Chapter 10: Thermochemistry

# 10.3 Enthalpy

### Learning Outcomes

• State the first law of thermodynamics
• Define enthalpy and explain its classification as a state function
• Write and balance thermochemical equations
• Calculate enthalpy changes for various chemical reactions
• Explain Hess’s law and use it to compute reaction enthalpies

Thermochemistry is a branch of chemical thermodynamics, the science that deals with the relationships between heat, work, and other forms of energy in the context of chemical and physical processes. As we concentrate on thermochemistry, we need to consider some widely used concepts of thermodynamics.

Substances act as reservoirs of energy, meaning that energy can be added to them or removed from them. Energy is stored in a substance when the kinetic energy of its atoms or molecules is raised. The greater kinetic energy may be in the form of increased translations (travel or straight-line motions), vibrations, or rotations of the atoms or molecules. When thermal energy is lost, the intensities of these motions decrease and the kinetic energy falls. The total of all possible kinds of energy present in a substance is called the internal energy (U), sometimes symbolized as E.

As a system undergoes a change, its internal energy can change, and energy can be transferred from the system to the surroundings, or from the surroundings to the system. Energy is transferred into a system when it absorbs heat (q) from the surroundings or when the surroundings do work (w) on the system. For example, energy is transferred into room-temperature metal wire if it is immersed in hot water (the wire absorbs heat from the water), or if you rapidly bend the wire back and forth (the wire becomes warmer because of the work done on it). Both processes increase the internal energy of the wire, which is reflected in an increase in the wire’s temperature. Conversely, energy is transferred out of a system when heat is lost from the system, or when the system does work on the surroundings.

The relationship between internal energy, heat, and work can be represented by the equation:

$\Delta U=q+w$

as shown in Figure 10.3.1. This is one version of the first law of thermodynamics, and it shows that the internal energy of a system changes through heat flow into or out of the system (positive $q$ is heat flow in; negative $q$ is heat flow out) or work done on or by the system. The work, $w$, is positive if it is done on the system and negative if it is done by the system.

A type of work called expansion work (or pressure-volume work) occurs when a system pushes back the surroundings against a restraining pressure, or when the surroundings compress the system. An example of this occurs during the operation of an internal combustion engine. The reaction of gasoline and oxygen is exothermic. Some of this energy is given off as heat, and some does work pushing the piston in the cylinder. The substances involved in the reaction are the system, and the engine and the rest of the universe are the surroundings. The system loses energy by both heating and doing work on the surroundings, and its internal energy decreases. (The engine is able to keep the car moving because this process is repeated many times per second while the engine is running.) We will consider how to determine the amount of work involved in a chemical or physical change in the chapter on thermodynamics.

As discussed, the relationship between internal energy, heat, and work can be represented as $\Delta U=q+w$. Internal energy is a type of quantity known as a state function (or state variable), whereas heat and work are not state functions. The value of a state function depends only on the state that a system is in, and not on how that state is reached. If a quantity is not a state function, then its value does depend on how the state is reached. An example of a state function is altitude or elevation. If you stand on the summit of Mt. Kilimanjaro, you are at an altitude of 5895 m, and it does not matter whether you hiked there or parachuted there. The distance you traveled to the top of Kilimanjaro, however, is not a state function. You could climb to the summit by a direct route or by a more roundabout, circuitous path (Figure 10.3.2). The distances traveled would differ (distance is not a state function) but the elevation reached would be the same (altitude is a state function).

Chemists ordinarily use a property known as enthalpy (H) to describe the thermodynamics of chemical and physical processes. Enthalpy is defined as the sum of a system’s internal energy (U) and the mathematical product of its pressure (P) and volume (V):

$H=U+PV$

Since it is derived from three state functions (U, P, and V), enthalpy is also a state function. Enthalpy values for specific substances cannot be measured directly; only enthalpy changes for chemical or physical processes can be determined. For processes that take place at constant pressure (a common condition for many chemical and physical changes), the enthalpy change (ΔH) is:

$\Delta H=\Delta U+P\Delta V$

The mathematical product PΔV represents work (w), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetic signs of ΔV and w will always be opposite:

$P\Delta V=-w$

Substituting this equation and the definition of internal energy into the enthalpy-change equation yields:

$\begin{array}{rll}\Delta H&=&\Delta U+P\Delta V\\&=&{q}_{\text{p}}+w-w\\&=&{q}_{\text{p}}\end{array}$

where ${q}_{\text{p}}$ is the heat of reaction under conditions of constant pressure.

And so, if a chemical or physical process is carried out at constant pressure with the only work done caused by expansion or contraction, then the heat flow (qp) and enthalpy change (ΔH) for the process are equal.

The heat given off when you operate a Bunsen burner is equal to the enthalpy change of the methane combustion reaction that takes place, since it occurs at the essentially constant pressure of the atmosphere. On the other hand, the heat produced by a reaction measured in a bomb calorimeter is not equal to ΔH, because the closed, constant-volume metal container prevents expansion work from occurring. Chemists usually perform experiments under normal atmospheric conditions, at constant external pressure with q = ΔH, which makes enthalpy the most convenient choice for determining heat.

The following conventions apply when we use ΔH:

• A negative value of an enthalpy change, ΔH < 0, indicates an exothermic reaction; a positive value, ΔH > 0, indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its ΔH is changed (a process that is endothermic in one direction is exothermic in the opposite direction).
• Chemists use a thermochemical equation to represent the changes in both matter and energy. In a thermochemical equation, the enthalpy change of a reaction is shown as a ΔH value following the equation for the reaction. This ΔH value indicates the amount of heat associated with the reaction involving the number of moles of reactants and products as shown in the chemical equation. For example, consider this equation:

${\ce{H}}_{2}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{H}}_{2}\ce{O}\left(l\right)\Delta\text{H}=-286\text{ kJ}$

This equation indicates that when 1 mole of hydrogen gas and $\frac{1}{2}$ mole of oxygen gas at some temperature and pressure change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released to the surroundings. If the coefficients of the chemical equation are multiplied by some factor, the enthalpy change must be multiplied by that same factor (ΔH is an extensive property):

$\begin{array}{l}\left(\text{two-fold increase in amounts}\right)\\ 2{\ce{H}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{H}}_{2}\ce{O}\left(l\right)\Delta\text{H}=2\times \left(-286\text{ kJ}\right)=-572\text{ kJ}\\ \left(\text{two-fold decrease in amounts}\right)\\ \frac{1}{2}{\ce{H}}_{2}\left(g\right)+\frac{1}{4}{\ce{O}}_{2}\left(g\right)\rightarrow\frac{1}{2}{\ce{H}}_{2}\ce{O}\left(l\right)\Delta\text{H}=\frac{1}{2}\times \left(-286\text{ kJ}\right)=-143\text{ kJ}\end{array}$

• The enthalpy change of a reaction depends on the physical state of the reactants and products of the reaction (whether we have gases, liquids, solids, or aqueous solutions), so these must be shown. For example, when 1 mole of hydrogen gas and $\frac{1}{2}$ mole of oxygen gas change to 1 mole of liquid water at the same temperature and pressure, 286 kJ of heat are released. If gaseous water forms, only 242 kJ of heat are released.

${\ce{H}}_{2}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{H}}_{2}\ce{O}\left(g\right)\Delta\text{H}=-242\text{ kJ}$

### Example 10.3.1: Measurement of an Enthalpy Change

When 0.0500 mol of $\ce{HCl}$(aq) reacts with 0.0500 mol of $\ce{NaOH}$(aq) to form 0.0500 mol of $\ce{NaCl}$(aq), 2.9 kJ of heat are produced. What is ΔH, the enthalpy change, per mole of acid reacting, for the acid-base reaction run under the conditions described in Example 10.2.3 of Calorimetry?

$\ce{HCl}\left(aq\right)+\ce{NaOH}\left(aq\right)\rightarrow\ce{NaCl}\left(aq\right)+{\ce{H}}_{2}\ce{O}\left(l\right)$

Show Solution

For the reaction of 0.0500 mol acid ($\ce{HCl}$), q = -2.9 kJ. This ratio $\dfrac{-2.9\text{ kJ}}{0.0500\text{mol}\text{HCl}}$ can be used as a conversion factor to find the heat produced when 1 mole of $\ce{HCl}$ reacts:

$\Delta\text{H}=1\cancel{\text{mol}\ce{HCl}}\times \dfrac{-2.9\text{ kJ}}{0.0500\cancel{\text{mol}\ce{HCl}}}=-58\text{ kJ}$

The enthalpy change when 1 mole of HCl reacts is −58 kJ. Since that is the number of moles in the chemical equation, we write the thermochemical equation as:

$\ce{HCl}\left(aq\right)+\ce{NaOH}\left(aq\right)\rightarrow\ce{NaCl}\left(aq\right)+{\ce{H}}_{2}\ce{O}\left(l\right)\Delta\text{H}=-58\text{ kJ}$

Be sure to take both stoichiometry and limiting reactants into account when determining the ΔH for a chemical reaction.

### Example 10.3.2: Another Example of the Measurement of an Enthalpy Change

A gummy bear contains 2.67 g sucrose, $\ce{C_{12}H_{22}O_{11}}$. When it reacts with 7.19 g potassium chlorate, $\ce{KClO3}$, 43.7 kJ of heat are produced. Determine the enthalpy change for the reaction

${\ce{C}}_{12}{\ce{H}}_{22}{\ce{O}}_{11}\left(aq\right)+8{\ce{KClO}}_{3}\left(aq\right)\rightarrow 12{\ce{CO}}_{2}\left(g\right)+11{\ce{H}}_{2}\ce{O}\left(l\right)+8\ce{KCl}\left(aq\right)$

Show Solution

We have $2.67\cancel{\text{g}}\times \dfrac{1\text{ mol}}{342.3\cancel{\text{g}}}=0.00780\text{ mol }{\ce{C}}_{12}{\ce{H}}_{22}{\ce{O}}_{11}$ available, and $7.19\cancel{\text{g}}\times \dfrac{1\text{ mol}}{122.5\cancel{\text{g}}}=0.0587\text{ mol }{\ce{KClO}}_{3}$ available. Since $0.0587\text{ mol }{\ce{KClO}}_{3}\times \dfrac{1\text{ mol }{\ce{C}}_{12}{\ce{H}}_{22}{\ce{O}}_{11}}{8\text{ mol }{\ce{KClO}}_{3}}=0.00734\text{ mol }{\ce{C}}_{12}{\ce{H}}_{22}{\ce{O}}_{11}$ is needed, $\ce{C_{12}H_{22}O_{11}}$ is the excess reactant and $\ce{KClO3}$ is the limiting reactant.

The reaction uses 8 mol $\ce{KClO3}$, and the conversion factor is $\dfrac{-43.7\text{ kJ}}{0.0587\text{ mol }{\ce{KClO}}_{3}},$ so we have

$\Delta\text{H}=8\text{mol}\times \dfrac{-43.7\text{ kJ}}{0.0587\text{ mol }{\ce{KClO}}_{3}}=-5960\text{ kJ}$

The enthalpy change for this reaction is $-5960\text{ kJ}$, and the thermochemical equation is:

${\ce{C}}_{12}{\ce{H}}_{22}{\ce{O}}_{11}+8{\ce{KClO}}_{3}\rightarrow 12{\ce{CO}}_{2}+11{\ce{H}}_{2}\ce{O}+8\ce{KCl}\Delta\text{H}=-5960\text{ kJ}$

Enthalpy changes are typically tabulated for reactions in which both the reactants and products are at the same conditions. A standard state is a commonly accepted set of conditions used as a reference point for the determination of properties under other different conditions. For chemists, the IUPAC standard state refers to materials under a pressure of 1 bar and solutions at 1 M, and does not specify a temperature. Many thermochemical tables list values with a standard state of 1 atm. Because the ΔH of a reaction changes very little with such small changes in pressure (1 bar = 0.987 atm), ΔH values (except for the most precisely measured values) are essentially the same under both sets of standard conditions. We will include a superscripted “o” in the enthalpy change symbol to designate standard state. Since the usual (but not technically standard) temperature is 298.15 K, this temperature will be assumed unless some other temperature is specified. Thus, the symbol $\left(\Delta{H}^{\circ}\right)$ is used to indicate an enthalpy change for a process occurring under these conditions. (The symbol ΔH is used to indicate an enthalpy change for a reaction occurring under nonstandard conditions.)

The enthalpy changes for many types of chemical and physical processes are available in the reference literature, including those for combustion reactions, phase transitions, and formation reactions. As we discuss these quantities, it is important to pay attention to the extensive nature of enthalpy and enthalpy changes. Since the enthalpy change for a given reaction is proportional to the amounts of substances involved, it may be reported on that basis (i.e., as the ΔH for specific amounts of reactants). However, we often find it more useful to divide one extensive property (ΔH) by another (amount of substance), and report a per-amount intensive value of ΔH, often “normalized” to a per-mole basis. (Note that this is similar to determining the intensive property specific heat from the extensive property heat capacity, as seen previously.)

## Standard Enthalpy of Combustion

Standard enthalpy of combustion $\left(\Delta{H}_{C}^{\circ }\right)$ is the enthalpy change when 1 mole of a substance burns (combines vigorously with oxygen) under standard state conditions; it is sometimes called “heat of combustion.” For example, the enthalpy of combustion of ethanol, -1366.8 kJ/mol, is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 °C and 1 atmosphere pressure, yielding products also at 25 °C and 1 atm.

${\ce{C}}_{2}{\ce{H}}_{5}\ce{OH}\left(l\right)+3{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{CO}}_{2}+3{\ce{H}}_{2}\ce{O}\left(l\right)\qquad\Delta{H}^{\circ }=-\text{1366.8}\text{ kJ}$

Enthalpies of combustion for many substances have been measured; a few of these are listed in Table 10.3.1. Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline.

Table 10.3.1. Standard Molar Enthalpies of Combustion
Substance Combustion Reaction Enthalpy of Combustion, $\Delta{H}_{c}^{\circ }$ $\left(\frac{\text{kJ}}{\text{mol}}\text{ at }25^{\circ}\text{ C}\right)$
carbon $\ce{C}\left(s\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)$ −393.5
hydrogen ${\ce{H}}_{2}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{H}}_{2}\ce{O}\left(l\right)$ −285.8
magnesium $\ce{Mg}\left(s\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow\ce{MgO}\left(s\right)$ −601.6
sulfur $\ce{S}\left(s\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{SO}}_{2}\left(g\right)$ −296.8
carbon monoxide $\ce{CO}\left(g\right)+\frac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)$ −283.0
methane ${\ce{CH}}_{4}\left(g\right)+2{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)+2{\ce{H}}_{2}\ce{O}\left(l\right)$ −890.8
acetylene ${\ce{C}}_{2}{\ce{H}}_{2}\left(g\right)+\frac{5}{2}{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{CO}}_{2}\left(g\right)+{\ce{H}}_{2}\ce{O}\left(l\right)$ −1301.1
ethanol ${\ce{C}}_{2}{\ce{H}}_{5}\ce{OH}\left(l\right)+2{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)+3{\ce{H}}_{2}\ce{O}\left(l\right)$ −1366.8
methanol ${\ce{CH}}_{3}\ce{OH}\left(l\right)+\frac{3}{2}{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)+2{\ce{H}}_{2}\ce{O}\left(l\right)$ −726.1
isooctane ${\ce{C}}_{8}{\ce{H}}_{18}\left(l\right)+\frac{25}{2}{\ce{O}}_{2}\left(g\right)\rightarrow 8{\ce{CO}}_{2}\left(g\right)+9{\ce{H}}_{2}\ce{O}\left(l\right)$ −5461

### Example 10.3.3: Using Enthalpy of Combustion

As Figure 10.3.3 suggests, the combustion of gasoline is a highly exothermic process. Let us determine the approximate amount of heat produced by burning 1.00 L of gasoline, assuming the enthalpy of combustion of gasoline is the same as that of isooctane, a common component of gasoline. The density of isooctane is 0.692 g/mL.

Show Solution

Starting with a known amount (1.00 L of isooctane), we can perform conversions between units until we arrive at the desired amount of heat or energy. The enthalpy of combustion of isooctane provides one of the necessary conversions. Table 1 gives this value as −5460 kJ per 1 mole of isooctane ($\ce{C8H_{18}}$).

Using these data,

$1.00\cancel{\text{L }{\ce{C}}_{8}{\ce{H}}_{18}}\times \dfrac{1000\cancel{\text{mL }{\ce{C}}_{8}{\ce{H}}_{18}}}{1\cancel{\text{L }{\ce{C}}_{8}{\ce{H}}_{18}}}\times \dfrac{0.692\cancel{\text{g }{\ce{C}}_{8}{\ce{H}}_{18}}}{1\cancel{\text{mL }{\ce{C}}_{8}{\ce{H}}_{18}}}\times \dfrac{1\cancel{\text{mol}{\ce{C}}_{8}{\ce{H}}_{18}}}{114\cancel{\text{g}{\ce{C}}_{8}{\ce{H}}_{18}}}\times \dfrac{-5460\text{ kJ}}{1\cancel{\text{mol}{\ce{C}}_{8}{\ce{H}}_{18}}}=-3.31\times {10}^{4}\text{ kJ}$

The combustion of 1.00 L of isooctane produces 33,100 kJ of heat. (This amount of energy is enough to melt 99.2 kg, or about 218 lbs, of ice.)

Note: If you do this calculation one step at a time, you would find:

$\begin{array}{l}\\ 1.00\text{L }{\ce{C}}_{8}{\ce{H}}_{18}\rightarrow 1.00\times {10}^{3}\text{mL }{\ce{C}}_{8}{\ce{H}}_{18}\\ 1.00\times {10}^{3}\text{mL }{\ce{C}}_{8}{\ce{H}}_{18}\rightarrow 692\text{g}{\ce{C}}_{8}{\ce{H}}_{18}\\ 692\text{g}{\ce{C}}_{8}{\ce{H}}_{18}\rightarrow 6.07\text{mol}{\ce{C}}_{8}{\ce{H}}_{18}\\ 692\text{g}{\ce{C}}_{8}{\ce{H}}_{18}\rightarrow-3.31\times {10}^{4}\text{ kJ}\end{array}$

### Emerging Algae-Based Energy Technologies (Biofuels)

As reserves of fossil fuels diminish and become more costly to extract, the search is ongoing for replacement fuel sources for the future. Among the most promising biofuels are those derived from algae (Figure 10.3.4). The species of algae used are nontoxic, biodegradable, and among the world’s fastest growing organisms. About 50% of algal weight is oil, which can be readily converted into fuel such as biodiesel. Algae can yield 26,000 gallons of biofuel per hectare—much more energy per acre than other crops. Some strains of algae can flourish in brackish water that is not usable for growing other crops. Algae can produce biodiesel, biogasoline, ethanol, butanol, methane, and even jet fuel.

According to the US Department of Energy, only 39,000 square kilometers (about 0.4% of the land mass of the US or less than $\frac{1}{7}$ of the area used to grow corn) can produce enough algal fuel to replace all the petroleum-based fuel used in the US. The cost of algal fuels is becoming more competitive—for instance, the US Air Force is producing jet fuel from algae at a total cost of under \$5 per gallon. The process used to produce algal fuel is as follows: grow the algae (which use sunlight as their energy source and $\ce{CO2}$ as a raw material); harvest the algae; extract the fuel compounds (or precursor compounds); process as necessary (e.g., perform a transesterification reaction to make biodiesel); purify; and distribute (Figure 10.3.5).

## Standard Enthalpy of Formation

A standard enthalpy of formation $\Delta{H}_{\text{f}}^{\circ}$ is an enthalpy change for a reaction in which exactly 1 mole of a pure substance is formed from free elements in their most stable states under standard state conditions. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess’s law.

The standard enthalpy of formation of $\ce{CO2}$(g) is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:

$\ce{C}\left(s\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)\qquad\Delta{H}_{\text{f}}^{\circ }=\Delta{H}_{298}^{\circ }=-393.5\text{ kJ}$

starting with the reactants at a pressure of 1 atm and 25°C (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of $\ce{CO2}$, also at 1 atm and 25°C. For nitrogen dioxide, $\ce{NO2}$(g), $\Delta{H}_{\text{f}}^{\circ }$ is 33.2 kJ/mol. This is the enthalpy change for the reaction:

$\frac{1}{2}{\ce{N}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{NO}}_{2}\left(g\right)\qquad\Delta{H}_{\text{f}}^{\circ }=\Delta{H}_{\text{298}}^{\circ }=\text{+33.2}\text{ kJ}$

A reaction equation with $\frac{1}{2}$ mole of $\ce{N2}$ and 1 mole of $\ce{O2}$ is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, $\ce{NO2}$(g).

You will find a table of standard enthalpies of formation of many common substances in Standard Thermodynamic Properties for Selected Substances. These values indicate that formation reactions range from highly exothermic (such as −2984 kJ/mol for the formation of $\ce{P4O_{10}}$) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, $\ce{C2H2}$). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under standard conditions, which is 1 atm for gases and 1 M for solutions.

### Example 10.3.4: Evaluating an Enthalpy of Formation

Ozone, $\ce{O3}$(g), forms from oxygen, $\ce{O2}$(g), by an endothermic process. Ultraviolet radiation is the source of the energy that drives this reaction in the upper atmosphere. Assuming that both the reactants and products of the reaction are in their standard states, determine the standard enthalpy of formation, $\Delta{H}_{\text{f}}^{\circ}$ of ozone from the following information:

$3{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{O}}_{3}\left(g\right)\qquad\Delta{H}_{298}^{\circ }=\text{+286}\text{ kJ}$

Show Solution

$\Delta{H}_{\text{f}}^{\circ }$ is the enthalpy change for the formation of one mole of a substance in its standard state from the elements in their standard states. Thus, $\Delta{H}_{\text{f}}^{\circ }$ for $\ce{O3}$(g) is the enthalpy change for the reaction:

$\dfrac{3}{2}{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{O}}_{3}\left(g\right)$

For the formation of 2 mol of $\ce{O3}$(g), $\Delta{H}_{\text{298}}^{\circ }=\text{+286 kJ}$. This ratio, $\left(\dfrac{286\text{ kJ}}{2\text{ mol }{\ce{O}}_{3}}\right)$, can be used as a conversion factor to find the heat produced when 1 mole of $\ce{O3}$(g) is formed, which is the enthalpy of formation for $\ce{O3}$(g):

$\Delta\text{H}^{\circ }\text{for }1\text{ mole of }{\ce{O}}_{3}\left(g\right)=1\cancel{\text{ mol }{\ce{O}}_{3}}\times \dfrac{286\text{ kJ}}{2\cancel{\text{ mol }{\ce{O}}_{3}}}=143\text{ kJ}$

Therefore, $\Delta{H}_{\text{f}}^{\circ }\left[{\ce{O}}_{3}\left(g\right)\right]=\text{+143}\text{ kJ/mol}.$

### Example 10.3.5: Writing Reaction Equations for $\Delta{H}_{\text{f}}^{\circ }$

Write the heat of formation reaction equations for:

1. $\ce{C2H5OH}$(l)
2. $\ce{Ca3(PO4)2}$(s)
Show Solutions

Remembering that $\Delta{H}_{\text{f}}^{\circ }$ reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:

1. $2\ce{C}\left(s,\text{graphite}\right)+3{\ce{H}}_{2}\left(g\right)+\dfrac{1}{2}{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{C}}_{2}{\text{H}}_{5}\text{OH}\left(l\right)$
2. $3\ce{Ca}\left(s\right)+\dfrac{1}{2}{\ce{P}}_{4}\left(s\right)+4{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{Ca}}_{3}{({\ce{PO}}_{4}\text{)}}_{2}\left(s\right)$

Note: The standard state of carbon is graphite, and phosphorus exists as $\ce{P}_{4}$.

### Key Concepts and Summary

If a chemical change is carried out at constant pressure and the only work done is caused by expansion or contraction, q for the change is called the enthalpy change with the symbol ΔH, or $\Delta{H}_{\text{298}}^{\circ }$ for reactions occurring under standard state conditions. The value of ΔH for a reaction in one direction is equal in magnitude, but opposite in sign, to ΔH for the reaction in the opposite direction, and ΔH is directly proportional to the quantity of reactants and products. Examples of enthalpy changes include enthalpy of combustion, enthalpy of fusion, enthalpy of vaporization, and standard enthalpy of formation. The standard enthalpy of formation, $\Delta{H}_{\text{f}}^{\circ },$ is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar (standard state). Many of the processes are carried out at 298.15 K.

#### Key Equations

• $\Delta U=q+w$
• $\Delta{H}_{\text{reaction}}^{\circ }=\sum n\times \Delta{H}_{\text{f}}^{\circ }\text{(products)}-\sum n\times \Delta{H}_{\text{f}}^{\circ }\left(\text{reactants}\right)$

### Try It

1. Explain how the heat measured in Example 10.2.3 of Calorimetry differs from the enthalpy change for the exothermic reaction described by the following equation:
1. $\ce{HCl}\left(aq\right)+\ce{NaOH}\left(aq\right)\rightarrow\ce{NaCl}\left(aq\right)+{\ce{H}}_{2}\ce{O}\left(l\right)$
2. Calculate the enthalpy of solution (ΔH for the dissolution) per mole of NH4NO3 under the conditions described in Figure 10.2.6 in Calorimetry.
3. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions?
4. How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions?
5. When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions?
6. Does the standard enthalpy of formation of $\ce{H2O}$(g) differ from ΔH° for the reaction ${\ce{2H}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{H}}_{2}\ce{O}\left(g\right)?$
7. How many kilojoules of heat will be released when exactly 1 mole of manganese, $\ce{Mn}$, is burned to form $\ce{Mn3O4}$(s) at standard state conditions?
8. The following sequence of reactions occurs in the commercial production of aqueous nitric acid:
1. $\begin{array}{cc}4{\ce{NH}}_{3}\left(g\right)+5{\ce{O}}_{2}\left(g\right)\rightarrow 4\ce{NO}\left(g\right)+6{\ce{H}}_{2}\ce{O}\left(l\right) & \Delta\text{H}=-907\text{ kJ} \\ 2\ce{NO}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{NO}}_{2}\left(g\right) & \Delta\text{H}=-113\text{ kJ} \\ 3{\ce{NO}}_{2}+{\ce{H}}_{2}\ce{O}\left(l\right)\rightarrow 2{\ce{HNO}}_{2}\left(aq\right)+\ce{NO}\left(g\right) & \Delta\text{H}=-139\text{ kJ}\end{array}$
9. Determine the total energy change for the production of one mole of aqueous nitric acid by this process.
10. Calculate the standard molar enthalpy of formation of NO(g) from the following data:
1. $\begin{array}{ll}{\ce{N}}_{2}\left(g\right)+2{\ce{O}}_{2}\rightarrow 2{\ce{NO}}_{2}\left(g\right) & \Delta{H}_{298}^{\circ }=66.4\text{ kJ}\\ \ce{2NO}\left(g\right)+{\ce{O}}_{2}\rightarrow 2{\ce{NO}}_{2}\left(g\right) & \Delta{H}_{298}^{\circ }=-114.1\text{ kJ}\end{array}$
11. Using the data in Standard Thermodynamic Properties for Selected Substances, calculate the standard enthalpy change for each of the following reactions:
1. $\ce{Si}\left(s\right)+2{\ce{F}}_{2}\left(g\right)\rightarrow{\ce{SiF}}_{4}\left(g\right)$
2. $2\ce{C}\left(s\right)+2{\ce{H}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CH}}_{3}{\ce{CO}}_{2}\ce{H}\left(l\right)$
3. ${\ce{CH}}_{4}\left(g\right)+{\ce{N}}_{2}\left(g\right)\rightarrow\ce{HCN}\left(g\right){\ce{+NH}}_{3}\left(g\right)$
4. ${\ce{Cs}}_{2}\left(g\right)+3{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{CCl}}_{4}\left(g\right){\ce{+S}}_{2}{\ce{Cl}}_{2}\left(g\right)$
12. Calculate ΔH for the process ${\ce{Hg}}_{2}{\ce{Cl}}_{2}\left(s\right)\rightarrow 2\ce{Hg}\left(l\right)+{\ce{Cl}}_{2}\left(g\right)$ from the following information:
1. $\begin{array}{ll}\\ \ce{Hg}\left(l\right)+{\ce{Cl}}_{\ce{2}}\left(g\right)\rightarrow{\ce{HgCl}}_{\ce{2}}\left(s\right) & \Delta H=-224\text{ kJ}\\ \ce{Hg}\left(l\right)+{\ce{HgCl}}_{\ce{2}}\left(s\right)\rightarrow{\ce{Hg}}_{\ce{2}}{\ce{Cl}}_{2}\left(s\right) & \Delta H=-41.2\text{ kJ}\end{array}$
13. Calculate the enthalpy of combustion of butane, $\ce{C4H_{10}}$(g) for the formation of $\ce{H2O}$(g) and $\ce{CO2}$(g). The enthalpy of formation of butane is −126 kJ/mol.
14. The enthalpy of combustion of hard coal averages −35 kJ/g, that of gasoline, 1.28 × 105 kJ/gal. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is 0.692 g/mL (the same as the density of isooctane).
Show Selected Solutions
1.  The enthalpy change of the indicated reaction is for exactly 1 mol $\ce{HCL}$ and 1 mol $\ce{NaOH}$; the heat in the example is produced by 0.0500 mol $\ce{HCl}$ and 0.0500 mol $\ce{NaOH}$.
2. The molar mass of $\ce{NH4NO3}$ is 80.0423 g/mol. From the example, 1000 J is required to dissolve 3.21 g of $\ce{NH4NO3}$. One mole under the same conditions would require

$\dfrac{80.0432\cancel{\text{g}}{\text{ mol}}^{-1}}{3.21\cancel{\text{g}}}\times 1000\text{ J}=25\text{ kJ}{\text{ mol}}^{-1}.$

(The heat of solution is positive because the process is endothermic.)

3. The heat of combustion is -1301.1 as given in Table 10.3.1.

heat released = 4.00 mol × (-1301.1 kJ/mol) = 5204.4 kJ

4. The value of ΔHcomb = -5461 kJ/mol. To produce 100 kJ requires:

$\dfrac{100\text{ kJ}}{5461\text{ kJ}{\text{mol}}^{-1}}=1.83\times {10}^{-2}\text{ mol}.$

5. The molar mass of $\ce{CH4}$ is 16.04 g/mol. Find the mole of $\ce{CH4}$ present:

$\dfrac{2.50\cancel{\text{g}}}{16.04\cancel{\text{g}}{\text{mol}}^{-1}}=0.15586\text{ mol}$

$\Delta{H}_{\text{comb}}=\dfrac{125\text{ kJ}}{0.15586\text{mol}}=802\text{ kJ}{\text{mol}}^{-1}.$

6. No. The standard enthalpy of formation can be determined for anything, including $\ce{H2O}$(g), and water does not have to be liquid in this case, it’s the gas-phase water that is the substance for which the heat of formation is to be found. However, the heat of this reaction is defined for two moles of $\ce{H2O}$(g), thus the heat of formation is half of the heat of the reaction.
7. This process requires 3 mol of Mn. For 1 mol, $\frac{1}{3}\left(-1378.83\text{ kJ}\right)=459.6\text{ kJ}.$
8. Enough material must be produced in each stage to proceed with the next. So $\frac{1}{2}$ of reaction 1 produces the $\ce{NO}$ required in reaction 2. But $\frac{3}{2}$ units of reaction 2 are required to provide enough $\ce{NO2}$ for reaction 3. Up to this stage, the heat produced is $\frac{3}{2}\left[\frac{1}{2}\left(-907\right)+\left(-113\right)\right]=\frac{3}{2}\left(-566.55\right)=-850\text{ kJ}$ to have the material to proceed with reaction 3. Therefore, from beginning to end, –850 + (–139) = –989 kJ are released. The question asks for the enthalpy change for 1 mole. Therefore, division of the last answer by 2 gives -495 kJ/mol.
9. Hess’s law can be applied to the two equations by reversing the direction of the second equation. The first equation is a formation reaction and is so indicated by writing $\Delta{H}_{298}^{\circ }$.

$\begin{array}{l}{\ce{N}}_{2}\left(g\right)+2{\ce{O}}_{2}\left(g\right)\rightarrow 2{\ce{NO}}_{2}\left(g\right)\Delta{H}_{298}^{\circ }=66.4\text{ kJ}\\ 2{\ce{NO}}_{\ce{2}}\left(g\right)\rightarrow 2\ce{NO}\left(g\right)+{\ce{O}}_{2}\left(g\right)\Delta H^{\circ }=114.1\text{ kJ}\end{array}$

${\ce{N}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow 2\ce{NO}\left(g\right)\Delta{H}^{\circ }=180.5\text{ kJ}$

This is the heat of formation of 2 mol of $\ce{NO}$. For 1 mol,

$\Delta{H}_{298}^{\circ }=\frac{180.5\text{ kJ}}{2}=90.3{\text{ mol}}^{-1}\text{of}\ce{ NO}.$

10. The standard enthalpy changes are as follows:
1. $\ce{Si}\left(s\right)+2{\ce{F}}_{2}\left(g\right)\rightarrow{\ce{SiF}}_{4}\left(g\right)$
$\begin{array}{rll} \\ \Delta{H}_{\text{reaction}}^{\circ }&=&\Delta{H}_{\text{products}}^{\circ }-\Delta{H}_{\text{reactants}}^{\circ }\\ &=&\Delta{H}_{{\ce{SiF}}_{4}\left(g\right)}^{\circ }-\Delta{H}_{\ce{Si}\left(s\right)}^{\circ }-2\Delta{H}_{{\ce{F}}_{2}\left(g\right)}^{\circ }\\ &=&-1614.9-\left(0\right)-2\left(0\right)=-1615.0\text{ kJ}{\text{mol}}^{-1}\end{array}$
2. $2\ce{C}\left(s\right)+2{\ce{H}}_{2}\left(g\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CH}}_{3}{\ce{CO}}_{2}\ce{H}\left(l\right)$
$\begin{array}{rll} \\ \Delta{H}_{\text{reaction}}^{\circ }&=&\Delta{H}_{\text{products}}^{\circ }-\Delta{H}_{\text{reactants}}^{\circ }\\ &=&\Delta{H}_{{\ce{CH}}_{3}{\ce{CO}}_{2}\ce{H}\left(l\right)}^{\circ }-2\Delta{H}_{\ce{C}\left(s\right)}^{\circ }-2\Delta{H}_{{\ce{H}}_{2}\left(g\right)}^{\circ }-\Delta{H}_{{\ce{O}}_{2}\left(g\right)}^{\circ }\\ &=&-484.5-2\left(0\right)-2\left(0\right)-\left(0\right)=-484.3\text{ kJ}{\text{mol}}^{-1}\end{array}$
3. ${\text{CH}}_{\text{4}}\left(g\right)\rightarrow\ce{C}\left(s\right)+2{\ce{H}}_{2}\left(g\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Delta{H}_{1}^{\circ }=-\left(-74.6\text{ kJ}\right)$
$\begin{array}{lll} \\ \frac{\ce{1}}{\ce{2}}{\ce{H}}_{\ce{2}}\left(g\right)+\ce{C}\left(s\right)+\frac{1}{2}{\ce{N}}_{2}\left(g\right)\rightarrow\ce{HCN}\left(g\right)\hfill & \hfill & \Delta{H}_{2}^{\circ }=135.5\text{ kJ}\hfill \\ \underline{\frac{\text{1}}{\ce{2}}{\ce{N}}_{\ce{2}}\left(g\right)+\frac{3}{2}{\ce{H}}_{2}\left(g\right)\rightarrow{\ce{NH}}_{3}\left(g\right)}\hfill & \hfill & \underline{\Delta{H}_{3}^{\circ }=-45.9\text{ kJ}}\\{{\ce{CH}}_{4}\left(g\right)+{\ce{N}}_{2}\left(g\right)\rightarrow\ce{HCN}\left(g\right)+{\text{NH}}_{3}\left(g\right)}\hfill & \hfill &{\Delta H^{\circ }=164.2\text{ kJ}}\end{array}$
4. ${\ce{CS}}_{\ce{2}}\left(g\right)\rightarrow\ce{C}\left(s\right)+2S\left(s\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta{H}_{1}^{\circ }=-\left(116.9\text{ kJ}\right)$
$\begin{array}{lll} \\ \ce{C}\left(s\right)+2{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{CCl}}_{\ce{4}}\left(g\right)\hfill & \hfill & \Delta{H}_{2}^{\circ }=-95.7\text{ kJ}\hfill \\ \underline{2\ce{S}\left(s\right)+{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{S}}_{2}{\ce{Cl}}_{2}\left(g\right)}\hfill & \hfill & \underline{\Delta{H}_{3}^{\circ }=-19.50\text{ kJ}}\\{{\ce{CS}}_{2}\left(g\right)+3{\ce{Cl}}_{2}\left(g\right)\rightarrow{\ce{CCl}}_{4}\left(g\right)+{\ce{S}}_{\ce{2}}{\ce{Cl}}_{2}\left(g\right)}\hfill & \hfill &{\Delta H^{\circ }=-232.1\text{ kJ}}\end{array}$
11. Reverse the direction of both equations and add the new equations and enthalpies.

$\begin{array}{lll}{\text{HgCl}}_{2}\rightarrow\text{Hg}\left(l\right)+{\text{Cl}}_{2}\left(l\right)\hfill & \hfill & \Delta H=224\text{ kJ}\hfill \\ \underline{{\text{Hg}}_{\text{2}}{\text{Cl}}_{2}\left(s\right)\rightarrow\text{Hg}\left(l\right)+{\text{HgCl}}_{2}\left(s\right)}\hfill & \hfill & \underline{\Delta H=41.2\text{ kJ}}\\{{\text{Hg}}_{\text{2}}{\text{Cl}}_{2}\left(s\right)\rightarrow 2\text{Hg}\left(l\right)+{\text{Cl}}_{2}\left(g\right)}\hfill & \hfill &{\Delta H=265\text{ kJ}} \end{array}$

12. The enthalpy can be found through the following steps:

$\begin{array}{llll}\text{Step}1:\hfill & 4\left[\ce{C}\left(s\right)+{\ce{O}}_{2}\left(g\right)\rightarrow{\ce{CO}}_{2}\left(g\right)\right]\hfill & \hfill & \Delta H^{\circ }=4\left(-394\text{ kJ}{\text{mol}}^{-1}\right)\hfill \\ \text{Step}2:\hfill & 5\left[{\ce{H}}_{\ce{2}}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow{\ce{H}}_{\ce{2}}\ce{O}\left(g\right)\right]\hfill & \hfill & \Delta H^{\circ }=5\left(-242\text{ kJ}{\text{mol}}^{-1}\right)\hfill \\ \text{Step}\text{3:}\hfill & \underline{{\ce{C}}_{\ce{4}}{\ce{H}}_{10}\rightarrow 4\ce{C}\left(s\right)+5{\ce{H}}_{2}\left(g\right)}\hfill & \hfill & \underline{\Delta H^{\circ }=+126\text{ kJ}{\text{mol}}^{-1}}\hfill \\ \text{Sum:}\hfill & {\ce{C}}_{\ce{4}}{\ce{H}}_{10}+\frac{13}{2}{\ce{O}}_{2}\left(g\right)\rightarrow 4{\ce{CO}}_{2}\left(g\right)+5{\ce{H}}_{2}\ce{O}\left(l\right)\hfill & \hfill & \Delta{H}_{298}^{\circ }=-2660\text{ kJ}{\text{mol}}^{-1}\hfill \end{array}$

13. the amount of heat produced by burning of 1.0 gallon of gasoline is:

q = 1.0 gallon × (–1.28 × 105 kJ/gal) =–1.28 × 105 kJ
Mass × (–35 kJ/g) =–1.28 × 105 kJ
Mass = 3657 g or 3.7 kg

## Glossary

chemical thermodynamics: area of science that deals with the relationships between heat, work, and all forms of energy associated with chemical and physical processes

enthalpy (H): sum of a system’s internal energy and the mathematical product of its pressure and volume

enthalpy change (ΔH): heat released or absorbed by a system under constant pressure during a chemical or physical process

expansion work (pressure-volume work): work done as a system expands or contracts against external pressure

first law of thermodynamics: internal energy of a system changes due to heat flow in or out of the system or work done on or by the system

hydrocarbon: compound composed only of hydrogen and carbon; the major component of fossil fuels

internal energy (U): total of all possible kinds of energy present in a substance or substances

standard enthalpy of combustion $\text{(}\Delta{H}_{\text{c}}^{\circ }\text{)}$: heat released when one mole of a compound undergoes complete combustion under standard conditions

standard enthalpy of formation $\text{(}\Delta{H}_{\text{f}}^{\circ }\text{)}$: enthalpy change of a chemical reaction in which 1 mole of a pure substance is formed from its elements in their most stable states under standard state conditions

standard state: set of physical conditions as accepted as common reference conditions for reporting thermodynamic properties; 1 bar of pressure, and solutions at 1 molar concentrations, usually at a temperature of 298.15 K

state function: property depending only on the state of a system, and not the path taken to reach that state