1. Chapter 17 Thermochemistry 17.4 Calculating Heats of Reaction
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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2. How much heat is released when a diamond changes into graphite?
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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3. Hess’s Law
Hess’s Law
How can you calculate the heat of reaction when it cannot be directly measured?
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4. 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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5. 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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6. C(s, diamond) → C(s, graphite)
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.
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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7. C(s, diamond) → C(s, graphite)
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 = kJ
When you reverse a reaction, you must also change the sign of ΔH.
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8. C(s, diamond) → C(s, graphite)
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 = 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 = kJ
C(s, diamond) → C(s, graphite)
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9. C(s, diamond) → C(s, graphite)
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.
C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ
CO2(g) → C(s, graphite) + O2(g) ΔH = kJ
C(s, diamond) → C(s, graphite) ΔH = –1.9 kJ
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10. C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ
Hess’s Law
C(s, diamond) + O2(g) → CO2(g) ΔH = –395.4 kJ
CO2(g) → C(s, graphite) + O2(g) ΔH = kJ
C(s, diamond) → C(s, graphite) ΔH = –1.9 kJ
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11. CHEMISTRY & YOU
How can you determine ΔH for the conversion of diamond to graphite without performing the reaction?
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12. 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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13. 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.
Carrying out the reaction in the laboratory as written is virtually impossible.
C(s, graphite)+ O2(g) → CO(g) ΔH = ? 1 2
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14. 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:
C(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ
CO2(g) → CO(g) + O2(g) ΔH = kJ
C(s, graphite)+ O2(g) → CO(g) ΔH = –110.5 kJ 1 2
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15. C(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ
Hess’s Law
C(s, graphite) + O2(g) → CO2(g) ΔH = –393.5 kJ
CO2(g) → CO(g) + O2(g) ΔH = kJ
C(s, graphite)+ O2(g) → CO(g) ΔH = –110.5 kJ 1 2
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16. 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. 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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18. Standard Heats of Formation
How can you calculate the heat of reaction when it cannot be directly measured?
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19. 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 kPa.
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20. 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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21. Standard Heats of Formation (ΔHf°) at 25°C and 101.3 kPa
Interpret Data
Standard Heats of Formation (ΔHf°) at 25°C and kPa
Substance
ΔHf° (kJ/mol)
Al2O3(s)
–1676.0
F2(g)
0.0
NO(g)
90.37
Br2(g)
30.91
Fe(s)
NO2(g)
33.85
Br2(l)
Fe2O3(s)
–822.1
NaCl(s)
–411.2
C(s, diamond)
1.9
H2(g)
O2(g)
C(s, graphite)
H2O(g)
–241.8
O3(g)
142.0
CH4(g)
–74.86
H2O(l)
–285.8
P(s, white)
CO(g)
–110.5
H2O2(l)
–187.8
P(s, red)
–18.4
CO2(g)
–393.5
I2(g)
62.4
S(s, rhombic)
CaCO3(s)
–1207.0
I2(s)
S(s, monoclinic)
0.30
CaO(s)
–635.1
N2(g)
SO2(g)
–296.8
Cl2(g)
NH3(g)
–46.19
SO3(g)
–395.7
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22. 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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23. 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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24. 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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25. 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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26. Calculating the Standard Heat 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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27. Analyze List the knowns and the unknown.
Sample Problem 17.8
Analyze List the knowns and the unknown. 1
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
UNKNOWN
ΔHf°CO(g) = –110.5 kJ/mol
ΔHf°O2(g) = 0 kJ/mol (free element)
ΔHf°CO2(g) = –393.5 kJ/mol
ΔH° = ? kJ
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28. Calculate Solve for the unknown.
Sample Problem 17.8
Calculate Solve for the unknown. 2
First write the balanced equation.
2CO(g) + O2(g) → 2CO2(g)
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29. Calculate Solve for the unknown.
Sample Problem 17.8
Calculate Solve for the unknown. 2
Find and add ΔHf° of all the reactants.
ΔHf°(reactants) = 2 mol CO(g)  ΔHf°CO(g) + 1 mol O2(g)  ΔHf°O2(g)
= 2 mol CO(g)  mol O2(g) 
= –221.0 kJ
–110.5 kJ
2 mol CO(g)
0 kJ
1 mol O2(g)
Remember to take into account the number of moles of each reactant and product.
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30. Calculate Solve for the unknown.
Sample Problem 17.8
Calculate Solve for the unknown. 2
Find ΔHf° of the product in a similar way.
ΔHf°(products) = 2 mol CO2(g)  ΔHf°CO2(g)
= 2 mol CO2(g) 
= –787.0 kJ
–393.5 kJ
1 mol CO2(g)
Remember to take into account the number of moles of each reactant and product.
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31. Calculate Solve for the unknown.
Sample Problem 17.8
Calculate Solve for the unknown. 2
Calculate ΔH° for the reaction.
ΔH° = ΔHf°(products) – ΔHf°(reactants)
= (–787.0 kJ) – (–221.0 kJ)
= –566.0 kJ
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32. Evaluate Does the result make sense?
Sample Problem 17.8
Evaluate Does the result make sense? 3
The ΔH° is negative, so the reaction is exothermic.
This outcome makes sense because combustion reactions always release heat.
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33. 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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34. CH4(g) + Cl2(g) → C(s, diamond) + 4HCl(g)
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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35. CH4(g) + Cl2(g) → C(s, diamond) + 4HCl(g)
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 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° = ΔHf°(products) – ΔHf°(reactants) = –367.3 kJ – (–74.86 kJ) = –292.4 kJ
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36. 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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37. 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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38. 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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39. END OF 17.4
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