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Standard Enthalpies of Formation

Tabulated enthalpy changes called standard enthalpy of formation (∆Hº f
) can also be used to determine enthalpies of reaction (Nelson textbook pp. 799-800)

Standard enthalpy of formation is the quantity of energy associated with the formation of one mole of a substance from its elements in their standard states
C
(s)
+ O
2(g)
 CO
2(g)
∆Hº f
= -393.5 kJ/mol
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How to write formation equations
1. Write one mole of product in the state that has been specified.
2. Write the reactant elements in their standard states.
Remember,



Most metals are monoatomic solids
(Mg
(s)
, Ca
(s)
, Fe
2(g)
, H
(s)
, Au
2(g)
)
(s)
, Na
(s)
)
Some nonmetals are diatomic gases
(N
2(g)
, O
Halogen family show a variety of states
(F
2(g)
, Cl
2(g)
, Br
2(l)
, I
2(s)
)
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3. Choose equation coefficients for the reactants to give a balanced equation yielding one mole of product.
Example 1
Write the equation for the formation of liquid water directly from its elements.
H
2(g)
+ ½ O
2(g)
 H
2
O
(l)
∆Hº f
= -285.8 kJ
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Example 2
Write the equation for the formation of solid calcium carbonate directly from its elements.
Ca
(s)
+ C
(s)
+ 3/2 O
2(g)
 CaCO
3(s)
∆Hº f
= -1206.9 kJ
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Using Standard Enthalpies of Formation

Generalization to all elements in their standard states  ∆Hº f for Elements, the standard enthalpy of formation of an element already in its standard state is zero

Thus, the ∆Hº f are all zero for Fe
(s)
, O
2(g)
, and Br
2(l)
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
We can apply Hess’s Law to predicting the energy changes for many reactions

The enthalpy change for any given equation equals the sum of the enthalpies of formation of the products MINUS the sum of the enthalpies of formation of the reactants
∆H° r
= ∑ n∆Hº f (products)
- ∑ n∆Hº f (reactants)
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Using Standard Enthalpies of Formation to
Calculate Enthalpy Changes
∆H° r
= ∑ n∆Hº f (products)
- ∑ n∆Hº f (reactants)
Determine the enthalpy change for the complete combustion of methane, CH
4(g)
CH
4(g)
+ 2O
2(g)
 CO
2(g)
+ 2H
2
O
(g)
∆H =
[1mol (∆Hº of CO
- [1mol (∆Hº f of CH
) + 2mol (∆Hº
4(g) f of H
) + 2mol (∆Hº f
2
O of O
(g)
)]
2(g)
)]
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Substitute the standard enthalpies of formation from Appendix C to get the following calculation.
∆H° r
=
[1mol (-393.5 kJ/mol) + 2mol (-241.8 kJ/mol)]
– [1mol (-74.8 kJ/mol) + 2mol (0 kJ/mol)]
= -802.3 kJ/mol of CH
4(g)
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∆H° r
= ∑ n∆Hº f (products)
- ∑ n∆Hº f (reactants)
Practice
Use standard enthalpies of formation to calculate the standard enthalpy change for the oxidation of ammonia represented by the following balanced equation:
4 NH
3(g)
+ 5 O
2(g)
 6 H
2
O
(g)
+ 4 NO
(g)
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Target Equation: CH
4(g)
+ 2O
2(g)
 CO
2(g)
+ 2H
2
O
(g)
Consider the equations for the formation of each compound that is involved in the reaction of methane with oxygen.
(1) H
2(g)
(2)
(3)
C
+ ½ O
2(g)
 H
2
O
C
(s)
(s)
+ O
+ 2H
2(g)
2(g)
(g)
 CO
2(g)
 CH
4(g)
∆Hº f
∆Hº f
∆Hº f
= -241.8 kJ
= -393.5 kJ
= -74.6 kJ
There is no equation for the formation of oxygen, because oxygen is an element in its standard state.
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p. 332 Practice UC # 1 p. 335 Practice UC # 2, 3, 4 p. 335 Correct Answers
4(a) 205.7 kJ
4(b) -41.2 kJ
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