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CHM 101 – Chapter Five
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The Nature of Energy
The First Law of Thermodynamics
Enthalpy
Enthalpies of Reaction
Calorimetry
Hess’ Law
Enthalpies of Formation
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CHM 101 – Chapter Five
The Nature of Energy
Chemical reactions frequently involve large exchanges of energy
Sometimes, the reaction releases large amounts of energy to
the surroundings in the form of heat and light.
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CHM 101 - Reeves
CHM 101 – Chapter Five
The Nature of Energy
Chemical reactions frequently involve large exchanges of energy
Less commonly, chemical reactions absorb energy from the surroundings.
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CHM 101 – Chapter Five
Forms of Energy
Kinetic Energy: Energy associated with motion:
Potential Energy: Energy associated with position:
The energy is dissipated as
heat. The temperature of the
water at the bottom of the falls
is higher than at the top
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CHM 101 – Chapter Five
Kinetic Energy & Units of Energy
What is the kinetic energy of a 1500 lb
automobile traveling at 100 mph (45 m/s)?
The energy unit Joule is defined as:
Thus, the energy of the automobile is
Department of Chemistry and Biochemistry
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CHM 101 – Chapter Five
The First Law of Thermodynamics
The total amount of energy in the universe is constant.
Energy cannot be created or destroyed.
If no energy crosses the boundary, the total energy of
the system cannot change.
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CHM 101 – Chapter Five
The First Law of Thermodynamics
system
surroundings
Energy that crosses the boundary from the surroundings into the system
increases the energy of the system and is considered positive.
Energy can only cross the boundary in the form of heat (q) or work (w).
Heat flowing into the system or work done on the system is defined as
positive.
Heat flowing out of the system or work done by the system is defined as
negative.
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CHM 101 - Reeves
CHM 101 – Chapter Five
The First Law of Thermodynamics
Thus the First Law is:
What is the change in the energy of a system if 1 kJ of
heat is absorbed by the system and 500J of work are
done by the system?
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CHM 101 – Chapter Five
Functions of State
A function of state is one that depends only on the state of the system,
as defined by its contents and conditions such at temperature and
pressure. Internal energy (E) is a function of state.
The change of a state variable such as DE depends only on the initial
and final states of the system, NOT on the path.
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CHM 101 – Chapter Five
Work
• Work is defined as force times distance in the direction of
the force. Since work done by the system is negative:
• If the system expands or contracts against a pressure P,
then "PV" work will result. Consider the expansion of the
cylinder below.
A
Dh
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CHM 101 – Chapter Five
Enthalpy
Work may take many forms including mechanical, electrical, etc. In this
course, we will not explicitly consider systems that involve work in these
forms.
Thus, if a process takes place at constant volume, DV = 0
Then, the change in DE can be determined directly by measuring the
heat that flows in a constant volume process.
However, most processes in chemistry occur at constant pressure, where
PV work can occur. Thus we define enthalpy (H) as H  E  PV Then
at constant pressure:
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CHM 101 – Chapter Five
Enthalpy
Like internal energy (E), enthalpy (H) is a state function.
Thus, to carry out the reverse reaction, 890 kJ must be put
into the system, an endothermic process.
CO2(g) + 2H2O(l)
CH4(g) + 2 O2(g)
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DH = 890 kJ
CHM 101 - Reeves
CHM 101 – Chapter Five
Enthalpy
Determine the enthalpy change when 1.5 g of CH4 is burned
in excess oxygen.
CH4(g) + 2 O2(g)
CO2(g) + 2H2O(l)
DH = -890 kJ
What mass of oxygen would be consumed if this reaction
released 100 kJ of heat?
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CHM 101 – Chapter Five
Calorimetry
If a process such as a chemical reaction is carried out at constant
pressure, the heat (q) is a direct measure of the enthalpy change (DH)
If the system is surrounded by a water bath (surroundings), the flow of
heat will result in a temperature change of the water bath.
5 kg
Tini = 25 oC
Water Bath
10 kg water
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CHM 101 – Chapter Five
Calorimetry
Heat and temperature are NOT the same; the amount of heat
(q) required to induce a temperature change (DT) in a
substance depends on:
1. The mass (m) of the substance,
2. The size of the temperature change (DT), and
3. The specific heat capacity (CP) or specific heat of the
substance.
q  mCP DT
CP is the amount of heat required to raise the temperature of
one gram of the substance by one degree centigrade.
J
The specific heat capacity of water is 4.184 g K
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CHM 101 – Chapter Five
Calorimetry
If 200 g of copper metal (CP = 0.39 gJK ) at 100oC is
immersed in 100 g of water (CP = 4.184 gJK ) at 20oC in an
insulated container, what it the final temperature?
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CHM 101 – Chapter Five
Calorimetry
J
If 200 g of copper metal (CP = 0.39 g o C ) at 100oC is immersed
J
in 100 g of water (CP = 4.184 g o C ) at 20oC in an insulated
container, what it the final temperature?
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CHM 101 – Chapter Five
Calorimetry
If 1.00 g of hydrogen gas were burned in an open container
surrounded by a 10.0 kg water bath originally at 25.00 oC, the
water temperature would rise to 28.42 oC. What is the
enthalpy change for the combustion of hydrogen?
2 H2(g) + O2(g)
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2 H2O(l)
CHM 101 - Reeves
CHM 101 – Chapter Five
Hess’ Law
When two moles of carbon is burned in limited
oxygen, carbon monoxide is formed. How might the
enthalpy change of this reaction be determined?
The enthalpy change cannot be measured directly
because combustion in limited oxygen produces
both CO(g) and CO2(g).
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CHM 101 – Chapter Five
Hess’ Law
2 C(s) + O2(g)
2 CO(g)
DH = ?
While this reaction cannot be carried out cleanly,
two related combustion reactions can be done:
1) The combustion of carbon in excess oxygen
produces carbon dioxide exclusively.
2) The combustion of carbon monoxide produces
carbon dioxide exclusively.
Department of Chemistry and Biochemistry
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CHM 101 – Chapter Five
Hess’ Law
2 C(s) + O2(g)
2 CO(g)
DH = ?
Hess’ Law: If two or more equations can be added together
to produce a desired reaction, the enthalpy of the desired
reaction will be the sum of their enthalpies.
C(s) + O2(g)
CO(g) + 12 O2(g)
CO2(g)
CO2(g)
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DH = -394 kJ
DH = -283 kJ
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CHM 101 – Chapter Five
Hess’ Law
The combustion of nitrogen monoxide forms nitrogen dioxide. Calculate
the heat of combustion of nitrogen monoxide given the following:
N2  g   O2  g   2NO  g 
N2  g   2O2  g   2NO2  g 
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DH  180.74kJ
DH  67.68kJ
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CHM 101 – Chapter Five
Enthalpy of Formation
Hess' Law can be formalized by defining Enthaply of
Formation, which is the enthalpy change (DHf0) that occurs
when one mole of a substance is formed from its elements
in their standard state at 25oC.
The Standard state of an element is its most stable form at
25oC. For example, carbon
is a solid (graphite),
oxygen
and hydrogen
are both diatomic gases.
Thus for methane (CH4) the enthalpy of formation is
represented by the equation
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CHM 101 – Chapter Five
Enthalpy of Formation
Enthalpies of Formation have been determined for a large
number of compounds, and are found in tabular form in
chemistry reference texts
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CHM 101 – Chapter Five
Enthalpy of Formation
Using Hess’ Law, it can be shown the Enthalpies of
Formation (DH 0f ) can be combined to determine the enthalpy
change (DH rxn ) for any chemical reaction. Thus:
DHrxn   nprod DH 0f  prod    nreac DH 0f reac
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CHM 101 – Chapter Five
Enthalpy of Formation
Calculate the enthalpy change (DH) the reaction of gaseous
hydrogen peroxide (H2O2) with hydrogen to form liquid water.
H2O2(g) + H2(g)
2 H2O(l)
Appendix C: Thermodynamic Quantities
for Selected Compounds at 298K
Compound DHof kJ/mol)
H2O(l)
-285.8
H2O(g)
-241.8
H2O2(l)
-187.8
H2O2(g)
-136.1
= -435.5kJ
Department of Chemistry and Biochemistry
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CHM 101 – Chapter Five
Enthalpy of Formation
To see how enthalpies of formation can be applied, consider the
combustion of methane:
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(l)
First, break the reactants down into their elements:
Department of Chemistry and Biochemistry
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CHM 101 – Chapter Five
Enthalpy of Formation
CH4(g) + 2 O2(g)
CO2(g) + 2 H2O(l)
Then, make the products from those elements:
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