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# Problems involving Heat of Reaction

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### Example

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