Download Chapter 6 Thermochemistry Section 6.1 The Nature of Energy and

Chapter 6
Thermochemistry
This chapter develops for the student the concepts of thermochemistry. Upon completion of Chapter 6,
the student should be able to:
1. Define and explain the following terms:
• Energy
• Radiant energy
• Thermal energy
• Chemical energy
• Potential energy
• Thermochemistry
• Open system
• Closed system
• Isolated system
• Endothermic
• Exothermic
• Enthalpy (∆H)
• Calorimetry
• Heat capacity
• Specific heat
2. Classify common processes as endothermic or exothermic.
3. Use thermochemical equations and stoichiometry to determine amount of heat lost or gained in a
chemical reaction.
4. Perform calculations involving specific heat, mass and temperature change.
5. Sketch the main components of a constant-volume bomb calorimeter.
6. Determine heats of reactions given experimental data collected in a calorimetry experiment.
7. Calculate standard enthalpy of reactions given the standard enthalpy of formations for products and
reactants.
8. Apply Hess’s law to a multi-step process to determine standard enthalpy of reaction.
9. Describe heat of solution, lattice energy, heat of hydration, heat of dilution, system, surrounding, and
internal energy.
10. Classify properties of materials as state functions or non-state functions.
11. Restate the First Law of Thermodynamics.
12. Recall the sign conventions for work and heat used in the textbook.
13. Apply heat and work relationships to gas-phase problems.
14. Define H in term of E, P, and V.
15. Calculate change in internal energy (∆E) given thermochemical equations.
Section 6.1
The Nature of Energy and Types of Energy
Energy, the ability to do work, takes many forms. They include kinetic, potential, radiant, thermal, and
chemical energy. Students sometimes confuse energy and temperature. If we place a pot of water on the
kitchen stove and turn the dial to high, we observe that the water’s temperature increases (using a
thermometer) as the water warms up. This process continues until we reach the boiling point of the water.
Once this temperature is reached, the water no longer increases in temperature, but we are still adding
energy to the system because the dial on the stove still reads “high”. What this shows is that the temperature
of the water and the energy pumped into it are proportional until the boiling point is reached. At that point, the
energy is used to change the physical state of the liquid water to steam without changing the temperature of
either. Using the concept of conservation of energy, we know that the energy provided by the stove at the
boiling point of water must be used to break the forces of attraction that hold the water molecules together in
the liquid phase.
Section 6.2
Energy Changes in Chemical Reactions
Heat is defined as the transfer of thermal energy from a hot body to a cold body. It is a process and is
not energy. It is incorrect to refer to heat energy. The study of transfer of energy that occurs during chemical
reactions is known as thermochemistry. In thermochemistry, we have the system and the surroundings that
make up the universe. We also refer to open, closed, or isolated systems. An open system allows for the
transfer of both energy and mass; a closed system allows the transfer of only energy; and an isolated system
does not allow the transfer of either energy or mass.
Exothermic processes give off energy. The prefix exo- refers to external, thus exothermic means energy
that is given off. The opposite of exothermic is endothermic which means energy is absorbed. In an
exothermic reaction, energy can be thought of as one of the reaction products. Energy can be thought of as a
reactant in endothermic reactions. Students sometimes think that since energy is added to the system in an
endothermic reaction, the temperature of the system should go up. Their logic is that if energy is added to the
system, then the temperature should rise. In fact, the energy is removed from the surroundings; thus a
cooling effect occurs and is incorporated in the products as energy stored in the chemical bonds. Exothermic
reactions seem to be easier for students since, if energy is released, then the temperature of the surroundings
should rise.
Section 6.3
Enthalpy
Since most of the reactions that we do are open to the atmosphere, we are usually interested in
constant-pressure processes. The energy that is transferred in a constant-pressure process is called enthalpy
and is symbolized by H. H is an extensive property and is also a state function which means that ∆H, the
change in enthalpy, is independent of the path taken in going from state one to state two. ∆H is equal to H for
the products minus H for the reactants. If ∆H is negative, the reaction is exothermic (energy is released) and
if ∆H is positive, the reaction is endothermic (energy is absorbed).
One method to assist your students in this is to consider the following sets of reactions. Assume the
following reaction occurs:
A+B
C
Assume further that reactants A and B contain a total of 100 units of energy and product C contains
80 units of energy. Therefore, because the law of conservation of energy must hold, 20 units of energy must
be released. If energy is released, then the reaction must be exothermic. For this reaction
∆ H = H products - H reactants = 80 - 100 = - 20
The important point is that ∆H<0 or is negative. Therefore, ∆H for exothermic reactions must be
negative. The converse is true for endothermic reactions. For example, assume the following reaction
occurs:
F+E
G
Assume that the reaction is endothermic and that the sum of the energy in F and E is 70 units and
the energy of G is 120 units. Thus 50 units of energy must be supplied for this reaction to proceed. The
reaction is endothermic.
∆ H = H products - H reactants = 120 - 70 = + 50
Here ∆H is >0 which is always true for endothermic reactions.
We could think of the exothermic reaction as
A+B
C + energy
where energy is a product for the exothermic reaction and
energy + F + E
G
where energy is a reactant for the endothermic reaction.
The melting of ice is
H2O(s)
H2O(ℓ)
The heat that is required to do this is 6.01 kJ for one mole and is known as the heat of fusion of water.
Therefore, ∆H = +6.01 kJ. The reverse process
H2O(ℓ)
H2O(s)
would have ∆H = -6.01 kJ. Therefore, the melting of ice is endothermic because heat is removed from the
surroundings while the freezing of water releases heat to the surroundings and is exothermic.
The process of converting liquid water to steam
H2O(ℓ)
H2O(g)
has ∆H = +44.0 kJ for one mole and is known as the heat of vaporization. The reverse process
H2O(g)
H2O(ℓ)
is exothermic. It is for this reason that steam burns are often so severe. The conversion of the steam to liquid
water releases a large amount of energy to the surroundings resulting in a burn to exposed skin.
Section 6.4
Calorimetry
Specific heat and heat capacity are often confused. Specific heat is defined as the amount of heat
required to raise the temperature of one gram of material one degree Celsius. It has units of joules per gram
degree Celsius. Heat capacity is the amount of energy required to raise a given quantity of material one
degree Celsius. It has the units of joules per degree Celsius. Just looking at the units, it is easy to see the
specific heat times mass is equal to heat capacity.
Your author introduces both constant-volume bomb calorimeters and constant-pressure calorimeters.
Most students will not experience bomb calorimetry since it requires fairly sophisticated equipment.
In example 6.4, your author shows that the heat of neutralization for HCℓ with NaOH is
-56.2 kJ/mol. This is a good opportunity to review the concept of net ionic equations since in fact -56.2 kJ/mol
is the ∆H for the reaction of
H+(aq) + OH—(aq)
H2O(ℓ)
Therefore, any neutralization reaction that results in this net ionic equation will always have the same ∆H.
This concept may not be obvious to your students. The reverse of this reaction is given in Table 6.2 and has
a ∆H equaling +56.2 kJ/mol.
Section 6.5
Standard Enthalpy of Formation and Reaction
The most stable form of oxygen at 25oC is molecular oxygen, O2, and not ozone, O3. Your students will
likely accept that because they have probably heard of the ozone layer and how it is being destroyed by
chlorofluorocarbons. They may have a more difficult time accepting that graphite is more stable than diamond
because they are likely aware that graphite will burn but diamonds "last forever". What needs to be explained
is that by stability we are referring to thermodynamic stability and not chemical reactivity.
Perhaps the following analogy may assist in explaining the difference between thermodynamic stability
and chemical reactivity. With respect to potential energy, a ten-pound box sitting on top of a cliff overlooking a
canyon is not as stable as a similar box sitting one inch above the canyon floor. If the box on the top of the
cliff is sitting firmly on the ground while the box at the bottom of the canyon is balancing on the head of a pin,
certainly the box at the bottom of the canyon is the more reactive of the two boxes. The box with the lowest
energy is the more reactive. The analogy to be made is the box on top of the cliff corresponds to diamond
(higher energy, less reactive) while the box on the pin corresponds to graphite (lower energy, more reactive).
The change in enthalpy for a reaction, ∆H, can be determined experimentally using calorimetry or
indirectly by using Hess’s law. Example 6.5 demonstrates the traditional method of using Hess’s law to
determine the ∆ H
o
f
for C2H2. Some students become confused or intimidated by all of the chemical
formulas. They may be more comfortable if each of the chemicals is assigned a single letter variable and then,
after writing the various equations, solve for the desired equation. For example, using the equations labeled
a, b, and c in example 6.5, let us define the following:
C (graphite) = A
O2(g) = B
CO2(g) = C
H2(g) = D
H2O(ℓ) = E
C2H2(g) = F
Reaction a in example 6.5 becomes
aa)
Reaction b
A+B=C
∆ H °rxn = - 393.5 kJ
bb)
D +1/ 2 B = E
∆ H°rxn = - 285.8 kJ
cc)
2F + 5B = 4C + 2E
∆ H°rxn = - 2598.8 kJ
Reaction c
The desired reaction for the formation of C2H2 from its elements is
2A + D = F
dd)
∆ H °rxn = ?
The task is then to algebraically change equations aa, bb, and cc to yield equation dd. The first step is to note that
equation dd has F on the right-hand side so equation cc must be multiplied by -1/2 to yield
ee)
- 1 / 2(2 F + 5 B = 4 C + 2 E)
- F - 5/ 2B = - 2C - E
2C + E = F + 5/ 2B
Equation dd has 2A on the left thus equation aa is doubled.
ff)
2A + 2B = 2C
- 1 / 2(∆ H°rxn )
(- 1 / 2)(-2598.8 kJ)
(2) ∆Horxn
(2) (--393.5 kJ)
Equation dd has D on the left thus equation bb will be used unchanged. Equation dd is obtained if ee, ff, and bb
are added
2C + E = F + 5/2B
∆Horxn = 1299.4 kJ
2A + 2B = 2C
∆Horxn = --787 kJ
D + 1/2B = E
∆Horxn = --285.8 kJ
2A + D = F
∆Horxn = 226.6 kJ
or
2C (graphite) + H2 (g)
C2H2 (g) ∆Horxn = + 226.6 kJ
This method is more time consuming, but once a student gains confidence in solving Hess’s law problems with this
method, he/she may feel more comfortable using the more traditional method shown in example 6.5.
Section 6.6
Heat of Solution and Dilution
Students are willing to accept exothermic heats of solution because many students have experienced
solutions that generate heat upon mixing. For example, they may have experienced the heat generated when
sulfuric acid and water are mixed. Endothermic reactions are less common but many ammonium salts are
endothermic when mixed with water. Cold packs used by athletes are examples of endothermic reactions that
occur upon mixing.
Section 6.7
Introduction to Thermodynamics
Properties of systems that are independent of how they were achieved are known as state functions.
Energy, volume, pressure, and temperature are all state functions. On the other hand, heat and work are not state
functions because they depend upon the path taken to reach the final state. The first law of thermodynamics states
that energy can be converted from one form to another but cannot be created or destroyed. We know it as the law
of conservation of energy. This can be expressed as
∆E = q + w
It is important to understand the sign convention for heat and work. The convention for q is positive for
endothermic processes and negative for exothermic reactions. Work is positive for work done on the system by
the surroundings and negative for work done by the system on the surroundings. The sign convention for q is the
same as for ∆H for exothermic and endothermic reactions. If we think of work being done on the system as a
“reactant”, then it follows that the sign on w will be positive as q is positive for endothermic reactions. For systems
where work is done by the system, work can be considered as a “product” and thus is similar to exothermic
reactions and the sign would be negative.
If we recall that R = 0.0821 L•atm / mol•K = 8.314 J/mol•K, then we have 1 L•atm = 101.3 J. The terms
L•atm may not be easily recognized as units for energy, but indeed they are.
We defined change in enthalpy as
∆ H = ∆ E + ∆(PV)
and also
∆E=q+w
We get
∆ H = q + w + ∆ (PV)
If we substitute w = -P∆V and hold pressure constant we get
∆H = qp
Since most of the common reactions that we encounter are performed at constant pressure, the heat that we
observe, either lost or gained, is a direct measure of ∆Hrxn.