FIRST
Check to see if contributing equations add to give the desired equation.
If so, no adjustments are necessary and the enthalpies can simply be added
Example
SECOND (if contributing equations need adjustment before combining)
Identify contributing equations having reactants/products in the overall equation.
Reverse contributing equations where substances in the overall equation on the wrong side of the equation.
Example continued
THIRD
Change the sign of the enthalpy change for any contributing equations that have been reversed.
Example continued
FOURTH
Multiply contributing equations having reactants/products in the overall equation by factors to make the coefficients on these reactants/products the same as they are in the overall equation.
Example continued
FIFTH
Multiply the enthalpy changes for contributing equations by the same factor.
Example nearly finished
SIXTH
Check that equations sum to desired equation and add the enthalpy changes.
Check to see if contributing equations add to give the desired equation.
If so, no adjustments are necessary and the enthalpies can simply be added
Example
desired:
S(s) + 3/2 O2(g)
SO3(g)
Δ3H° = Δ1H° + Δ2H°
contributing:
(1) S(s) + O2(g) SO2(g)
ΔH° = Δ1H°
(2)SO2(g) + ½O2(g) SO3(g)
ΔH° = Δ2H°
S(s) + 3/2 O2(g)
SO3(g)Δ3H° = Δ1H° + Δ2H°
contributing:
(1) S(s) + O2(g)
ΔH° = Δ1H°
(2)
SO3(g)ΔH° = Δ2H°
Identify contributing equations having reactants/products in the overall equation.
Reverse contributing equations where substances in the overall equation on the wrong side of the equation.
Example continued
desired:
Fe2O3(s) + CO(g) 2FeO(s) + CO2(g)
ΔrH = ??? kJ mol–1
contributing:
(1) Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
ΔrH = -23.4 kJ mol–1 as is
(2) FeO(s) + CO(g) Fe(s) + CO2(g)
ΔrH = -10.9 kJ mol–1 reverse
Fe2O3(s) + CO(g)
2FeO(s) + CO2(g)ΔrH = ??? kJ mol–1
contributing:
(1) Fe2O3(s) + 3CO(g)
2Fe(s) + 3CO2(g)ΔrH = -23.4 kJ mol–1 as is
(2) FeO(s) + CO(g)
Fe(s) + CO2(g)ΔrH = -10.9 kJ mol–1 reverse
THIRD
Change the sign of the enthalpy change for any contributing equations that have been reversed.
Example continued
desired:
Fe2O3(s) + CO(g) 2FeO(s) + CO2(g)
ΔrH = ??? kJ mol–1
contributing:
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
ΔrH = -23.4 kJ mol–1
Fe(s) + CO2(g) FeO(s) + CO(g)
Fe2O3(s) + CO(g)
2FeO(s) + CO2(g)ΔrH = ??? kJ mol–1
contributing:
Fe2O3(s) + 3CO(g)
2Fe(s) + 3CO2(g)ΔrH = -23.4 kJ mol–1
Fe(s) + CO2(g)
FeO(s) + CO(g)ΔrH = +10.9 kJ mol–1
FOURTH
Multiply contributing equations having reactants/products in the overall equation by factors to make the coefficients on these reactants/products the same as they are in the overall equation.
Example continued
desired:
Fe2O3(s) + CO(g) 2FeO(s) + CO2(g)
ΔrH = ??? kJ mol–1
contributing:
Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)
ΔrH = -23.4 kJ mol–1
2Fe(s) + 2CO2(g) 2FeO(s) + 2CO(g)
Fe2O3(s) + CO(g)
2FeO(s) + CO2(g)ΔrH = ??? kJ mol–1
contributing:
Fe2O3(s) + 3CO(g)
2Fe(s) + 3CO2(g)ΔrH = -23.4 kJ mol–1
2Fe(s) + 2CO2(g)
2FeO(s) + 2CO(g)ΔrH = +10.9 kJ mol–1 times 2
FIFTH
Multiply the enthalpy changes for contributing equations by the same factor.
Example nearly finished
desired:
Fe2O3(s) + CO(g)2FeO(s) + CO2(g)
ΔrH = -1.6 kJ mol–1
contributing:
Fe2O3(s) +3CO(g) 2Fe(s) + 3CO2(g)
ΔrH = -23.4 kJ mol–1
2Fe(s) + 2CO2(g) 2FeO(s) + 2CO(g)
ΔrH = +21.8 kJ mol–1
Fe2O3(s) + CO(g)
2FeO(s) + CO2(g)ΔrH = -1.6 kJ mol–1
contributing:
Fe2O3(s) +
ΔrH = -23.4 kJ mol–1
2FeO(s) + ΔrH = +21.8 kJ mol–1
SIXTH
Check that equations sum to desired equation and add the enthalpy changes.