# 22.2: Enthalpy

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• define and understand enthalpy of reaction.
• calculate the enthalpy of reaction, \begin{align*}\triangle H_{\text{rxn}}\end{align*}.
• define and understand \begin{align*}\triangle H_f\end{align*}.
• define Hess’s law.
• calculate \begin{align*}\triangle H_{\mathrm{rxn}}\end{align*} 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 \begin{align*}\triangle H\end{align*} or heat of reaction.

\begin{align*} \triangle H_{\text{rxn}}= \triangle H_{\text{products}} - \triangle H_{\text{reactants}}\end{align*}

Remember that for an endothermic reaction, the value of \begin{align*}\triangle H\end{align*} is positive, therefore the \begin{align*}\triangle H_{\mathrm{products}}\end{align*} must be greater than \begin{align*}\triangle H_{\mathrm{reactants}}\end{align*}. 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 \begin{align*}\triangle H\end{align*} will be positive.

The opposite is true for exothermic reactions. For exothermic reactions, the value of \begin{align*}\triangle H\end{align*} 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 \begin{align*}\triangle H\end{align*} will be negative.

Example:

Using the diagram below, answer questions 1-4.

1. Which letter represents the activation energy for the reaction? What is its value?
2. Which letter represents the change in enthalpy of the reaction or the \begin{align*}\triangle H\end{align*}? What is its value?
3. Is the reaction endothermic or exothermic? How can you tell?
4. What does the letter E represent?

Solution:

1. The activation energy is represented by letter D. It has a value of \begin{align*}250 - 120 = 130 \ \mathrm{kJ}\end{align*}.
2. The enthalpy change \begin{align*}(\triangle H)\end{align*} is represented by letter C. It has a value of \begin{align*}35 - 120 = -85 \ \mathrm{kJ}\end{align*}.
3. The reaction is exothermic because \begin{align*}\triangle H\end{align*} is negative and the products are lower than the reactants on the potential energy diagram.
4. 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, \begin{align*}\triangle H_f\end{align*}, 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 \begin{align*}\triangle H_f\end{align*} for the formation of one mole of \begin{align*}\mathrm{NH}_{3(g)}\end{align*}:

\begin{align*}\frac{1} {2} \mathrm{N}_{2(g)} + \frac{3} {2} \mathrm{H}_{2(g)} \rightarrow \mathrm{NH}_{3(g)} \ \ \ \ \ \triangle H_f = -46.1\ \text{kJ/mol}\end{align*}

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 \begin{align*}\triangle H_f\end{align*}.

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 \begin{align*}\triangle H\end{align*} to be a \begin{align*}\triangle H_f\end{align*}. In comparison, the equation below does not qualify to be a formation reaction.

\begin{align*}2\ \mathrm{H}_{2(g)} + \mathrm{O}_{2(g)} \rightarrow 2 \ \mathrm{H}_2\mathrm{O}_{(l)} \ \ \ \ \ \triangle H = -571.6 \ \text{kJ}\end{align*}

The equation above shows the formation of two moles of water, so it does not represent \begin{align*}\triangle H_f\end{align*}. In order to represent the \begin{align*}\triangle H_f\end{align*} for water, we must divide the equation by two:

\begin{align*}\mathrm{H}_{2(g)} + \frac{1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{H}_2\mathrm{O}_{(l)} \ \ \ \ \ \triangle H_f = -285.8 \ \text{kJ} \end{align*}

If we were to reverse this reaction, look at what would happen to the value of \begin{align*}\triangle H\end{align*}:

\begin{align*}\mathrm{H}_2\mathrm{O}_{(l)} \rightarrow \mathrm{H}_{2(g)} + \frac{1} {2} \mathrm{O}_{2(g)} \ \ \ \ \ \triangle H = 285.8\ \text{kJ} \end{align*}

The equation no longer represents the heat of formation because the equation represents a decomposition reaction. However, look at the value of \begin{align*}\triangle H\end{align*}. Since the equation was reversed, the sign of the value of \begin{align*}\triangle H\end{align*} was also reversed. \begin{align*}285.8 \ \mathrm{kJ}\end{align*} 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 \begin{align*}285.8 \ \mathrm{kJ}\end{align*} is required. The \begin{align*}\triangle H\end{align*} for the forward reaction will be exactly the opposite of the \begin{align*}\triangle H\end{align*} for the reverse reaction.

You can use the values of \begin{align*}\triangle H_f\end{align*} found in Table below of standard heats of formation to find the enthalpy of a reaction (or \begin{align*}\triangle H_{\text{rxn}}\end{align*}).

Standard Enthalpy of Formation for Some Selected Compounds
Name of Compound Formation Reaction Standard Enthalpy of Formation, \begin{align*}\triangle H_f^o\end{align*} (\begin{align*}\mathrm{kJ/mol}\end{align*} of product)
aluminum oxide \begin{align*}2 \ \mathrm{Al}_{(s)} + \frac {3} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{Al}_2\mathrm{O}_{3(s)}\end{align*} \begin{align*}-1669.8\end{align*}
ammonia \begin{align*}\frac {1} {2} \mathrm{N}_{2(g)} + \frac {3} {2} \mathrm{H}_{2(g)} \rightarrow \mathrm{NH}_{3(g)}\end{align*} \begin{align*} -46.1\end{align*}
carbon dioxide \begin{align*}\mathrm{C}_{(s)} + \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)}\end{align*} \begin{align*}-393.5\end{align*}
carbon monoxide \begin{align*}\mathrm{C}_{(s)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{(g)}\end{align*} \begin{align*}-110.5\end{align*}
copper(I) oxide \begin{align*}\mathrm{Cu}_{(s)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{CuO}_{(s)}\end{align*} \begin{align*}-156\end{align*}
iron(III) oxide \begin{align*}2 \ \mathrm{Fe}_{(s)} + \frac {3} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{Fe}_2\mathrm{O}_{3(s)}\end{align*} \begin{align*}-822.2\end{align*}
magnesium oxide \begin{align*}\mathrm{Mg}_{(s)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{MgO}_{(s)}\end{align*} \begin{align*}-602\end{align*}
methane \begin{align*}\mathrm{C}_{(s)} + 2 \ \mathrm{H}_{2(g)} \rightarrow \mathrm{CH}_{4(g)}\end{align*} \begin{align*}-74.8\end{align*}
nitrogen monoxoide \begin{align*}\frac {1} {2} \mathrm{N}_{2(g)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{NO}_{(g)}\end{align*} \begin{align*}+90\end{align*}.
nitrogen dioxide \begin{align*}\frac {1} {2} \mathrm{N}_{2(s)} + \mathrm{O}_{2(g)} \rightarrow \mathrm{NO}_{2(g)}\end{align*} \begin{align*}+34\end{align*}
sodium chloride \begin{align*}\mathrm{Na}_{(s)} + \frac {1} {2} \mathrm{Cl}_{2(g)} \rightarrow \mathrm{NaCl}_{(s)}\end{align*} \begin{align*} -411\end{align*}
sulfur dioxide \begin{align*}\mathrm{S}_{(s)} + \mathrm{O}_{2(g)} \rightarrow \mathrm{SO}_{2(g)}\end{align*} \begin{align*}-297\end{align*}
sulfur trioxide \begin{align*} \mathrm{S}_{(s)} + \frac {3} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{SO}_{3(g)}\end{align*} \begin{align*}-393.2\end{align*}
water (gaseous) \begin{align*}\mathrm{H}_{2(g)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{H}_2\mathrm{O}_{(g)}\end{align*} \begin{align*}-241.8\end{align*}
water (liquid) \begin{align*}\mathrm{H}_{2(g)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{H}_2\mathrm{O}_{(l)}\end{align*} \begin{align*}-285.8\end{align*}

Consider the following equation:

\begin{align*}\mathrm{CH}_{4(g)} + 2 \ \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} + 2 \ \mathrm{H}_2\mathrm{O}_{(g)} \ \ \ \ \ \triangle H =?\end{align*}

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

\begin{align*}\triangle H_f (\mathrm{CH}_{4(g)}) = -74.8 \ \text{kJ/mol}\end{align*}
\begin{align*} \triangle H_f (\mathrm{O}_{2(g)}) = 0\ \text{kJ/mol}\end{align*}
\begin{align*}\text{(Note: all elements in their natural state have a} \ \triangle H_f = 0 \ \text{kJ/mol})\end{align*}
\begin{align*}\triangle H_f (\mathrm{CO}_{2(g)}) = -393.5 \ \text{kJ/mol}\end{align*}
\begin{align*}\triangle H_f (\mathrm{H}_2\mathrm{O}_{(g)}) = -241.8\ \text{kJ/mol}\end{align*}

We also know that \begin{align*}\triangle H_{\mathrm{rxn}} = \triangle H_{\mathrm{products}} - \triangle H_{\mathrm{reactants}}\end{align*}

Therefore, for this reaction:

\begin{align*}\triangle H_{rxn} = [\triangle H_f (\mathrm{CO}_{2(g)}) + 2 \triangle H_f(\mathrm{H}_2\mathrm{O}_{(g)})] - [\triangle H_f(\mathrm{CH}_{4(g)}) + 2 \triangle H_f (\mathrm{O}_{2(g)})]\end{align*}

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 \begin{align*}\triangle H_{\mathrm{rxn}}\end{align*} because we have all of the required values for \begin{align*}\triangle H_{\mathrm{products}}\end{align*} and \begin{align*}\triangle H_{\mathrm{reactants}}\end{align*}.

\begin{align*}\triangle H_{\text{rxn}} = [(-393.5 \ \text{kJ/mol}) + (2)(-241.8 \ \text{kJ/mol})] - [(-74.8 \ \text{kJ/mol}) + (2)(0 \ \text{kJ/mol})]\end{align*}
\begin{align*}\triangle H_{\text{rxn}} = [-877.1\ \text{kJ/mol}] - [-74.8\ \text{kJ/mol}] \end{align*}
\begin{align*}\triangle H_{\text{rxn}} = -802.3\ \text{kJ/mol}\end{align*}

Rewriting the equation we see:

\begin{align*}\mathrm{CH}_{4(g)} + 2 \ \mathrm{O}_{2(g)} \rightarrow \mathrm{CO}_{2(g)} + 2 \ \mathrm{H}_2\mathrm{O}_{(g)} \ \ \ \ \ \triangle H = -802.3 \ \text{kJ/mol}\end{align*}

Example:

Calculate the value of \begin{align*}\triangle H_{\mathrm{rxn}}\end{align*} for the following reaction.

\begin{align*}2 \ \mathrm{Al}_{(s)} + \mathrm{Fe}_2\mathrm{O}_{3(s)} \rightarrow \mathrm{Al}_2\mathrm{O}_{3(s)} + 2 \ \mathrm{Fe}_{(l)}\end{align*}

Solution:

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

\begin{align*}\triangle H_f (\mathrm{Al}_{(s)}) = 0 \ \text{kJ/mol}\end{align*}
\begin{align*}\text{(Note: all elements in their natural state have a} \ \triangle H_f = 0\ \text{kJ/mol})\end{align*}
\begin{align*}\triangle H_f (\mathrm{Fe}_2\mathrm{O}_{3(s)}) = -822.2 \ \text{kJ/mol}\end{align*}
\begin{align*}\triangle H_f (\mathrm{Al}_2\mathrm{O}_{3(s)}) = -1669.8 \ \text{kJ/mol}\end{align*}
\begin{align*}\triangle H_f (\mathrm{Fe}_{(l)}) = -12.4 \ \text{kJ/mol}\end{align*}

Therefore, for this reaction:

\begin{align*}\triangle H_{rxn} = [\triangle H_f (\mathrm{Al}_2\mathrm{O}_{3(s)}) + 2 \times \triangle H_f (\mathrm{Fe}_{(l)})] - [2 \times \triangle H_f (\mathrm{Al}_{(s)}) + \triangle H_f (\mathrm{Fe}_2\mathrm{O}_{3(s)})]\end{align*}
\begin{align*}\triangle H_{rxn} = [-1669.8 \ \text{kJ/mol} + (2)(-12.4 \ \text{kJ/mol})] - [(2)(0 \ \text{kJ/mol}) + (-822.2 \ \text{kJ/mol})] \end{align*}
\begin{align*}\triangle H_{rxn} = [-1694.6 \ \text{kJ/mol}] - [-822.2\ \text{kJ/mol}]\end{align*}
\begin{align*}\triangle H_{rxn} = -872.4\ \text{kJ/mol}\end{align*}

Rewriting the equation:

\begin{align*}2 \ \mathrm{Al}_{(s)} + \mathrm{Fe}_2\mathrm{O}_{3(s)} \rightarrow \mathrm{Al}_2\mathrm{O}_{3(s)} + 2 \ \mathrm{Fe}_{(l)} \ \ \ \ \ \triangle H_{\text{rxn}} = -872.4 \ \text{kJ/mol}\end{align*}

## Hess’s Law of Heat Summation

The first method shown for finding \begin{align*}\triangle H_{\mathrm{rxn}}\end{align*} is to subtract the \begin{align*}\triangle H\end{align*} (reactants) from the \begin{align*}\triangle H\end{align*} (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 \begin{align*}\triangle H_f\end{align*}. Other times, there may be side reactions happening, and there is a need for a more indirect method for calculating the value of the \begin{align*}\triangle H_{\mathrm{rxn}}\end{align*}. In the middle of the 1800s, Germain Hess developed a method for determining the \begin{align*}\triangle H_{\mathrm{rxn}}\end{align*} 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, \begin{align*}\triangle H\end{align*}, 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:

\begin{align*}\mathrm{S}_{(s)} + \frac{3} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{SO}_{3(g)} \ \ \ \ \ \triangle H_f = -393.2\ \text{kJ/mol}\end{align*}

Calculate \begin{align*}\triangle H\end{align*} for the reaction:

\begin{align*}2 \ \mathrm{S}_{(s)} + 3 \ \mathrm{O}_{2(g)} \rightarrow 2 \ \mathrm{SO}_{3(g)}\end{align*}

Notice that the second equation is the first equation multiplied by two. Therefore, we can simply multiply the \begin{align*}\triangle H\end{align*} by the same factor, that is, two.

\begin{align*}2 \ \mathrm{S}_{(s)} + 3 \ \mathrm{O}_{2(g)} \rightarrow 2 \ \mathrm{SO}_{3(g)} \ \ \ \ \ \triangle H_f = -786.4 \ \text{kJ}\end{align*}

If we were to reverse the direction of a reaction, then the sign of the \begin{align*}\triangle H\end{align*} must also be reversed. Look at the example below that illustrates this possibility.

Given the following equation:

\begin{align*}2 \ \mathrm{NO}_{(g)} + \mathrm{O}_{2(g)} \rightarrow 2 \ \mathrm{NO}_{2(g)} \ \ \ \ \ \triangle H = -114 \ \text{kJ}\end{align*}

What is the value of \begin{align*}\triangle H\end{align*} for the reaction \begin{align*}2 \ \mathrm{NO}_{2(g)} \rightarrow 2 \ \mathrm{NO}_{(g)} + \mathrm{O}_{2(g)}\end{align*} ?

Notice when looking at the two equations, the second equation is the reverse of the first equation. Therefore, the \begin{align*}\triangle H\end{align*} value will change signs (or be multiplied by -1).

\begin{align*}2 \ \mathrm{NO}_{2(g)} \rightarrow 2 \ \mathrm{NO}_{(g)} + \mathrm{O}_{2(g)} \ \ \ \ \ \triangle H = +114 \ \text{kJ}\end{align*}

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 \begin{align*}\triangle H's\end{align*}. Consider the equation below:

\begin{align*}\mathrm{CuO}_{(s)} + \mathrm{H}_{2(g)} \rightarrow \mathrm{Cu}_{(s)} + \mathrm{H}_2\mathrm{O}_{(g)} \ \ \ \ \ \text{Reaction 1}\end{align*}

Suppose we wish to know the \begin{align*}\triangle H\end{align*} for this reaction but necessary \begin{align*}\triangle H_f\end{align*} values are not available. We do, however, have the following two equations available:

\begin{align*}\mathrm{CuO}_{(s)} \rightarrow \mathrm{Cu}_{(s)} + \frac {1} {2} \mathrm{O}_{2(g)} \ \ \ \ \ \triangle H_2 = +155\ \text{kJ} \ \ \ \ \ \text{Reaction 2}\end{align*}
\begin{align*}\mathrm{H}_{2(g)} + \frac {1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{H}_2\mathrm{O}_{(g)} \ \ \ \ \ \triangle H_3 = -242\ \text{kJ} \ \ \ \ \ \text{Reaction 3}\end{align*}

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

\begin{align*} \begin{array}{rllclcl} \mathrm{CuO}_{(s)} & \rightarrow & \mathrm{Cu}_{(s)} + \frac {1} {2} \mathrm{O}_{2(g)} & & \triangle H_2 = +155 \ \text{kJ} & & \text{Reaction 2}\\ \mathrm{H}_{2(g)} + \frac {1} {2} \mathrm{O}_{2(g)} & \rightarrow & \mathrm{H}_2\mathrm{O}_{(g)} & & \triangle H_3 = -242 \ \text{kJ} & &\text{Reaction 3}\\ \hline \mathrm{CuO}_{(s)} + \mathrm{H}_{2(g)} & \rightarrow & \mathrm{Cu}_{(s)} + \mathrm{H}_2\mathrm{O}_{(g)} & & \triangle H_1 = ? \ \text{kJ} & &\text{Reaction 1} \end{array} \end{align*}

Hess's Law tells us that since Reactions 2 and 3 add to give Reaction 1, the sum of \begin{align*}\triangle H_2\end{align*} and \begin{align*}\triangle H_3\end{align*} will be equal to \begin{align*}\triangle H_1\end{align*} (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 \begin{align*}\triangle H_1\end{align*} as the heats of formation of its products minus the heats of formation of its reactants in the normal way.

\begin{align*}\triangle H_1 = [\triangle H_f (Cu_{(s)}) + \triangle H_f (H_2O_{(g)})] - [\triangle H_f (CuO_{(s)}) + \triangle H_f (H_{2(g)})]\end{align*}
\begin{align*}\triangle H_1 = \triangle H_f (Cu_{(s)}) + \triangle H_f (H_2O_{(g)}) - \triangle H_f (CuO_{(s)}) - \triangle H_f (H_{2(g)})\end{align*}

We can do the same for \begin{align*}\triangle H_2\end{align*} and \begin{align*}\triangle H_3\end{align*} and then add the expressions for \begin{align*}\triangle H_2\end{align*} and \begin{align*}\triangle H_3\end{align*} together.

\begin{align*}\triangle H_2 = \triangle H_f (\mathrm{Cu}_{(s)}) + \triangle H_f \left ( \frac {1} {2} \mathrm{O}_{2(g)} \right ) - \triangle H_f (\mathrm{CuO}_{(s)})\end{align*}
\begin{align*}\triangle H_3 = \triangle H_f (\mathrm{H}_2\mathrm{O}_{(g)}) - \triangle H_f (\mathrm{H}_{2(g)}) - \triangle H_f \left ( \frac {1} {2} \mathrm{O}_{2(g)}\right )\end{align*}
\begin{align*}\triangle H_2 + \triangle H_3 = \triangle H_f (\mathrm{Cu}_{(s)}) + \triangle H_f \left ( \frac {1} {2} \mathrm{O}_{2(g)}\right ) - \triangle H_f (\mathrm{CuO}_{(s)}) + \triangle H_f (\mathrm{H}_2\mathrm{O}_{(g)}) - \triangle H_f (\mathrm{H}_{2(g)}) - \triangle H_f \left ( \frac {1} {2} \mathrm{O}_{2(g)}\right )\end{align*}

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.

\begin{align*}\triangle H_2 + \triangle H_3 = \triangle H_f (\mathrm{Cu}_{(s)}) - \triangle H_f (\mathrm{CuO}_{(s)}) + \triangle H_f (\mathrm{H}_2\mathrm{O}_{(g)} - \triangle H_f (\mathrm{H}_{2(g)})\end{align*}

You can see that this is equivalent to the expression for \begin{align*}\triangle H_1\end{align*} above. Therefore, \begin{align*}\triangle H_1\end{align*} is equal to the sum of \begin{align*}\triangle H_2\end{align*} and \begin{align*}\triangle H_3.\end{align*}

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 \begin{align*}\triangle H_f = 0 \;\mathrm{kJ/mol}.\end{align*}
• Hess’s Law states that if multiple reactions are combined, the enthalpy \begin{align*}(\triangle H)\end{align*} 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 \begin{align*}\triangle H\end{align*} is also reversed.
• If you multiply an equation by a factor, the \begin{align*}\triangle H\end{align*} is also multiplied by that same factor.

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

## Review Questions

1. Define the Hess’s Law and the need to use this method.
2. Draw a potential energy diagram to represent the reaction: \begin{align*}\mathrm{S}_{8(s)} + 8 \ \mathrm{Cl}_{2(g)} \rightarrow 8 \ \mathrm{SCl}_{2(s)} \ \ \triangle H = -376 \ \text{kJ}\end{align*}.
3. Which of the following does not have a \begin{align*}\triangle H_f = 0?\end{align*}
1. \begin{align*}\mathrm{H}_2\mathrm{O}_{(l)}\end{align*}
2. \begin{align*}\mathrm{O}_{2(g)}\end{align*}
3. \begin{align*}\mathrm{H}_{2(g)}\end{align*}
4. \begin{align*}\mathrm{Fe}_{(s)}\end{align*}
4. Which statement would describe an endothermic reaction?
1. The potential energy of the reactants is greater than the potential energy of the products.
2. The potential energy of the reactants is less than the potential energy of the products.
3. Energy is released in the chemical reaction.
4. The energy required to break bonds is more than the energy produced when bonds are formed.
5. Given the reaction \begin{align*}2 \ \mathrm{HCl}_{(g)} \rightarrow \mathrm{H}_{2(g)} + \mathrm{Cl}_{2(g)}, \ \triangle H = 185 \ \mathrm{kJ}\end{align*}, what would be the \begin{align*}\triangle H\end{align*} for the following reaction: \begin{align*}\frac{1}{2} \mathrm{H}_{2(g)} + \frac{1}{2} \mathrm{Cl}_{2(g)} \ \rightarrow\ \mathrm{HCl}_{(g)}\end{align*}?
1. \begin{align*}185\end{align*} kJ
2. \begin{align*}-185\end{align*} kJ
3. \begin{align*}92.5\end{align*} kJ
4. \begin{align*}-92.5\end{align*} kJ
6. Which of the following reactions represents that for a \begin{align*}\triangle H_f?\end{align*}
1. \begin{align*}4 \ \mathrm{Fe}_{(s)} + 3 \ \mathrm{O}_{2(g)} \rightarrow 2 \ \mathrm{Fe}_2\mathrm{O}_{3(s)}\end{align*}
2. \begin{align*}\mathrm{SO}_{2(g)} + \frac{1} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{SO}_{3(g)}\end{align*}
3. \begin{align*}2 \ \mathrm{Al}_{(s)} + \frac{3} {2} \mathrm{O}_{2(g)} \rightarrow \mathrm{Al}_2\mathrm{O}_{3(s)}\end{align*}
4. \begin{align*}\frac{1} {2} \mathrm{C}_4\mathrm{H}_{10(g)} + \frac{1} {2} \mathrm{H}_{2(g)} \rightarrow \mathrm{C}_2\mathrm{H}_{6(g)} \end{align*}
7. Hydrogen sulfide can mix with carbon dioxide to make a very smelly liquid, carbon disulfide. Given that the enthalpies of formation for \begin{align*}\mathrm{CO}_{2(g)}\end{align*}, \begin{align*}\mathrm{H}_2S_{(g)}\end{align*}, \begin{align*}CS_{2(l)}\end{align*}, and \begin{align*}\mathrm{H}_2{O}_{(l)}\end{align*} are -393.5 kJ/mol, -20.6 kJ/mol, 116.7 kJ/mol, and -285.8 kJ/mol, respectively, calculate \begin{align*}\triangle H_{\mathrm{rxn}}\end{align*}.
8. Ethene is a common compound used in the production of plastics for plastic bottles. Calculate the \begin{align*}\triangle H_{\mathrm{rxn}}\end{align*} for ethene: \begin{align*}2 \ \mathrm{C}_{(g)} + 2 \ \mathrm{H}_{2(g)} \rightarrow \mathrm{C}_2\mathrm{H}_{4(g)}\end{align*}.

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