Using Standard Molar
Enthalpies of Formation
SCH4U0

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



In a formation reaction, a substance is formed from elements in their standard states.
From what elements is water formed?
H
2(g)
+ ½O
2 (g)
 H
2
O
(l)
ΔH f o = -285.8 kJ

The enthalpy change of a formation reaction is called the standard molar enthalpy of
formation, ∆H˚ f
.

Definition of ∆H˚ f

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

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

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


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

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.

CH
4(g)
+ 2O
2(g)
→ C0
2(g)
+ 2H
2
O
(g)
∆H˚ = [(n∆H˚ f
of C0
2(g)
) + 2(n∆H˚ f
of H
2
O
(g)
)] – [(n∆H˚ f
of CH
4(g)
) +
2(n∆H˚ f
of O
2(g)
)]
∆H˚ = [(-393.5 kJ/mol) + 2(-241.8 kJ/mol )] – [(-74.4 kJ/mol) +
2(0 kJ/mol)]
= -802.7 kJ/mol of CH
4

How does this method of adding heats of formation relate to Hess’s law?
(1) H
2(g)
+ ½ O
2(g)
→ H
2
O
2(g)
∆H˚ f
= -241.8 kJ
(2) C
(s)
+ O
2(g)
→ CO
2(g)
∆H˚
f
= -393.5 kJ
(3) C
(s)
+ 2H
2(g)
→ CH
4(s)
∆H˚
f
=-74.4 kJ
2 x (1) 2H
2(g)
+ O
2(g)
→ 2H
2
O
2(g)
∆H˚ f
= 2(-241.8) kJ
(2) C
(s)
+ O
2(g)
→ CO
2(g)
∆H˚
f
= -393.5 kJ
-1 x (3) CH
4(s)
→ C
(s)
+ 2H
2(g)
∆H˚
f
=-1(-74.4) kJ
CH
4(g)
+ 2O
2(g)
→ 2H
2
O
(g)
+ CO
2(g)
∆H˚
f
= -802.7 kJ

Note

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.

Homework Questions
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 p.332 # 1 p. 335 # 2 -5 p. 339 # 1, 2