Chemical energetics

1. Chemical Energetics

Friday, 10 April 2026
Chemical Energetics
REVISION

2. What is enthalpy?

Friday, 10 April 2026
What is enthalpy?
When a chemical reaction occurs there is normally a change in energy
The greek letter ‘delta’
means ‘change in’. So this
is the change in enthalpy
The enthalpy change of a
reaction is the heat
change in a reaction at
constant pressure.
It is given the
symbol ΔH and has
the units of kJmol-1
⦵
ΔH
Standard conditions
1. 100kPa pressure
2. 298K (25°C)
This symbol
means the
substance was in
the standard
state under
standard
conditions.

3. Endothermic and Exothermic

Friday, 10 April 2026
Endothermic and Exothermic
Reactions can give out heat energy or take it in.
Endothermic Reactions
Exothermic Reactions
Reactions that absorb energy from the surroundings
Reactions that release energy to the surroundings
enthalpy
products
ΔH POSITIVE
Products higher in
energy than
reactants. ΔH is
POSITIVE
enthalpy
reactants
ΔH NEGATIVE
reactants
Reaction progress
CaCO3(s) CaO(s) + CO2(g) ΔcH = +178kJmol-1
Example – thermal decomposition
of calcium carbonate
Products lower in
energy than
reactants. ΔH is
NEGATIVE
products
Reaction progress
C2H6(g) + 3.5O2(g) 3H2O(l) + 2CO2(g) ΔcH = -1560kJmol-1
Example – combustion of ethane

4. Bond breaking and making

Friday, 10 April 2026
Bond breaking and making
Bonds are broken and made during a reaction
Bonds Broken
Bonds Made
To break a bond energy needs to be absorbed
When bonds are formed energy is released
Bonds are broken in the reactants and this is an
endothermic process so ΔH is positive.
Bonds are made when products are being produced
and this is an exothermic process so ΔH is negative
enthalpy
products
ΔH POSITIVE
reactants
Reaction progress
If more energy is
needed to break
bonds than energy
given out when
bonds are formed
the reaction will be
ENDOTHERMIC
enthalpy
reactants
ΔH NEGATIVE
products
Reaction progress
If more energy is
released when
bonds are formed
than what was
needed to break
initial bonds the
reaction will be
EXOTHERMIC

5. Mean Bond Enthalpy

Friday, 10 April 2026
Mean Bond Enthalpy
Bonds of the same type don’t all have the same amount of energy
There are 4 C-H bonds in methane
(CH4) all with slightly different
bond enthalpies.
The energy needed to break the
first C-H will be different to the
energy needed to break the
second C-H bond.
Enthalpy Changes using mean bond enthalpy
Enthalpy Change = Total energy to break bonds – Total energy released forming bonds
Calculate energy change for combustion of methane
CH4 + 2O2 CO2 + 2H2O
Bond
We can measure the enthalpy of breaking
all 4 bonds and divide by 4 to get a mean
bond enthalpy of a C-H bond.
Total = 1662kJ for 4 C-H bonds broken in CH4
Mean bond enthalpy of C-H – 1662/4 =
415.5kJmol-1
C=O C-H O=O O-H
Mean bond enthalpy (kJmol-1)
Break (endothermic)
4 × C-H = 4 × 435 = 1740
2 × O=O = 2 × 498 = 996
Total = 2736kJmol-1
805 435
Forming (exothermic)
4 × O-H = 4 × 464 = 1856
2 × C=O = 2 × 805 = 1610
Total = 3466kJmol-1
498
464
Enthalpy change
= 2736 – 3466 =
-730kJmol-1

6. Calorimetry

Friday, 10 April 2026
Calorimetry
Calorimetry is used to work out the enthalpy change of combustion
The energy from the fuel is
transferred into the water although
some is lost to the surroundings
too.
The energy transferred can be calculated
using -
q=mc∆T (see next slide)
We can go further and work out
enthalpy change too.
A fuel is burned to raise the
temperature of the water by a
specific amount. Generally we
weigh the fuel before and after
burning to work out the mass of
fuel burned.
A lid is placed on top to
prevent heat loss and wind
shields placed by the flame
to prevent a draught
moving the flame.

7. Calorimetry

Friday, 10 April 2026
Calorimetry
Energy from calorimetry experiment can be calculated
q = mcΔT
Temperature
change (K)
Heat energy lost Mass of water Specific heat
or solution (g) capacity of water
or gained (J)
(4.18Jg-1K-1)
100g of water was heated from 23°C to 57°C by
1.8g of ethanol. Calculate the energy transferred
and hence the enthalpy change of the fuel.
q = 100g × 4.18 × 34
q = 14212 J OR 14.212 kJ
Moles of Ethanol = Mass/Mr
Moles of ethanol = 1.8 / 46
Moles = 0.039
Enthalpy = q / moles
Enthalpy = -14.212 / 0.039
Enthalpy = -364.4 kJmol-1
Calculate energy
first!
Calculate moles
of ethanol to
work out
enthalpy

8. Calorimetry

Friday, 10 April 2026
Calorimetry
Energy from calorimetry experiment of solutions can be calculated
Polystyrene cup to
prevent heat loss
Add acid first – measure
temperature.
Then add alkali/solid, stir and
measure temperature change
25.0cm3 of 1 moldm-3 HCl had a temperature of
20°C. 25cm3 of 1 moldm-3 of NaOH was added
and this raised the temperature to a maximum
of 26°C.
Calculate the enthalpy of neutralisation for
hydrochloric acid.
q = 50g × 4.18 × 6
q = 1254J OR 1.254 kJ
Total mass of
both liquids
mixed
Moles of HCl = Conc × Vol
Moles of HCl = 1 × 25×10-3
Moles = 0.0250
We must convert
to dm3 by ÷1000.
We can also add
‘×10-3’
Enthalpy = q / moles
Enthalpy = -1.254 / 0.0250
Enthalpy = -50.16 kJmol-1
Calculate moles
of HCl to work
out enthalpy

9. Hess’s Law

Friday, 10 April 2026
Hess’s Law
Germain Hess came up with a law to work out enthalpy changes you can’t
find out by doing an experiment
HESS’S LAW
THE TOTAL ENTHALPY
CHANGE OF A REACTION
IS INDEPENDENT OF THE
ROUTE TAKEN

10. Hess’s Cycle - Formation

Friday, 10 April 2026
Hess’s Cycle - Formation
If you are given formation data you draw a formation cycle
SUBSTANCE
ENTHALPY OF
FORMATION (ΔfH) kJmol-1
CH3OH
-234
CO2
-394
H2O
-286
ΔcH / ΔrH
REACTANTS
ΔfH
The data is formation
KEY
so we draw a
ΔfH – Enthalpy of Formation
formation cycle
ΔcH – Enthalpy of Combustion
ΔrH – Enthalpy of reaction
Set out your cycle like
this. Balance your
equations too!
PRODUCTS
ΔfH
ELEMENTS IN
THEIR STANDARD
Remember! Substitute
STATES
the FORMATION
symbols for numbers.
Multiply by the number
of moles in the
equation

11. Hess’s Cycle - Formation

Friday, 10 April 2026
Hess’s Cycle - Formation
Go with the arrow – keep the sign the same. Go against the arrow you
change the sign
Calculate the enthalpy of combustion of
methanol burning completely in oxygen to
make carbon dioxide and water.
SUBSTANCE
ENTHALPY OF
FORMATION (ΔfH) kJmol-1
CH3OH
-234
CO2
-394
H2O
-286
To work out ΔcH we need to take an alternative
route. So we go against the arrow via A and
with the arrow via B
ΔcH
CH3OH(l) + 1.5O2(g) CO2(g) + 2H2O(l)
A
ΔfH CH3OH = -234
ΔfH O2 = 0
B
ΔfH CO2 = -394
ΔfH H2O = 2 × -286
= -572
C(s) + 2O2(g) + 2H2(g)
ΔfH TOTAL = -234
ΔfH TOTAL = -966
ΔcH = +234 - 966 = -732 kJmol-1

12. Hess’s Cycle - Combustion

Friday, 10 April 2026
Hess’s Cycle - Combustion
If you are given combustion data you draw a combustion cycle
SUBSTANCE
ENTHALPY OF
COMBUSTION (ΔCH)
kJmol-1
C
-394
H2
-286
C5H12
-3509
ΔfH / ΔrH
REACTANTS/ELEMENTS PRODUCTS
ΔcH
The data is combustion
KEY
so we draw a combustion
ΔfH – Enthalpy of Formation
cycle
ΔcH – Enthalpy of Combustion
ΔrH – Enthalpy of reaction
Set out your cycle like
this. Balance your
equations too!
COMBUSTION
PRODUCTS –
CO2 + H2O
ΔcH
Remember!
Substitute the
COMBUSTION
symbols for
numbers.
Multiply by the
number of moles
in the equation

13. Hess’s Cycle - Combustion

Friday, 10 April 2026
Hess’s Cycle - Combustion
Go with the arrow – keep the sign the same. Go against the arrow you
change the sign
Calculate the enthalpy of formation of pentane
from their elements in their standard states.
SUBSTANCE
ENTHALPY OF COMBUSTION
(ΔCH) kJmol-1
ΔfH
5C(s) + 6H2(g) + 8O2(g) C5H12(l) + 8O2(g)
C
-394
H2
-286
ΔcH C = 5×-394
= -1970
ΔcH H2 = 6×-286
= -1716
C5H12
-3509
ΔcH TOTAL = -3686
To work out ΔfH we need to take an alternative
route. So we go with the arrow via A and
against the arrow via B
A
B
ΔcH C5H12 = -3509
5CO2(g) + 6H2O(l)
ΔcH TOTAL = -3509
ΔfH = -3686 + 3509 = -177kJmol-1