HESS`S LAW
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 The enthalpy changes of some reactions cannot be measured
directly.
 So instead we can use an indirect approach, which would give you
the same value
 This indirect method of working out an enthalpy change which
cannot be measured is called HESS`S LAW
HESS`S LAW – states that the total enthalpy change for a reaction
is independent of the route taken as long as the initial and final
conditions are the same
NOTE: Hess’s law is a version of the first law of thermodynamics,
which is that energy is always conserved
EXAMPLES 1
Consider the reactions shown in the diagram.
Lets imagine that DH cannot be measured directly
Instead we could measure the DH of reactions a
and b
Because the overall reaction gives the same result
we can use reactions a and b to give the value of
DH
Since both the arrows in blue are in the same
direction we can write that:
DH + b = a
∴ DH = a - b
EXAMPLE 2
Again this is another type of example.
Lets imagine that we cannot measure DH directly but
we can measure the enthalpy changes a, c and d
Then using the equation below we can work out DH:
a + DH = c + d
NOTE –the two black arrow are in the same
direction and the two blue arrows are in the same
direction
So ∴
DH = c + d - a
SUMMARY - Often Hess’s law cycles are used to measure the enthalpy change for a
reaction that cannot be measured directly by experiments. Instead alternative reactions
are carried out that can be measured experimentally.
Using Hess’s law to determine enthalpy changes from enthalpy
changes of formation.
The following thermochemical cycle can be used to determine the DHOr having
been given the DH0f values in the question
2.
1.
3.
Arrows 1 and 2 are going in the same direction ∴ we can write the expression, using
Hess`s Law:
1 + 2 = 3
i.e
𝜮 DHOf(reactants) + DHOreaction = 𝜮 DHOf(products)
So rearranging this equation
DHOreaction = 𝜮 DHOf(products) - 𝜮 DHOf(reactants)
CHECKING YOUR CALCULATION: The following equation can be
used to check your final answer:
REMEMBER//
THAT THE DHOf Of ELEMENTS = 0
Example 1: What is the enthalpy change for this reaction ?
Al2O3 + 3Mg  3MgO + 2Al
Given that:
∆Hf(MgO)= -601.7 kJ mol-1
∆Hf(Al2O3) = -1675.7 kJ mol-1
REMEMBER: THE ∆HOf of ELEMENTS IS ALWAYS ZERO
ANSWER
Al2O3 (s) + 3Mg (s)
𝛴 ΔHOf (reactants)
= -1675.7 + 0
DHOr
3MgO (s) + 2Al (s)
2.
1.
3.
𝛴 ΔHOf (products)
= 3(-601.7) + 0
2Al (s) + 3Mg (s) + 1.5O2 (g)
So,
∴1+2=3
-1675.7 + DHOr = 3(-601.7)
DHOr = 3(-601.7) + 1675.7
= -1294kJmol-1
The calculation can now be checked using the equation:
∆HOr = 3(-601.7) – (-1675.7)
= -1294kJmol-1
Example 2 - Using the following data to calculate the heat of combustion of propene.
Given that:
∆Hf C3H6(g) = +20 kJ mol-1
∆Hf CO2(g)= –394 kJ mol-1
∆Hf H2O(g)= –242 kJ mol-1
The equation for the reaction is:
C3H6 + 4½O2  3CO2 + 3H2O
ANSWER:
C3H6 + 4½O2
+20 + 0
1.
DHOr
2.
3CO2 + 3H2O
3.
3(-394) + 3(-242)
∴1+2=3
20 + DHOr = 3(-394) + 3(-242)
DHO = 3(-394) + 3(-242) -20
3C(s) + 3H2(g) + 4½O2
= -1928kJmol-1
The calculation can now be checked using the equation:
∆HOr = (3(–394) – 3(–242 )) - (+20)
= -1928kJmol-1
Using Hess’s law to determine enthalpy changes from
enthalpy changes of combustion & formation combined
The following thermochemical cycle can be used to determine the DH having
been given DH0c & DHOf values in the question
OR DHOf
2.
1.
3.
Arrows 2 and 3 are going in the same direction ∴ we can write the expression, using
Hess`s Law:
2 + 3 = 1
Example 1. Using the following combustion data to calculate the heat
of reaction:
CO (g) + 2H2 (g)  CH3OH (g)
Given:
∆HOc(CO(g)) = -283 kJ mol-1
∆HOc(H2(g))= –286 kJ mol-1
∆HOc(CH3OH(g))= –671 kJ mol-1
∆HOr
CO(g) + 2H2(g)
Σ∆HOc(reactants)
= (-283) + (-286)
CH3OH(g)
2.
3.
1.
Σ∆HO(products)
= -671
CO2(g) + 2H2O(l)
So,
∴1=2+3
(-283) + (-286) = ∆HOr + -671
∆HOr = -671 + 283 + 286
= -102kJmol-1
Example 2. Using the following combustion data to calculate the heat of formation of
propene
3C(s) + 3H2 (g)  C3H6 (g)
∆HOc (C(s)) = -393kJ mol-1
∆HOc(H2(g)) = –286 kJ mol-1
∆HOc(C3H6(g)) = –2058 kJ mol-1
3C(s) + 3H2(g)
3(-393) + 3(-286)
∆HOc
2.
1.
C3H6(g)
3.
3CO2(g) + 3H2O(l)
∴1=2+3
3(-393) + 3(-286) = ∆HOc + (-2058)
∆HOc = 3(-393) + 3(-286) + 2058
∆HOc = +21kJmol-1
-2058
BOND ENTHALPIES
To break a covalent bond energy MUST be put in. This is an ENDOTHERMIC
process
BOND DISSOCIATION ENTHALPY – Is defined as the enthalpy change
required to break one mole of a covalent bond with all species in a gaseous
state
Mean Bond energies
Definition: The Mean bond energy is
the enthalpy needed to break the
covalent bond into gaseous atoms,
averaged over different molecules
NOTE: These values are positive because energy is required to
break a bond.
The definition only applies when the substances start and end in
the gaseous state.
We use values of mean bond energies because every single bond in a compound has a
slightly different bond energy. E.g. In CH4 there are 4 CH bonds. Breaking each one
will require a different amount of energy. However, we use an average value for the
C-H bond for all hydrocarbons.
Enthalpies of reaction that have been calculates using mean bond
enthalpies are not as accurate as they might be because the values
used are averages and not the specific ones for that compound.
CALCULATING BOND ENTHALPIES
There are two methods which can be used to calculate the enthalpy
change of a reaction using bond enthalpy values:
1. THE MOST COMMON METHOD
IN GENERAL
DHr = Σ LHS ENERGIES + Σ RHS ENERGIES
REMEMBER – MAKE THE RHS NEGATIVE
2. USING A HESS`S LAW CYCLE
2.
1.
∴ Using Hess`s
law:
1=2+3
3.
EXAMPLE 1 – Using the following mean bond enthalpy data to calculate the heat of
combustion of propene
METHOD 1
6(C-H) = 6(412)
1(C-C) = 1(348)
1(C=C) = 1(612)
4½(O=O) = 4.5(496)
6(C=O) = 6(743)
6(O-H)) = 6(463)
5664
-7236
DHr = Σ LHS ENERGIES + Σ RHS ENERGIES
∆HOr = (5664) + (-7236)
= -1572kJmol-1
METHOD 2 –
DHr
2.
3.
1.
3C(g) + 6H(g) + 9O(g)
1=2+3
5664 = DH + (7236)
DH = 5664 – 7236
= -1572kJmol-1