# Problems involving Heat of Reaction

### Example

Take this reaction, the combustion of acetylene:

**2C _{2}H_{2}(g) + 5O_{2}(g) —> 4CO_{2}(g) + 2H_{2}O(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:

ΔHº_{f} CO_{2} = -393.5 kJ/mole

Multiply this value by the stoichiometric coefficient, which in this case is equal to 4 mole.

v_{p}ΔH^{º}_{f} CO_{2} = 4 mol (-393.5 kJ/mole)

= -1574 kJ

ΔH^{º}_{f} H_{2}O = -241.8 kJ/mole

The stoichiometric coefficient of this compound is equal to 2 mole. So,

v_{p}ΔH^{º}_{f} H_{2}O = 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 v_{p}ΔHº_{f}(products)) = (-1574 kJ) + (-483.6 kJ) = -2057.6 kJ

**Now, find the enthalpies of the reactants:**

ΔHº_{f} C_{2}H_{2} = +227 kJ/mole

Multiply this value by the stoichiometric coefficient, which in this case is equal to 2 mole.

v_{p}ΔHº_{f} C_{2}H_{2} = 2 mol (+227 kJ/mole)

= +454 kJ

ΔHº_{f} O_{2} = 0.00 kJ/mole

The stoichiometric coefficient of this compound is equal to 5 mole. So,

v_{p}ΔHº_{f} O_{2} = 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 v_{r}ΔHº_{f}(reactants)) = (+454 kJ) + (0.00 kJ) = +454 kJ

**The sum of the reactants and products can now be inserted into the formula:**

ΔHº = Δ v_{p}ΔHº_{f}(products) - ? v_{r}ΔHº_{f}(reactants)

= -2057.6 kJ - +454 kJ

= -2511.6 kJ