Chemistry 20 H
Chemistry: The Central Science
Chapter 5 Thermochemistry
Objectives
1. Define terms related to energy of systems. (5.1)
2. Discuss the relationship between kinetic energy and potential energy. (5.1)
3. Discuss the relationship between work and heat. (5.1)
4. Discuss the first law of thermodynamics and its’ implications in chemical systems. (5.2)
5. Discuss what is meant by an endothermic or an exothermic process. (5.2)
6. Discuss what is meant by a state function. (5.2)
7. Discuss the concept of enthalpy and explain it in terms of energy, pressure and volume. (5.3)
8. Given a physical or chemical reaction, be able to determine the sign of the enthalpy change. (5.3)
9. Discuss the relationship between the magnitude of enthalpy and the amounts of reactant consumed in a chemical process. (5.4)
10. Perform stoichiometric thermochemistry calculations (5.4)
11. Discuss the principle of calorimetry and perform heat calculations. (5.5)
12. Discuss Hess’ law and use it to calculate the change in enthalpy. (5.6)
13. Use the enthalpies of formation to determine the ΔH for a multistep chemical reaction. (5.7)
14. Examine foods and fuels as sources of energy and discuss some related health and social issues. (5.8)
Vocabulary
See Summary and Key terms, pages 203 - 204
Section 5.1 - The Nature of Energy o Thermodynamics is the study of energy and its transformations. o Thermochemistry is the study of the relationships between chemical reactions and energy changes involving heat. o Definitions: o Energy is the capacity to do work or to transfer heat. o Work is energy used to cause an object with mass to move.
 Work = force x distance
 w = F x d o Heat is the energy used to cause the temperature of an object to increase. o Force is any kind of push or pull exerted on an object.
 The most familiar force is the pull of gravity.
 Force = mass x acceleration
 F = m x a
1

Kinetic Energy and Potential Energy o Kinetic energy is the energy of motion:
E k

1 mv
2
2 o Potential energy is the energy an object possesses by virtue of its position or composition. o Electrostatic energy is an example: o It arises from interactions between charged particles.
E el
 kQ
1
Q
2 d o Potential energy can be converted into kinetic energy.
 An example is a ball of clay dropped off a building.
Units of Energy o SI unit is the joule, J. o 1 Joule equal 1 N·m (1 kg·m 2 ·s -2 ) o Traditionally, we use the calorie as a unit of energy. o 1 cal = 4.184 J (exactly) o The nutritional Calorie, Cal = 1,000 cal = 1 kcal.
Complete questions 5.11, 5.12 and 5.13
System and Surroundings o A system is the part of the universe we are interested in studying. o Surroundings are the rest of the universe (i.e., the surroundings are the portions of the universe that are not involved in the system). o Example: If we are interested in the interaction between hydrogen and oxygen in a cylinder, then the H
2
and O
2
in the cylinder form a system. o A closed system is one where no inputs or outputs occur. o These systems are usually set up at a specific temperature and pressure in order to keep them as stable as possible. o Such systems are often studied to determine the effects of making changes in one component of system on the system as a whole.
Section 5.2 - The First Law of Thermodynamics o The first law of thermodynamics states that energy cannot be created or destroyed. o The first law of thermodynamics is the law of conservation of energy. o That is, the energy of system + surroundings is constant. o Thus, any energy transferred from a system must be transferred to the surroundings (and vice versa).
2

Internal Energy o The total energy of a system is called the internal energy: o It is the sum of all the kinetic and potential energies of all components of the system. o Absolute internal energy cannot be measured, only changes in internal energy. o Change in internal energy, ΔE = E final
– E initial
. o Example: o A mixture of H
2(g)
and O
2(g)
has a higher internal energy than H
2
O
(g)
therefore: o Going from H
2(g)
and O
2(g)
to H
2
O
(g)
results in a negative change in internal energy, indicating that the system has lost energy to the surroundings:
 2 H
2(g)
+ O
2(g)
 2 H
2
O
(g)
ΔE < 0 o Going from H
2
O
(g)
to H
2(g)
and O
2(g)
results in a positive change in internal energy, indicating that the system has gained energy from the surroundings:
 2 H
2
O
(g)
 2 H
2(g)
+ O
2(g)
ΔE > 0
Relating ΔE to Heat and Work o From the first law of thermodynamics: o When a system undergoes a physical or chemical change, the change in internal energy is given by the heat added to or liberated from the system plus the work done on or by the system:
 ΔE = q + w o Heat flowing from the surroundings to the system is positive, q > 0. o Work done by the surroundings on the system is positive, w > 0.
3

Endothermic and Exothermic Processes o An endothermic process is one that absorbs heat from the surroundings. o An endothermic reaction feels cold. o ΔE > 0 o An exothermic process is one that transfers heat to the surroundings. o An exothermic reaction feels hot. o ΔE < 0
To summarize:
• For q :
– if positive (+) the system
– if negative (-) the system
• For w :
heat.
heat.
– if positive (+) work is done
– if negative (-) work is done
• For 
:
– if positive (+) there is a
– if negative (-) there is a
.
of energy by system.
of energy by system.
Complete questions 5.25 and 5.26
State Functions o A state function depends only on the initial and final states of a system. o Example: The altitude difference between Vancouver and Saskatoon does not depend on whether you fly or drive, only on the elevation of the two cities above sea level. o Similarly, the internal energy of 50 g of H
H
2
O
(l)
from 1
2
O
(l)
at 25°C does not depend on whether we cool 50 g of
H
2
O
(l)
at 0°C to 25°. o A state function does not depend on how the internal energy is used. o Example: A battery in a flashlight can be discharged by producing heat and light. The same battery in a toy car is used to produce heat and work. The change in internal energy of the battery is the same in both cases.
4

Section 5.3 - Enthalpy o Chemical changes may involve the release or absorption of heat. o Many also involve work done on or by the system. o Work is often either electrical or mechanical work. o Mechanical work done by a system involving expanding gases is called pressure-volume work or P-V work. o The heat transferred between the system and surroundings during a chemical reaction carried out under constant pressure is called enthalpy, H. o Again, we can only measure the change in enthalpy, ΔH. o Mathematically, o ΔH = H final
– H initial
= ΔE + PΔV o w = –PΔV; ΔE = q + w o ΔH = ΔE + PΔV = (qp + w) – w = qp o For most reactions PΔV is small thus ΔH = ΔE o Heat transferred from surroundings to the system has a positive enthalpy o ΔH > 0 for an endothermic reaction o Heat transferred from the system to the surroundings has a negative enthalpy o ΔH < 0 for an exothermic reaction o Enthalpy is a state function.
A Closer Look at Energy, Enthalpy, and P-V Work o Consider: o A cylinder has a cross-sectional area A. o A piston exerts a pressure, P = F/A, on a gas inside the cylinder. o The volume of gas expands through ΔV while the piston moves a height Δh = h o The magnitude of work done = F x Δh = P x A x Δh = P x ΔV. o Since work is being done by the system on the surroundings, then
 w = –PΔV. f
– h i
. o Using the first law of thermodynamics,
 ΔE = q – PΔV. o If the reaction is carried out under constant volume,
 ΔV = 0 and ΔE = qv. o If the reaction is carried out under constant pressure, o ΔE = qp – PΔV, or o qp = ΔH = ΔE + PΔV o and ΔE = ΔH – PΔV
Section 5.4 - Enthalpies of Reaction o For a reaction, ΔH rxn
= H products
– H reactants
. o The enthalpy change that accompanies a reaction is called the enthalpy of reaction or heat of reaction
(ΔH rxn
). o Consider the thermochemical equation for the production of water: o 2 H
2(g)
+ O
2(g)
 2 H
2
O
(g)
ΔH = –483.6 kJ o The equation tells us that 483.6 kJ of energy are released to the surroundings when water is formed. o ΔH noted at the end of the balanced equation depends on the number of moles of reactants and products associated with the ΔH value.
5

o These equations are called thermochemical equations. o Enthalpy diagrams are used to represent enthalpy changes associated with a reaction. o In the enthalpy diagram for the combustion of H
2(g)
, the reactants, 2H
2(g) than the products 2H
2
O
(g)
; this reaction is exothermic. o Enthalpy is an extensive property.
+ O
2(g)
, have a higher enthalpy o Therefore, the magnitude of enthalpy is directly proportional to the amount of reactant consumed. o Example: If one mol of CH
4
is burned in oxygen to produce CO
2
and water, 890 kJ of heat are released to the surroundings. If two mol of CH
4
are burned, then 1780 kJ of heat are released. o The sign of ΔH depends on the direction of the reaction. o The enthalpy change for a reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction. o Example: CH
4(g) o But CO
2(g)
+ 2 H
+ 2 O
2
O
(l)
2(g)
 CH
 CO
2(g)
4(g)
+ 2 H
2
O
(l)
+ 2 O
2(g)
ΔH = –890 kJ,
ΔH = +890 kJ. o Enthalpy change depends on state. o 2 H
2
O
(g)
 2 H
2
O
(l)
Complete questions 5.36 to 5.40, and 5.45 to 5.46
Section 5.5 - Calorimetry o Calorimetry is a measurement of heat flow. o A calorimeter is an apparatus that measures heat flow.
Heat Capacity and Specific Heat o Heat capacity is the amount of energy required to raise the temperature of an object by 1°C. o Molar heat capacity is the heat capacity of 1 mol of a substance.
ΔH = –88 kJ
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o Specific heat, or specific heat capacity, is the heat capacity of 1 g of a substance. o Heat = specific heat x grams of substance x change of temperature o q = C x m x ΔT. o Be careful of the sign of q.
Constant-Pressure Calorimetry o The most common technique is to use atmospheric pressure as the constant pressure. o Recall ΔH = qp. o The easiest method is to use a coffee cup calorimeter. o q soln
= (specific heat of solution) x (grams of solution) x ΔT = –q rxn o For dilute aqueous solutions, the specific heat of the solution will be close to that of pure water.
Bomb Calorimetry (Constant-Volume Calorimetry) o Reactions can be carried out under conditions of constant volume instead of constant pressure. o Constant volume calorimetry is carried out in a bomb calorimeter. o The most common type of reaction studied under these conditions is combustion. o If we know the heat capacity of the calorimeter, C q rxn
= –C cal x ΔT. cal
, then the heat of reaction, o Since the reaction is carried out under constant volume, q relates to ΔE.
Remember the heat calculations you learned in Grade 10.
Do questions 5.47 to 5.56
Section 5.6 - Hess’s Law
Hess’s Law: If a reaction is carried out in a series of steps, ΔH for the reaction is the sum of ΔH for each of the steps. o The total change in enthalpy is independent of the number of steps. o Total ΔH is also independent of the nature of the path. o Note that ΔH is sensitive to the states of the reactants and the products. o Hess’s law allows us to calculate enthalpy data for reactions that are difficult to carry out directly
For instance, carbon and oxygen can react to form carbon monoxide:
(1) C
(s)
+ ½ O
2 (g)
 CO
(g)
Δ H = - 110 kJ
Carbon monoxide can react further with oxygen to form carbon dioxide:
(2) CO
(g)
+ ½ O
2 (g)
 CO
2 (g)
ΔH = - 294 kJ
A third possible reaction involves carbon and oxygen reacting to form carbon dioxide directly:
(3) C
(s)
+ O
2 (g)
 CO
2 (g)
ΔH = - 394 kJ
7

You will note that the heat of reaction from (1) plus the heat of reaction from (2) exactly equals the heat of reaction from (3). This makes sense when you consider what is going on; chemically, reactions (1) and (2) perform in two steps what reaction (3) does in one step. Note what happens when reactions (1) and (2) are added up:
C
(s)
+ ½ O
2 (g)
 CO
(g)
ΔH = - 110 kJ
CO
(g)
+ ½ O
2 (g)
 CO
2 (g)
C
(s)
+ O
2 (g)
 CO
2 (g)
ΔH = - 294 kJ
ΔH = - 394 kJ
This result is stated by Hess's Law:
This law allows us to take several reactions from the heats of formation table, add them up and get ΔH values for reactions that do not appear directly on the table.
Section 5.7 - Enthalpies of Formation o Hess’s law states that if a reaction is carried out in a number of steps, ΔH for the overall reaction is the sum of the ΔH’s for each of the individual steps. o If a compound is formed from its constituent elements, then the enthalpy change for the reaction is called the enthalpy of formation, ΔH f
. o Standard state (standard conditions) refer to the substance at: o 1 atm and 25°C (298 K). o Standard enthalpy, ΔH°, is the enthalpy measured when everything is in its standard state. o Standard enthalpy of formation of a compound, ΔH° f
, is the enthalpy change for the formation of 1 mol of compound with all substances in their standard states. o If there is more than one state for a substance under standard conditions, the more stable state is used.
Example: When dealing with carbon we use graphite because graphite is more stable than diamond or C
60
. o The standard enthalpy of formation of the most stable form of an element is zero.
Using Enthalpies of Formation to Calculate Enthalpies of Reaction
Example 1: 2 Al
(s)
+ 3 CuO
(s)
 3 Cu
(s)
+ Al
2
O
3(s)
1.
Find an equation on the Heat of Formation Table to match equation.
of the compounds in the given
2 Al
(s)
+ 1½ O
2(g)
 Al
2
O
3(s)
ΔH = - 1676.0 kJ/mol
Cu
(s)
+ ½ O
2(g)
 CuO
(s)
ΔH = - 155.0 kJ/mol
8

2.
Flip the CuO equation to make the compound a reactant as well:
Cu
(s)
+ ½ O
2(g)
 CuO
(s)
ΔH = - 155.0 kJ/mol
Becomes
CuO
(s)
 Cu
(s)
+ ½ O
2(g)
ΔH = + 155.0 kJ/mol
3.
Multiply the CuO equation by 3 to have the same number of molecules as in the original equation.
2 Al
(s)
+ 1½ O
2(g)
 Al
2
O
3(s)
ΔH = - 1676.0 kJ/mol
3 x CuO
(s)
 Cu
(s)
+ ½ O
2(g)
ΔH = 3 x (+ 155.0 kJ/mol)
Gives
2 Al
(s)
+ 1½ O
2(g)
 Al
2
O
3(s)
ΔH = - 1676.0 kJ/mol
3 CuO
(s)
 3 Cu
(s)
+ 1½ O
2(g)
ΔH = + 465.0 kJ/mol
4.
Add the two equations together. You treat the equations just like you would in math; add the material to the left of the arrow together and the material on the right of the arrow together. The enthalpy changes are also added:
2 Al
(s)
+ 1½ O
3 CuO
(s)
 3 Cu
2(g)
 Al
(s)
2
O
3(s)
+ 1½ O
2(g)
ΔH = - 1676.0 kJ/mol
ΔH = + 465.0 kJ/mol
2 Al
(s)
+ 1½ O
2(g)
+ 3 CuO
(s)
 Al
2
O
3(s)
+ 3 Cu
(s)
+ 1½ O
2(g)
ΔH = - 1211 kJ
You will notice that there are similar terms on each side of the reaction equation. These can be treated again just like mathematical terms and cancelled out where you can:
2 Al
(s)
+ 3 CuO
(s)
 Al
2
O
3(s)
+ 3 Cu
(s)
ΔH = - 1211 kJ
You know you are right if the net equation you end up with is the same as the original equation you started with.
9

Example 2: 2 C
2
H
6(g)
+ 7 O
2(g)
 4 CO
2(g)
+ 6 H
2
O
(l)
1.
Find an equation on the Heat of Formation Table to match equation.
of the compounds in the given
2 C
(s)
+ 3 H
2(g)
 C
2
H
6(g)
ΔH = - 84.0 kJ/mol
C
H
(s)
+ O
2(g)
 CO
2(g)
+ ½ O
2(g)
2(g)
 H
2
O
(l)
ΔH = - 394.0 kJ/mol
ΔH = - 286.0 kJ/mol
2.
Flip the C
2
H
6
equation to make the compound a reactant:
2 C
(s)
+ 3 H
2(g)
 C
2
H
6(g)
ΔH = - 84.0 kJ/mol
Becomes
C
2
H
6(g)
 2 C
(s)
6 x H
Gives
+ 3 H
2(g)
2(g)
+ ½ O
2(g)
 H
2
O
(l)
ΔH = + 84.0 kJ/mol
3.
Multiply each equation to have the same number of molecules as in the original equation.
2 x C
2
H
6(g)
 2 C
(s)
+ 3 H
2(g)
ΔH = 2 x(+ 84.0 kJ/mol)
4 x C
(s)
+ O
2(g)
 CO
2(g)
ΔH = 4 x (- 394.0 kJ/mol)
ΔH = 6 x (- 286.0 kJ/mol)
2 C
2
H
6(g)
 4 C
(s)
+ 6 H
2(g)
4 C
(s)
+ 4 O
2(g)
 4 CO
2(g)
6 H
2(g)
+ 3 O
2(g)
 6 H
2
O
(l)
ΔH = + 168.0 kJ/mol
ΔH = - 1576.0 kJ/mol
ΔH = - 1716.0 kJ/mol
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4. Add the equations together.
2 C
2
H
6(g)
 4 C
(s)
+ 6 H
2(g)
4 C
(s)
+ 4 O
2(g)
 4 CO
2(g)
6 H
2(g)
+ 3 O
2(g)
 6 H
2
O
(l)
ΔH = + 168.0 kJ/mol
ΔH = - 1576.0 kJ/mol
ΔH = - 1716.0 kJ/mol
2 C
2
H
6(g)
+ 4 C
(s)
+ 4 O
2(g)
+ 6 H
2(g)
+ 3 O
2(g)
 4 C
(s)
+ 6 H
2(g)
+ 4 CO
2(g)
+ 6 H
2
O
(l)
ΔH = - 3124 kJ
Cancel like terms:
2 C
2
H
6(g)
+ 7 O
2(g)
 4 CO
2(g)
+ 6 H
2
O
(l)
ΔH = - 3124 kJ
Example 3: SiO
2(s)
+ C
(s)
 CO
2(g)
+ Si
(s)
Where the heat of formation of SiO
2(s)
= - 861 kJ/mol
1.
Find an equation on the Heat of Formation Table to match equation.
of the compounds in the given
C
(s)
+ O
2(g)
 CO
2(g)
ΔH = - 394.0 kJ/mol
The equation for SiO its elements:
2
is not on the table, but you are given the heat of formation of the substance from
Si
(s)
+ O
2(g)
 SiO
2(s)
ΔH = - 861.0 kJ/mol
2.
Flip the SiO
2
equation to make the compound a reactant as well:
Si
(s)
+ O
2(g)
 SiO
2(s)
ΔH = - 861.0 kJ/mol
Becomes
SiO
2(s)
 Si
(s)
+ O
2(g)
ΔH = + 861.0 kJ/mol
3.
Since there is only one of each molecule in the original equation, no multiplication is necessary.
11

4. Add the two equations together. You treat the equations just like you would in math; add the material to the left of the arrow together and the material on the right of the arrow together. The enthalpy changes are also added:
C
(s)
+ O
2(g)
 CO
2(g)
SiO
2(s)
 Si
(s)
+ O
2(g)
ΔH = - 394.0 kJ/mol
ΔH = + 861.0 kJ/mol
C
(s)
+ O
2(g)
+ SiO
2(s)
 CO
2(g)
+ Si
(s)
+ O
2(g)
ΔH = + 467 kJ
Cancel like terms:
C
(s)
+ SiO
2(s)
 CO
2(g)
+ Si
(s)
ΔH = + 467 kJ
Complete questions 5.67 to 5.75
Section 5.8 - Foods and Fuels o Fuel value is the energy released when 1 g of substance is burned. o The fuel value of any food or fuel is a positive value that must be measured by calorimetry.
Foods o Fuel value is usually measured in Calories (1 nutritional Calorie, 1 Cal = 1000 cal). o Most energy in our bodies comes from the oxidation of carbohydrates and fats. o In the intestines, carbohydrates are converted into glucose, C
6
H
12
O
6
, or blood sugar. o In the cells glucose reacts with O
2
in a series of steps which ultimately produce CO
2
, H
2
O, and energy:
 C
6
H
12
O
6 (s)
+ 6 O
2 (g)
 6 CO
2 (g)
+ 6 H
2
O
(l)
ΔH° f
= –2803 kJ o Fats, for example tristearin, react with O2 as follows:
 2 C
57
H
110
O
6 (s)
+ 163 O
2 (g)
 114 CO
2 (g)
+ 110 H
2
O
(l)
ΔH° f
= –75,250 kJ. o Fats contain more energy than carbohydrates. Fats are not water soluble. Therefore, fats are good for energy storage.
12

Fuels o In Canada we use about 10 000 PJ of energy per year (about 8 x 10 5 kJ of fuel per person per day). o Approximately 70% of this energy comes from petroleum and natural gas. o The remainder of the energy comes from coal, nuclear, and hydroelectric sources. o In 2003, Canada generated 567 Terrawatt-hours of electricity. This was generated by Hydroelectric (58%), coal (16%), nuclear (12%), natural gas (6%) , and fuel oil (3%), while other sources such as biomass, solar energy and wind generators made up the remaining 2%. The total installed capacity of Canada was
111000 MW in 2000. o Coal, petroleum, and natural gas are fossil fuels. They are not renewable. o Natural gas consists largely of carbon and hydrogen. Compounds such as CH
4 typical constituents.
, C
2
H
6
, C
3
H
8
and C
4
H
10
are o Petroleum is a liquid consisting of hundreds of compounds. Impurities include S, N, and O compounds. o Coal contains high molecular weight compounds of C and H. In addition, compounds containing S, O, and
N are present as impurities that form air pollutants when burned in air. o Syngas (synthesis gas): a gaseous mixture of hydrocarbons produced from coal by coal gasification.
Other Energy Sources o Nuclear energy is the energy released in the splitting or fusion of nuclei of atoms. o Fossil fuels and nuclear energy are nonrenewable sources of energy. o Renewable energy sources include: o solar energy. o wind energy. o geothermal energy. o hydroelectric energy. o biomass energy. o These are virtually inexhaustible and will become increasingly important as fossil fuels are depleted.
Complete questions 5.67 to 5.75
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