Chemistry 1011
TOPIC
Thermochemistry
TEXT REFERENCE
Masterton and Hurley Chapter 8
Chemistry 1011 Slot 5
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8.5 Enthalpies of Formation
YOU ARE EXPECTED TO BE ABLE TO:
•
Relate DH to enthalpies of formation.
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Enthalpies of Formation
• Heats of Formation, or Enthalpies of Formation,
are used to provide a concise collection of
thermochemical data
• For consistency, enthalpy of formation data are
recorded for reactions that take place under
standard conditions
• Standard conditions are
– Constant pressure of 1 atmosphere
– Fixed temperature of 25oC
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Standard Molar Enthalpy of
Formation
• The enthalpy change when one mole of a
compound is formed from its elements at
1 atm and 25oC
• The elements must be in their stable states
at this pressure and temperature
1/ N
2 2(g)
+ 3/2H2(g)  NH3(g); DHof = -46.1kJ
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Notes About DHof
• Enthalpies of formation are listed in most data
books and are shown in Table 8.3 of the text (page
221)
• Most enthalpies of formation are negative. This
means that the formation of a compound from its
elements is exothermic
• There are no entries in a table of enthalpies of
formation for elemental species such as Br2, O2,
N2, etc. DHof for an element is zero
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Using Heats of Formation
• Enthalpies of formation can be used to
determine the Standard Enthalpy Change
for a reaction, DHo
• One way is to apply Hess’ Law:
• Use DHof for CaCO3, CaO and CO2 to
determine the standard heat of reaction for
CaCO3(s)  CaO(s) + CO2(g)
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1. Ca(s) + C(s) + 3/2O2(g)  CaCO3(s) ; DH = -1206.9kJ
2. Ca(s) + 1/2O2(g)  CaO(s) ; DH = -635.1kJ
3. C(s) + O2(g)  CO2(g) ; DH = -393.5kJ
–
–
–
Reverse equation 1
Add equation 2
Add equation 3
CaCO3(s)  Ca(s) + C(s) + 3/2O2(g) ; DH = +1206.9kJ
Ca(s) + 1/2O2(g)  CaO(s) ; DH = -635.1kJ
C(s) + O2(g)  CO2(g) ; DH = -393.5kJ
CaCO3(s)  CaO(s) + CO2(g); DH = +178.3kJ
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An Alternative Process
• The standard enthalpy change for a reaction is
equal to the sum of the standard enthalpies of
formation of the products minus the sum of the
standard enthalpies of formation of the reactants
DHo = SDHof products – SDHof reactants
• Coefficients of reactants and products in the
balanced equation for the reaction must be taken
into account
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Textbook Example 8.8
• Calculate DHo for the combustion of one
mole of propane, C3H8
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
DHo = [3 DHof CO2(g) + 4 DHof H2O(l)] - [DHof C3H8(g) + 5 DHof O2(g)]
DHo = [3 x -393.5 + 4 x -285.8] – [1 x –103.8 + 5 x 0]
DHo = -2219.9kJ
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Enthalpies of Formation of Ions in
Solution
• It is possible to construct a table to show the
heats of formation of ions in solution
• However, all ionic solution processes
involve both a + and a – ion
HCl(g)  H+(aq) + Cl-(aq)
• To get around this problem, DHof for the H+(aq)
ion is set at zero
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Determining Enthalpies of Formation
of Ions in Solution
• When HCl(g) is added to water
HCl(g)  H+(aq) + Cl-(aq); DHo = -74.9kJ
• Using the summation method:
-74.9kJ = [DHof H+(aq) + DHof Cl-(aq)] - [DHof HCl(g) ]
-74.9kJ = 0 + DHof Cl-(aq) + 92.3kJ (from Table 8.3)
DHof Cl-(aq) = -74.9kJ - 92.3kJ = -167.2kJ
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Using Enthalpies of Formation of Ions
in Solution
• Example 8.10
• Calculate DHo for the reaction:
2H+(aq) + CO32-(aq)  CO2(g) + H2O(l)
• Use the summation method
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