Energy
The law of conservation of energy states that energy can’t be created
or destroyed, however energy can change forms.
• Chemical reactions and changes of state involve energy changes.
• Exothermic reactions occur when energy is released to the
surroundings.
• The net strength of the bonds in the products is less than the
reactants.
• Endothermic reactions occur when energy is absorbed from the
surroundings.
• The net strength of the bonds in the products is greater than the
reactants.
• Exchange of heat energy between the system and the surroundings is
called the enthalpy change (∆H)
∆H = Hreactants – Hproducts
Energy Profile Diagrams
Exothermic
Endothermic
Thermochemical equations
• The ∆H value is written to the right of the equation.
• The ∆H value is measured in kJ mol–1
• Combustion of methane – exothermic = -ve ∆H
CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O(g) ΔH = –890 kJ mol–1
• If the reaction was reversed what happens to ∆H?
• Just change the symbol to a positive as it is an endothermic
reaction.
CO2 (g) + 2H2O(g) →CH4 (g) + 2O2 (g) ΔH = + 890 kJ mol–1
Thermochemical Equations – changing the mole ratio, what happens to ΔH?
Another example of a thermochemical equation is the combustion of
methanol (CH3OH):
CH3OH(l) + 3/2O2(g) → CO2(g) + 2H2O(g)
ΔH = –726 kJ mol–1 (1 mole of methanol)
• What happens when there are two moles of methanol reacting?
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g) ΔH = –1452 kJ mol–1 (Double ΔH )
• What happens when there are three moles of methanol reacting?
3CH3OH(l) + 4 ½ O2(g) → 3CO2(g) + 6H2O(g) ΔH = – 2178 kJ mol–1 (Triple ΔH )
Thermochemical Equations – what happens to ΔH when changing states?
Activation energy
• Activation energy (Ea) is the energy that must be absorbed to break the
bonds in the reactants so that a chemical reaction can proceed.
H2O(s)  H2O(l)
ΔH = +6.00 kJ mol–1
• Both endothermic and exothermic reactions require activation energy.
H2O(l)  H2O(g)
ΔH = +40.7 kJ mol–1
More energy is needed break the hydrogen bonds in liquid water
to form a water vapour (gas).
Less energy is required to break the bonds between water
molecules in ice to form liquid water.
Hess’s Law: Calculate the H of the following reaction 2Fe + 1.5O2  Fe2O3 given the following 2
equations
1. Fe2O3 + 3CO  2Fe + 3CO2 H= – 26.74 kJ
2. CO + ½ O2  CO2
H= – 282.96 kJ
Step 1: need Fe2O3 to be on the
product side so flip reaction 1
Reverse reaction causes symbol of H to become the
opposite charge
2Fe + 3CO2  Fe2O3 + 3CO
H= +26.74 kJ
Step 2: the final equation has no CO so
this must be removed from both
equations.
Multiple equation x 3
CO + ½ O2  CO2 H= – 282.96 kJ
Step 3: Add equations and ΔH’s
together and cancel down common
molecules
3CO +1.5O2  3CO2 H=-848.88 kJ
2Fe + 3CO2  Fe2O3 + 3CO H=+26.74 kJ
3CO +1.5 O2  3CO2 H= - 848.88 kJ
3CO +1.5O2 + 2Fe + 3CO2  3CO2 +Fe2O3+ 3CO
H= - 848.88 kJ +26.74 kJ
Final answer
2Fe + 1.5O2  Fe2O3 H= -822.14kJ
Hess’s Law using Energy profile diagrams
The energy profile diagrams below relates to the following two reactions
NO(g)  ½ N2(g) + ½ O2(g)
H= – 90.37 kJ
½ N2(g) + O2(g)  NO2(g)
H= + 33.8 kJ
Question: Calculate the enthalpy change ΔH for the reaction
NO(g) + ½ O2(g) NO2(g)
Step 1: determine if any
equations need to be flipped.
NO
Step 2: Does adding the
equations together get you the
final equation?
Yes, add equations together and ΔH to get final answer
NO(g) + ½ N2(g) + O2(g)  ½ N2(g) + ½ O2(g) + NO2(g)
H= 33.8 kJ + – 90.37 kJ
Final Answer: NO(g) + ½ O2(g) NO2(g) H= -56.57 kJ
Drawing energy profiles
Given that NO2(g) + CO(g)  NO(g) + CO2(g) ΔH = - 225 kJ/mol and
the activation energy of the reaction is 116 kJ/mol draw an
appropriate energy profile.
Heat of Combustion
• The heat of combustion of a fuel is defined as the enthalpy
change that occurs when a specified amount (e.g. 1 g, 1 L or 1 mol)
of the fuel burns completely in oxygen.
• It is usually measured at Standard Lab Conditions (SLC) of 298K
(25⁰C) and 100kPa.
• The units for ∆H can either be kJ mol-1 , kJ g-1 , kJ L-1 or
MJ/tonne.
• NOTE: Only pure substances can have their ∆H measure in kJ mol1 as they have a known molar mass.
Unpure substance that don’t have a known molecule formula eg.
Coal ∆H is measured in kJ g-1
Formula for calculating energy kj mol-1 released for different fuels
Enthalpy of combustion
Use
∆H = q/n
Where:
∆H = Enthalpy of combustion
q = energy in Joules
n = mol
CALCUALTING ENERGY GIVEN DENISTY OF A FUEL AND VOLUME
The molar heat of combustion of ethanol is given as -1364 kJ mol-1.
If the density of the fuel is 0.79 g ml-1, calculate the energy evolved in
MJ if 1.00 L of the fuel is burnt.
Step 1: Write a thermochemical equation
C2H5OH (l) + 3O2(g) à 3H2O (g) + 2CO2 (g) ΔH= -1364 kJ mol-1
Step 2: Use Denisty formula to work out mass of fuel.
Denisty = g/ml g = 0.79 x 1000 ml = 790 g
Step 3: mole (n) = m/Mr = 790/46 = 17.17 mol
Step 4: 17.17 x 1364 = 23,425J = 23.4 MJ
Thermochemical equations and states
•
Methane, ethane, ethene, propane, propene, butane, butene are
all gases (g) at SLC (25°C) (1 – 4 C straight chain hydrocarbons)
•
Pentane, pentene, hexane, hexene, heptane, heptane, octane,
octene are all liquids (l) at SLC (25°C). (5 – 8C straight chain
hydrocarbons)
NOTE: Hydrocarbons above C16H34 are semi=solid waxes
•
Fuels should never be listed as aqueous in thermochemical
equations.
•
Mixing water with fuels prevents them combusting properly so no
writing (aq) for fuels !
Balancing Thermochemical equations
Odd numbered alkanes
• Methane, propane, pentane, heptane all
require 1 mole of fuel to balance the
thermochemical equation.
• E.g. Pentane
• C5H12 (l) + 8O2 (g)  5CO2(g) +6H2O (l)
1 mole
therefore ΔH = -3520 kJ mol-1
Balancing Thermochemical equations
Even numbered alkanes
• Ethane, butane, hexane and octane all
require 2 mole of fuel to balance the
thermochemical equation.
• E.g. butane
• 2C4H10 (l) + 8O2 (g)  5CO2(g) +6H2O (l)
1 mole
therefore ΔH = -3520 kJ mol-1
Writing balance thermochemical equations for complete combustion
E.g. butane
Tips.
1. Write the formula for the fuel reacting with
oxygen gas.
2. C4H10 (g)
3. The products carbon dioxide and water.