Hess suggested that the sum of the enthalpies (?H) of the steps of a reaction will equal the enthalpy of the overall reaction. - PowerPoint PPT Presentation

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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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Hess s Law (4-4) Hess suggested that the sum of the enthalpies ( H) of the steps of a reaction will equal the enthalpy of the overall reaction. – PowerPoint PPT presentation

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Title: Hess suggested that the sum of the enthalpies (?H) of the steps of a reaction will equal the enthalpy of the overall reaction.


1
Hesss Law (4-4)
  • Hess suggested that the sum of the enthalpies
    (?H) of the steps of a reaction will equal the
    enthalpy of the overall reaction.

2
Why Does It Work?
  • If you turn an equation around, you change the
    sign
  • If H2(g) 1/2 O2(g) H2O(g) DH-241.8 kJ
  • then, H2O(g) H2(g) 1/2 O2(g) DH
    241.8 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 483.6 kJ

3
  • Find the heat of formation (?H) of
  • 2 Al (s) 3 CuO (s) ? 3 Cu (s) 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 (s) ?H -157.3kJ
  • 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 (s)?3Cu(s) 3/2 O2 (g) ?H(3)157.3kJ
  • ___________________________________
  • 2 Al (s) 3 CuO (s) ? 3 Cu (s) Al2O3 (s)
  • ?H - 1204.1 kJ
  • ( oxygen can be cancelled because it exists on
    both sides of the reactions)

4
  • 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.

5
  • 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.

6
  • 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

7
  • 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)

8
  • We can manipulate the equations by
  • a) Reversing equation 1
  • b) Multiplying equation 2 by 4
  • c) Multiplying equation 3 by 3

9
  • ?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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