Grade 11 IB CHEMISTRY
IB Core Objective
5.1.1 Define the terms exothermic reaction,
endothermic reaction and standard enthalpy
change of reaction (Ho)
Define: Give the precise meaning of a word,
phrase or physical quantity.
5.1.1 Define the terms exothermic reaction, endothermic
reaction and standard enthalpy change of reaction (Ho)
 Thermochemistry = study of energy changes in
chemical reactions
 Most chemical reactions absorb or evolve energy
(usually as heat, sometimes as light and mechanical
energy)
 Energy is measured in Joules (J)
5.1.1 Define the terms exothermic reaction, endothermic
reaction and standard enthalpy change of reaction (Ho)
• Enthalpy (H, heat content) = the total energy of a
•
•
•
•
system
Some is stored as chemical potential energy in the
chemical bonds
Energy is absorbed to BREAK bonds
Energy is released when bonds are MADE
The potential energy of the bonds changes in
chemical reactions  this is ENTHALPY CHANGE
5.1.1 Define the terms exothermic reaction, endothermic
reaction and standard enthalpy change of reaction (Ho)
This quantity, H, is called the enthalpy of
reaction, or the heat of reaction.
5.1.1 Define the terms exothermic reaction, endothermic
reaction and standard enthalpy change of reaction (Ho)
1st Law of Thermodynamics:
Energy can not be created nor destroyed, but
it can change form
 Energy lost = Energy gained
The
represents
 System + Surroundings = Universe (Which
is System
constant
)
the chemical reaction
Reactions
 Endothermic: Energy is used by the system
 Exothermic: Energy is produced by the system
5.1.1 Define the terms exothermic reaction, endothermic
reaction and standard enthalpy change of reaction (Ho)
 The system includes the
molecules we want to
study
(here,
the
hydrogen and oxygen
molecules).
 The surroundings are
everything else (here,
the cylinder and piston).
5.1.1 Define the terms exothermic reaction, endothermic
reaction and standard enthalpy change of reaction (Ho)
The change in enthalpy,
H, is the enthalpy of
the products minus the
enthalpy
of
the
reactants:
H = Hproducts − Hreactants
5.1.1 Define the terms exothermic reaction, endothermic
reaction and standard enthalpy change of reaction (Ho)
 The change in enthalpy ΔH = the net
energy after bonds are broken and
remade
 ΔH = (Negative) More energy was
created than used. EXOTHERMIC
 ΔH = (Positive) More energy was
used than created. ENDOTHERMIC
IB Core Objective
 5.1.2 State that combustion and neutralization are
exothermic processes.
Combustion:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ∆H = - 882 kJ mol -1
Neutralization:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H= -57.3 kJ mol-1
IB Core Objective
 5.1.3 Apply the relationship between temperature change,
enthalpy change and the classification of a reaction a
endothermic or exothermic.
What happens to the temperature when the reaction is
exothermic?
A: Temperature goes up.
What happens to the enthalpy change?
A: It is a negative value.
If the same moles of substance were reacted in a larger container
of water, would the temperature and enthalpy values stay the
same?
A: Enthalpy value would stay the same, the temperature
however would not go up as much.
IB Core Objective
 5.1.4 Deduce, from an enthalpy level diagram, the
relative stabilities of reactants and products, and the
sign of the enthalpy change for the reaction.
5.1.4 Deduce, from an enthalpy level diagram, the relative stabilities of
reactants and products, and the sign of the enthalpy change for the reaction.
 Exothermic
Products are more stable
Enthalpy
ΔH is Negative
Heat energy is
RELEASED into the
SURROUNDINGS
Enthalpy
 Endothermic
ΔH is Positive
Reactants are more
stable
Heat energy is
ABSORBED into
the SYSTEM
Standard State
 Enthalpy is affected by multiple factors (concentration,
pressure, state of reactants, temperature))
θ (Theta) Means
under SATP
 ΔHθ Under SATP (Standard Ambient Temperature
and Pressure)
 P = 101.3kPa
 T = 25oC or 298 K
 C = 1mol/dm3 (Concentration for aqueous solutions aq)
 Standard state (physical state of element under these
conditions)
Different from STP
(Temperature is not 0 oC)
IB Core Objective
 5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
Specific Heat Capacity
 The energy required to heat 1g of substance 1 degree
Celsius
 Specific Heat Capacity of water
 1 g of H2O to heat 1 degree C requires 4.18 J
 Therefore the specific heat capacity of water is 4180
J•kg-1•K-1

4.18 Jg-1 K-1
Typically we use this
for labs due to the small
quantities used.
5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
Calorimetry
 To measure the energy of a system is hard
 To measure the energy of surroundings is easy
 q = mcΔT
Energy must have come
from the system
 q = Energy change in the surroundings
 m = mass of surroundings (g)
 c = Specific Heat Capacity of surrounding substance
(Sometimes s is used instead of c)
 ΔT = Temperature change
For us, this will
be the mass of
water used
5.2.1 Calculate the heat energy change when the
temperature of a pure substance is changed.
Questions
How much heat is needed to warm 250 g of water from
22°C to near its boiling point 98°C?
A: 7.9 x 104 J or 79kJ.
Calculate the quantity of heat used when you mix 50cm3 of
1.0M HCl and 50cm3 of NaOH in a coffee-cup calorimeter,
and the temperature increases from 21.0°C to 27.5°C.
Assume the density is 1.00 g/cm3.
A: 2.7 kJ
Calculate the enthalpy change for the above reaction in
kJ/mol.
A: -54 kJ/mol
Questions
 Excess solid Magnesium is added to a 100g of a 2M
solution of Copper(II) Sulphate. The temperature
increased from 20.0oC to 65.0oC. (Since the solution
is largely water we will assume specific heat capacity
as 4.18 Jg-1K-1.
 What is the enthalpy change for the reaction?
 Don’t forget it must be calculated as a ratio of a full
mol.
 Push for answer
 ΔH = -94.1 kJ/mol.
IB Core Objective
 5.2.2 Design suitable experimental procedures for
measuring the heat energy changes of reactions.
 5.2.4 Evaluate the results of experiments to
determine enthalpy changes.
5.2.2 Design suitable experimental procedures for measuring the heat energy
changes of reactions.
5.2.4 Evaluate the results of experiments to determine enthalpy changes.
For liquids, a calorimeter should be well insulated and
the heat capacity should be low.
Calorimetry: Technique used to measure the enthalpy
associated with a particular change.
Calorimetry depends on the assumption that no heat is
gained from or lost to the surroundings.
Even with well insulated calorimeters, heat exchange
with surroundings is a major source of error.
5.2.2 Design suitable experimental procedures for measuring the heat energy
changes of reactions.
5.2.4 Evaluate the results of experiments to determine enthalpy changes.
 In combustion experiments,
where burning gas is used to
heat liquid in a calorimeter,
the error is quite large.
 Temperature rises are much
less than expected and thus
∆H values are less than
literature values.
IB Core Objective
 5.2.3 Calculate the enthalpy change for a reaction
using experimental data on temperature changes,
quantities of reactants and mass of water.
5.2.2 Design suitable experimental procedures for measuring
the heat energy changes of reactions.
 Question: 20.0 cm3 of 2 mol dm-3 aqueous sodium
hydroxide is added to 30.0 cm3 of hydrochloric acid
of the same concentration, the temperature increases
by 12.0 C. For dilute aqueous solutions, we can
assume the density is 1.00 g cm-3.
 What is ∆H?
 A: -62.7 kJ mol-1
IB Core Objective
 5.3.1 Determine the enthalpy change of a reaction
that is the sum of two or three reactions with known
enthalpy changes.
 Students should be able to use simple enthalpy
cycles and enthalpy level diagrams and to
manipulate equations. Students will not be required
to state Hess’s law.
5.3.1 Determine the enthalpy change of a reaction that is the
sum of two or three reactions with known enthalpy changes.
Hess’s Law
 Heat of the whole = the sum of the parts
 Reaction that take a direct route or multiple step route
make no difference with Enthalpy.
ANALOGY
 I just bought a $1200 TV, I could either pay 12 equal
instalments of $100, pay 2 instalments of $600 or pay it all
$1200 at once. How I do it doesn’t matter, in the end I
still pay $1200.
 In enthalpy terms: ∆H1 = ∆H2 + ∆H3
5.3.1 Determine the enthalpy change of a reaction that is the
sum of two or three reactions with known enthalpy changes.
ΔHf =
CO2 -395 kJ mol-1
ΔHf =
H2O -287 kJ mol-1
ΔHc =
C2H2 -1301 kJ mol-1
C2H2(g)
2C (s) + H2(g)
2½O2(g)
A: 224 kJ mol-1
 Calculate ΔHf for C2H2(g)
2CO2(g) + H2O(l)
IB Core Objective
 5.4.1 Define the term average bond enthalpy.
 Enthalpies are a measure of the strength of a
covalent bond: the stronger the bond, the more
tightly the atoms are joined together.
 Breaking of a chemical bond requires energy, and is
thus and endothermic process.
 Average bond enthalpy: The bond enthalpy for a
compound will be affected by surrounding bonds,
therefore we use the average bond enthalpy.
IB Core Objective
 5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic.
Energy 5.4.2 Explain, in terms of average bond enthalpies,
why some reactions are exothermic and others are
endothermic.
H
463
kJ/mol
H
463
kJ/mol
436
kJ/mol
½
O
O
 It takes energy to
496
break
bonds
kJ/mol
 Energy is produced upon bond formation
 ΔH = Energy IN (Bonds broken) – Energy OUT(Bonds formed)
 ΔH = 1(H-H 436) + ½ (O=O 496) – 2(O-H 463)
 ΔH = 684 – 926
 ΔH = -242 (Energy is left over, Exothermic)
5.4.2 Explain, in terms of average bond enthalpies, why some
reactions are exothermic and others are endothermic.
 Sometimes Bond ΔH
energy
is broken
effected
by
= {Bonds
– Bonds
surrounding bonds so anformed}
average must
be used.
 Calculate the Enthalpy of change for the
following:
 CH4(g) + O2(g)  CO2(g) + H2O(g)
 Compare to enthalpy of combustion
value (Data book)
Bond
Bond
Enthalpy
kJ/mol
H-H(g)
436
Cl-Cl(g)
242
F-F(g)
158
H-Cl(g)
431
H-F(g)
562
Bond
Avg. Bond
Enthalpy
kJ/mol
C-C(g)
348
C=C(g)
612
C=C(g) (in
518
benzene)
Why the
difference?
C-H(g)
412
C=O(g)
743
O-H(g)
463
N-H(g)
388
O=O(g)
496
5.4.2 Explain, in terms of average bond enthalpies, why
some reactions are exothermic and others are endothermic.
 Given that the enthalpy change for the reaction
N2(g) + 3Cl2(g) → 2NCl3(g) is +688kJ/mol
Calculate the enthalpy of the N-Cl bond, given that the
bond enthalpies in the nitrogen molecule and the
chlorine molecule are 944kJ/mol and 242kJ/mol.
A: 164 kJ/mol
IB HL Objective
 15.1.1 Define and apply the terms standard state,
standard enthalpy change of formation (∆Hfѳ) and
standard enthalpy change of combustion (∆Hcѳ)
 You have already learned about standard state.
What are the conditions?
 A: Form normally found at 298K, 101.3 kPa, 1 mol
dm-3
15.1.1 Define and apply the terms standard state, standard enthalpy change
of formation (∆Hfѳ) and standard enthalpy change of combustion (∆Hcѳ)
 Hθc = Standard enthalpy change of combustion
 Enthalpy change when one mole of compound is burned in
excess oxygen under standard conditions. (Always
exothermic).
 Hθf = Standard enthalpy change of formation
 Amount of energy released or absorbed in the formation of
one mol of compound from elements in their normal states.
 By using the definition of standard state, the enthalpy of
formation of an element in the standard state is zero.
IB HL Objective
 15.1.2 Determine the enthalpy change of a reaction
using standard enthalpy changes of formation and
combustion.
∆Hѳ = ∑∆Hfѳ(products) - ∑∆Hfѳ(reactants)
15.1.2 Determine the enthalpy change of a reaction using
standard enthalpy changes of formation and combustion.
For the following reaction, find the enthalpy of
formation:
2 C(graphite) + 3H2(g) + ½O2(g) → C2H5OH(l)
A: ∆Hfѳ = -277 kJ mol-1 (All the reactants are in their
elemental, standard state, so they would be zero!)
Find the enthalpy of formation for the following reaction
15.1.2 Determine the enthalpy change of a reaction using
standard enthalpy changes of formation and combustion.
Find the enthalpy of formation for the following reaction:
NH4NO3 (s) → N2O(g) +2H2O(l)
The enthalpies for formation for the above compounds
are: NH4NO3 (s) :-366 kJ mol-1, N2O(g) +82 kJ mol-1,
and H2O(l) -285 kJ mol -1
A: -122 kJ mol-1
When we have learned more about Hess’s Law, we
will return to this later…
15.1.2 Determine the enthalpy change of a reaction using
standard enthalpy changes of formation and combustion.
Find the formation of combustion of benzene using
enthalpies of formation values.
First, balance the equation:
C6H6(l) + O2(g) → CO2(g) + H2O(g)
A: C6H6(l) +152 O2(g) → 6CO2(g) + 3H2O(l)
The enthalpies of formation are: CO2(g): -393.5 kJ mol1, H2O(g): -285.8 kJ, benzene (find in your data
booklet), O2???
A: -3267 kJ mol-1 (Compare this answer with the data
booklet).
IB HL Objective
 15.2.1 Define and apply the terms lattice enthalpy and
electron affinity.
Lattice enthalpy: Enthalpy change to convert one mole
of a solid ionic compound into gaseous ions or vice
versa.
Electron affinity: Enthalpy change when one mole of
gaseous atoms or anions gain electrons to form a
mole of negatively charged gaseous ions.
15.2.1 Define and apply the terms lattice enthalpy and electron
affinity.
 Standard enthalpy change of atomization (also
known as standard enthalpy of vaporization). This
is the enthalpy required to change one mole of atoms
from the standard state to the gaseous state.
Na(s) → Na(g) ∆Hѳ +103kJ mol-1
Exothermic or endothermic?
15.2.1 Define and apply the terms lattice enthalpy and electron
affinity.
First ionization energy (remember this?)
• Na(g) → Na+(g) + e- ∆Hѳ = +494 kJ mol-1
Enthalpy atomization of Cl
• ½Cl2(g) → Cl(g) ∆Hѳ = +121 kJ mol-1
First electron affinity of Cl
Cl(g) + e- → Cl-(g) ∆Hѳ = -364 kJ mol-1
IB HL Objective
 15.2.3 Construct a Born-Haber cycle for Group 1 and
2 oxides and chlorides, and use it to calculate an
enthalpy change.
15.2.3 Construct a Born-Haber cycle for Group 1 and 2 oxides
and chlorides, and use it to calculate an enthalpy change.
Na+(g) + Cl-(g)
Na+ (g)
Na(g)
Cl-(g)
+ Cl (g)
Na(s) + ½Cl2(g)
ΔHof =
NaCl(S)
15.2.3 Construct a Born-Haber cycle for Group 1 and 2 oxides
and chlorides, and use it to calculate an enthalpy change.
Affinity
 Formation of NaCl(s) from its gaseous elements.




1) Na(s)  Na(g)
2) ½Cl2(g)  Cl(g)
3) Cl(g) + e-  Cl(g)4) Na(g)  Na+(g)
ΔHoat = 103 kJ/mol
ΔHoat = ½(242) kJ/mol
ΔHo = -364 kJ/mol
ΔHoat = 500 kJ/mol
 Find the enthalpy of formation. Literature value for
lattice enthalpy can be found in the data booklet.
 A: -430 kJ/mol
15.2.3 Construct a Born-Haber cycle for Group 1 and 2 oxides
and chlorides, and use it to calculate an enthalpy change.
 Draw a Born-Haber cycle for the formation of




magnesium oxide, and calculate the enthalpy of
formation.
The enthalpy of atomization of magnesium is +150
kJ/mol, and for oxygen it is +249 kJ/mol.
The second ionization energy of magnesium is +1450
kJ/mol.
Use the data booklet to find other relevant
information.
A: -547 kJ/mol
IB HL Objective
 15.2.2 Explain how the relative sizes and the charges
of ions affect the lattice enthalpies of different ionic
compounds.
 The relative value of the theoretical lattice enthalpy
increases with higher ionic charge and smaller ionic
radius due to increased attractive forces.
15.2.2 Explain how the relative sizes and the charges of ions
affect the lattice enthalpies of different ionic compounds.
 The greater the charge on the ions, the greater the





electrostatic attraction and thus greater the lattice
enthalpy, and vice versa.
The larger the ions, then the greater the separation of
the charges and the lower the lattice enthalpy, and
vice versa.
Lattice enthalpy of MgO > NaCl. Why?
A: Increased ionic charge.
Lattice enthalpy of KBr<NaCl. Why?
A: KBr has larger ions.
IB HL Objective
 15.2.4 Discuss the difference between theoretical
and experimental lattice enthalpy values of ionic
compounds in terms of their covalent character.
 A significant difference between the two values
indicates covalent character.
15.2.4 Discuss the difference between theoretical and
experimental lattice enthalpy values of ionic compounds in terms
of their covalent character.
 Energy cycles such as Hess’s Law and the Born-Haber




cycle are used to find unknown values, provided that the
other values are known.
The Born-Haber cycle assumes that the salt formed has
100% ionic character.
Look at Table 13 in your data booklet. There are two
tables showing different enthalpy values. What do they
label them as?
Lattice enthalpies can be calculated experimentally by
using the Born-Haber cycle.
They can also be calculated theoretically from the size,
charge, and packing of the ions.
15.2.4 Discuss the difference between theoretical and
experimental lattice enthalpy values of ionic compounds in
terms of their covalent character.
 When the electronegativity difference is lower, will the




compound exhibit greater or less ionic character?
A: Less.
When the electronegativity difference is less, there will be
a greater amount of covalent bonding, and this will result
in a higher experimental value for lattice enthalpy than
theoretical.
Look at electronegativity differences between NaCl and
AgI. Which would you guess has a greater difference in
experimental vs. theoretical lattice enthalpy values?
A: AgI
IB HL Objective
 15.3.1 State and explain the factors that increase the
entropy in a system.
 Systems naturally tend towards disorder.
 Entropy: A measure of the degree of disorder or
randomness in a system.
15.3.1 State and explain the factors that increase the entropy
in a system.
 Entropy values for gas>liquid>solid.
 Gas pressure increases, then entropy decreases
(reduces volume for gas particles to move in).
 When a solid or liquid dissolves in a solvent,
entropy increases.
 Increase the number of moles increases the entropy.
 Heating a substance increases the entropy, since this
increases the movement of particles.
IB HL Objective
 15.3.2 Predict whether the entropy change (∆S) for a
given reaction or process is positive or negative.
 ΔS = ∑S(Products) – ∑S(Reactants)
 So if the products have more disorder than the
reactants, would the entropy change be positive or
negative?
 A: Positive
15.3.2 Predict whether the entropy change (∆S) for a given
reaction or process is positive or negative.
 Consider the following equation:
NH4Cl(s) → NH3(g) + HCl(g)
State ∆S as positive or negative, and explain why.
A: Positive. (∆S=+285 J K-1 mol-1) More moles on
product side, so more disorder. Also, solid has
become a gas, so more disorder.
IB HL Objective
 15.3.3 Calculate the standard entropy change for a
reaction (∆Sѳ) using standard entropy values (∆Sѳ).
 ΔS = ∑S(Products) – ∑S(Reactants)
15.3.3 Calculate the standard entropy change for a reaction
(∆Sѳ) using standard entropy values (∆Sѳ).
 Calculate the entropy change for the combustion of
methane (CH4).
 First, write the equation.
 Make a prediction if it will be positive or negative,
and explain why.
 Next, plug in the values and calculate. You can find
the entropy value for methane in your data booklet.
O2 is 205, CO2 is 214, and H2O is 70.
 A: -242 J K-1 mol-1
IB HL Objective
 15.4.1 Predict whether a reaction or process will be
spontaneous by using the sign of ∆Gѳ.
 A reaction will be spontaneous when ∆Gѳ has a negative
value.
 Spontaneous: Once started, a reaction will continue
without any extra energy having to be added.
 Remember redox? An electrochemical cell can create
energy through a spontaneous reactions. So if Eѳcell is
positive, what would ∆Gѳ be?
 A: Negative
15.4.1 Predict whether a reaction or process will be
spontaneous by using the sign of ∆Gѳ.
 Is the combustion of coal spontaneous at room
temperature?
 Yes. The activation energy is different than
spontaneous reactions. You have to get the coal
going, but once you do, it continues to react.
 What about the combustion of diamond at room
temperature?
 Yes, although it has a very high activation energy.
Once this is reached, it will be a spontaneous
reaction.
IB HL Objective
 15.4.2 Calculate ∆Gѳ for a reaction using the
equation
∆Gѳ = ∆Hѳ – T∆Sѳ
and by using values of the standard free energy
change of formation, ∆Gfѳ.
15.4.2 Calculate ∆Gѳ for a reaction using the equation ∆Gѳ = ∆Hѳ – T∆Sѳ and by
using values of the standard free energy change of formation, ∆Gfѳ.
 Let’s go back to the combustion of diamond at room
temperature (298K):
C (s) + O2(g) → CO2(g)
∆Hѳ = -395.4 kJ mol-1.
∆Sѳ = +6.6 J K-1 mol-1.
What is ∆Gѳ?
A: -397.4 kJ mol-1
Would this reaction be spontaneous at all temperatures?
Why or why not? Refer to the equation!
A: If ∆Hѳ is negative and ∆Sѳ is positive, then ∆Gѳ will always be
negative.
IB HL Objective
 15.4.3 Predict the effect of a change in temperature
on the spontaneity of a reaction, using standard
entropy and enthalpy changes and the equation:
∆Gѳ = ∆Hѳ – T∆Sѳ
• If both ∆Sѳ and ∆Hѳ are positive, then spontaneity
depends on T. (Spontaneous at higher
temperatures).
• If both ∆Sѳ and ∆Hѳ are negative, then spontaneity
depends on T. (Spontaneous at lower temperatures).