Unit 1 Mod 1 Energetics page 1 of 12

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!

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

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.

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!!!!

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 moleculecontains 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 CH4
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

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.

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 ??

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

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.

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

Practice Questions

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