# Hess* Law - lets-learn ```Hess’ Law
Learning Goal: I will be able to use Hess’ Law
to add enthalpy changes for the individual steps
leading to a product in order to determine an
overall enthalpy change for the reaction. I will
also be able to use standard enthalpy (ΔH)
values from tables in order to determine the
standard enthalpy change for a reaction
Review
Ex 1)When white phosphorus burns in the air, it
produces phosphorus (V) oxide.
White phosphorus can be used as a weapon as it burns fiercely and can ignite cloth, fuel, ammunition and other combustibles. It is
also a highly efficient smoke-producing agent, used in smoke grenades to mask movement, position or the origin of fire from the
enemy.
Ex 2) Carbon disulfide burns in air, producing
carbon dioxide and sulfur dioxide.
It is used as an insecticide for the fumigation of grains, nursery stock, in fresh fruit conservation and as a soil
disinfectant against insects and nematodes. It is also a solvent for fats, resins, rubber, and asphalt.
Hess’s Law of Heat of Summation
• It states that for a chemical equation that can
be written as the sum of two or more steps,
the enthalpy change for the overall equation
is the sum of the enthalpy changes for the
individual steps.
A  B + C
H = x
B + C  D
H = y
A  D
H = ?
What will the change in altitude be if you take the path on the right? Path
on the left? Is there a difference?
Enthalpy diagram illustrating Hess’s
law
Example # 1
• For example, suppose you are given the following
data:
S(s )  O 2 (g )  SO 2 (g ); H  -297 kJ
o
2SO 3 (g )  2SO 2 (g )  O 2 (g ); H  198 kJ
o
• Could you use these data to obtain the enthalpy
change for the following reaction?
2S(s )  3O 2 (g )  2SO 3 (g ); H o  ?
• If we multiply the first equation by 2 and
reverse the second equation, they will sum
together to become the third.
Techniques for Manipulating
Equations
1.Reverse an equation
• the products become the reactants, and reactants become
the products
• the sign of the ΔH value must be changed
2.Multiply each coefficient
• all coefficients in an equation are multiplied by the same
integer or fraction
• the value of ΔH must also be multiplied by the same
number
Example 1
• What is the change in enthalpy for the reaction of
2NO2(g) N2O4(g) if:
2NO2(g)N2(g) + 2O2(g)
N2(g) + 2O2(g)N2O4(g)
ΔH = -84.8kJ
ΔH = 12kJ
Standard Enthalpies of Formation
• The term standard state refers to the
standard thermodynamic conditions chosen for
substances when listing or comparing
thermodynamic data: 1 atmosphere pressure
and the specified temperature (usually 25 oC).
– The enthalpy change for a reaction in
which reactants are in their standard states
is denoted Ho (“delta H naught”).
• The standard enthalpy of formation of a substance,
denoted Hfo, is the enthalpy change for the
formation of one mole of a substance in its
standard state from its component elements in
their standard state.
– Note that the standard enthalpy of formation for a
pure element in its standard state is zero.
– i.e. O2(g), Na(s)  these things cannot be broken
down further
• The law of summation of heats of formation states
that the enthalpy of a reaction is equal to the total
formation energy of the products minus that of the
reactants.
H  
o
o
nH f (products ) 

o
mH f (reactants)
S is the mathematical symbol meaning “the sum of”, n is
the coefficients of the substances in the chemical
equation.
Example # 2
• Large quantities of ammonia are used to prepare nitric acid
according to the following equation:
4NH 3 (g )  5O 2 (g )  4NO(g )  6H 2O(g )
What is the standard enthalpy change for this reaction? (Refer to
tables for data)
Learning Check
Determine ∆H˚r for the following reaction using the enthalpies of formation
that are provided.
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
∆H˚f of C2H5OH(l): –277.6 kJ/mol
∆H˚f of CO2(g): –393.5 kJ/mol
∆H˚f of H2O(l): –285.8 kJ/mol
Homework
• Pg 324 Q2, 5, 7
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