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Take this reaction, the combustion of acetylene:
2C2H2(g) + 5O2(g) —> 4CO2(g) + 2H2O(g)
1) The first step is to make sure that the equation is balanced and correct. Remember, the combustion of a hydrocarbon requires oxygen and results in the production of carbon dioxide and water.
2) Next, locate a table of Standard Enthalpies of Formation to look up the values for the components of the reaction (Table 7.2, Petrucci Text)
3) First find the enthalpies of the products:
ΔHºf CO2 = -393.5 kJ/mole
Multiply this value by the stoichiometric coefficient, which in this case is equal to 4 mole.
vpΔHºf CO2 = 4 mol (-393.5 kJ/mole)
= -1574 kJ
ΔHºf H2O = -241.8 kJ/mole
The stoichiometric coefficient of this compound is equal to 2 mole. So,
vpΔHºf H2O = 2 mol ( -241.8 kJ/mole)
= -483.6 kJ
Now add these two values in order to get the sum of the products
Sum of products (Σ vpΔHºf(products)) = (-1574 kJ) + (-483.6 kJ) = -2057.6 kJ
Now, find the enthalpies of the reactants:
ΔHºf C2H2 = +227 kJ/mole
Multiply this value by the stoichiometric coefficient, which in this case is equal to 2 mole.
vpΔHºf C2H2 = 2 mol (+227 kJ/mole)
= +454 kJ
ΔHºf O2 = 0.00 kJ/mole
The stoichiometric coefficient of this compound is equal to 5 mole. So,
vpΔHºf O2 = 5 mol ( 0.00 kJ/mole)
= 0.00 kJ
Add these two values in order to get the sum of the reactants
Sum of reactants (Δ vrΔHºf(reactants)) = (+454 kJ) + (0.00 kJ) = +454 kJ
The sum of the reactants and products can now be inserted into the formula:
ΔHº = Δ vpΔHºf(products) - ? vrΔHºf(reactants)
= -2057.6 kJ - +454 kJ
= -2511.6 kJ