compounds in
the reaction
B. two other reactions with known heats of
reaction
C. specific heat capacities for each compound in
the reaction
D. density for each compound in the reaction
17
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17.4 Calculating Heats of Reaction > Standard Heats
of Formation
Standard Heats of Formation
How can you calculate the heat of
reaction when it cannot be directly
measured?
18
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17.4 Calculating Heats of Reaction > Standard Heats
of Formation
Enthalpy changes generally depend on the
conditions of the process.
• Scientists specify a common set of conditions
as a reference point.
• These conditions, called the standard state,
refer to the stable form of a substance at
25°C and 101.3 kPa.
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17.4 Calculating Heats of Reaction > Standard Heats
of Formation
The standard heat of formation (ΔHf°) of
a compound is the change in enthalpy that
accompanies the formation of one mole of a
compound from its elements with all
substances in their standard states.
• The ΔHf° of a free element in its standard
state is arbitrarily set at zero.
• Thus, ΔHf° = 0 for the diatomic molecules
H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(l), and
I2(s).
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17.4 Calculating Heats of Reaction >
Interpret Data
Standard Heats of Formation (ΔHf°) at 25°C and 101.3 kPa
Substance
ΔHf°
(kJ/mol)
Substance
Al2O3(s)
–1676.0
F2(g)
0.0
NO(g)
90.37
Fe(s)
0.0
NO2(g)
33.85
–822.1
NaCl(s)
Br2(g)
30.91
Br2(l)
0.0
Fe2O3(s)
C(s, diamond)
1.9
H2(g)
C(s, graphite)
0.0
–74.86
CH4(g)
ΔHf°
(kJ/mol)
0.0
H2O(g)
–241.8
O3(g)
142.0
H2O(l)
–285.8
P(s, white)
–187.8
P(s, red)
H2O2(l)
CO2(g)
–393.5
I2(g)
62.4
–1207.0
I2(s)
–635.1
N2(g)
Cl2(g)
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0.0
–411.2
O2(g)
–110.5
CaO(s)
ΔHf°
(kJ/mol)
0.0
CO(g)
CaCO3(s)
Substance
NH3(g)
0.0
–18.4
S(s, rhombic)
0.0
0.0
S(s, monoclinic)
0.30
0.0
SO2(g)
–296.8
–46.19
SO3(g)
–395.7
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17.4 Calculating Heats of Reaction > Standard Heats
of Formation
For a reaction that occurs at standard
conditions, you can calculate the heat
of reaction by using standard heats of
formation.
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17.4 Calculating Heats of Reaction > Standard Heats
of Formation
For a reaction that occurs at standard
conditions, you can calculate the heat
of reaction by using standard heats of
formation.
• Such an enthalpy change is called
the standard heat of reaction (ΔH°).
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17.4 Calculating Heats of Reaction > Standard Heats
of Formation
For a reaction that occurs at standard
conditions, you can calculate the heat
of reaction by using standard heats of
formation.
• The standard heat of reaction is the
difference between the standard
heats of formation of all the reactants
and products.
ΔH° = ΔHf°(products) – ΔHf°(reactants)
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17.4 Calculating Heats of Reaction >
Standard Heats
of Formation
This enthalpy diagram shows the
standard heat of formation of water.
• The enthalpy difference
between the reactants and
products, –285.8 kJ/mol, is
the standard heat of formation
of liquid water from the gases
hydrogen and oxygen.
• Notice that water has a lower
enthalpy than the elements
from which it is formed.
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17.4 Calculating Heats of Reaction >
Sample Problem 17.8
Calculating the Standard Heat
of Reaction
What is the standard heat of
reaction (ΔH°) for the reaction
of CO(g) with O2(g) to form
CO2(g)?
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17.4 Calculating Heats of Reaction >
Sample Problem 17.8
1 Analyze List the knowns and the unknown.
Balance the equation of the reaction of CO(g)
with O2(g) to form CO2(g). Then determine
ΔH° using the standard heats of formation of
the reactants and products.
KNOWNS
ΔHf°CO(g) = –110.5 kJ/mol
UNKNOWN
ΔH° = ? kJ
ΔHf°O2(g) = 0 kJ/mol (free
element)
ΔHf°CO2(g) = –393.5 kJ/mol
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17.4 Calculating Heats of Reaction >
Sample Problem 17.8
2 Calculate Solve for the unknown.
First write the balanced equation.
2CO(g) + O2(g) → 2CO2(g)
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17.4 Calculating Heats of Reaction >
Sample Problem 17.8
2 Calculate Solve for the unknown.
Find and add ΔHf° of all the reactants.
ΔHf°(reactants) = 2 mol CO(g)  ΔHf°CO(g) + 1 mol O2(g) 
ΔHf°O2(g)
–110.5 kJ
0 kJ
= 2 mol CO(g) 
2 mol CO(g)+ 1 mol O2(g)  1 mol O2(g)
= –221.0 kJ
Remember to take into
account the number of moles
of each reactant and product.
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17.4 Calculating Heats of Reaction >
Sample Problem 17.8
2 Calculate Solve for the unknown.
Find ΔHf° of the product in a similar way.
ΔHf°(products) = 2 mol CO2(g) 
ΔHf°CO2(g)
–393.5 kJ
= 2 mol CO2(g)  1 mol CO (g)
2
= –787.0 kJ
Remember to take into
account the number of moles
of each reactant and product.
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17.4 Calculating Heats of Reaction >
Sample Problem 17.8
2 Calculate Solve for the unknown.
Calculate ΔH° for the reaction.
ΔH° = ΔHf°(products) – ΔHf°(reactants)
= (–787.0 kJ) – (–221.0 kJ)
= –566.0 kJ
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17.4 Calculating Heats of Reaction >
Sample Problem 17.8
3 Evaluate Does the result make sense?
• The ΔH° is negative, so the reaction is
exothermic.
• This outcome makes sense because
combustion reactions always release heat.
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17.4 Calculating Heats of Reaction >
Standard Heats
of Formation
Standard heats of formation are used to
calculate the enthalpy change for the
reaction of carbon monoxide and oxygen.
• 2CO(g) + O2(g) → 2CO2(g)
• The diagram shows the
difference between
ΔHf°(product) and
ΔHf°(reactants) after taking
into account the number of
moles of each.
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17.4 Calculating Heats of Reaction >
Calculate the standard heat of reaction
for the following:
CH4(g) + Cl2(g) → C(s, diamond) + 4HCl(g)
ΔHf°(CH4(g)) = –74.86 kJ/mol
ΔHf°(C(s, diamond)) = 1.9 kJ/mol
ΔHf°(HCl(g)) = –92.3 kJ/mol
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17.4 Calculating Heats of Reaction >
Calculate the standard heat of reaction
for the following:
CH4(g) + Cl2(g) → C(s, diamond) + 4HCl(g)
ΔHf°(CH4(g)) = –74.86 kJ/mol
ΔHf°(C(s, diamond)) = 1.9 kJ/mol
ΔHf°(HCl(g)) = –92.3 kJ/mol
ΔHf°(reactants) = [1 mol CH4(g)  ΔHf°CH4(g)] + [1 mol Cl2  ΔHf°Cl2(g)]
= –74.86 kJ + 0.0 kJ = –74.86 kJ
ΔHf°(products) = [1 mol C(s)  ΔHf°C(s, diamond)] + [4 mol HCl 
ΔHf°HCl(g)]
= 1.9 kJ + (4  –92.3 kJ) = –367.3 kJ
ΔH°
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= ΔHf°(products) – ΔHf°(reactants)
= –367.3 kJ – (–74.86 kJ) = –
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17.4 Calculating Heats of Reaction > Key Concepts &
Key Equation
Hess’s law allows you to determine the
heat of reaction indirectly by using the
known heats of reaction of two or more
thermochemical equations.
For a reaction that occurs at standard
conditions, you can calculate the heat of
reaction by using standard heats of
formation.
ΔH° = ΔHf°(products) –
ΔHf°(reactants)
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17.4 Calculating Heats of Reaction > Glossary Terms
• Hess’s law of heat summation: if you add
two or more thermochemical equations to give
a final equation, then you also add the heats of
reaction to give the final heat of reaction
• standard heat of formation (ΔHf°): the
change in enthalpy that accompanies the
formation of one mole of a compound from its
elements with all substances in their standard
states at 25°C
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17.4 Calculating Heats of Reaction >
BIG IDEA
Matter and Energy
The heat of reaction can be calculated by
using the known heats of reaction of two
or more thermochemical equations or by
using standard heats of formation.
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17.4 Calculating Heats of Reaction >
END OF 17.4
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