5.1 Exothermic and endothermic reactions
If a reaction produces heat (increases the
temperature of the surroundings) it is exothermic.
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If the temperature of the reaction mixture
decreases (ie heat is absorbed) then the reaction is
endothermic.
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Exothermic -> a reaction which produces heat
(ΔH has a negative value by convention, -ve)
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Endothermic -> a reaction which absorbs heat
(ΔH has a positive value by convention, +ve)
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Enthalpy of reaction, ΔH:
The change in internal (chemical) energy (H)
in a reaction
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The most stable state is where all energy has
been released. When going to a more stable state,
energy will be released, and when going to a less
stable state, energy will be gained (from the
surroundings).
On an enthalpy level diagram, higher positions
will be less stable (with more internal energy)
therefore, if the product is lower, heat is released
(more stable, ΔH is -ve) but if it is higher, heat is
gained (less stable, ΔH is +ve).
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Formation of bonds causes an
energy release (exothermic).
Breaking of bonds requires
energy (endothermic).
Calculation of enthalpy changes

Change in energy = mass x specific
heat capacity x change in
temperature
E = m x c x ΔT
where m is the mass of water present
(kilograms), and c = 4.18 kJ Kg-1 K1.
Hess' Law
Hess' Law states that the total
enthalpy change between given
reactants and products is that same
regardless of any intermediate steps
(or the reaction pathway).
Example: Calculate the enthalpy of
formation of methane.
Example 1:
Data:
ΔH combustion (carbon) = -394kj
ΔH combustion (hydrogen) =-242kj
ΔHcombustion (methane) =-891kj
If ΔHf methane is represented by the equation:
C(s) + H2(g) → CH4 (g)
This equation can be constructed using the equations for combustion
of the reactants (carbon and hydrogen) and the product (methane)
C(s) + O2(g) → CO2(g)
-394k…..enthalpy 1
H2(g) + ½ O2(g) → H2O(l)
multiply this equation by 2 to get
2H2(g) + O2(g) → 2H2O(l)
-242kj…..enthalpy 2
-484kj …….enthalpy 3
Add the first and third equation together to get:
C(s) +2O2(g) +2H2(g) →CO2(g) + 2H2O(l) -878kj
enthalpy (1 + 3) = 4
Now take away the eq for the combustion of methane
CH4 (g) + 2O2(g) → CO2(g) + 2H2O(l)
-891kj
enthalpy 5
And after rearrangement ( take the CH4 to the right hand side) the
result is the equation for the formation of methane
C(s) + H2(g) → CH4 (g)
+13kj
enthalpy ( 4 - 5)
Example 1
Monoclinic sulphur is formed in volcanic regions by reaction between sulphur
dioxide and hydrogen sulphide according to the equation:
SO2 + 2H2S --> 2H2O + 3S
Draw an enthalpy diagram or cycle and calculate the standard enthalpy change for this
reaction.
Here are some values.
Standard enthalpy of formation (all in kJ)
H2O(l)
-286
H2S(g) - 20.2
Standard enthalpy of combustion of S (monoclinic) = -297.2kJ
equation 1:
equation 2:
equation 3:
H2 + ½ O2 --> H2O
H2 + S --> H2S
S + O2 --> SO2
-286 kJ
-20.2kJ
-297.2 kJ
multiply E2 by 2… (equation 4) H2 + 2S --> 2H2S
kJ
add equation 3 & 4….….. 2H2 + 3S + O2 --> SO2 + 2H2S
multiply E1 by 3…(equation 5) 2H2 + O2 --> 2H2O
subtract 4 from 5……-3S ---> 2H2O - SO2 - 2H2S
rearrange
SO2 + 2H2S --> 2H2O + 3S
-234.4 kJ
-40.4
-337.6 kJ
-572 kJ
-234.4 kJ
Born Haber Cycles
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Born Haber cycles are the application of
Hess' law to ionic systems. An ionic solid
consists of a giant structure of ions held
together in a giant lattice.
Application of Hess law tells us that the
enthalpy of formation of an ionic crystal is
equal to the sum of the energies of
formation of the ions plus the enthalpy of
the lattice.
it is a several step process that is best
represented by a diagram showing the
individual steps as endothermic
upwards and exothermic downwards.