Chapter 5 Energy Relationships in Chemistry: Thermochemistry
In order to study thermochemical changes, we first have to define
(a)
system  that specify part of the universe of interest to us.
(b)
surrounding  the part of the universe not contained in the system.
3 types of Systems
1. open system  exchanges mass and energy
2. closed system  exchanges energy but no mass
3. isolated system  no exchange of either mass or energy

In thermodynamics, we are interested in the studying the energy of a system and its
interconversion into heat and other kinds of energy. We define energy as the ability to do
work.
There are a number of different types of energy.
1. thermal energy - associated with the random motions of atoms and molecules of atoms
and molecules. The greater the thermal energy  the more quickly the molecules move.
2. heat energy - transfer of thermal energy between two bodies that are at different
temperature., i.e. heat flows from a hot body to a cold body!
3. chemical energy - energy stored within the structural units of chemical substance.
4. potential energy - the ability of an object to due work because of its position in a field of
force.
5. kinetic energy - the work that can be performed by a moving object
The Law of Conservation of Energy - energy is neither created nor destroyed in ordinary
chemical and physical processes; it is merely changed in form.
Energy Changes in Chemistry Reactions
First Law of Thermodynamics
 Chemical reactions either absorb or release energy. We are interested in the study of heat
changes during a chemical reaction  thermochemistry. We have already stated that for
the enthalpy of the system,
 exothermic reaction 
heat is released to the surroundings
 endothermic reaction heat is supplied to the system by the surroundings.
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 For the internal energy,
 exothermic process the total energy of the products is less than the total energy of
the reactants since heat was transferred to the surroundings.
 endothermic process total energy of products is greater than the total energy of
reactants (reason: heat supplied by surroundings).
 From the law of conservation of energy, we cannot create or destroy energy, but we can
convert one type of energy into another (energy of the universe is constant). In chemical
systems, we are interested in the conversion of heat into work.
E = q + w = Ef - Ei
E  the change in the internal energy

The internal energy change (E) has two components: the kinetic energy and the potential
energy.
 kinetic energy part
(a) molecular motions
(b) movement of electrons within molecules
 potential energy part
(a) attractive electron - nuclear inter.
(b) repulsive nuclear - nuclear inter.
(c) repulsive electron - election inter.

We can't accurately determine all these quantities, but since we are only interested in the
change in E, this doesn't matter.
Example
C (s) + O2 (g)  CO2 (g)
E = E[CO2 (g)] – E[C(s)] – E[O2(g)]

We already know something about the energy content of the system, since this reaction ha
a negative enthalpy change rxnH = -393.5 kJ/mol, this means that some energy is lost to
the surroundings.

From the first law, since the total energy of the universe is constant (1st law restated)
surrE + sysE = 0
or surrE = -sysE
i.e., the energy "lost" from the system is "gained" in the surroundings.

In our discussion of the first law, we will only concern ourselves with pressure volume
work (it should be noted that there are many other possible types of work).
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1.
2.
3.
4.
Work done by system on surroundings, w  Work done by surroundings on system, w  +
q <heat flows to surroundings (exothermic) q > 0, heat flows to system (endothermic) +

Definition of pressure - volume work  when a gas expands and it pushes a piston
upwards against a constant opposing force (constant pressure F/A), the work done by the
gas on the surroundings.
w = -Pop V = Pop (Vf-Vi)
We should be able to see that if the opposing pressure is greater, the gas has to "work
harder" to obtain the same volume change. If we allow the gas to expand in a vacuum,
w = 0.

Example
6.0 L of a gas is allowed to expand against a constant pressure of 12 atm (T = const.).
Calculate the work done by the gas on the surroundings if the gas volume
(a) doubles (Vf = 12.0L)
(b) triples (Vf = 18.0L)
(a) w = -Pop V
V = Vf - Vi = 12.0L - 6.0L = 6.0L
 w = -12atm * 6.0L
= -72 L atm
(note: conversion factor  L atm = 101.3 J)
w = -72 L atm * 101.3 J/L atm = -7.3 * 103 J = -7.3 kJ
(b) w = -PV = -12 atm * (18.0 L - 6.0 L) = -144 L atm
convert to J
w = -144 L atm * 101.3 J/L atm = -14.6 kJ
 The other component of the internal energy is q - heat. As a reminder, heat flows from a
hot object to a cool one.
 ASIDE: The thermodynamic functions, E, H, V, etc., are examples of system
properties that do not depend on the history of the systems (i.e., they are independent of
the path of the transformation). They are known as state functions, as their value is
determined by the difference in the thermodynamic variable specified at the final state and
the initial state of the system (i.e., at a specified temperature, pressure, and volume).
However, work (w) and q (heat) are not state functions. They are called path functions
i.e., the amount of work done or heat released depends on how the system changes states.
 This is easy to see with w, i.e., when the opposing pressure is larger, the gas (system) has
to "work" harder to obtain the same final volume (Vf similar).
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For a constant volume process, we see that E = q v. The internal energy change for a system
si equal to the heat evolved in the constant volume process. This definition is similar to what
we have seen for the enthalpy (i.e., H = q p) .
Example  An ideal gas contracts from 8.4 L to 4.2 L under a constant opposing pressure of
1.5 atm. The system evolves 830 J of heat during the contraction. Calculate E.
 Since the gas is contracting work is being done on the system (i.e. w is +). The magnitude
of w is:
w = -Pop (Vf - Vi)
=
-1.5 atm (4.2 L) * 101.3 J/ (L atm ) = 640 J
E = q + w; we are given q = -830 J (i.e. heat flows to the surroundings)
E = -830 J + 640 J = -190 J
The Measurement of Energy Changes - Calorimetry
 We define calorimetry as the measurement of heat and energy changes in chemical and
physical processes. Before we get into a discussion of the calorimeter, we must define a
few simple terms.
 specific heat capacity (s) the amount of energy (heat) required to raise 1 g of a substance
by 1C (units = J/gC)
 heat capacity (C) the amount of heat (energy) needed to raise the temperature of a given
mass of substance by 1C.
 We note from the definitions C  m s where m is the mass of the substance (in grams).
 For 1 mole of the substance, we have the molar heat capacity, Cm = M s where M is the
molar mass of the substance and s is its specific heat capacity. Hence, for water Cm =
4.184 J/ (g C) * 18.016 g/mole = 75.38 J / (C mole)
 What would be the molar heat capacity for ethanol?
For ethanol (1 mole = 46.08 g); s = 2.46 J/gC
2.46J
J
= 113 Error! Switch argument not
C  m s  46.08 g of C2 H5 OH
g C
C
specified.
 What if we wanted to raise the temperature of 1.00 g of water by 2C, how much heat
would we need?
qH2O = m s t = m s (tf – ti)
tf = final temperature of substance
ti = initial temperature of substance
What would be the temperature change if we needed 5.20 kJ of energy (heat) to 250.0 g of
H2O?
q = m s t
m = 250.0 g ; q = 5.20 kJ = 5.2 * 103 J;
s = 4.184 J/gC
3
t = q/ms = 5.20 * 10 J / (250.0 g * 4.184 J/gC) = 4.97C
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 if the initial temperature (ti = 20.00C), what would tf be?
t = 4.97C = 20.00C + 4.97C = 24.97C
 How do we measure the heat changes in a chemical reaction? We use a calorimeter.
 Calorimeter  a device which contains water and/or other substance with a known
capacity for absorbing energy (heat). The outside walls are insulated so that all the heat by
the process is absorbed by the calorimeter. A calorimeter is an example of an adiabatic
system, i.e., there is no exchange of energy (heat) permitted between the system and the
surroundings. Therefore, all energy changes take place within the calorimeter.
 In the calorimeter, the heat evolved by the reacting system is absorbed by the (dry)
components of the calorimeter (the calorimeter walls, the heater, the stirrer, the
thermometer, etc.) and any added fluid (usually water).
 For an adiabatic system, q tot = qrxn + qH2O + q cal = 0
-qrxn = qH2O + q cal
 We will discuss two types of calorimeters, the constant volume (bomb) calorimeter and the
constant pressure calorimeter. In the constant volume calorimeter, the heat flow of the
reaction is equal to the internal energy change (i.e., E = q v), whereas in the constant
pressure calorimeter, the heat flow of the reaction is equal to the enthalpy change (i.e., H
= q p).
Examples  3.00 g of C6H12O6 was burned in a constant volume calorimeter, filled with O2
(g) at a high pressure, according to the following reaction.
C6H12O6 (s) + O2 (g)  6 CO2 (g) + 6 H2O (l)
Given that the calorimeter contained 1.20 kg of water, and the temperature of the water rose
from 19.00C to 25.50C, calculate the internal energy change for the combustion of glucose.
(note: Ccal = 2.21 kJ/C)
Note: since we are dealing with a calorimeter,
q tot = qrxn + qH2O + q cal = 0
-qrxn = qH2O + q cal
We must first calculate the heat absorbed by each part of the calorimeter.
Calculate the heat capacity of the water?
CH2O = g (H2O) * s (H2O)
= 1200 g * 4.184 J/gC = 5016 J/C
q H2O = CH2O t = 5016 J/C * (25.50C – 19.00C)
= 32604 J
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Next, we must calculate how much energy (heat) was absorbed by the calorimeter.
qcal = Ccal t = 2.21 x 103 J/C * (25.50C – 19.00C)
= 14365 J
the heat released by the combustion reaction at constant volume
qrxn = -( qH2O + q cal)
= - (32604 J + 14365 J) = - 46969 J = 46.97 x 103 J = 47.0 kJ
What is the internal energy change per mole of glucose?
3.00 g of C6H12O6 combusted  47.0 kJ of heat released at constant volume.
Calculate the number of moles of glucose
n gluc = 3.00 g / (180.2 g/mole)
= 1.66 * 10-2 moles
comb = qrxn / ngluc = 47.0 kJ / 1.66 * 10-2 moles
= -2.82 * 103 kJ/mole
 H is measured at constant pressure, however, we have a constant volume process.
Therefore, we must convert the E values to H values!
H  E  PV  E  n g RT
where ng = the change in the number of moles of gaseous products – reactants, and
the value of R = 8.314 J / (K mole)
 In the above calorimetry example, calculate the enthalpy change for the combustion
processes from its internal energy change measured in the calorimeter.
H  E  PV  E  n g RT 
C6H12O6 (s) + O2 (g)  6 CO2 (g) + 6 H2O (l)
Note: in this case, ng = 6 –6 = 0, therefore, the internal energy change is equal to its
enthalpy change, i.e., combH = combE = -2.82 * 103 kJ/mole.
Another Example
For the reaction,
2 N2 (g) + O2 (g)  2 N2O (g)
The enthalpy change for the reaction is rxnH = 164 kJ. Calculate the internal
energy change E (T = 298.15 K)
Note: Again the internal energy change is related to the enthalpy change as follows.
E = H - ng RT V
For the above reaction,  ng = 2 – 3 = -1

E = H -  RT = 164 x 103 J + 1 * 8.314 J/ (K mole)* 298.2 K
= 166 kJ.
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We can see that for most reactions, the difference between the internal energy change and the
enthalpy change is small. In most of our constant volume (bomb calorimetry) problem, this
can be safely neglected.
Enthalpy of Formation and Reaction
 Note  we said previously that rxnH =  H (products) -  H (reactants)
 However, we cannot measure the absolute values of the enthalpies directly (the same is
also true of the internal energy). How do we get around estimating enthalpies (or heat
contents) of chemical systems.
The FORMATION REACTION (a "chemical thermodynamic reference point")
e.g. for CO and CO2
2 C (s) + O2 (g)  2 CO (g)
or
C (s) + ½ O2 (g)  CO (g)
 This is the reaction for the "formation" of CO from its constituent elements in their
standard states under standard conditions. The degree sign, either  or , indicates that the
thermodynamic property has been measured under standard conditions, i.e., P = 1.00 atm,
[aqueous species] = 1.00 mol/L, and T = temperature of interest (note the temperature used
in the thermodynamic data tables is 25C or 298 K).
 Define the formation reaction. The formation reaction is a reaction where 1.00 mole of a
compound is formed from its constituent elements in their standard states at that
temperature.
i.e., for the reaction 2 Na (s) + S (s) + 3/2 O2 (g)  Na2SO3 (s)
The enthalpy change for this reaction would be the ‘formation enthalpy of
Na2SO3 (s)’, symbolised fH.
 NOTE fH is "measurable".
 For CO (g)
C (s) + O2 (g)  CO (g)
fH (CO(g)) = -110.5 kJ/mol CO
 Let’s compare with CO2
 The formation reaction for CO2 (g)
C (s) + O2 (g)  CO2 (g)
fH = -393.5 kJ/mol
 We note that the enthalpy change for forming CO2 (g) from its constituent elements is
considerable larger than the enthalpy change for forming CO (g) from the same elements!
 Why?
= fH or
the enthalpy change (changes in heat content of
products - reactants) when 1 mol of CO is formed
 In terms of gas phase bond breaking and bond formation,
1st reaction  ½
2nd reaction 
O=O bond broken
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O=O broken
C=O bond formed
2 C=O formed
 We can use the formation enthalpies as our thermodynamic reference point, since the
formation of the elements from themselves is a null reaction!
 i.e., for O2 (We want the enthalpy of this reaction.
rxnH = H [CO (g)] - ½ H [O2 (g)] - H [C (s)]
 For any element in its stable form at a specified temperature, fH = 0 kJ/mol (i.e. O2 (g),
H2 (g), F2 (g), Br2 (g), C(s), Na(s) all have fH = 0).
i.e.,
fH(O2) = 0 but
fH(O3)  0
fH(Cl2) = 0 but
fH(Cl)  0
rxnH = fH CO2(g) - fH CO(g) = -283 kJ/mol (394-111)
IN GENERAL
rxnH =  fH(products) -  fHreactants)
Example 2: Calculate the rxnH for the combustion of 1 mol of "iso-octane" to form H2O(g)
and CO2(l)
Reaction:
want
Balanced = C8H18(l) + 25/2 O2(g)  8 CO2(g) 9 H2O(g)
fH CO2(g) = -394 kJ/mol
fH H2O(g) = -242 kJ/mol
fH C8H18(g) = -225 kJ/mol
rxnH =  fH (products - reactants)
= 8 * fH CO2(g) + 9fH H2O(g)
- fH C8H18(l)
= 8 * -344 + 9 * -242 -(-225) kJ/mol = -5.10 * 103 kJ/mol
Other important Enthalpy changes
Heat of dilution and solution.
 So far, we have been concerned with chemical reactions only. Many other important
processes have associate enthalpy changes. In a number of cases, dissolving a solute in a
solvent produce measurable heat effects.
 definition solnH = the heat absorbed or given off when a quantity of solute is dissolved in
a solvent. Represents the difference in enthalpy of the final solution and H's of original
component.
solnH = Hsoln - Hcomponent
(i.e., H component = H solid + H solvent)
 Note: as before, we can't measure absolute values of enthalpy, we can only measure the
difference between the solution state and the solid compound.
 For the process,
HCl (aq, 6 M)  HCl (aq, 1 M)
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 There will be a significant quantity of heat released when the acid solution is diluted. This
is known as the enthalpy of dilution. It represents the difference in the enthalpy of the
final solution and the original solution.
dil H = Hsoln, 2 – Hsol’n , 1
 This is why we must always do the following when preparing acid solutions.
"Always add acid to water, SLOWLY, with constant stirring"
 Look at the following process.
NaCl (s)  Na+ (g) + Cl- (g)
H = latticeH = 788 kJ/mol  the lattice energy
 Latent Heats
These are the enthalpy changes associated with phase transitions.
e.g., H2O (l)  H2O (g)
rxnH = vapH  the enthalpy of vapourization.
H2O (s)  H2O (l) rxnH = fusH  the enthalpy of fusion.
H2O (s)  H2O (g) rxnH = subH  the enthalpy of sublimation.
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