Hess

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Hesss Law
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• Hess suggested that the sum of the enthalpies
  (?H) of the steps of a reaction will equal the
  enthalpy of the overall reaction.

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Why Does It Work?

• If you turn an equation around, you change the
  sign
• If H2(g) 1/2 O2(g) H2O(g) DH-285.5 kJ
• then, H2O(g) H2(g) 1/2 O2(g) DH
  285.5 kJ
• also,
• If you multiply the equation by a number, you
  multiply the heat by that number
• 2 H2O(g) 2 H2(g) O2(g) DH 571.0 kJ

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• How do you get good at this?

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• Find the heat of formation (?H) of
• Al (s) 3 CuO (g) ? 3 Cu (g) Al2O3 (s)
• Using two equations from the sheet
• 2Al (s) 3/2 O2 (g) ? Al2O3 (s) ?H-1676kJ
• Cu(s) 1/2 O2 (g) ? CuO (g) ?H -155kJ

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• Need to flip the second equation and change the
  sign on ?H and multiply it by 3
• 2Al (s) 3/2 O2 (g) ? Al2O3 (s) ?H-1676kJ
• 3CuO (g)?3Cu(s) 3/2 O2 (g) ?H(3)155kJ
• ___________________________________
• Al (s) 3 CuO (g) ? 3 Cu (g) Al2O3 (s)
• ?H - 1211 kJ
• ( oxygen can be cancelled because it exists on
  both sides of the reactions)

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• Calculating Heats of Reaction using Hess's Law
• 1) Write the overall equation for the reaction if
  not given.
• 2) Manipulate the given equations for the steps
  of the reaction so they add up to the overall
  equation.
• 3) Add up the equations canceling common
  substances in reactant and product.
• 4) Add up the heats of the steps heat of
  overall reaction.

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• H2O(g) C(s) ? CO(g) H2(g) ?H?
• These are the equations chosen from the sheet
• 1) H2(g) 1/2O2(g) ?H2O(g) ?H -242.kJ
• 2) C(s) 1/2 O2(g) ? CO(g) ?H -110. kJ
• Note that the H2O (g) is a reactant and in step
  1 it is a product. Thus step one needs to be
  reversed and the sign of the heat.

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• H2O(g) ? H2(g) ½ O2(g) ?H 242.kJ
• C(s) ½ O2(g) ? CO(g) ?H -110. kJ
• _________________________________
• H2O(g) C(s) ? CO(g) H2(g) ?H132 kJ

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• Calculate the heat of reaction for the following
  equation
• C3H8(g) 5 O2 (g) ---gt 3 CO2(g)4H2O (g)
• given the following steps in the reaction
  mechanism.
• 3C (s) 4 H2 (g) -------gt C3H8 (g)
• H2 (g) ½ O2 (g) -------gt H2O (g)
• C (s) O2 (g) --------gt CO2 (g)

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• We can manipulate the equations by
• a) Reversing equation 1
• b) Multiplying equation 2 by 2
• c) Multiplying equation 3 by 3

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• ?H
• C3H8 (g) -------gt 3C (s) 4 H2 (g) 103.8
• 4 H2 (g) 2O2 (g) -----gt 4H2O (g) -967.2
• 3C (s) 3O2 (g) -----gt 3CO2 (g) -1180.5
• C3H8(g) 5 O2 (g) ---gt 3 CO2(g)4H2O (g)
• ?H - 2043.9 kJ

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Standard Heats of Formation

• The DH for a reaction that produces 1 mol of a
  compound from its elements at standard
  conditions
• Standard conditions 25C and 1 atm.
• Symbol is

• The standard heat of formation of an element 0
• This includes the diatomics

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What good are they?

• The heat of a reaction can be calculated by
• subtracting the heats of formation of the
  reactants from the products

DHo
(
Products) -
(
Reactants)
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Examples

• CH4(g) 2 O2(g) CO2(g) 2 H2O(g)

• DH -393.5 2(-241.8) - -74.68 2 (0)
• DH - 802.4 kJ