Thermochemistry
The Nature of Energy
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Energy is defined as the ability to do work (a force exerted over a distance) or
produce heat.
Types of Energy -- Kinetic Energy – is the energy of motion (KE = ½ mv2).
There are three kinetic energies associated with the states of matter.
Vibrational motion – is the bond bending and stretching of molecules in the
solid state. As the temperature of the solid approaches the melting point of the
pure substance the vibrational motion or energy increases.
Rotational motion – is the tumbling of molecules in the liquid phase of
matter. This energy is gained when the solid completely melts. As the
temperature of the liquid approaches the boiling point the rotational motion
of the pure substance increases. In the liquid phase the pure substance has
both vibrational and rotational motion.
Translational motion – is straight-line motion. This energy is gained when
the pure substance completely vaporizes (liquid to gas phase). As the
temperature of the gas increases the translational motion increases. A gas has
vibrational, rotational and translational energy or motion.
The Nature of Energy
 Potential Energy – is energy due to position or composition.
 Chemical potential energy – energy due to composition.
Energy that is stored in the bonds of the substance. (CPE)
 Gravitational potential energy – energy due to position. (GPE
= mgh)
 Conservation of Energy -- states that energy can be
converted from one form to another but can be neither
created nor destroyed. THE ENERGY OF THE UNIVERSE IS
CONSTANT.
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The Nature of Energy
 ENERGY CHANGE IS INDEPENDENT OF THE PATHWAY,
WHEREAS WORK AND HEAT ARE BOTH DEPENDENT ON
THE PATHWAY.
 State Function – is a property of the system that changes independently
of its pathway. A trip from Chicago to Denver; the distance travelled is not
a state function while the elevation is a state function.
 In varying the surface of a hill from very smooth to very rough a
ball always loses the same amount of energy because of its
position. The way that this energy transfer is divided between
work (w) and heat (q), however, depends on the specific
conditions – the pathway. If more energy is expended (due to
the surface being rough, frictional heating is higher) less work is
done on the ball. If less energy is expended (due to the surface
being smooth, frictional heating is lower) more work is done on
the ball.
The Nature of Energy
Temperature and Heat
 Temperature – is the measure of the average kinetic energy of all the
molecules. The kinetic energy of warm water has the water molecules
moving around more rapidly than the water molecules in cold water.
 Heat – is the flow of energy due to a temperature difference. Energy
will always flow from the warmer substance (higher temperature) to
the cooler substance (colder temperature).
 Because the energy is constant the energy lost by the warmer substance
is equal to the energy gained by the cooler substance (assuming no
energy is lost to the environment).
 Mathematically:
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Energy Lost (EL) = Energy Gained (EG)
- EL =
EG
System Vs Surroundings
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To consider energy changes that accompany chemical reactions, definitions of system and
surrounding have to be established.
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The SYSTEM is part of the universe on which we wish to focus attention, the reactants and products
of the reaction.
 The SURROUNDINGS include everything else in the universe, the air in the room and anything else
other than the reactants and products.
System Vs Surroundings
 Sodium hydroxide dissolved in water in a beaker.
 System --Surroundings --
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Methanol is ignited in a 20.0 litre container.
System --Surroundings --Aluminum and iodine reaction (in an evaporating dish) is sped up by the addition of a water.
System --Surroundings ---
Exothermic Vs Endothermic
 Exothermic – this is a process that results in the
evolution of heat (exo- meaning ‘out of’). The burning
of a match, energy in the form of heat flows out of the
system.
 Endothermic – this is a process that absorbs energy
from the surroundings (endo –meaning within, inner).
The evaporation of water from a pond or a lake has
energy in the form of heat being absorbed from the air
surrounding the pond or lake.
Diagram -- Exothermic
Diagram – Endothermic
Thermodynamics
 The study of energy is called thermodynamics. The law of conservation of
energy is often called the first law of thermodynamics – the energy of the
universe is constant.
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 The internal energy (E) of a system is the sum of all the kinetic and potential
energies of all particles in the system. The internal energy (E) of the system can
be changed by a flow of work, heat or both.
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E = q + w
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  = means a change in the internal energy
 q = represents heat
 w = represents work
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 Thermodynamics quantities consist of a number (indicating magnitude) and a
sign (+, -) indicating the direction of flow. THE SIGN REFLECTS THE
SYSTEMS POINT OF VIEW.
Heat Vs Work
 HEAT
 If energy flows into the system in the form of heat
(endothermic) ‘q’ equals a ‘+x’, the positive sign indicates the
energy of the system is increasing.
 When energy flows out of the system in the form of heat
(exothermic) then ‘q’ is a ‘–x’, where the negative sign indicates
that the system’s energy is decreasing.
 WORK
 If the system does work on the surroundings (energy flows
out of the system), ‘w’ is negative.
 If the surroundings do work on the system (energy flows into
the system), ‘w’ is positive.
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Change in Internal Energy (ΔE)
Calorimetry
 Measuring Energy Changes
 The units for energy measurement are the calorie and
the joule. In the metric system the calorie is the
amount of energy (heat) required to raise the
temperature of one gram of water by one degree
Celsius. The calorie which is used to measure the
energy content of food is actually the kilocalorie (1000
calories) written with a capital C. The joule (J) is
defined in terms of the calorie used in chemistry
 1 cal = 4.184 J
Calorimetry – Determining Energy
Content
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Energy is dependent of the mass of a substance the greater the mass of the substance the more energy it contains, thus
energy is proportional to the mass. Energy is also dependent of the temperature change of the substance the greater
the temperature change the more energy the substance contains thus energy is proportional to the temperature
change.
Energy  mass (m)
Energy  temperature change (T)
Energy  (m) (T)
In thermodynamics energy is given the letter ‘Q’, thus:
Q  (m)(T)
To change a proportionality to an equality a constant must be introduced into the equation, thus:
Q = (m) (s) (T)
Q = energy ,m = mass of the substance ,s = specific heat capacity of the substance, T = temperature change for the
substance
The specific heat capacity of a substance is the amount of heat per unit mass that is required to raise the temperature
by one degree Celcius.
Q = (m)
(J)
(g)
(kJ)
(s)
(J/g oC)
(kg) (kJ/kg oC)
(T)
(oC)
(oC)
The equation works for both energy gained and energy lost.
Calorimeter
Calorimetry -- Questions
 Large beds of rocks are used in some solar-heated
homes to store heat. Assume that the specific heat of
the rocks is 0.0820 J/g K. What quantity of heat is
absorbed by 50.o kg of rocks if their temperature
increases by 12.0 o C?
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 What temperature change would these rocks undergo
if they emitted 520. kJ of heat?
Calorimetry -- Questions
 Determine the amount of energy (heat) in joules required
to raise the temperature of 10.5 grams of water from 15.0 oC
to 45.8 oC.
 What quantity of energy is needed to raise 7.89 grams of
aluminum (‘s’ = 0.900 J/g oC) from 22.0 oC to 78.1 oC? What
is the energy value in calories?
 If 55.9 joules of energy is applied to 6.00 grams of gold
(‘s’ = 0.126 J/goC), what is the temperature change for the
gold?
Calorimetry -- Questions
 When 30.0 grams of methane (CH4) undergoes
combustion it increases the temperature of 150.0
mL of water by 45.0oC. Answer the following
questions:
 Determine the amount of heat absorbed by the water.
 Determine the amount of heat released by the methane.
 Determine the heat released by the methane in kJ/mol.
Calorimetry -- Questions
 If 1.00 kg of hot water at 90.0 oC is mixed with 1.00 kg of
cold water at 10.0 oC, what is the final temperature of the
water?
 A 24.60 g sample of nickel is heated to 110.0oC and
then placed in a coffee cup calorimeter containing
125.0 g of water at a temperature of 23.00oC. After the
nickel cools, the final temperature of the metal and
water is 24.83oC. Assuming that no heat has escaped to
the surroundings or has been absorbed by the
calorimeter, calculate the specific heat of nickel.
Enthalpy – H
 Chemists like to know exactly how much energy is produced or
absorbed by a given reaction. Enthalpy is invented for this
purpose it is a special energy function designated by ‘H’. For a
reaction occurring under conditions of constant pressure, the
change in enthalpy (H) is equal to the energy that flows as
heat.
Hp = heat
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 Hp = change in enthalpy has occurred under conditions of
constant pressure.
 For exothermic processes the Δ H is negative (-) and for
endothermic processes the Δ H is positive (+).
Enthalpy – Questions
 When one mole of methane undergoes combustion at constant
pressure, 890. kJ of energy is released as heat. Calculate the H
for a process in which 25.0 grams of methane are burned at
constant pressure.
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 The reaction that occurs on unprotected parts of an vehicle made
of iron is;
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 4Fe + 3O2  2Fe2O3
ΔH = -1652.0 kJ
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 How much heat is released when 3.00 grams of Fe is exposed (in
the presence of excess oxygen)?
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Hess’ Law
 Enthalpy is a state function, that is the change in enthalpy for given
process is independent of the pathway for the process. Consequently,
in going from a particular set of reactants to a particular set of
products, the change in enthalpy is the same whether the reaction
takes place in one step or in a series of steps..
 Hess’ Law has been called the law of additivity of heats of reaction.
It is important to understand two characteristics of H for a reaction:
 If a reaction is reversed, the sign of H is also reversed.
 The magnitude of H is directly proportional to the quantities of
reactants and products, in the reaction. If the coefficients in a
balanced reaction are multiplied by an integer, the value of H is
multiplied by the same integer.
 If two or more reactions add up to a third reaction then their
heats are additive.
Hess’ Law – Questions
 Calculate the H for the following reaction: P4 + 5O2  P4O10, given
the two reaction below.
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 P4 + 3O2  P4O6
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 P4O6 + 2O2  P4O10
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H = -1640 kJ
H = -1344 kJ
 From the following information calculate the Hf of nitrogen
monoxide, NO.
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 4NH3 + 5O2  4NO + 6H2O
 4NH3 + 3O2  2N2 + 6H2O
H = -1170 kJ
H = -1530 kJ
Hess’ Law -- Question
Calculate the change in heat for the conversion of graphite to diamond.
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C(graphite) + O2(g) → CO2(g)
∆H = -393.5 kJ
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C(diamond) + O2(g) → CO2(g)
∆H = -395.4 kJ
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Calculate the ∆H for the reaction:
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2H3BO3(aq) → B2O3(s) + 3H2O(l)
using the following data
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H3BO3(aq) → HBO2(aq) + H2O(l)
∆H = -0.02 kJ
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2B2O3(s) + H2O(l) → H2B4O7(aq)
∆H = -17.5 kJ
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H2B4O7(aq) + H2O(l) → 4HBO2(aq)
∆H = -11.3 kJ
Exercise 5.65
Enthaplies of Formation
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(ΔH
f)
 Enthalpies of formation can be used to determine
enthalpies of reaction using the following formula.
 Enthalpy of formation is the energy absorbed or
released to produce one mole of a compound.
 C(s) + O2(g)  CO2(g) ΔHof = -393.5 kJ/mol
 2C(s) + 3H2(g) + 1/2O2(g)  C2H5OH(l) ΔHof = -277.7 kJ/mol
 N2(g) + 1/2O2(g)  N2O(g) ΔHof = 82.1 kJ/mol
 All element and elemental molecules have a heat of
formation (ΔHof ) of zero.
Enthalpy of Formation
o
(ΔH
f)
 ∆Hrxn = Σ np ∆Hof(products) - Σ nr ∆Hof(reactants)
 Or
 ∆Hrxn = SHOP - SHOR
 A themo-chemical data sheet is required for any
calculation.
Enthalpies of Formation –
Questions
 Determine the enthalpies of reaction (for the following
reactions) using enthalpies of formation from a
thermochemical data sheet.
 C2H5OH(l) + O2(g) → CO2(g) + H2O(l)
 10N2O(g) + C3H8(g) → 10N2(g) + 3CO2(g) + 4H2O(g)
Enthalpies of Formation -Question
 The heat of reaction (Hrxn) for the balanced reaction
below is given. Using the values found on your
thermochemical data sheet determine the heat of
formation value (Hof ) for glucose (C6H12O6).
 2CO2(g) + 2C2H5OH(l) → C6H12O6(s) Hrxn = -1462.0 kJ
 Exercise 5.71, 5.73
Bond Enthalpy
 The application of bond enthalpies is to give a quick
estimate of whether a reaction will be endothermic or
exothermic. Recall, that endothermic reactions have a
(+) ∆H value whereas exothermic reactions have a
(-) ∆H value. To estimate reaction enthalpy, an
application of Hess' Law is used that is the fact that
bond breaking is always endothermic and bond
formation is always exothermic. Typically, bond
enthalpies are used if the needed ΔHof values are
not readily available.
Bond Enthalpy – Calculation
 Phase 1 is the breaking of the bonds present in the
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reactants and Phase 2 is the formation of the bonds of
the products.
Write a balanced chemical equation and draw
Lewis structures for all molecules
Identify the bonds broken and formed and their
bond enthalpies (BE).
Calculate ∆H using:
ΔHrxn = Σ BE(bonds broken) – Σ BE(bonds formed)
Bond Enthalpy -- Calculation
 Write a balanced chemical equation and draw
Lewis structures for all molecules
 Identify the bonds broken (reactants) and formed
(products) and their bond enthalpies (BE).
 Calculate ∆H using:
 ΔHrxn = ΣBE(bonds broken) – ΣBE(bonds formed)
Bond Enthalpy Table
Bond Enthalpy -- Example
Bond Enthalpy – Example
 Determine the ΔH for the following reaction using
bond enthalpies.
 2CH3CHCH2 + 9O2  6CO2 + 6H2O
 Exercise 8.69