Unit 1 Mod 1 Energetics Revised

8/17/2019 Unit 1 Mod 1 Energetics Revised

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Unit 1 Mod 1 Energetics page 1 of 12Module 1 Energetics

Conditions needed for a reaction to occur

1. Particles MUST collide2. 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

EFFECTIVECOLLISIONS

Processes that occur in chemical reactions

1. Bonds are broken FIRST (energy is needed or absorbed) i.e. bond

breakage2. 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 chemicalreaction, 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. Notetemperature ALWAYS INCREASES in an exothermic reaction!


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Unit 1 Mod 1 Energetics page 2 of 12ENERGY 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 changein 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 thereactants 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


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Unit 1 Mod 1 Energetics page 3 of 12Relationship 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 isindirectly 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 kJmol-1 making oxygen a lot more reactive than nitrogen.

Factors affecting bond energy

1. Strength of covalent bond2. 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 canapproach 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.


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Unit 1 Mod 1 Energetics page 4 of 12Calculating 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 wasin 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!!!!


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Unit 1 Mod 1 Energetics page 5 of 12Calculating 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 = +1370Each 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 H2Ofrom the enthalpies of formation: CH4 -75 kJ/mol, CO2 -394, and H2O(l) -286kJ/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


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Unit 1 Mod 1 Energetics page 6 of 12How 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 thuscalculations 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.


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Unit 1 Mod 1 Energetics page 7 of 12M+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 secondelectron 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)

ExampleBelow 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


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Unit 1 Mod 1 Energetics page 8 of 12Checkpoint 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


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Unit 1 Mod 1 Energetics page 9 of 12How 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 ioniccompounds 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 ofMn+ with increasing charge increases the attractive force between the

ions and also increases the lattice energy.

Example:- The ionic radius of the Na+ and Ca

2+ 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.


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Unit 1 Mod 1 Energetics page 10 of 12Calculating enthalpy of formation of compounds using combustion

data

ExampleGiven

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


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Unit 1 Mod 1 Energetics page 11 of 12Practice Questions1.


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Unit 1 Mod 1 Energetics page 12 of 12

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Unit 1 Mod 1 Energetics Revised