<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

17.4: Hess’s Law

Difficulty Level: At Grade Created by: CK-12

Lesson Objectives

• Use Hess’s law of heat summation to add chemical reactions together in such a way as to produce a desired final equation. Calculate the enthalpy change for that final reaction.
• Define the standard heat of formation of a compound.
• Use known values for standard heats of formation to calculate a previously unknown standard heat of reaction.

Lesson Vocabulary

• heat of combustion
• Hess’s law of heat summation
• standard heat of formation

Recalling Prior Knowledge

• What is a combustion reaction?
• How does a compound compare to an element?

Calorimetry is an experimental technique used to directly measure a heat of reaction in a laboratory setting. However, many reactions for which it may be desirable to know the heat of reaction are too difficult to perform in a controlled manner. In this lesson, you will learn about two indirect methods that will allow you to find the enthalpy change for almost any chemical reaction.

It is sometimes very difficult or even impossible to measure the enthalpy change for a reaction directly in the laboratory. Some reactions take place extremely slowly, making a direct measurement unfeasible. In other cases, a given reaction may be an intermediate step in a series of reactions. Some reactions may be difficult to isolate because multiple side reactions may occur at the same time. Fortunately, it is possible to measure the enthalpy change for a reaction by an indirect method. Hess’s law of heat summation states that if two or more thermochemical equations are added together to give a final equation, then the heats of reaction for those equations can also be added together to give a heat of reaction for the final equation.

An example will illustrate how Hess’s law can be used. Acetylene (C2H2) is a gas that burns at an extremely high temperature (3300°C) and is used in welding (Figure below). On paper, acetylene gas can be produced by the reaction of solid carbon (graphite) with hydrogen gas. However, this is not an especially easy reaction for which to measure the total enthalpy change. In contrast, enthalpy changes for combustion reactions are relatively easy to measure. The heat of combustion is the heat released when one mole of a substance is completely reacted with oxygen gas.

The heats of combustion for carbon, hydrogen, and acetylene are shown below along with the balanced equation for each process.

1. \begin{align*}\text{C}{(s, \ \text{graphite})}+\text{O}_{2}{(g)} \rightarrow \text{CO}_{2}{(g)} \ \ \ \ \ \ \ \ \ \ \ \Delta \text{H}=-393.5 \ \text{kJ}\end{align*}
2. \begin{align*}\text{H}_{2}{(g)}+\dfrac{1}{2}\text{O}_{2}{(g)} \rightarrow \text{H}_2\text{O}{(l)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Delta \text{H}=-285.8 \ \text{kJ}\end{align*}
3. \begin{align*}\text{C}_2\text{H}_{2}{(g)}+\dfrac{5}{2}\text{O}_{2}{(g)} \rightarrow 2\text{CO}_{2}{(g)}+\text{H}_2\text{O}{(l)} \ \ \ \Delta \text{H}=-1301.1 \ \text{kJ}\end{align*}

The combustion reactions are written with fractional coefficients for O2 because the heats of combustion that are found in a table are for the combustion of 1 mol of the given substance. To use Hess’s law, we need to determine how the three equations above can be manipulated so that they can be added together to result in the desired equation (the formation of acetylene from carbon and hydrogen).

In order to do this, we will go through the desired equation, one substance at a time – choosing the combustion reaction from the equations numbered 1-3 above that contains that substance. It may be necessary to either reverse a combustion reaction or multiply it by some factor in order to make it “fit” to the desired equation. The first reactant is carbon, and the in the equation for the desired reaction, the coefficient of the carbon is a 2. We will therefore start by writing the first combustion reaction with all of its coefficients doubled. Because we are doubling the coefficients, we also need to double the value of ΔH, since we are now looking at the heat released when two moles of graphite are combusted instead of just one.

\begin{align*}2\text{C}{(s, \ \text{graphite})}+2\text{O}_{2}{(g)} \rightarrow 2\text{CO}_{2}{(g)} \ \ \ \ \ \ \Delta \text{H}=2(-393.5) = -787.0 \ \text{kJ}\end{align*}

The second reactant is hydrogen, and its coefficient is a 1, as it is in the second combustion reaction. Therefore, that reaction will be used as written.

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

The product of the reaction is C2H2, and its coefficient is also a 1. In combustion reaction #3, the acetylene is a reactant. Therefore, we will reverse reaction 3. When the reactants and products are reversed, ΔH for the resulting equation has the same numerical value but the opposite sign.

\begin{align*}2\text{CO}_{2}{(g)}+\text{H}_2\text{O}{(l)} \rightarrow \text{C}_2\text{H}_{2}{(g)}+\dfrac{5}{2}\text{O}_{2}{(g)} \ \ \ \ \Delta \text{H}=1301.1 \ \text{kJ}\end{align*}

Now, these three reactions can be summed together. Any substance that appears in equal quantities as a reactant in one equation and a product in another equation cancels out algebraically. The values for the enthalpy changes are likewise added.

\begin{align*} 2\text{C}{(s,\ \text{graphite})}+\cancel{2\text{O}_{2}{(g)}} & \rightarrow \cancel{2\text{CO}_{2}{(g)}} & & \Delta \text{H}=2(-393.5)=-787.0 \ \text{kJ} \\ \text{H}_{2}{(g)}+\cancel{\frac{1}{2}\text{O}_{2}{(g)}} & \rightarrow \cancel{\text{H}_2\text{O}{(l)}} & & \Delta \text{H}=-285.8 \ \text{kJ} \\ \cancel{2\text{CO}_{2}{(g)}}+\cancel{\text{H}_2\text{O}{(l)}} & \rightarrow \text{C}_2\text{H}_{2}{(g)}+\cancel{\dfrac{5}{2}\text{O}_{2}{(g)}} & & \Delta \text{H}=1301.1 \ \text{kJ} \\ \hline 2\text{C}{(s,\ \text{graphite})}+\text{H}_{2}{(g)} & \rightarrow \text{C}_2\text{H}_{2}{(g)} & & \Delta \text{H}=228.3 \ \text{kJ} \end{align*}

So the heat of reaction for the combination of carbon with hydrogen to produce acetylene is 228.3 kJ. When one mole of acetylene is produced, 228.3 kJ of heat are absorbed, making the reaction endothermic.

Standard Heat of Formation

A relatively straightforward chemical reaction is one in which elements are combined to form a compound. For example, sodium and chlorine react to form sodium chloride (as seen in the video below), and hydrogen and oxygen combine to form water. Like other reactions, these are accompanied by either the absorption or release of heat. The standard heat of formation (ΔHf°) is the enthalpy change associated with the formation of one mole of a compound from its elements in their standard states. The standard conditions for thermochemistry are 25°C and 101.3 kPa. Therefore, the standard state of an element is its state at 25°C and 101.3 kPa. For example, iron is a solid, bromine is a liquid, and oxygen is a gas under those conditions. The standard heat of formation of an element in its standard state is by definition equal to zero. ΔHf° = 0 for H2(g), N2(g), O2(g), F2(g), Cl2(g), Br2(l), and I2(g), because these elements are most commonly found in their pure form as diatomic molecules. The graphite form of solid carbon is defined as its standard state, so it has a ΔHf° of 0. Because diamond is not the standard state of carbon, it has a non-zero value for ΔHf°. Some standard heats of formation are listed in Table below.

Elemental sodium (Na) and chlorine gas (Cl2) react to form sodium chloride, releasing 411 kJ of heat for every mole of NaCl produced. This spectacular reaction can be seen in the following video: http://www.youtube.com/watch?v=Ftw7a5ccubs (0:42).

Standard Heats of Formation of Selected Substances
Substance ΔHf° (kJ/mol) Substance ΔHf° (kJ/mol)
Al2O3(s) −1669.8 H2O2(l) −187.6
BaCl2(s) −860.1 KCl(s) −435.87
Br2(g) 30.91 NH3(g) −46.3
C (s, graphite) 0 NO(g) 90.4
C (s, diamond) 1.90 NO2(g) 33.85
CH4(g) −74.85 NaCl(s) −411.0
C2H5OH(l) −276.98 O3(g) 142.2
CO(g) −110.5 P(s, white) 0
CO2(g) −393.5 P(s, red) −18.4
CaO(s) −635.6 PbO(s) −217.86
CaCO3(s) −1206.9 S(rhombic) 0
HCl(g) −92.3 S(monoclinic) 0.30
CuO(s) −155.2 SO2(g) −296.1
CuSO4(s) −769.86 SO3(g) −395.2
Fe2O3(s) −822.2 H2S(g) −20.15
H2O(g) −241.8 SiO2(s, quartz) −910.7
H2O(l) −285.8 ZnCl2(s) −415.89

An application of Hess’s law allows us to use standard heats of formation to indirectly calculate the heat of reaction for any reaction that occurs at standard conditions. An enthalpy change that occurs specifically under standard conditions is called the standard enthalpy (or heat) of reaction and is given the symbol ΔH°. The standard heat of reaction can be calculated by using the following equation.

\begin{align*}\mathrm{\Delta H^{\circ} = \Sigma n \Delta H^{\circ}_f (products)- \Sigma n \Delta H^{\circ}_f (reactants)}\end{align*}

The symbol Σ is the Greek letter sigma and means “the sum of.” The standard heat of reaction is equal to the sum of all the standard heats of formation of the products minus the sum of all the standard heats of formation of the reactants. The symbol “n” signifies that each heat of formation must first be multiplied by its coefficient in the balanced equation. Sample Problem 17.8 illustrates the use of this equation.

Sample Problem 17.8: Calculating Standard Heat of Reaction

Calculate the standard heat of reaction (ΔH°) for the reaction of nitrogen monoxide gas with oxygen to form nitrogen dioxide gas.

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

Known

• ΔHf° for NO(g) = 90.4 kJ/mol
• ΔHf° for O2(g) = 0 (pure element in its standard state)
• ΔHf° for NO2(g) = 33.85 kJ/mol

Unknown

• ΔH° = ? kJ

First, write the balanced equation for the reaction. Then, apply the equation to calculate the standard heat of reaction from the standard heats of formation.

Step 2: Solve.

The balanced equation is: \begin{align*}2\text{NO}(g)+\text{O}_2(g) \rightarrow 2\text{NO}_2(g)\end{align*}

Applying the equation from the text:

ΔHf° = [2 mol NO2 (33.85 kJ/mol)] - [2 mol NO (90.4 kJ/mol) + 1 mol O2 (0 kJ/mol)] = -113 kJ

The standard heat of reaction is −113 kJ.

The reaction is exothermic, which makes sense, because combustion reactions typically release heat.

Practice Problem
1. Calculate ΔH° for the following reactions.
1. \begin{align*}\text{CaCO}_3(s) \rightarrow \text{CaO}(s)+\text{CO}_2(g)\end{align*}
2. \begin{align*}7\text{O}_2(g)+4\text{NH}_3(g) \rightarrow 4\text{NO}_2(g)+6\text{H}_2\text{O}(l)\end{align*}

Lesson Summary

• Hess’s law of heat summation allows chemical reactions to be added together in such a way as to produce a final desired equation. The heats of reaction for each individual reaction are also added, resulting in the enthalpy change for the final reaction.
• The standard heat of formation is the enthalpy change that occurs when a compound is formed from its elements in their standard states.
• The standard enthalpy change for any reaction is equal to the sum of the heats of formation of the products minus the sum of the heats of formation of the reactants.

Lesson Review Questions

Reviewing Concepts

1. What are two reasons why the heat of reaction may not be able to be determined directly?
2. What are standard conditions for thermochemical problems?
3. What is the standard heat of formation of an element in its standard state?
4. For which of the following is the standard heat of formation equal to zero?
1. F(g)
2. F2(g)
3. He(l)
4. Hg(l)
5. CO(g)
6. Co(s)
5. In general, the more negative the value of the ΔHf° for a compound, the more stable the compound. Explain why this is true. Use Table above to choose the more stable substance from each pair below.
1. CO(g) or CO2(g)
2. graphite or diamond
3. Al2O3 or Fe2O3
4. O2(g) or O3(g)

Problems

1. Write the balanced chemical equation for the formation of one mole of solid sodium nitrite, NaNO2, from its elements in their standard states.
2. Determine ΔH for the reaction below.
3. Given the heats of reaction for the following combustion reactions: Calculate the enthalpy of formation of methanol from its elements, according to the following equation:
4. Find the enthalpy change for the reaction below for the formation of phosphorus pentachloride from its elements. Use the thermochemical equations below.
5. Use standard heats of formation from Table above to calculate the standard heat of reaction for each of the following.
1. \begin{align*}2\text{H}_2\text{O}_{2}{(l)} \rightarrow 2\text{H}_2\text{O}{(l)}+\text{O}_{2}{(g)}\end{align*}
2. \begin{align*}\text{Fe}_2\text{O}_{3}{(s)}+3\text{CO}{(g)} \rightarrow 2\text{Fe}{(s)}+3\text{CO}_{2}{(g)}\end{align*}
3. \begin{align*}\text{C}_2\text{H}_5\text{OH}{(l)}+3\text{O}_{2}{(g)} \rightarrow 2\text{CO}_{2}{(g)}+3\text{H}_2\text{O}{(l)}\end{align*}

Points to Consider

Chemical reactions proceed at a wide variety of speeds or rates. Some reactions happen almost instantly, as soon as the reactants come into contact with one another, while other reactions may take years to reach completion.

• How do chemical reactions occur on the molecular level?
• What factors influence the rate of a chemical reaction?

Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

Show Hide Details
Description
Tags:
Subjects: