Formula for “Reversible, Isothermal Work”
Remember that the amount of work by the system on the surroundings varies according to how
one goes from point (Pinitial,Vinitial) to (Pfinal,Vfinal).
Remember that for a gas:
P·V = n R T
and
P=nRT/V
And that
w = - P · V
The symbol signifies any change but in calculus the infintesimal change that we have in
reversible cases is represented by the roman letter d.
d w = - P · dV
Substituting for P:
d w = - (n R T / V) · dV
Calculus gives a final equation for reversible isothermal work done by a gas on the
surroundings:
w = - (n R T) ln (Vfinal / Vinitial)
Do Self-Test 6.2B
The First Law of Thermodynamics
Energy is neither created nor destroyed. Energy is conserved.
Any energy lost by the system is gained by the surroundings.
U = q + w
Self-test 6.6B
Chapter 5 Thermochemistry 1
State Functions
A state function is a property of a system that can be determined by specifying its final and initial
conditions (in terms of temperatue, pressure, etc).
The value of a state function does not depend on the particular history of the sample, only its
present condition.
The change in the state function depends only on the initial and final states of the system, not
how the change occurs.
Can use an “energy diagram”.
Work, w and heat, q are not state functions because they depend on the path taken
Heat at constant pressure
Consider the case where no non-expansion work is done and where the volume of the system
does not change. Since V=0, w=0 and all the heat q involved in the change is thus equal to the
internal energy.
U = q + w = qv
(const vol, w=0)
This means that when a chemical reaction is performed in a constant volume calorimeter (a
“bomb”calorimeter), the heat measured is the internal energy, U. Because the heat capacity, C
equals q/T,
CV = U /T
Chapter 5 Thermochemistry 2
Heat at constant pressure
Enthalpy
Heat transferred under constant-pressure conditions is called enthalpy, H.
It is related to the internal energy by
H = U + PV
Consider the case for a change when all the work done by the system on the surroundings is
expansion under constant pressure.
H = U + PV
but the internal energy equals the heat and work:
U = qp - PexV
H = (qp - PexV) + PV
The change in enthalpy, ∆H, equals the heat, qp, gained or lost by the system when the process
occurs under constant pressure:
H  H final  Hinitial  q p

If ∆H is positive then the system has gained heat from surroundings, and is an endothermic
process.

If ∆H is negative, the system has released heat to the surroundings, and is an exothermic
process.
Enthalpy, H, is a state function; heat in general, q, is not.
Since heat transferred at constant pressure is identified as the change in enthalpy, ∆H, we can
define the heat capacity at constant pressure, CP as
Chapter 5 Thermochemistry 3
CP = H /T
The heat involved in a reaction at constant pressure can be measured with a simple apparatus
called the “coffee cup” calorimeter:
[Section 6.10 of textbook will not be covered in this course.]
Enthalpies of Physical Change
The molecules in a solid are vibrating in place. Temperature is a measure of the average kinetic
energy that the molecules. As the temperature rises, more kinetic energy is added and the
molecules vibrate more. At one particular temperature, the molecules begin to tumble past each
other. This breaking loose to tumble around each other (still in contact with each other) requires
more energy. This is called the latent energy. At constant pressure, the molar enthalpy change
that accompanies melting (fusion) is called the enthalpy of fusion, ∆Hfus.
∆Hfus = H (liquid) – H (solid)
Chapter 5 Thermochemistry 4
Fusion is the “act of making fluid” and is the opposite of freezing. The enthalpy of freezing is
equal and opposite to ∆Hfus.
The amount of energy needed to vaporize a liquid into a gas is much more than that needed to
melt (fuse). This is because the molecule in a gas are no longer in contact. The difference in
molar enthalpy between the vapor and liquid states of a substance is called the enthalpy of
vaporization, ∆Hvap = H(gas) – H (liquid).
Sublimation is the direct conversion of a solid into a gas. Frost disappearing is an example, as is
the disappearance of “dry ice”, CO2(s). The enthalpy of sublimation,
∆Hsub = H(gas) – H (solid) = ∆Hfus + ∆Hvap
Self-test 6.10B
Chapter 5 Thermochemistry 5
Heating Curves
Chapter 5 Thermochemistry 6
Enthalpies of Reaction
H  Hproducts  Hreactants 
heat of reaction, ∆Hrxn
2H2 g  O2 g   2H2Og
H  483.6 kJ
enthalpy – viewed as a measure of how much heat stored as potential energy in the system or as
‘heat content.’
Guidelines
1. Enthalpy is an extensive property. Magnitude of ∆H is directly proportional to amount of
reactant consumed in process.
CH4 g   2O2 g   CO2 g  2H2Ol 
H   890 kJ
The combustion of 2 mol of CH4 would release –1780 kJ.
2 (CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) H = +890 kJ )
2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(l) H = +1780 kJ
2. The enthalpy change for a reaction is equal in magnitude but opposite in sign to ∆H for
reverse reaction.
Chapter 5 Thermochemistry 7
CO2 g   2H2 Ol  CH4 g   2O2 g 
H   890 kJ
3. The enthalpy change for a reaction depends on the state of the reactants and products.
2H2O(g)  2H2O(l)
H = - 88 kJ
So adding the two equations ( the H2O(l) on both sides cancel out):
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g)
H = +890 kJ
2H2O(g)  2H2O(l)
H = - 88 kJ
--------------------------------------------------------------------CO2(g) + 2H2O(g)  CH4(g) + 2O2(g)
H = +802 kJ
4. These are thermochemical equations and the H quantity is stoichiomentric
Example:
When one mole of ethylene, C2H6(g) is combusted to CO2(g) and H2O(g), 316.3 kcal of heat are
liberated under constant pressure. If 78.6g CO2(g) are produced from this reaction, calculate the
enthalpy change for this reaction.
Chapter 5 Thermochemistry 8
Step 1: Balance the thermochemical equation
C2H6(g) + 3O2(g)  2CO2(g) + 2H2O(g) H = -316.3 kcal
Step 2: Calculate the molar relation
78.6g CO2 · 1 mol CO2 · -316.3 kcal
44.0 g CO2
2 mol CO2
=
- 283. kcal
Self-test 6.11B
Hess’s Law
Hess’s law states that if a reaction is carried out in a series of steps, ∆H for the reaction will be
equal to the sum of the enthalpy changes for the individual steps.
Hess pointed out that the heat absorbed (or evolved) in a given chemical reaction is the same
whether the process takes one step or in several steps.
CO(g) + 1/2 O2
2
C(graph) + O2(g)
H2
H3
H1
1
Chapter 5 Thermochemistry 9
3
CO2
We can burn graphite with excess O2 to get CO2 and measure the heat evolved, but it is
technically very difficult to burn graphite and get only CO. We take advantage of the fact that
CO burns to give CO2.
H2 = H1 - H3
H1
C + O2  CO2
CO2  CO + ½ O2 -H3
--------------------------------H2
C + ½ O2  CO
Tool Box 6.1
Do Self-tests 6.13,14B
Enthalpies of Formation
Enthalpy of formation, ∆Hf,(or heat of formation), the formation of a compound from its
constituent elements.
The standard enthalpy of formation of a compound,
H f , is the change in enthalpy for the
reaction that forms 1 mol of the compound from its elements, with all substances in their
standard states.
The most stable form of the element is used. E.g. O2 not O3, C(graphite not diamond)
The standard enthalpy of formation of the most stable form of any element is zero.
Using Enthalpies of Formation to Calculate Enthalpies of Reaction
We can use Hess's law to calculate the standard enthalpy change for any reaction for which we
know the
H f
values for all reactants and products.
Chapter 5 Thermochemistry 10
5 ↓ (missing from eqn)
example:
C3 H8 g  O2 g  3CO2 g  4 H2 Ol 
Hrxn   nH f  products   mH f reactants 
From Appendix 2:
H f
Molecule
(kJ/mol)
C3H8(g)
-2220.
O2(g)
0
CO2(g)
-393.51
H2O(g)
-241.82
H2O(l)
-285.83
(not all tables give values for elements in most stable form)
Horxn = [3mol(-393.51kJ/mol) + 4mol(-285.83kJ/mol)] –
[1mol(-2220.kJ/mol) + 5mol(0kJ/mol)] = - 104. kJ
Self Tests 6.15,16B
Section C of notes will include 6.19,20,21
Chapter 5 Thermochemistry 11