Download Chapter 5 Outline 1213 full

AP Chemistry
Chapter 5. Thermochemistry
Chapter 5. Thermochemistry
5.1 The Nature of Energy
•
Thermodynamics = study of energy and its transformations.
•
Thermochemistry = the study of the relationships between chemical reactions and energy changes
involving heat.
 Definitions:
•
Energy is the capacity to do work or to transfer heat.
•
Work is energy used to cause an object with mass to move. w = F  d
•
Heat is the energy used to cause the temperature of an object to increase.
•
A force is any kind of push or pull exerted on an object. The most familiar force is the pull of gravity.
Kinetic Energy and Potential Energy
•
Kinetic energy = energy of motion:
Ek 
•
1 2
mv
2
Potential energy = the energy an object possesses by virtue of its position or composition.
• Electrostatic energy (Ed) is an example
• It arises from interactions between charged particles.  = 8.99 x 109 J-m/C2
Eel 
•
kQ1Q2
r2
Potential energy can be converted into kinetic energy. Example: A ball of clay dropped off a building.
Units of Energy
•
SI unit is the joule, J.
•
From E k 
•
•
2
1 2
mv , 1J  1kg x m
2
s2
Traditionally, we use the calorie as a unit of energy.
• 1 cal = 4.184 J (exactly)
The nutritional Calorie, Cal = 1,000 cal.
System and Surroundings
•
A system is the part of the universe we are interested in studying.
•
Surroundings are the rest of the universe (i.e., the surroundings are the portions of the universe
not involved in the system).
•
Example: If we are interested in the interaction between hydrogen and oxygen in a cylinder, then the H2 and
O2 in the cylinder form a system.
-1-
AP Chemistry
Chapter 5. Thermochemistry
Transferring Energy: Work and Heat
•
From physics:
• Force = a push or pull on an object.
•
Work = the energy used to move an object against a force.
w=F x d
•
Heat = the energy transferred from a hotter object to a colder one.
•
Energy = the capacity to do work or to transfer heat.

Internal Energy = total energy of a system

Absolute internal energy cannot be measured.

Change in internal energy: E = Efinal - Einitial
Sample Exercise 5.1 (p. 169)
A bowler lifts a 5.4-kg (12-lb.) bowling ball from ground level to a height of 1.6 m (5.2 ft.) and then drops the
ball back to the ground.
a) What happens to the potential energy of the bowling ball as it is raised from the ground?
b) What quantity of work, in J, is used to raise the ball? (85 J)
c) After the ball is dropped, it gains kinetic energy. If we assume that all of the work in part (b) is converted
into kinetic energy at the point of impact with the ground, what is the speed of the ball at the
point of impact? (5.6 m/s)
(Note: The force due to gravity is F = m x g, where m is the mass of the object and g is the gravitational
constant; g = 9.8 m/s2.)
Practice Exercise 5.1
What is the kinetic energy, in J, of
a) an Ar atom moving with a speed of 650 m/s?
(5.6 m/s)
b) a mole of Ar atoms moving with a speed of 650 m/s (Hint: 1 amu = 1.66 x 10-27 kg)?
(8.4 x 103 J)
-2-
AP Chemistry
Chapter 5. Thermochemistry
5.2 The First Law of Thermodynamics
•
•
The first law of thermodynamics: Energy cannot be created or destroyed.
The first law of thermodynamics = the law of conservation of energy.
• That is, the energy of (system + surroundings) is constant.
• Thus any energy transferred from a system must be transferred to the surroundings (and vice versa).
Internal Energy
•
•
The total energy of a system = internal energy.
= sum of all the kinetic and potential energies of all components of the system.
Absolute internal energy cannot be measured, only changes in internal energy.
•
Change in internal energy, E = Efinal – Einitial.
•
•
Example: A mixture of H2(g) and O2(g) has a higher internal energy than H2O(g).
Going from H2(g) and O2(g) to H2O(g) results in a negative change in internal energy, indicating that the
system has lost energy to the surroundings:
H2(g) + O2(g)  2H2O(g)
E < 0
Going from H2O(g) to H2(g) and O2(g) results in a positive change in internal energy, indicating that the
system has gained energy from the surroundings:
2H2O  H2(g) + O2(g)
E > 0
•
Figure 5.4 Changes in internal energy.
a) When a system loses energy, that energy is released to the surroundings. The loss of energy is
represented by an arrow that points downward between the initial and final states of the system. In this
case, the energy change of the system, E = Efinal – Einitial, is negative.
b) When a system gains energy, that energy is gained from the surroundings. In this case, the gain of
energy is represented by an arrow that points upward between the initial and final states of the system,
and the energy change of the system is positive.
Notice that in both (a) and (b) that the vertical arrow originates at the initial state and has its head at the
final state.
-3-
AP Chemistry
Chapter 5. Thermochemistry
Figure 5.5 Energy diagram for the interconversion of H2(g), O2(g), and H2O(l)
A system composed of H2(g) and O2(g) has a greater internal energy than one composed of H2O(l). The system
loses energy (E < 0) when H2 and O2 are converted to H2O. It gains energy (E > 0) when H2O is decomposed
into H2 and O2.
Relating E to Heat and Work
•
•
•
From the first law of thermodynamics:
• 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:
E = q + w
Heat flowing from the surroundings to the system is positive, q > 0.
Work done by the surroundings on the system is positive, w > 0.
Figure 5.6 Sign conventions for heat and work
Heat, q, gained by a system and work, w done on a system are both positive quantities.
Both increase the ineternal energy, E, of the system, causing E to be a positive quantity.
-4-
AP Chemistry
Chapter 5. Thermochemistry
Sample Exercise 5.2 (p. 171)
Two gases, A(g) and B(g), are confined in a cylinder-and-piston arrangement like that in the Figure below.
Substances A and B react to form a solid product: A(g) + B(g)  C(s). As the reaction occurs, the system
loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the
volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work
on the system. What is the change in the internal energy of the system?
(-670 J)
Practice Exercise 5.2
Calculate the change in the internal energy of the system for a process in which the system absorbs 140 J of heat
from the surroundings and does 85 J of work on the surroundings.
(+55 J)
Endothermic and Exothermic Processes
•
An endothermic process is one that absorbs heat from the surroundings.
• An endothermic reaction feels cold.
•
An exothermic process is one that transfers heat to the surroundings.
• An exothermic reaction feels hot.
State Functions
•
A state function depends only on the initial and final states of a system.
• e.g. The altitude difference between Denver and Chicago does not depend on whether you fly or drive,
only on the elevation of the two cities above sea level.
• Similarly, the internal energy of 50 g of H2O(l) at 25oC does not depend on whether we cool 50 g of
H2O(l) from 100 oC to 25 oC or heat 50 g of H2O(l) at 0 oC to 25 oC
-5-
AP Chemistry
Chapter 5. Thermochemistry
Figure 5.8 Internal energy, E, a state function
E depends only on the present state of the system and not on the path by which it arrived at that state.
The internal energy of 50 g of water at 25oC is the same whether the water is cooled from a higher temperature
to 25oC or warmed from a lower temperature to 25oC.
•
A state function does not depend on how the internal energy is used.
• 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.
Figure 5.9 Internal energy is a state function, but heat and work are not.
The amounts of heat and work transferred between the system and the surroundings depend on the way in which
the system goes from one state to another. (a) A battery shorted out by a wire loses energy to the surroundings
only as heat; no work is performed by the system. (b) A battery discharged through a motor loses energy as
work (to make the fan turn) and also loses energy as heat. Now, however, the amount of heat lost is much less
than in (a). The value of ΔE is the same for both processes even though the values of q and w in (a) are different
from the values of q and w in (b).
-6-
AP Chemistry
Chapter 5. Thermochemistry
5.3 Enthalpy
•
•
Chemical changes may involve release or absorption of heat.
Many also involve work done on or by the system.
• Work is often either electrical or mechanical work.
• Mechanical work done by a system involving expanding gases is called pressure-volume work or P-V
work.
Figure 5.10 A system that does work on its surroundings.
(a) An apparatus for studying the reaction of zinc metal with hydrochloric acid at constant pressure. The piston
is free to move up and down in its cylinder to maintain a constant pressure equal to atmospheric pressure inside
the apparatus. Notice the pellets of zinc in the L-shaped arm on the left. When this arm is rotated, the pellets will
fall into the main container and the reaction will begin. (b) When zinc is added to the acid solution, hydrogen
gas is evolved. The hydrogen gas does work on the surroundings, raising the piston against atmospheric pressure
to maintain constant pressure inside the reaction vessel.
•
•
•
The heat transferred between the system and surroundings during a chemical reaction carried out under
constant pressure is called enthalpy, H.
Again, we can only measure the change in enthalpy, H.
Mathematically,
H = Hfinal – Hinitial = E + PV)= E + PV
E = q + w; w = –PV
 H = E + PV = q + w – w = q
• For most reactions PV is small thus H = E = q
•
Heat transferred from surroundings to the system has a positive enthalpy (i.e., H > 0 for an endothermic
reaction).
•
Heat transferred from the system to the surroundings has a negative enthalpy (i.e., H < 0 for an exothermic
reaction).
•
Enthalpy is a state function.
-7-
AP Chemistry
Chapter 5. Thermochemistry
Figure 5.11 Endothermic and exothermic processes
a) If the system absorbs heat (endothermic process), H will be positive (DH > 0).
b) If the system loses heat (exothermic process), H will be negative (DH < 0).
Sample Exercise 5.3 (p. 176)
Indicate the sign of the enthalpy change, H, in each of the following processes carried out under atmospheric
pressure, and indicate whether the process is endothermic or exothermic.
a) an ice cube melts;
b) 1 g of butane (C4H10) is combusted in sufficient oxygen to give complete combustion of CO2 and H2O.
Practice Exercise 5.3
Suppose we confine 1 g of butane and sufficient oxygen to completely combust it in a cylinder like that in
Figure 5.13. The cylinder is perfectly insulating, so no heat can escape to the surroundings. A spark initiates
combustion of the butane, which forms carbon dioxide and water vapor. If we used this apparatus to measure
the enthalpy change in the reaction, would the piston rise, fall, or stay the same?
-8-
AP Chemistry
Chapter 5. Thermochemistry
5.4 Enthalpies of Reaction
•
•
For a reaction, Hrxn = H(products) – H(reactants).
The enthalpy change that accompanies a reaction is called the enthalpy of reaction or heat of reaction
(Hrxn).
•
Consider the thermochemical equation for the production of water:
2H2(g) + O2(g)  2H2O(g)
H = –483.6 kJ
• The equation tells us that 483.6 kJ of energy are released to the surroundings when water is formed.
• H noted at the end of the balanced equation depends on the number of moles of reactants and products
associated with the H value.
• These equations are called thermochemical equations.
•
•
Enthalpy diagrams are used to represent enthalpy changes associated with a reaction.
In the enthalpy diagram for the combustion of H2(g), the reactants, 2H2(g) + O2(g), have a higher enthalpy
than the products 2H2O(g); this reaction is exothermic.
1. Enthalpy is an extensive property.
• Therefore, the magnitude of enthalpy is directly proportional to the amount of reactant consumed.
• Example: If one mol of CH4 is burned in oxygen to produce CO2 and water, 890 kJ of heat is released to
the surroundings. If two mol of CH4 is burned, then 1780 kJ of heat is released.
H = -890 kJ
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
(Equation 5.18)
Figure 5.15 H for a reverse reaction
Reversing a reaction changes the sign but not the magnitude of the enthalpy change: H2 – H1
-9-
AP Chemistry
Chapter 5. Thermochemistry
2. The sign of H depends on the direction of the reaction.
• The enthalpy change for a reaction is equal in magnitude but opposite in sign to H for the reverse
reaction.
• Example: CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = –890 kJ,
• But CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
H = +890 kJ.
3. Enthalpy change depends on state.
• 2H2O(g)  2H2O(l)
H = –88 kJ
Sample Exercise 5.4 (p. 179)
How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system?
(Use the information given in Equation 5.18.)
(-250 kJ)
Practice Exercise 5.4
Hydrogen peroxide can decompose to water and oxygen by the following reaction:
2 H2O2(l)  2 H2O(l) + O2(g)
H = -196 kJ
Calculate the value of q when 5.00 g of H2O2(l) decomposes at constant pressure.
(-14.4 kJ)
5.5 Calorimetry
•
•
Calorimetry is a measurement of heat flow.
Calorimeter is an apparatus that measures heat flow.
Heat Capacity and Specific Heat
•
•
•
Heat capacity is the amount of energy required to raise the temperature of an object by 1oC.
• Molar heat capacity is the heat capacity of 1 mol of a substance.
• Specific heat, or specific heat capacity is the heat capacity of 1 g of a substance.
Heat, q = (specific heat) x (grams of substance) x T.
Be careful of the sign of q.
- 10 -
AP Chemistry
Chapter 5. Thermochemistry
Sample Problem 5.5 (p. 181)
a) How much heat is needed to warm 250 g of water from 22oC to near its boiling point, 98oC? The
specific heat of water is 4.18 J/g-K.
(7.9 x 104 J)
b) What is the molar heat capacity of water?
(75.2 J/mol-K)
Practice Problem 5.5
a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of
the rocks is 0.82 J/g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature
increases by 12.0oC.
(4.9 x 105 J)
b) What temperature change would these rocks undergo if they emitted 450 kJ of heat?
(11K = 11oC decrease)
Constant-Pressure Calorimetry
•
•
•
•
Most common technique: use atmospheric pressure as the constant pressure.
Recall H = qp.
Easiest method: use a coffee cup calorimeter.
qsoln = (specific heat of solution) x (grams of solution) x T = –qrxn
For dilute aqueous solutions, the specific heat of the solution will be close to that of pure water.
- 11 -
AP Chemistry
Chapter 5. Thermochemistry
Figure 5.17 Coffee-cup calorimeter
This simple apparatus is used to measure heat-accompanying reactions at constant pressure.
Sample Exercise 5.6 (p.182)
When a student mixes 50. mL of 1.0 M HCl and 50. mL of 1.0 M NaOH in a coffee-cup calorimeter, the
temperature of the resultant solution increases from 21.0oC to 27.5oC. Calculate the enthalpy change for the
reaction, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the
solution is 100. mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g-K.
(-54 kJ/mol)
Practice Exercise 5.6
When 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter,
the temperature of the mixture increases from 22.30oC to 23.11oC. The temperature increase is caused by the
following reaction:
AgNO3(aq) + HCl(aq)  AgCl(s) + HNO3(aq)
Calculate H for this reaction, assuming that the combined solution has a mass of 100.0 g and a specific heat of
4.18 J/g-oC.
(-68 kJ/mol)
- 12 -
AP Chemistry
Chapter 5. Thermochemistry
Bomb Calorimetry (Constant-Volume Calorimetry)
•
•
•
•
•
Reactions can be carried out under conditions of constant volume instead of constant pressure.
Constant volume calorimetry is carried out in a bomb calorimeter.
The most common type of reaction studied under these conditions is combustion.
If we know the heat capacity of the calorimeter, Ccalorimeter, then the heat of reaction,
qrxn = –Ccalorimeter x T.
Since the reaction is carried out under constant volume, q relates to E.
Figure 5.18 Bomb calorimeter
Sample Exercise 5.7 (p. 184)
Methylhydrazine (CH6N2) is commonly used as a liquid rocket fuel. The combustion of methylhydrazine with
oxygen produces N2(g), CO2(g), and H2O(l):
2 CH6N2(l) + 5 O2(g)  2 N2(g) + 2 CO2(g) + 6 H2O(l)
When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter
increases from 25.00oC to 39.50oC. In a separate experiment the heat capacity of the bomb calorimeter is
measured to be 7.794 kJ/oC. What is the heat of reaction for the combustion of a mole of CH6N2 in this
calorimeter?
(-1.30 x 103 kJ/mol CH6N2)
- 13 -
AP Chemistry
Chapter 5. Thermochemistry
Practice Exercise 5.7
A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/oC. The
temperature increases from 23.10oC to 24.95oC. Calculate the heat of combustion of
a) lactic acid per gram (-15.2 kJ/g) and
b) per mole (-1370 kJ/mol)
5.6 Hess’s Law
•
•
•
•
•
•
Hess’s Law: If a reaction is carried out in a series of steps, H for the reaction is the sum of H for each of
the steps.
The total change in enthalpy is independent of the number of steps.
Total H is also independent of the nature of the path.
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
H = –802 kJ
2H2O(g)  2H2O(l)
H = –88 kJ
______________________________________________________________
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
H = –890 kJ
Therefore, for the reaction CH4(g) + 2O2(g)  CO2(g) + 2H2O(l), H= –890 kJ.
Note that H is sensitive to the states of the reactants and products.
Hess’s law allows us to calculate enthalpy data for reactions which are difficult to carry out directly: C(s) +
O2(g) produces a mixture of CO(g) and CO2(g).
Figure 5.20 An enthalpy diagram comparing a one-step and a two-step process for a reaction
The enthalpy change of the direct reaction on the left equals the sum of the two steps on the right. That is, H
for the overall reaction equals the sum of the H values for the two steps shown.
- 14 -
AP Chemistry
Chapter 5. Thermochemistry
Sample Exercise 5.8 (p. 186)
The enthalpy of combustion of C to CO2 is -393.5 kJ/mol C, and the enthalpy of combustion of CO to
CO2 is -283.0 kJ/mol CO:
1)
2)
H1 = -393.5 kJ
C(s) + O2(g)  CO2(g)
H2 = -283.0 kJ
CO(g) + ½ O2(g)  CO2(g)
Using these data, calculate the enthalpy of combustion of C to CO:
3)
H3 = ?
C(s) + ½ O2(g)  CO(g)
(-110.5 kJ)
Practice Exercise 5.8
Carbon occurs in two forms, graphite and diamond. The enthalpy of combustion of graphite is -393.4 kJ/mol
and that of diamond is -395.4 kJ:
C(graphite) + O2(g)  CO2(g)
H1 = -393.5 kJ
C(diamond) + O2(g)  CO2(g)
H2 = -395.4 kJ
Calculate H for the conversion of graphite to diamond:
C(graphite)  C(diamond)
H3 = ?
(+1.9 kJ)
- 15 -
AP Chemistry
Chapter 5. Thermochemistry
Sample Exercise 5.9 (p. 187)
Calculate H for the reaction
2 C(s) + H2(g)  C2H2(g)
Given the following reactions and their respective enthalpy changes:
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(l)
H = -1299.6 kJ
C(s) + O2(g)  CO2(g)
H = -393.5 kJ
H2(g) + ½ O2(g)  H2O(l)
H = -285.8 kJ
(226.8 kJ)
Practice Exercise 5.9
Calculate H for the reaction
NO(g) + O(g)  NO2(g)
given the following reactions and their respective enthalpy changes:
NO(g) + O3(g)  NO2(g) + O2(g)
H = -198.9 kJ
O3(g)  3/2 O2(g)
H = -142.3 kJ
O2(g)  2 O(g)
H = 495.0 kJ
(-304.1 kJ)
- 16 -
AP Chemistry
Chapter 5. Thermochemistry
5.7 Enthalpies of Formation
•
Hess’s law states that if a reaction is carried out in a number of steps, H for the overall reaction is the sum
of the Hs for each of the individual steps.
•
Consider the formation of CO2(g) and 2H2O(l) from CH4(g) and 2O2(g).
• If the reaction proceeds in one step:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l),
then H1 = –890 kJ.
• However, if the reaction proceeds through a CO intermediate:
CH4(g) + 2O2(g)  CO(g) + 2H2O(l) + ½O2(g)
H2 = –607 kJ
CO(g) + 2H2O(l) + ½ O2(g)  CO2(g) + 2H2O(l)
H3 = –283 kJ,
Then H for the overall reaction is:
H2 + H3 = –607 kJ – 283 kJ = –890 kJ = H1
•
•
If a compound is formed from its constituent elements, then the enthalpy change for the reaction is called
the enthalpy of formation, Hf.
Standard state (standard conditions) refer to the substance at:
• 1 atm and 25 oC (298 K).
•
Standard enthalpy, Ho, is the enthalpy measured when everything is in its standard state.
•
Standard enthalpy of formation of a compound, Hof is the enthalpy change for the formation of 1 mol of
compound with all substances in their standard states.
•
∆H˚f for selected substances are tabulated in Appendix C.
• A large majority of ∆H˚f values tabulated in Appendix C are negative, meaning that most
formation reactions are exothermic.
If there is more than one state for a substance under standard conditions, the more stable one is used.
Example: When dealing with carbon we use graphite because graphite is more stable than diamond or C60.
•
•
The standard enthalpy of formation of the most stable form of an element is zero.
- 17 -
AP Chemistry
Chapter 5. Thermochemistry
Sample Exercise 5.10 (p. 190)
For which of the following reactions at 25oC would the enthalpy change represent a standard enthalpy of
formation? For those where it does not, what changes would need to be made in the reaction conditions?
a) 2 Na(s) + ½ O2(g)  Na2O(s)
b) 2 K(l) + Cl2(g)  2 KCl(s)
c) C6H12O6(s)  6 C(diamond) + 6 H2(g) + 3 O2(g)
Practice Exercise 5.10
Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl4).
Using Enthalpies of Formation to Calculate Enthalpies of Reaction
•
•
•
•
Use Hess’s law!
Example: Calculate H for
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
We start with the reactants, decompose them into elements, then rearrange the elements to form products.
The overall enthalpy change is the sum of the enthalpy changes for each step.
• Decomposing into elements (note O2 is already elemental, so we concern ourselves with C3H8):
C3H8(g)  3C(s) + 4H2(g)
H1 = –Hof [C3H8(g)]
• Next we form CO2 and H2O from their elements:
3C(s) + 3O2(g)  3CO2(g)
H = 3 Hof CO2(g)]
4H2(g) + 2O2(g)  4H2O(l)
H3 = 4Hof [H2O(l)]
• We look up the values and add:
Horxn = –1(–103.85 kJ) + 3(–393.5 kJ) + 4(–285.8 kJ) = –2220 kJ
In general:
Horxn = nHof (products) – mHof (reactants)
•
Where n and m are the stoichiometric coefficients.
Figure 5.22 An enthalpy diagram relating the
enthalpy change for a reaction to enthalpies of
formation.
For the combustion of propane gas, C3H8(g), the reaction is
C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l).
We can imagine this reaction as occurring in three steps.
First, C3H8(g) is decomposed to its elements,
so H1 = Hf[C3H8](g). Second, 3 mol CO2(g) are formed, so
H2 = 3 H[CO2](g). Finally 4 mol H2O(l) are formed, so
H3 = 4 H[H2O](l).
Hess’s law tells us that Hrxn = H1 + H2 + H3.
- 18 -
AP Chemistry
Chapter 5. Thermochemistry
Sample Exercise 5.11 (p. 192)
a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g) and
H2O(l).
(-3267 kJ)
b) Compare the quantity of heat produced by combustion of 1.00 g propane (C3H8) to that produced by
1.00 g benzene.
(C3H8(g): -50.3 kJ/g; C6H6(l): -41.8 kJ/g)
Practice Exercise 5.11
Using the standard enthalpies of formation listed in Table 5.3, calculate the enthalpy change for the combustion
of 1 mol of ethanol:
C2H5OH(l) + 3 O2(g)  2 CO2(g) + 3 H2O(l)
(-1367kJ)
- 19 -
AP Chemistry
Chapter 5. Thermochemistry
Sample Exercise 5.12 (p. 192)
The standard enthalpy change for the reaction
CaCO3(s)  CaO(s) + CO2(g)
is 178.1 kJ. From the values for the standard enthalpies of formation of CaO(s) and CO2(g) given in Table 5.3,
calculate the standard enthalpy of formation of CaCO3(s).
(-1207.1 kJ/mol)
Practice Exercise 5.12
Given the following standard enthalpy of reaction, use the standard enthalpies of formation in Table 5.3 to
calculate the standard enthalpy of formation of CuO(s):
CuO(s) + H2(g)  Cu(s) + H2O(l)
Ho = -129.7 kJ
(-156.1 kJ/mol)
Sample Integrative Exercise (p. 186) (complete on binder paper)
Trinitroglycerin, C3H5N3O9, (usually referred to simply as nitroglycerin) has been widely used as an explosive.
Alfred Nobel used it to make dynamite in 1866. Rather surprisingly, it also is used as a medication, to relieve
angina (chest pains resulting from partially blocked arteries to the heart), by dilating the blood vessels. The
enthalpy of decomposition at 1 atmosphere pressure of trinitroglycerin to form nitrogen gas, carbon dioxide gas,
liquid water, and oxygen gas at 25oC is -1541.4 kJ/mol.
a) Write a balanced chemical equation for the decomposition of trinitroglycerin.
b) Calculate the standard heat of formation of trinitroglycerin.
c) A standard dose of trinitroglycerin for relief of angina is 0.60 mg. Assuming that the sample is
eventually completely combusted in the body (not explosively, though!) to nitrogen gas, carbon
dioxide gas, and liquid water, what number of calories is released?
d) One common form of trinitroglycerin melts at about 3oC. From this information and the formula for the
substance, would you expect it to be a molecular or ionic compound? Explain.
e) Describe the various conversions of forms of energy when trinitroglycerin is used as an explosive to
break rockfaces in highway construction.
- 20 -