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Energy & Rates of
Unit 3
is the study of heat change in chemical
Chapter 5
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 5 Key Concepts: After completing this chapter you
will be able to…..
• compare the energy changes of different substances, perform
calorimetry calculations, and describe the relationship between
reactions using enthalpy terms.
• represent thermochemical equations using different methods, and
determine whether a reaction is exothermic or endothermic
• calculate and use bond energies to estimate the enthalpy change
of a reaction.
• define and solve problems using Hess’s law.
• write formation reactions and calculate enthalpy changes using
standard enthalpy of formation values.
• describe current and future energy sources and explain their
advantages and disadvantages
5.1: Energy: is the capacity to do work
Thermal energy is the energy associated with
the random motion of atoms and molecules
Chemical energy is the energy stored within the
bonds of chemical substances
Nuclear energy is the energy stored within the
collection of neutrons and protons in the atom
Electrical energy is the energy associated with
the flow of electrons
Potential energy is the energy available by virtue
of an object’s position
Energy Changes in Chemical Reactions
Heat is the transfer of thermal energy between two bodies that
are at different temperatures.
Temperature is a measure of the thermal energy.
thermal energy: the total quantity of kinetic and
potential energy in a substance
Temperature = Thermal Energy
greater thermal energy
Law of Conservation of Energy: Energy cannot be created
or destroyed
The system is the specific part of the universe that is of
interest in the study.
Exchange: mass & energy
Exothermic process is any process that gives off heat –
transfers thermal energy from the system to the surroundings.
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic process is any process in which heat has to be
supplied to the system from the surroundings.
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
Nuclear Energy
All nuclear reactions are exothermic.
Per unit of mass, nuclear reactions release much more energy than
exothermic chemical reactions.
Two main types of Reactions: Fusion and Fission
A fusion reaction occurs when nuclei of small atomic mass
combine to form larger molecule.
2 H + 3 H → 4 He + 1 n
In Fission, large nuclei with high atomic mass are split to smaller,
lighter mass by collision with a neutron:
1 n → 92 Kr +
See P. 290 Table 1
1 n + 1.9 x 10 10 kJ/mol
5.2: Calorimetry and Enthalpy
The heat capacity (C) of a substance is the amount of heat (q)
required to raise the temperature of a given quantity (m) of the
substance by one degree Celsius.
The specific heat (c) of a substance is the amount of heat (q)
required to raise the temperature of one gram of the substance
by one degree Celsius.
C = mc
Heat (q) absorbed or released:
q = mcDt
q = CDt
Dt = tfinal - tinitial
Calorimetry: the measurement of heat into or out of a
system for chemical and physical processes.
the heat released = the heat absorbed
The device used to measure the absorption or
release of heat in chemical or physical processes is
called a “Calorimeter”
a “Calorimeter”
Calorimetry Calculations
When analyzing data obtained using a calorimeter, make
these assumptions:
• Any thermal energy transferred from the calorimeter to
outside environment is negligible.
• Any thermal energy absorbed by the calorimeter itself is
• All dilute, aqueous solutions have the same density (1.00
g/mL) and specifi c heat capacity (4.18 J/(g∙°C)) as water
• thermal energy absorbed or released by a
chemical system is given the symbol q
qwater = mcDt
Calculations Involving Thermal Energy Transfer
Thermal Energy transfer by metal
How much heat is given off when an
869 g iron bar cools from 940C to
c of Fe = 0.444 J/g • 0C
m = 869 g
Dt = tfinal – tinitial = 50C – 940C = -890C
q = mcDt
= 869 g x 0.444 J/g • 0C x –890C
= -34,000 J
No heat enters or leaves!
Calculations of metal in water
A student places 50.0 mL of liquid water at 21.00 °C into a coffeecup calorimeter. She places a sample of gold at 100 °C into the
The final temperature of the water is 21.33 °C. The specific heat
capacity of water is 4.18 J/g∙°C and the density of water, d, is 1.00
g/mL. Calculate the quantity of thermal energy, q , absorbed by
the water in the calorimeter
Given: VH2O=50.0 mL; c=4.18 J/(g∙°C); dH2O=1.00 g/mL;
Tinitial=21.00 °C;Tfinal=21.33 °C
R:quantity of thermal energy absorbed, q
d =m/v
m= vd, m =50mL x 1.00g/mL
= 50 g
∆t = 21.33oC – 21.00oC
q = mc∆t
=50 g x 4.18 J x 0.33oC
g. oC
= 69 J
Determining specific heat capacity of a substance
Using the value of q above, calculate the specific heat capacity, c, of
gold if its mass is 6.77 g. Final temperature of the gold is same as
that of water in the calorimeter.
Given:mAu =6.77 g; q = 69 J; t initial =100 oC; t final = 21.33oC
∆t =21.33 oC – 100oC
= - 78.67 oC
Required: cAu
q = mc∆t
Rearrage and solve for c
Thermal Energy Transfer during a Neutralization reaction
A 50.0 mL sample of a 1.0 mol/L aqueous solution of hydrochloric
acid, HCl(aq), was mixed with 50.0 mL of a 1.0 mol/L aqueous
solution of sodium hydroxide, NaOH(aq), at 25.0 °C in a
calorimeter. After the solutions were mixed by stirring, the
temperature was 31.9 °C.
(a) Determine the quantity of thermal energy transferred by the
reaction to the water, q , Assume that the specific heat capacity and
density of both solutions is the same as that of liquid water
(b) State whether the reaction was endothermic or exothermic.
Given: Total volume = 100 mL; c water= 4.18J/g. oC
dwater = m =1g/mL
q = mc∆t
m = vd; msolution = 100 g
∆t = 31.9 oC – 25.0 oC
= 6.9 oC;
=100 g x 4.18 J x 6.9oC
g. oC
= 2.9 x103 J
= 2.9 kJ
Enthalpy (H):
Total amount of thermal energy in a substance
Enthalpy Change (∆H) :
Energy changes in Physical and Chemical Reactions
The enthalpy change of the chemical system is equal to the flow
of thermal energy in and out of the system,
∆H surrounding = |q system|
∆H surrounding > 0, Endothermic reaction
∆H surrounding < 0, Exothermic reaction
Energy Changes during Exothermic & Endothermic
Reaction in an Open System
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
DH < 0
Hproducts > Hreactant
DH > 0
Molar Enthalphy Change
Energy changes involving 1 mole of a substances… (∆H)
Types of molar Enthalpies
Solution (∆Hsol)
NaBr(s) →Na(aq) + Br(aq)
Combustion (∆Hcomb) CH4(g) + 2O2(aq) → CO2(aq) + H2O(l)
Vapourization (∆Hvap) CH3OH(l) → CH3OH(g)
Freezing (∆Hfr)
H2O(l) → H2O(s)
Neutralization (∆Hneut) 2NaOH(aq) + H2SO4(aq) →Na2SO4(aq) +
Formation (∆Hf) C(s) +2H2(g) + 1/2O2(g) → CH3OH(l)
∆H = n∆Hr
Where n is the amount and ∆Hr is the molar enthalpy change of the
Molar Enthalpy Calculations
Calculate ∆H for Vaporization Reactions
Ethanol, CH3CH2OH(l), is used to disinfect the skin prior to an
injection. If a 1.00 g sample of ethanol is spread across the skin and
evaporated, what is the expected enthalpy change? The molar enthalpy
of vaporization of ethanol is 38.6 kJ/mol.
Methanol=1.00 g; ∆Hvap=38.6 kJ/mol Methanol= 24 +6 +16
R = ∆H?
n = 1 g x 1 mol/46 g
A= ∆H = n∆Hvap
= 0.022
∆H = 0.022 mol x 38kJ/mol
= 0.836 KJ
Representing Molar Enthalpy Changes
1. The equations we use to represent energy changes are called
thermochemical equations.
2. Endothermic enthalpy changes are reported as positive values;
H2O(l) → H2 + 1/2O2(g) ∆Hdecomp. = + 285.8 KJ/mol H2O
3. Exothermic enthalpy changes are reported as negative values
H2(l) + 1/2O2 (g) → H2O(l)
∆Hcomb = -285.8 KJ/mol H2
Energy changes can be communicated in four different ways.
Three of them are thermochemical equations and one uses diagram.
These are:
1. By including an energy value as a term in the thermochemical equation
e.g., CH3OH(l) + 3/2 O2 → CO2(g) + 2H2O(g) + 726 KJ(Exothermic Reaction)
2. By writing a chemical equation and stating its enthalpy change
e.g., CH3OH(l) + 3/2O2(g)→CO2(g)+ 2H2O(g) ∆H = -726 KJ
(Exothermic Reaction)
3. By stating the molar enthalpy of a specific reaction
e.g., ∆Hcombustion or ∆Hc = - 726 KJ/mol CH3OH
4. By drawing a chemical potential energy diagram (next slides)
Fractions are convenient in many thermochemical equations.
They apply to fractions of moles of substances e.g., 3/2 mol
represents 1.5 mol rather than fractions of actual molecules
Potential Energy Diagram
Is DH negative or positive?
System absorbs heat
DH > 0
6.01 kJ are absorbed for every
1 mole of ice that melts at 00C and 1 atm.
Thermochemical Equations
H2O (s)
H2O (l)
DH = 6.01 kJ
Is DH negative or positive?
System gives off heat
DH < 0
890.4 kJ are released for every 1 mole of methane that is
combusted at 250C and 1 atm.
Thermochemical Equations
CH4 (g) + 2O2 (g)
CO2 (g) + 2H2O (l)
DH = -890.4 kJ
Calculations in Thermochemical Equations
The stoichiometric coefficients always refer to the number
of moles of a substance
H2O (s)
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l)
H2O (l)
H2O (s)
DH = -6.01 kJ
If you multiply both sides of the equation by a factor n,
then DH must change by the same factor n.
2H2O (s)
2H2O (l)
DH = 2 x 6.01 = 12.0 kJ
The physical states of all reactants and products must be
specified in thermochemical equations.
H2O (s)
H2O (l)
H2O (l)
DH = 6.01 kJ
H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
266 g P4 x
P4O10 (s)
1 mol P4
123.9 g P4
DHc = -3013 kJ/mol
3013 kJ
= 6470 kJ
1 mol P4
5.3:Bond energies
Bonds in compounds are broken when energies are supplied.
The energy required to break a chemical bond is called bond
dissociation energy.
Bond dissociation energies have positive values
Bond dissociation energies are reported as average bond energies
because it depends on types and # of bonds in the molecule.
E.g. bond energy for methane: 413KJ/mol
Calculated as follows:CH4 →CH3 + H---- (435KJ/mol)
CH3 → CH2 + H----(453KJ/mol)
CH2 → CH + H ----(425KJ/mol)
CH → C + H ----- (339KJ/mol)
Total = 1652
This value is convenient for calculating
Av. = 413
Multiple Bonds and Bond Energies
• Multiple bonds are generally stronger than single bonds.
c≡c (839 KJ/mol); c=c (614 KJ/mol); c-c (347 KJ/mol)
• As the # of bonds increases, the bond length decreases, the
more the energy required to break it. E.g.
Bond length (pm)
Bond energy(KJ/mol)
Enthalpy and Bond Energies
• During a chemical reaction, the bonds in the reactants must first
break, energy must be added —an endothermic process.
• Bond breaking have positive signs
• Making new bonds in the products releases energy—an exothermic
Process, energy terms associated with bond making have negative
ΔH=sum of energies required + Sum of energies released
to Break old bonds
In the formation of new bonds
(positive values)
(negative values)
This leads to the equation
ΔH= Σn x D (bonds broken) – Σn x D(bonds formed)
Energy required
energy released
Σ (sigma) means “the sum of,” n is the amount (in moles) of a
particular bond type, and D is the bond energy per mole of
D is obtained from reference tables
Steps in using bond energies to predict ∆H for a reaction
1. Determine how many of each type of bond must be broken in
the reactants.
2. Determine the number of bonds of each type that form in the
3. Use bond energy data from Table 1 (page 307) to calculate the
total energy required to break the reactant bonds, followed by
the total energy released by the formation of product bonds.
4. The energy change, ∆H, of the reaction is the difference
between these two sums
Calculate the enthalpy change for the reaction in which
hydrogen gas, H2 (g), is combined with fluorine gas, F2(g), to
produce 2 moles of hydrogen fluoride gas, HF(g). This reaction
is represented by the balanced chemical equation
H2(g)+F2(g) → 2HF(g)
for H2(g): nH–H = 1 mol; DH–H = 432 kJ/mol;
for F2(g): nF–F = 1 mol; DF–F = 154 kJ/mol;
Analysis: ∆H = Σn x D bonds broken –Σn x D bonds formed
ΔH = (nH-FDH-F + nF-F DF-F) – (nH-FDH-F
1 mol each of H–H and F–F bonds are broken
The bonds formed are 2 mol of H–F bonds
∆H= (nH-HDH-H + nF-FDF-F) – nH-FDH-F
(1 mol x 432KJ) + (1 mol x 154 KJ) - (2 mol x 565 KJ
∆H = -544 KJ
The enthalpy change for the reaction of 1 mol hydrogen gas
and 1 mol fluorine gas to ptoduce 2 mol. Hydrogen fluoride
is -544 KJ
Bond Energies Review
Video Presentation
Hess’s Law
5.5: Standard enthalpy of formation
Standard enthalpy of formation (DH0f) is the heat change
that results when one mole of a compound is formed from
its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its most
stable form is zero.
DH0 (O2) = 0
DH0 (O3) = 142 kJ/mol
DH0f (C, graphite) = 0
DHf0 (C, diamond) = 1.90 kJ/mol
0 ) is the enthalpy of
The standard enthalpy of reaction (DHrxn
a reaction carried out at 1 atm.
aA + bB
cC + dD
DH0rxn = [ cDH0f (C) + dDH0f (D) ] - [ aDH0f (A) + bDH0f (B) ]
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
Hess’s Law: When reactants are converted to products, the
change in enthalpy is the same whether the reaction takes
place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get
there, only where you start and end.)
Calculate the standard enthalpy of formation of CS2 (l)
given that:
C(graphite) + O2 (g)
CO2 (g) DH0rxn = -393.5 kJ
S(rhombic) + O2 (g)
CS2(l) + 3O2 (g)
SO2 (g)
DH0rxn = -296.1 kJ
CO2 (g) + 2SO2 (g)
0 = -1072 kJ
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic)
CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)
2S(rhombic) + 2O2 (g)
+ CO2(g) + 2SO2 (g)
CO2 (g) DH0rxn = -393.5 kJ
2SO2 (g) DH0rxn = -296.1x2 kJ
CS2 (l) + 3O2 (g)
0 = +1072 kJ
C(graphite) + 2S(rhombic)
CS2 (l)
0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJ
Benzene (C6H6) burns in air to produce carbon dioxide and
liquid water. How much heat is released per mole of
benzene combusted? The standard enthalpy of formation
of benzene is 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g)
12CO2 (g) + 6H2O (l)
DH0rxn = S nDH0f (products) - S mDHf0 (reactants)
DH0rxn = [ 12DH0f (CO2) + 6DH0f (H2O)] - [ 2DH0f (C6H6)]
DH0rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ
- 6534.9 kJ
= - 3267.44 kJ/mol C6H6
2 mol
The enthalpy of solution (DHsoln) is the heat generated or
absorbed when a certain amount of solute dissolves in a
certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be
used for melting ice?
NaCl, KCl, NH4Cl, NH4NO3
Which substance(s) could be
used for a cold pack?
LiCl & CaCl2