Standard Molar Enthalpies
of Formation
ΔHf
o
Focus Questions
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1) What are formation reactions?
2) What is standard molar enthalpy of
formation?
3) How do we write a formation equation?
4) How do we calculate the ∆H using the
standard molar enthalpy of formations?
5) How does this method relate to Hess’s
Law?
Formation Reactions
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In a formation reaction, a substance is
formed from elements in their standard states.
From what elements is water formed?
H2(g) + ½O2 (g)  H2O (l) ΔHfo = -285.8 kJ
The enthalpy change of a formation reaction is
called the standard molar enthalpy of
formation, ∆H˚f.
Definition of ∆H˚f
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The standard molar enthalpy of
formation is the quantity of energy that
is absorbed or released when one mole
of a compound is formed directly from
its elements in their standard states.
Formation Equations and ∆H˚f
Note
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The standard enthalpies of formation of
most compounds are negative.
By definition, the enthalpy of formation
of an element in its standard state is
zero
Writing a formation equation
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Always write the elements in their
standard state (l, g, or s).
A formation equation should also show
the formation of exactly one mole of
the compound of interest.
Calculating Enthalpy Changes
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You can calculate the enthalpy change of a
chemical reaction by adding the heats of
formation of the products and
subtracting the heats of formation of
the reactants.
∆H˚ = Σ(n∆H˚f products) - Σ(n∆H˚f reactants)
Note
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As usual, you need to begin with a
balanced chemical equation.
If a given reactant or product has a
molar coefficient that is not 1, you need
to multiply its ∆H˚f by the same molar
coefficient.
Try Solving using Standard enthalpy of formation equation…
CH4(g) + 2O2(g) → C02(g) + 2H2O(g)
∆H˚ = [(n∆H˚f of C02(g) ) + 2(n∆H˚f of H2O(g) )] – [(n∆H˚f of CH4(g)) +
2(n∆H˚f of O2(g) )]
∆H˚ = [(-393.5 kJ/mol) + 2(-241.8 kJ/mol )] – [(-74.4 kJ/mol) +
2(0 kJ/mol)]
= -802.7 kJ/mol of CH4
How does this method of adding
heats of formation relate to Hess’s
law? Solve previous question using Hess’s Law…
(1) H2(g) + ½ O2(g) → H2O2(g)
∆H˚f= -241.8 kJ
(2)
C(s) + O2(g) → CO2(g)
∆H˚ f = -393.5 kJ
(3)
C(s) + 2H2(g) → CH4(s)
∆H˚ f =-74.4 kJ
2 x (1) 2H2(g) + O2(g) → 2H2O2(g)
∆H˚f= 2(-241.8) kJ
(2) C(s) + O2(g) → CO2(g)
∆H˚ f = -393.5 kJ
-1 x (3)
CH4(s) → C(s) + 2H2(g)
CH4(g) + 2O2(g) → 2H2O(g) + CO2(g)
∆H˚ f =-1(-74.4) kJ
∆H˚ f = -802.7 kJ
Note
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It is important to realize that, in most
reactions, the reactants do not actually break
down into their elements and then react to
form products.
Since there is extensive data about enthalpies
of formation, however, it is useful to calculate
the overall enthalpy change this way.
Challenging Question:
When octane burns in a car engine, heat
is released to the air and to the metal of
the engine, but a significant portion is
absorb by the liquid in the cooling
system-an aqueous solution of ethylene
glycol. What mass of octane is completely
burned to cause the heating of 20kg of
ethylene glycol from 10oC to 70oC?
Assume water is produced as a gas and
that all the heat flows into the coolant. (
ethylene glycol 3.5J/goC)