# 4.9: Stoichiometric Calculations and Enthalpy Changes


There is a growing concern about damage to the environment done by emissions from manufacturing plants. Many companies are taking steps to reduce these harmful emissions by adding equipment that will trap the pollutants. In order to know what equipment (and the quantity) to order, studies are done to measure the amount of product currently produced. Since pollution is often both particulate and thermal, energy changes need to be determined in addition to the amounts of products released.

## Stoichiometric Calculations and Enthalpy Changes

Chemistry problems that involve enthalpy changes can be solved by techniques similar to stoichiometry problems. Refer again to the combustion reaction of methane. Since the reaction of $$1 \: \text{mol}$$ of methane released $$890.4 \: \text{kJ}$$, the reaction of $$2 \: \text{mol}$$ of methane would release $$2 \times 890.4 \: \text{kJ} = 1781 \: \text{kJ}$$. The reaction of $$0.5 \: \text{mol}$$ of methane would release $$\frac{890,4 \: \text{kJ}}{2} = 445.2 \: \text{kJ}$$. As with other stoichiometry problems, the moles of a reactant or product can be linked to mass or volume.

##### Example $$\PageIndex{1}$$

Sulfur dioxide gas reacts with oxygen to form sulfur trioxide in an exothermic reaction, according to the following thermochemical equation.

$2 \ce{SO_2} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{SO_3} \left( g \right) + 198 \: \text{kJ}\nonumber$

Calculate the enthalpy change that occurs when $$58.0 \: \text{g}$$ of sulfur dioxide is reacted with excess oxygen.

###### Step 1: List the known quantities and plan the problem.
• Mass $$\ce{SO_2} = 58.0 \: \text{g}$$
• Molar mass $$\ce{SO_2} = 64.07 \: \text{g/mol}$$
• $$\Delta H = -198 \: \text{kJ}$$ for the reaction of $$2 \: \text{mol} \: \ce{SO_2}$$
###### Unknown

The calculation requires two steps. The mass of $$\ce{SO_2}$$ is converted to moles. Then the moles of $$\ce{SO_2}$$ is multiplied by the conversion factor of $$\left( \frac{-198 \: \text{kJ}}{2 \: \text{mol} \: \ce{SO_2}} \right)$$.

###### Step 2: Solve.

$\Delta H = 58.0 \: \text{g} \: \ce{SO_2} \times \frac{1 \: \text{mol} \: \ce{SO_2}}{64.07 \: \text{g} \: \ce{SO_2}} \times \frac{-198 \: \text{kJ}}{2 \: \text{mol} \: \ce{SO_2}} = 89.6 \: \text{kJ}\nonumber$

The mass of sulfur dioxide is slightly less than $$1 \: \text{mol}$$. Since $$198 \: \text{kJ}$$ is released for every $$2 \: \text{mol}$$ of $$\ce{SO_2}$$ that reacts, the heat released when about $$1 \: \text{mol}$$ reacts is one half of 198. The $$89.6 \: \text{kJ}$$ is slightly less than half of 198. The sign of $$\Delta H$$ is negative because the reaction is exothermic.