Hess's Law - Example

Example

Given:

  • B2O3 (s) + 3H2O (g) → 3O2 (g) + B2H6 (g) (ΔH = 2035 kJ/mol)
  • H2O (l) → H2O (g) (ΔH = 44 kJ/mol)
  • H2 (g) + (1/2)O2 (g) → H2O (l) (ΔH = -286 kJ/mol)
  • 2B (s) + 3H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol)

Find the ΔHf of:

  • 2B (s) + (3/2) O2 (g) → B2O3 (s)

After the multiplication and reversing of the equations (and their enthalpy changes), the result is:

  • B2H6 (g) + 3O2 (g) → B2O3 (s) + 3H2O (g) (ΔH = -2035 kJ/mol)
  • 3H2O (g) → 3H2O (l) (ΔH = -132 kJ/mol)
  • 3H2O (l) → 3H2 (g) + (3/2) O2 (g) (ΔH = 858 kJ/mol)
  • 2B (s) + 3H2 (g) → B2H6 (g) (ΔH = 36 kJ/mol)

Adding these equations and canceling out the common terms on both sides, we get

  • 2B (s) + (3/2) O2 (g) → B2O3 (s) (ΔH = -1273 kJ/mol)

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