The total enthaply change in a reaction can be found from the enthalpies of formation of the reactants and products. Their relationship is simple: the total change is just the sum of the enthalpies of the products minus the sum of the enthalpies of the reactants.

DH^o = S DH_f^o products - S DH_f^o reactants

(The S means "sum over") It is necessary to take into account the coefficients in front of the products and reactants in the balanced equation: If you have the reaction

aA + bB -> cC + dD

then the sum of the enthalpies of the products (S DH_f^o products) and the sum of the enthalpies of the reactants (S DH_f^o reactants) are

S DH_f^o products = cDH_f^oC + dDH_f^oD

S DH_f^o reactants = aDH_f^oA + bDH_f^oB

Example 1: Hydrogen gas is burned in oxygen gas to form water vapor at 25^oC. What is the change in enthalpy in this reaction?

Solution: First, write down the balanced equation for the reaction

2H₂(g) + O₂(g) -> 2H₂O(g)

Next, look up the enthalpies of formation of the products and reactants

Substance	DHfo
H2(g)	0.00 kJ/mol*
O2(g)	0.00 kJ/mol*
H2O(g)	-241.8 kJ/mol

*Remember, by the definition of standard molar enthalpy of formation, it is zero of any element in its most stable state at 1 atm and 25^oC.

S DH_f^o reactants = 2*DH_f^oH₂ + DH_f^oO₂ = 0.00 kJ/mol
S DH_f^o products = 2*DH_f^oH₂O = 2*(-241.8) = -483.6 kJ/mol

DH^o = S DH_f^o products - S DH_f^o reactants = -483.6 - 0 = -483.6 kJ

The final thermochemical reaction is

2H₂(g) + O₂(g) -> 2H₂O(g)      DH = -483.6 kJ

Solution: First, write down the balanced equation for the reaction

2CH₃OH(l) + 3O₂(g) -> 4H₂O(g) + 2CO₂(g)

Next, look up the enthalpies of formation of the products and reactants

Substance	DHfo
CH3OH(l)	-238.7 kJ/mol
O2(g)	0.00 kJ/mol
CO2(g)	-393.5 kJ/mol
H2O(g)	-241.8 kJ/mol

S DH_f^o reactants = 2*DH_f^oCH₃OH + DH_f^oO₂ = 2*(-238.7) + 3*0.00 = -477.4 kJ/mol
S DH_f^o products = 4*DH_f^oH₂O + 2*DH_f^oCO₂= 4*(-241.8) +2*(-393.5)= -1754.2 kJ/mol

DH^o = S DH_f^o products - S DH_f^o reactants = -1754.2 - (-477.4) = -1276.8 kJ/mol

The final thermochemical reaction is

2CH₃OH(l) + 3O₂(g) -> 4H₂O(g) + 2CO₂(g)      DH = -1276.8 kJ

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