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# 17.14: Heat of Combustion

Difficulty Level: At Grade Created by: CK-12
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Practice Heat of Combustion

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#### What is gasohol?

In efforts to reduce gas consumption from oil, ethanol is often added to regular gasoline.  It has a high octane rating and burns more slowly than regular gas.  This “gasohol” is widely used in many countries.  It produces somewhat lower carbon monoxide and carbon dioxide emissions, but does increase air pollution from other materials.

### Molar Heat of Combustion

Many chemical reactions are combustion reactions.  It is often important to know the energy produced in such a reaction so we can determine which fuel might be the most efficient for a given purpose. The molar heat of combustion (He) is the heat released when one mole of a substance is completely burned.

Typical combustion reactions involve the reaction of a carbon-containing material with oxygen to form carbon dioxide and water as products.  If methanol is burned in air, we have:

CH3OH+O2CO2+2H2OHe=890 kJ/mol\begin{align*}\text{CH}_3\text{OH} + \text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \quad \text{He}= 890 \text{ kJ/mol}\end{align*}

In this case, one mole of oxygen reacts with one mole of methanol to form one mole of carbon dioxide and two moles of water.

It should be noted that inorganic substances can also undergo a form of combustion reaction:

2Mg+O22 MgO\begin{align*}2\text{Mg} + \text{O}_2 \rightarrow 2 \ \text{MgO}\end{align*}

In this case there is no water and no carbon dioxide formed.  These reactions are generally not what we would be talking about when we discuss combustion reactions.

#### Sample Problem: Calculation of Heat of Combustion

Heats of combustion are usually determined by burning a known amount of the material in a bomb calorimeter with an excess of oxygen.  By measuring  the temperature change, the heat of combustion can be determined.

A 1.55 gram sample of ethanol is burned and produced a temperature increase of 55°C in 200 grams of water.  Calculate the molar heat of combustion.

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

Known

• mass of ethanol = 1.55 grams
• molar mass of ethanol = 46.1 g/mol
• mass of water = 200 grams
• cp water:4.18 J/gC\begin{align*}c_{p} \ \text{water}: 4.18 \text{ J/g}^\circ \text{C}\end{align*}
• temperature increase = 55°C

Unknown

• He of ethanol

Step 2: Solve.

amount of ethanol used: 1.55 g46.1 g/mol=0.0336 moles\begin{align*}\frac{1.55 \text{ g}}{46.1 \text{ g/mol}} = 0.0336 \ \text{moles}\end{align*}

energy generated: 4.184 J/gC×200 g×55C=46024 J=46.024 kJ\begin{align*}4.184 \text{ J/g}^\circ \text{C} \times 200 \text{ g} \times 55^\circ \text{C} = 46024 \text{ J} = 46.024 \text{ kJ}\end{align*}

molar heat of combustion: 46.024 kJ0.0336 moles=1369 kJ/mol\begin{align*}\frac{46.024 \text{ kJ}}{0.0336 \text{ moles}}=1369 \text{ kJ/mol}\end{align*}

The burning of ethanol produces a significant amount of heat.

### Summary

• The molar heat of combustion is defined.
• Calculations using the molar heat of combustion are described.

### Review

1. Write the reaction for the combustion of ethanol in oxygen?
2. What would be the limiting reagent in the reaction?
3. If you made an error in weighing the ethanol and added less than you planned on, would the heat of combustion result be higher or lower?

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