# 22.2: Enthalpy

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

## Lesson Objectives

The student will:

- define and understand enthalpy of reaction.
- calculate the enthalpy of reaction, .
- define and understand .
- define Hess’s law.
- calculate using Hess’s law.

## Vocabulary

- enthalpy of formation
- Hess's law

## Introduction

The change in enthalpy for a reaction can be determined by three methods. First, the enthalpy of the reaction can be found by finding the difference between the enthalpies of the products and reactants in the lab using a calorimeter. Second, the change in enthalpy for a reaction can also be calculated using the heats of formation of all the reactants and products. Thirdly, the change in enthalpy can be calculated using the mathematical application of Hess's Law, which will be introduced in this lesson.

## The Energy Content of a System

Enthalpy has been defined previously as the measure of the total internal energy of a system. The difference between the enthalpy of the reactants and the enthalpy of the products is called the change in enthalpy. When reactions take place in an open system, such as a beaker or a container on a counter, the pressure in the system is constant because the pressure is the atmospheric pressure in the room. The change in enthalpy for reactions occurring under constant pressure is also called the or heat of reaction.

Remember that for an endothermic reaction, the value of is positive, therefore the must be greater than . To illustrate this, look at the potential energy diagram below.

In the figure above, the enthalpy of the reactants is lower than the energy of the products. Therefore, energy must be input into the reaction, and the value of the will be positive.

The opposite is true for exothermic reactions. For exothermic reactions, the value of is negative. Therefore, the enthalpy of the products must be less than enthalpy of the reactants. Notice that in the exothermic reaction below, the energy of the reactants is higher than the energy of the products. Therefore, the value of the will be negative.

**Example:**

Using the diagram below, answer questions 1-4.

- Which letter represents the activation energy for the reaction? What is its value?
- Which letter represents the change in enthalpy of the reaction or the ? What is its value?
- Is the reaction endothermic or exothermic? How can you tell?
- What does the letter E represent?

**Solution:**

- The activation energy is represented by letter D. It has a value of .
- The enthalpy change is represented by letter C. It has a value of .
- The reaction is exothermic because is negative and the products are lower than the reactants on the potential energy diagram.
- Letter E represents the activation energy for the reverse reaction.

## Enthalpy of Formation

A formation reaction is a reaction in which exactly one mole of a product is formed from its elements. The **enthalpy of formation**, , is the energy required to form one mole of a substance from its constituent elements at standard temperature and pressure. The equation below represents the for the formation of one mole of :

We can find the values for enthalpies of formation using a table of standard molar enthalpies found in the *CRC Handbook of Chemistry and Physics*, online, or in most chemistry textbooks. The values found in these sources are the values of .

In the equation below, exactly one mole of ammonia is formed from its elements and that qualifies the reaction as a formation reaction and its to be a . In comparison, the equation below does not qualify to be a formation reaction.

The equation above shows the formation of two moles of water, so it does not represent . In order to represent the for water, we must divide the equation by two:

If we were to reverse this reaction, look at what would happen to the value of :

The equation no longer represents the heat of formation because the equation represents a decomposition reaction. However, look at the value of . Since the equation was reversed, the sign of the value of was also reversed. was released when one mole of water formed from its elements, so to decompose that mole of water back into its elements, an input of is required. The for the forward reaction will be exactly the opposite of the for the reverse reaction.

You can use the values of found in **Table** below of standard heats of formation to find the enthalpy of a reaction (or ).

Name of Compound | Formation Reaction | Standard Enthalpy of Formation, ( of product) |
---|---|---|

aluminum oxide | ||

ammonia | ||

carbon dioxide | ||

carbon monoxide | ||

copper(I) oxide | ||

iron(III) oxide | ||

magnesium oxide | ||

methane | ||

nitrogen monoxoide | . | |

nitrogen dioxide | ||

sodium chloride | ||

sulfur dioxide | ||

sulfur trioxide | ||

water (gaseous) | ||

water (liquid) |

Consider the following equation:

From the table of standard heats of formation, we know that:

We also know that

Therefore, for this reaction:

Note that the heat of formation for gaseous water and that for oxygen gas are to be multiplied by two because there are two moles of oxygen gas and two moles of gaseous water in the combustion reaction.

We can now calculate the value for because we have all of the required values for and .

Rewriting the equation we see:

**Example:**

Calculate the value of for the following reaction.

**Solution:**

From the table of standard heats of formation we know that:

Therefore, for this reaction:

Rewriting the equation:

## Hess’s Law of Heat Summation

The first method shown for finding is to subtract the (reactants) from the (products). Sometimes, however, this method is not always possible. Compounds may not be easily produced from their elements, so there is not an available value for the . Other times, there may be side reactions happening, and there is a need for a more indirect method for calculating the value of the . In the middle of the 1800s, Germain Hess developed a method for determining the indirectly. **Hess’s law** states in any series of reactions that start with the same reactants and end with the same products, the net change in energy must be the same. This means that if multiple reactions are combined, the enthalpy change, , of the combined reaction is equal to the sum of all the individual enthalpy changes. So how does this work? It can be as straight forward as multiplying a reaction by a number or rearranging the reactions. Consider the following example.

Given the following equation:

Calculate for the reaction:

Notice that the second equation is the first equation multiplied by two. Therefore, we can simply multiply the by the same factor, that is, two.

If we were to reverse the direction of a reaction, then the sign of the must also be reversed. Look at the example below that illustrates this possibility.

Given the following equation:

What is the value of for the reaction ?

Notice when looking at the two equations, the second equation is the reverse of the first equation. Therefore, the value will change signs (or be multiplied by -1).

The most useful of Hess’s law, and the critical part of his definition, is the ability to add multiple reactions to obtain a final reaction and subsequently add the . Consider the equation below:

Suppose we wish to know the for this reaction but necessary values are not available. We do, however, have the following two equations available:

If we add these two equations by the normal addition of equations process, the result is exactly the same as Reaction 1:

Hess's Law tells us that since Reactions 2 and 3 add to give Reaction 1, the sum of and will be equal to (for Reaction 1). (Note: Since one-half mole of oxygen gas appears on each side of the equation, it cancels out).

It is relatively easy to demonstrate the truth of this statement mathematically. We can express as the heats of formation of its products minus the heats of formation of its reactants in the normal way.

We can do the same for and and then add the expressions for and together.

If you can sort through that mess of symbols, you will see that the heat of formation for one-half mole of oxygen gas is added in one place and subtracted in another place. Therefore, those cancel and can be removed from the equation.

You can see that this is equivalent to the expression for above. Therefore, is equal to the sum of and

This video serves a blackboard lecture showing the concepts involved with an example of using Hess's Law **(7e)**: http://www.youtube.com/watch?v=j4-UrAaAy3M (9:37).

## Lesson Summary

- All elements in their natural state have a
- Hess’s Law states that if multiple reactions are combined, the enthalpy of the combined reaction is equal to the sum of all the individual enthalpies.
- If we were to reverse a reaction, the sign of the is also reversed.
- If you multiply an equation by a factor, the is also multiplied by that same factor.

## Further Reading / Supplemental Links

For more practice using Hess's law, visit the following website.

## Review Questions

- Define the Hess’s Law and the need to use this method.
- Draw a potential energy diagram to represent the reaction: .
- Which of the following does not have a
- Which statement would describe an endothermic reaction?
- The potential energy of the reactants is greater than the potential energy of the products.
- The potential energy of the reactants is less than the potential energy of the products.
- Energy is released in the chemical reaction.
- The energy required to break bonds is more than the energy produced when bonds are formed.

- Given the reaction , what would be the for the following reaction: ?
- kJ
- kJ
- kJ
- kJ

- Which of the following reactions represents that for a
- Hydrogen sulfide can mix with carbon dioxide to make a very smelly liquid, carbon disulfide. Given that the enthalpies of formation for , , , and are -393.5 kJ/mol, -20.6 kJ/mol, 116.7 kJ/mol, and -285.8 kJ/mol, respectively, calculate .
- Ethene is a common compound used in the production of plastics for plastic bottles. Calculate the for ethene: .