Unit 1 Mod 1 Energetics
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Module 1 Energetics
Conditions needed for a reaction to occur
1. Particles MUST collide
2. Particles must collide with the CORRECT ORIENTATION
3. Particles must collide with a certain MINIMUM AMOUNT OF
ENERGY (ACTIVATION ENERGY)
These three conditions are collectively called EFFECTIVE
COLLISIONS
Processes that occur in chemical reactions
1. Bonds are broken FIRST (energy is needed or absorbed) i.e. bond
breakage
2. Bonds are formed AFTERWARDS (energy is released) i.e. bond
formation
Note BOTH processes ALWAYS OCCUR in a chemical reaction!!!
In some cases, more energy is needed than is released in a chemical
reaction, these reactions which absorb MORE energy than is released are
called ENDOTHERMIC reactions. Examples of endothermic reactions
are:- dissolving potassium nitrate solid in water, dissolving sodium
thiosulphate solid in water and photosynthesis. Note temperature
ALWAYS DECREASES in an endothermic reaction!
In some cases, more energy is released than is absorbed in a chemical
reaction, these reactions which release MORE energy than is absorbed
are called EXOTHERMIC reactions. Examples of exothermic reactions
are:- acid-base reactions, combustion reactions and respiration. Note
temperature ALWAYS INCREASES in an exothermic reaction!
Unit 1 Mod 1 Energetics
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ENERGY PROFILE DIAGRAMS
The change in the amount of energy contained in a substance between
the beginning of a reaction and the end of a reaction is called the change
in energy / enthalpy and is given the symbol ΔH
The symbol Ea represents the activation energy of the reaction. Note the
activation energy starts from the reactants to the TOP of the HILL! All
reactions have their own UNIQUE Ea !
Note in exothermic reactions, the products have LESS energy than the
reactants and in endothermic reactions, they have MORE energy than
the reactants.
Checkpoint A
1. Write either “exo” or “endo” for the processes that you consider to be
exothermic or endothermic respectively.
a) reaction between sodium hydroxide and hydrochloric acid
b) N2 + 3H2  2NH3 ΔH = - 92 kJ mol-1
c) The burning of gasoline
d) The dissolving of potassium nitrate in water
Unit 1 Mod 1 Energetics
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Relationship between bond energy and reactivity of molecules
Remember bonds must first be broken in a chemical reaction. If the
bonds can be broken easily, the reaction would occur quickly.. If the
bonds cannot be broken easily, the reaction would occur slowly.
The term “bond energy” refers to the amount of energy required to
dissociate a molecule into its respective atoms.
Bond energy is directly related to the strength of the covalent bond and is
indirectly related to its reactivity.
For example, in the nitrogen molecule, the bond energy is 945 kJ mol-1.
This is a very HIGH value making nitrogen gas a very UNREACTIVE
molecule. However the bond energy of an oxygen molecule is 498 kJ
mol-1 making oxygen a lot more reactive than nitrogen.
Factors affecting bond energy
1. Strength of covalent bond
2. Size of the atoms in the molecule
3. Degree of orbital overlap in the covalent bond
Note: a triple bond is stronger than a double bond and a double bond is
stronger than a single bond
All three factors are related. The smaller the atom, the more extensive
the degree of overlap with another atom ( because of its small size, it can
approach another atom closer before their electron clouds interact). The
better the degree of overlap, the stronger the covalent bond formed. The
stronger the covalent bond, the higher the bond energy.
Unit 1 Mod 1 Energetics
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Calculating the enthalpy change of a reaction using enthalpies of
formation of substances
ΔH reaction = ΔH formation products – ΔH formation reactants
Example
Standard enthalpies of formation are: C2H5OH(l) -228, CO2 -394, and
H2O(l) -286 kJ/mol. Calculate the enthalpy of the reaction,
C2H5OH (l) + 3 O2 (g) 2 CO2 (g) + 3 H2O (l)
Please note the enthalpy of formation of an element is ALWAYS
equal to ZERO!!! Remember that!!
Answer ΔH reaction = (3 x -286) + (2 x -394) – (-228)
= -1418 kJ mol-1
Definition of enthalpy of formation
The enthalpy change when ONE mole of a compound is formed from
its elements in their standard states.
Writing equations representing enthalpy of formation of compounds
Example 1 Write the equation representing the enthalpy of formation of
ethane C2H6.  2C (s) + 3H2 (g)  C2H6
Example 2 Write the equation representing the enthalpy of formation of
carbon monoxide  C (s) + ½O2 (g)  CO
Note in example 2, a fraction was used in order to ensure the product was
in the quantity of one mole.
Remember when writing equations to represent the enthalpy of formation
of compounds, it must be written and balanced in respect to ONE mole
of the product!!!!
Unit 1 Mod 1 Energetics
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Calculating enthalpy of reaction using bond energies
Example Calculate the enthalpy of reaction for 2 H2 + O2  2 H2O
Bond energies: H2 436 kJ/mol, O2 498, HO 463 kJ/mol
Each H2 contains one covalent bond and there are 2 molecules, so for
2H2 = +(2 x 436) . Oxygen = +498 .
Therefore sum of energy required (for bond breakage)
= +872 + 498 = +1370
Each water molecule contains 2 OH bonds and there are 2 water
molecules therefore 4 OH bonds are formed
thus sum of energy released (for bond formation) = -(4 x 463) = -1852
ΔH reaction = sum of endothermic process and exothermic processes
= + 1370 – 1852 = -482 kJ mol-1
Checkpoint B
1. Calculate the enthalpy of the reaction CH4 + 2 O2  CO2 + 2 H2O
from the enthalpies of formation: CH4 -75 kJ/mol, CO2 -394, and H2O(l) -286
kJ/mol.
2. Calculate the heat of reaction for H2 + Cl2  2 HCl
Bond energies: H2 436 kJ/mol, O2 498, HO 463, Cl2 243, HCl 432 kJ/mol
3. Write the equation for the enthalpy of formation of CH 4
4. Calculate the enthalpy of reaction below using the bond energies
(all in kJ mol-1)
C-H +412 , O=O +496, C=O +803 (in carbon dioxide), H-O +463
Unit 1 Mod 1 Energetics
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How to calculate the enthalpy of formation or lattice enthalpy for ionic
compounds via application of Hess’s Law and Born-Haber cycles
Lattice enthalpies cannot be determined experimentally and thus
calculations are used.
Definition of Hess’s Law: The standard enthalpy change of a reaction is
independent of the route taken from reactants to products.
Born-Haber cycle of a simple M+X- ionic compound
Born-Haber cycles for all ionic compounds would be similar to the one above
except for a few insertions. But first here are the labels for each step
Step 1 – enthalpy of atomisation / vapourisation of solid
(usually endothermic so arrow is pointed up)
Step 2 – first ionisation energy of gaseous metal atoms
Step 3 – bond dissociation energy of halogen molecule
Step 4 – first electron affinity of halogen atom
(usually exothermic so the arrow points downward)
Step 5 – lattice energy of compound in question
(definition of lattice energy in Born- Haber cycle is defined as
an output of energy. The amount of energy released when 1
mole of an ionic compound is formed from its respective ions in
their gaseous states.
Unit 1 Mod 1 Energetics
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M+X- compounds would be like NaCl, or LiCl or LiF
However there would be slight changes for compounds like M2+X2- , for
example the insertion of a second ionisation energy and a second
electron affinity or other similar changes.
The formula used for Born-Haber cycles for ionic compounds 
Enthalpy of lattice energy = -(step 1) – (step 2) – (step 3) - (step 4) +
(enthalpy of formation)
Example
Below is a Born-Haber cycle for the formation of sodium fluoride
Sample calculation based on diagram above
Enthalpy of formation = - 411 kJ
Step 1 + 108 kJ
Step 2 + 496 kJ
Step 3 + 122 kJ
Step 4 - 349
Step 5 ??
Enthalpy of lattice enthalpy = -(108) – (496) –
(122) – (-349) + (-411)
Therefore lattice energy = - 788 kJ
Unit 1 Mod 1 Energetics
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Checkpoint B
1. Draw a Born-Haber cycle for the compound LiCl and using the data
below calculate the lattice energy of LiCl
2. Determine the lattice energy of CaF2 using the following data
Unit 1 Mod 1 Energetics
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How to calculate the enthalpy of solution for ionic compounds via
application of Hess’s Law and Born-Haber cycles
Note Born-Haber cycles can be used for the dissolving of ionic
compounds called the enthalpy of solution. Below is a Born-Haber cycle
for the dissolving of any ionic compound
Enthalpy of solution = ΔHlattice + ΔHhydration
Effect of ionic charge and radius on lattice energy
As the charge on Mn+ increases there is a greater attractive force between
the ions and lattice energies increase. In addition, the decrease in size of
Mn+ with increasing charge increases the attractive force between the
ions and also increases the lattice energy.
Example:- The ionic radius of the Na+ and Ca2+ ions are very similar.
However the lattice energy of CaCl2 is about 3 times that of NaCl.
Effect of ionic charge and radius on lattice energy
The smaller and the greater the charge on the ion, the more hydration
energy would be released resulting in a more exothermic enthalpy of
solution. e.g. NaCl has ΔH solution = + 3.9 kJ mol-1 while LiCl has ΔH
solution = - 37.2 kJ mol-1.
Unit 1 Mod 1 Energetics
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Calculating enthalpy of formation of compounds using combustion
data
Example
Given
 C + O2  CO2 (g) ΔH = -394 kJ mol-1 eqn 1
 CO + ½O2  CO2 ΔH = -283 kJ mol-1 eqn 2
Find the ΔHf of: C(s) + ½O2 (g) → CO(g)
Rearrange the equations in order to get the equation in question
C + O2  CO2 (g)
CO2  CO + ½O2
If you remove the compounds on either side, the result is the equation in
question  C(s) + ½O2 (g) → CO(g)
Then the enthalpy of formation is sum of values
 -394 + 283 = -110.5 kJ mol-1
OR via Enthalpy diagram
 Then the enthalpy of formation is sum of values
 -394 + 283 = -110.5 kJ mol-1
END OF ENERGETICS
Unit 1 Mod 1 Energetics
Practice Questions
1.
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Unit 1 Mod 1 Energetics
2.
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