Thermochemistry

1
Thermochemistry

• Lesson 4
• Hesss Law

2
Enthalpy in Chemical Reactions

• From experimental evidence we have determined
  that the change of enthalpy in chemical reactions
  is independent of the path taken.
• This means that whether a reaction occurs in one
  step or five, if the products are the same in the
  end, the enthalpy change will be the same overall.

3
Nitrogen Dioxide Example

• For example, we have determined that
• N2 (g) 2 O2 (g) ? 2 NO2 (g) ?H 68 kJ.
• But we can do this reaction in two steps as well
• N2 (g) O2 (g) ? 2 NO (g) ?H 180 kJ
• 2 NO (g) O2 (g) ? 2 NO2 (g) ?H -112 kJ
• If we add these two enthalpies together
• (?H 180 kJ ?H -112 kJ) we get the same
  value of 68 kJ.

4
Potential Energy Diagram
5
Hesss Law

• The enthalpy change for the conversion of
  reactants to products is the same whether the
  conversion occurs in one step or several steps.
• Rules for using Hesss Law
• If you reverse a chemical reaction, you must also
  reverse the sign of ?H.
• The magnitude of ?H id directly proportional to
  the number of moles of reactants and products in
  a reaction. If the coefficients in a balanced
  equation are multiplied by a factor, the value of
  ?H is multiplied by the same factor.

6
Example 1

• Graphite and diamond are two forms of solid
  carbon. Use the data below to calculate the
  enthalpy change for the conversion of graphite
  into diamond. Cgraphite (s) ? Cdiamond (s)
• GIVEN
• Cgraphite (s) O2 (g) ? CO2 (g) ?H -394 kJ
• Cdiamond (s) O2 (g) ? CO2 (g) ?H -396 kJ

7
Example 2

• Ethane gas dehydrogenates to ethene and hydrogen
  
• C2H6 (g) ? C2H4 (g) H2 (g). Determine the
  change in enthalpy for this reaction using the
  following thermochemical equations
• C2H6 (g) 3.5 O2 (g) ? 2 CO2 (g) 3 H2O (l) ?H
  -1559 kJ
• C2H4 (g) 3 O2 (g) ? 2 CO2 (g) 2 H2O (l) ?H
  -1411 kJ
• 2 H2 (g) O2 (g) ? 2 H2O (l) ?H -572 kJ