# 17.9: Stoichiometric Calculations and Enthalpy Changes

**At Grade**Created by: CK-12

**Practice**Stoichiometric Calculations and Enthalpy Changes

**What will it cost?**

There is a growing concern about the damage to the environment from 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 how many) to order, studies need to be done to measure the amount of product currently produced. The 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 mol of methane released 890.4 kJ, the reaction of 2 mol of methane would release \begin{align*}2 \times 890.4 \text{ kJ} = 1781 \text{ kJ}\end{align*}. The reaction of 0.5 mol of methane would release \begin{align*}\frac{890.4 \text{ kJ}}{2}= 445.2 \text{ kJ}\end{align*}. As with other stoichiometry problems, the moles of a reactant or product can be linked to mass or volume.

#### Sample Problem: Calculating Enthalpy Changes

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

\begin{align*}2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightarrow 2\text{SO}_3(\text{g})+198 \text{ kJ}\end{align*}

Calculate the enthalpy change that occurs when 58.0 g of sulfur dioxide is reacted with excess oxygen.

*Step 1: List the known quantities and plan the problem*.

Known

- mass SO
_{2}= 58.0 g - molar mass SO
_{2}= 64.07 g/mol - \begin{align*}\Delta H = -198 \text{ kJ for the reaction of } 2 \text{ mol SO}_2 \end{align*}

Unknown

- \begin{align*}\Delta H = ? \text{ kJ} \end{align*}

The calculation requires two steps. The mass of SO_{2} is converted to moles. Then the mol SO_{2} is multiplied by the conversion factor of \begin{align*}\left(\frac{-198 \text{ kJ}}{2 \text{ mol SO}_2} \right)\end{align*}.

*Step 2: Solve*.

\begin{align*}\Delta H=58.0 \text{ g SO}_2 \times \frac{1 \text{ mol SO}_2}{64.07 \text{ g SO}_2} \times \frac{-198 \text{ kJ}}{2 \text{ mol SO}_2}=-89.6 \text{ kJ}\end{align*}

*Step 3: Think about your result*.

The mass of sulfur dioxide is slightly less than 1 mol. Since 198 kJ is released for every 2 mol of SO_{2} that reacts, the heat released when about 1 mol reacts is one half of 198. The 89.6 kJ is slightly less than half of 198. The sign of \begin{align*}\Delta H\end{align*} is negative because the reaction is exothermic.

### Summary

- Calculations of energy changes in enthalpy equations are described.

### Review

- What do you need to determine to solve enthalpy stoichiometry problems?
- If I react 1.75 moles of methane, how much energy will be involved?
- I ran a reaction producing sulfur dioxide and releasing 267.3 kJ of energy. How many moles of sulfur dioxide were involved in the reaction?

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### Image Attributions

- Perform calculations of enthalpy equations.

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