Chapter 5
Thermochemistry
Energy :capacity to do work or to transfer heat
Work: energy used to cause an object to move against a force .
Heat: energy used to cause the temperature of an object to increase.
Course: First
2013/2014
Kinetic energy and potential energy
Kinetic energy:
kinetic energy, is the energy of motion. It depends on mass and velocity of the object.
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Ek  m v 2
[5.1]
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Potential energy:
Potential energy is the “stored” energy that arises from the attractions and repulsions an
object experiences in relation to other objects. It depends on the force that is acting on the
object and has many forms.
interactions between charged particles. (electrostatic potential energy, Eel)
EP = mgh
potential energy can be converted into kinetic energy. For example, a cyclist poised at the top
of a hill. The potential energy of the cyclist is greater at the top of the hill than at the bottom.
The bicycle easily moves down the hill with increasing speed. As it does so, the potential
energy initially stored in it is converted into kinetic energy. The potential energy decreases as
the bicycle rolls down the hill, but its kinetic energy increases as the speed increases.
There are many forms of potential energy, one of the most important forms of potential
energy in chemistry is electrostatic potential energy, Eel, which arises from the interactions
between charged particles. This energy is proportional to the electrical charges on the two
interacting objects, Q1 and Q2, and inversely proportional to the distance, d, separating them:
Eel 
K Q1 Q2
d
[5.2]
Where K is the proportionality constant, k = 8.99  109 . m/C2
Equation 5.2 shows that the electrostatic potential energy goes to zero as d becomes
infinite; in other words, the zero of electrostatic potential energy is defined as infinite
separation of the charged particles.
FIGURE 5.3 illustrates how Eel behaves for charges of the same and different sign.
When and Q2 have the same sign the two charged particles repel each other, and a repulsive
force pushes them apart. In this case, Eel is positive, and the potential energy decreases as the
particles move farther and farther apart.
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When Q1 and Q2 have opposite signs, the particles attract each other, and an attractive force
pulls them toward each other. In this case, Eel is negative, and the potential energy increases
(becomes less negative).
FIGURE 5.3 Electrostatic potential
energy. At finite separation distances for
two charged particles, Eel is positive for like
charges and negative for opposite charges.
As the particles move farther apart,
their electrostatic potential energy
approaches zero.
Units of Energy:
The SI unit for energy is the joule, J.
1 J = 1 kg.m2/s2
Equation 5.1 shows that a mass of 2 kg moving at a speed of possesses a kinetic
energy of 1 J:
Because a joule is not a large amount of energy, we often use kilojoules (kJ) in discussing
the energies associated with chemical reactions.
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System and Surroundings:
System: is the limited and well defined part of the universe. (the portion of study)
Classes of systems: Open, closed, and isolated
Open system: both matter and energy can be exchanged with the surroundings.
(For example, uncovered boiling pot containing water on the stove)
Closed system: can exchange energy but not mass. (Hydrogen and oxygen in a piston)
Isolated system: neither matter nor energy can be exchanged. (thermos containing hot coffee)
Surroundings: All parts of the universe other than the system.
Work and Heat
Work (w) : Energy used to cause an object to move.
w= F × d where F is the force and d is the distance of motion.
Heat: is the energy transferred from a hotter object to colder one.
Practice exercise 5.2
What is the kinetic energy, in J, of (a) an Ar atom moving at a speed of 650 m/s , (b) a mole
of Ar atoms moving at650 m/s ? (Hint: 1 amu = 1.66  10-27 kg)
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THE FIRST LAW OF THERMODYNAMICS
Simply: The first law of thermodynamics summarizes the truth: energy is conserved.
Energy that is lost by system, is gained by the surroundings.
∆E = Efinal – Einitial
Where , ∆E is the change in internal energy.
The internal energy, E, of a system as the sum of all the kinetic and potential energies of the
components of the system. For the system in Figure 5.4, for example, the internal energy
includes not only the motions and interactions of the H2 and O2 molecules but also the
motions and interactions of the nuclei
and electrons. We generally do not know the numerical value of a system’s internal energy.
In thermodynamics, we are mainly concerned with the change in E (and, as we shall
see, changes in other quantities as well) that accompanies a change in the system.
Transferring Energy: Heat and Work
Energy changes can be expressed in two formal ways as ( work and heat) to cause motion of
objects ( work and heat)
A force is any push or pull exerted on an object. We define work, w, as the energy
transferred when a force moves an object. The magnitude of this work equals the product
of the force, F, and the distance, d, the object moves:
w= fd
[5.3]
The other way in which energy is transferred is as heat.
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Heat is the energy transferred from a hotter object to a colder one. A combustion reaction,
such as the burning of natural gas illustrated in Figure 5.1(b), releases the chemical energy
stored in the molecules of the fuel. If we define the substances involved in the reaction
as the system and everything else as the surroundings, we find that the released energy
causes the temperature of the system to increase. Energy in the form of heat is then
transferred from the hotter system to the cooler surroundings.
See sample exercise 5.1 and practice exercise
Relating ∆E to work and heat
Internal energy of the system may exchange energy with its surroundings as heat (q) or as
work (w).
The internal energy of a system changes in magnitude as heat is added to or removed from the
system or as work is done on or by the system.
Algebraic expression of the first law of thermodynamics: ∆E = q + w
Q = heat added to or removed from the system., w is the work done on or by the system.
When heat is added to a system or work is done on a system, its internal energy increases.
when work is done on the system by the surroundings, w has a positive value.
Conversely, both the heat lost by the system to the surroundings and the work done by the
system on the surroundings have negative values; they will lower the internal energy of the
system. They are energy withdrawals.
Sign conventions for q, w and ∆E: (table 5.1)
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Sample Exercise 5.2
Gases A(g) and B(g) are confined in a cylinder-and-piston arrangement like that in Figure 5.4
and react to form a solid product C(s): 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?
Solution
q = - 1150 J
w= + 480 J
∆E = ?
∆E = q + w
∆E = (-1150) + (480) = -670 J
Negative value of ∆E means : Energy is transferred from system to surroundings
Endothermic and exothermic processes
Endothermic: A process in which the system absorbs heat from surroundings.
Melting of ice: heat flows into the system from its surroundings
Exothermic process: A process in which the system release heat to the surroundings
combustion of gasoline, heat exits or flows out of the system into the surroundings.
State function a property of a System that is determined by the difference between the final and
initial states.
The state function isa property of a system that is determined by specifying the system’s condition,
or state (Final and Initial).
Internal Energy change (E) is a state function.
q and w are not state function.
5.3 Enthalpy
A system that consists of a gas confined to a container can be characterized by several
different properties. That are: The internal energy, E , the pressure of the gas, P, and the
volume of the container, V.
We can combine these three state functions E, P, and V to define a new state
function called enthalpy which is useful for discussing heat flow in processes that occur
under constant (or nearly constant) pressure. Enthalpy, which we denote by the symbol H, is
defined as the internal energy plus the product of the pressure and volume of the system:
H = E + PV
[5.6]
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The work involved in the expansion or compression of gases is called pressure volume
work (or P-V work).When pressure is constant in a process, as in our preceding
example, the sign and magnitude of the pressure-volume work are given by
w = -P V
[5.8]
Where P is the pressure, and V = Vfinal – V initial.
The negative sign in Equation 5.8 is necessary to conform to the sign conventions of Table
5.1.
In case of expansion: The volume of the system expands, and V is positive. Because the
expanding system does work on the surroundings, w is negative — energy leaves the system
as work.
In case of compression: the system is compressed, and V is negative (the volume decreases),
and Equation 5.8 indicates that w is positive, meaning work is done on the system by the
surroundings.
Sample exercise 5.3
Indicate the sign of the enthalpy change, H, in these processes carried out under atmospheric
pressure and indicate whether each 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
to CO2 and H2O.
Solution:
(a) The ice cube (system) absorbs heat from the surroundings as it melts  H is positive
and the process is endothermic.
(b) The system is the 1 g of butane and the oxygen required to combust it. The combustion of
butane in oxygen gives off heat,  H is negative and the process is exothermic.
5.4 Enthalpies of reaction (∆Hrxn)
The enthalpy change of the reaction is given by: ∆Hrxn = H products – H reactants
Guidelines for enthalpies of reactions:
1. Enthalpy is an extensive property. It depends on the amount of substance reacted or
produced.
2. The enthalpy change for a reaction is equal in magnitude, but opposite in sign, to the
enthalpy change of the reverse reaction. For example,
CO2 (g) + O2 (g)
→ CO2 (g) + 2 H2O (l)
CO2 (g) + 2 H2O (l) → CO2 (g) + O2 (g)
∆Hrxn= -890 kJ
∆Hrxn= 890 kJ
3. The enthalpy change for a reaction depends on the state of the reactants and
products.
Compare the following reactions:
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→ CO2 (g) + 2 H2O (l)
∆Hrxn= -890 kJ
And
CO2 (g) + O2 (g)
→ CO2 (g) + 2 H2O (g) ∆Hrxn= -802 kJ
So, it is important to specify the states of the reactants and products in the
thermochemical equations.
CO2 (g) + O2 (g)
Sample exercise 5.4
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.)
Solution:
From guideline 1: H is an extensive property.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
H = -890 kJ
Heat released  4.5 g CH 4 
1mol CH 4  890 kJ

  250 kJ
16 g CH 4 1mol CH 4
Solution
From the data given in the equation above
Two moles of H2O2 which correspond to 68 g H2O2, yield -196 kJ as heat, thus the amount of
5 g H2O2 will release 14.412 kJ of heat.
68 g H2O2
196 kJ
5 g H2O2
x ? kJ
X =  14.412 kJ
Standard Enthalpy of formation ∆Hf°
Standard enthalpy of formation of a substance is defined as the heat absorbed or released
when one mole of a substance is formed from its most stable elements in standard states. The
reason for choosing the elements is because substances are formed by the elements. The
standard enthalpy of formation is given by a symbol, ∆Hf°
The most stable elements are assigned zero enthalpy change, i.e., ∆Hf° = 0, because we do
not know how much heat is involved when the elements are formed.
Standard enthalpies of formation of some substances are given in the table below.
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Standard
enthalpy of
formation (∆Hf°)
(kJ/mol)
Substance
0.0
C graphite)
C (diamond)
1.90
Cl2 (g)
0.0
H2O(g)
-241.8
-285.83
H2O(l)
O2
O3
0.0
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CALORIMETRY
Heat capacity and Specific heat
Heat capacity: is the amount of heat required to raise the temperature of an object by 1 degree.
Specific heat: the amount of heat required to raise the temperature of 1 g of a substance, 1
degree.
Specific heat 
quantity of heat tranferred 
mass of subs tan ceT 
or , Cs 
q
m  T
Sample Exercise 5.5
(a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 °C (about room
temperature) to 98 °C (near its boiling point)? (b)What is the molar heat capacity of water?
Solution
Cs 
q
Or ,
m  T
q  Cs  m  T

J 
 250 g 76 K   7.6 10 4 J
From the equation: q =  4.18
g.K 


J   18.0 g 

  75.2 J / mol.K
Molar heat capacity of water   4.18
g . K   1 mol 

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Constant Pressure Calorimetry
(Coffee cup calorimeter)
Coffee cup calorimeter is often used in general chemistry labs, to illustrate ‫ لتوضح‬the
principles of calorimetry.
∆H is measured directly at constant pressure, i.e ∆H = qp
In case of exothermic reactions: the heat lost by reactants, is gained by the solution. The
oppposite will occur for endothermic reactions.
qsoln =  qrxn
qsoln = m×s×∆T=  qrxn
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Sample Exercise 5.6
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 21.0 °C to 27.5 °C.
Calculate the enthalpy change for the reaction in kJ/mol HCl, 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 1g/mL, and that its specific heat is 4.18 J/g.K.
Solution
Reaction :
HCl (aq) + NaOH (aq) 
H2O (l) + NaCl (aq)
Mass of solution = 100 mL × 1 g/mL = 100 g
Difference in temperature = ∆T = 27.5 °C – 21 °C = 6.5 °C = 6.5 K
qsoln = m×Cs×∆T=  qrxn
qrxn = - m×Cs×∆T=  100 ×4.18×6.5 =  2700 J
qp = -2700 J
∆H = qp = -2700 J (exothermic reaction)
∆H (per mol) = -2700/0.05 mol = -54 000 J/mol
Moles of acid =(1.0 M × 0.05 L) = 0.05 moles acid
Bomb calorimeter (Constant volume calorimeter)
Ignition ‫ حرق‬of the sample in the bomb calorimeter is initiated by passing an electric current through
a fine wire that is in contact with the sample.
When combustion occurs, the heat is released.
The heat is absorbed by the calorimeter contents causing a rise in the temperature of the calorimeter
contents.
qrxn =  Ccal × ∆T
where Ccal is the heat capacity of the calorimeter.
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Solution
T  39.5  C  25.0  C 14.50  C
q rxn   C cal  T   (7.794 kJ )(14.50  C )  113.0 kJ
Heat released by combustion 4 g (0.087 mol ) CH 6 N 2 is  113.0 kJ
 Heat released per mol of CH 6 N 2 combusted   113 / 0.087 1302.3 kJ / mol
Hess′ s Law
The enthalpy change of a reaction (∆Hrxn) is the same whether the reaction is occurred in one or in
several steps.
∆H does not depend on the path of reaction. (state function)
Because the enthalpy is a state function, it depends only on the amount of matter that undergoes
change and on the nature of initial state of reactants and products in the chemical reaction.
Hess′s law provides a useful means for calculating energy changes that are difficult to
measure directly.
Solution
Equation (1) – equation (2) will give the result ∆H3 = -393 + 383.0 = -110.5 kJ.
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Solution
Equation (1) – equation (2)

∆H3= + 1.9 kJ
Solution
Equation (1) – equation (2) – 1/2 equation (3)  ∆H = -304 kJ.
Enthalpies of formation
The magnitude of enthalpy change depends on the conditions of T, P, and the physical state
of the substance (g, l or s) in the chemical reaction.

: is defined as the enthalpy change when all
Standard enthalpy change of a reaction H rxn
reactants and products are in their standard states.
Standard enthalpy change of formation H f : is the enthalpy change when 1 mole of the


compound is formed from its elements in their standard states.
For example: formation of ethanol (C2H5OH)
2 C (graphite) + 3 H2 (g) + 1/2 O2 (g)→ C2H5OH(l) H f = -277.7 kJ
Standard enthalpy of formation of the element in its standard state = zero
H f (C graphite)=0
H f (gas, 1 atm, 25 ºC)=0
Note that: Standard enthalpy of formation of the most stable form of any substance (i.e in its
standard state) is zero.
H f (Carbon, graphite) = 0
H f (H2, g)= H f (O2, g) =Zero
) 2013-2014 : ‫ غير مقرر في هذا القصل (األول‬5.8 Foods and Fuels : SELF STUDY
Important problems on Chapter Five:
5 27, 29, 32, 37. 41, 42, 43. 45, 48, 51-58, 65, 69-80.
HW: 29, 42, 52, 56, 72, 76
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