Learning Objectives

By the end of this section, you will be able to:

  • 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 in this chapter, 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:

ΔU=q+w\Delta U=q+w

as shown in Figure 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 rectangular diagram is shown. A green oval lies in the center of a tan field inside of a gray box. The tan field is labeled “Surroundings” and the equation “Δ U = q + w” is written at the bottom of the diagram. Two arrows face into the green oval and are labeled “q subscript in” and “w subscript on” while two more arrows face away from the oval and are labeled “q subscript out” and “w subscript by.” The center of the oval contains the terms “Δ U < 0”, “System,” and “Δ U > 0.” Figure 1. The internal energy, U, of a system can be changed by heat flow and work. If heat flows into the system, qin, or work is done on the system, won, its internal energy increases, ΔU < 0. If heat flows out of the system, qout, or work is done by the system, wby, its internal energy decreases, ΔU > 0.

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.

This view of an internal combustion engine on How Stuff Works illustrates the conversion of energy produced by the exothermic combustion reaction of a fuel such as gasoline into energy of motion.
As discussed, the relationship between internal energy, heat, and work can be represented as ΔU = 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 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).

An aerial photo depicts a view of Mount Kilimanjaro. A straight, green arrow labeled X is drawn from the term “base,” written at the bottom of the mountain, to the term “Summit,” written at the top of the mountain. Another arrow labeled Y is draw from the base to the summit alongside the green arrow, but this arrow is pink and has three large S-shaped curves along its length. Figure 2. Paths X and Y represent two different routes to the summit of Mt. Kilimanjaro. Both have the same change in elevation (altitude or elevation on a mountain is a state function; it does not depend on path), but they have very different distances traveled (distance walked is not a state function; it depends on the path). (credit: modification of work by Paul Shaffner)

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):


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:

ΔH=ΔU+PΔV\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:

PΔV=wP\Delta V=-w

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

ΔH=ΔU+PΔV=qp+ww=qp\begin{array}{rll}\Delta H&=&\Delta U+P\Delta V\\&=&{q}_{\text{p}}+w-w\\&=&{q}_{\text{p}}\end{array}

where qp 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:

  1. 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:


    This equation indicates that when 1 mole of hydrogen gas and
    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):

    (two-fold increase in amounts)2H2(g)+O2(g)2H2O(l)ΔH=2×(286kJ)=572kJ(two-fold decrease in amounts)12H2(g)+14O2(g)12H2O(l)ΔH=12×(286kJ)=143kJ\begin{array}{l}\left(\text{two-fold increase in amounts}\right)\\ 2{\text{H}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow 2{\text{H}}_{2}\text{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}{\text{H}}_{2}\left(g\right)+\frac{1}{4}{\text{O}}_{2}\left(g\right)\rightarrow\frac{1}{2}{\text{H}}_{2}\text{O}\left(l\right)\Delta\text{H}=\frac{1}{2}\times \left(-286\text{kJ}\right)=-143\text{kJ}\end{array}

  2. 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
    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.


  3. A negative value of an enthalpy change, ΔH, indicates an exothermic reaction; a positive value of ΔH 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).

Example 1: Measurement of an Enthalpy Change

When 0.0500 mol of HCl(aq) reacts with 0.0500 mol of NaOH(aq) to form 0.0500 mol of 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 3 of Calorimetry?


Check Your Learning

When 1.34 g Zn(s) reacts with 60.0 mL of 0.750 M HCl(aq), 3.14 kJ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:


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

Example 2: Another Example of the Measurement of an Enthalpy Change

A gummy bear contains 2.67 g sucrose, C12H22O11. When it reacts with 7.19 g potassium chlorate, KClO3, 43.7 kJ of heat are produced. Determine the enthalpy change for the reaction

C12H22O11(aq)+8KClO3(aq)12CO2(g)+11H2O(l)+8KCl(aq){\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\left(aq\right)+8{\text{KClO}}_{3}\left(aq\right)\rightarrow 12{\text{CO}}_{2}\left(g\right)+11{\text{H}}_{2}\text{O}\left(l\right)+8\text{KCl}\left(aq\right)

Check Your Learning

When 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of FeCl2(s) and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of FeCl2(s) is produced?

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, we will use a subscripted “298” to designate this temperature. Thus, the symbol
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.)


  1. Explain how the heat measured in Example 3 of Calorimetry differs from the enthalpy change for the exothermic reaction described by the following equation:
  2. Using the data in the check your learning section of Example 3 of Calorimetry, calculate ΔH in kJ/mol of AgNO3(aq) for the reaction:
  3. Calculate the enthalpy of solution (ΔH for the dissolution) per mole of NH4NO3 under the conditions described in Figure 6 in Calorimetry.
  4. Calculate ΔH for the reaction described by the equation.
    Ba(OH)28H2O(s)+2NH4SCN(aq)Ba(SCN)2(aq)+2NH3(aq)+10H2O(l)\text{Ba}{\left(\text{OH}\right)}_{2}\cdot 8{\text{H}}_{2}\text{O}\left(s\right)+2{\text{NH}}_{4}\text{SCN}\left(aq\right)\rightarrow\text{Ba}{\left(\text{SCN}\right)}_{2}\left(aq\right)+2{\text{NH}}_{3}\left(aq\right)+10{\text{H}}_{2}\text{O}\left(l\right)
  5. Calculate the enthalpy of solution (ΔH for the dissolution) per mole of CaCl2.

Enthalpy of Combustion

Standard enthalpy of combustion
(ΔHC)\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.

C2H5OH(l)+3O2(g)2CO2+3H2O(l)ΔH298=1366.8kJ{\text{C}}_{2}{\text{H}}_{5}\text{OH}\left(l\right)+3{\text{O}}_{2}\left(g\right)\rightarrow 2{\text{CO}}_{2}+3{\text{H}}_{2}\text{O}\left(l\right)\Delta{H}_{298}^{\circ }=-\text{1366.8}\text{kJ}

Enthalpies of combustion for many substances have been measured; a few of these are listed in Table 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 1. Standard Molar Enthalpies of Combustion
Substance Combustion Reaction Enthalpy of Combustion,
ΔHc\Delta{H}_{c}^{\circ }
(kJmol at 25 C)\left(\frac{\text{kJ}}{\text{mol}}\text{ at }25^{\circ}\text{ C}\right)
carbon monoxide
C2H2(g)+52O2(g)2CO2(g)+H2O(l){\text{C}}_{2}{\text{H}}_{2}\left(g\right)+\frac{5}{2}{\text{O}}_{2}\left(g\right)\rightarrow 2{\text{CO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)
C8H18(l)+252O2(g)8CO2(g)+9H2O(l){\text{C}}_{8}{\text{H}}_{18}\left(l\right)+\frac{25}{2}{\text{O}}_{2}\left(g\right)\rightarrow 8{\text{CO}}_{2}\left(g\right)+9{\text{H}}_{2}\text{O}\left(l\right)

Example 3: Using Enthalpy of Combustion

As Figure 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.

A picture shows a large ball of fire burning on a road. A fire truck and fireman are shown in the foreground. Figure 3. The combustion of gasoline is very exothermic. (credit: modification of work by “AlexEagle”/Flickr)

Check Your Learning

How much heat is produced by the combustion of 125 g of acetylene?

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 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.

Three pictures are shown and labeled a, b, and c. Picture a shows a microscopic view of algal organisms. They are brown, multipart strands and net-like structures on a background of light violet. Picture b shows five large tubs full of a brown liquid containing these algal organisms. Picture c depicts a cylinder full of green liquid in the foreground and a poster in the background that has the title “From Field to Fleet.” Figure 4. (a) Tiny algal organisms can be (b) grown in large quantities and eventually (c) turned into a useful fuel such as biodiesel. (credit a: modification of work by Micah Sittig; credit b: modification of work by Robert Kerton; credit c: modification of work by John F. Williams)

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
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 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 5).

A flowchart is shown that contains pictures and words. Reading from left to right, the terms “Grow,” “Harvest,” “Extract,” “Process and purify,” and “Jet fuel gasoline diesel” are shown with right-facing arrows in between each. Above each term, respectively, are diagrams of three containers, three cylinders lying side-by-side, a pyramid-like container with liquid inside, a factory, and a fuel pump. In the space above all of the diagrams and to the left of the images is a diagram of the sun. Figure 5. Algae convert sunlight and carbon dioxide into oil that is harvested, extracted, purified, and transformed into a variety of renewable fuels.

Watch this video to learn more about the process of creating algae biofuel.


  1. Although the gas used in an oxyacetylene torch (Figure 6) is essentially pure acetylene, the heat produced by combustion of one mole of acetylene in such a torch is likely not equal to the enthalpy of combustion of acetylene listed in Table 1. Considering the conditions for which the tabulated data are reported, suggest an explanation.

    Two pictures are shown and labeled a and b. Picture a shows a metal railroad tie being cut with the flame of an acetylene torch. Picture b shows a chemical cold pack containing ammonium nitrate. Figure 6. (a) An oxyacetylene torch. (b) A cold pack. (credit a: modification of work by “Skatebiker”/Wikimedia commons)
  2. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions?
  3. How much heat is produced by combustion of 125 g of methanol under standard state conditions?
  4. How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions?
  5. What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions?
  6. 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?
  7. How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed?
    HCl(aq)+NaOH(aq)NaCl(aq)+H2O(l)ΔH298=58kJ\text{HCl}\left(aq\right)+\text{NaOH}\left(aq\right)\rightarrow\text{NaCl}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\Delta{H}_{298}^{\circ }=-58\text{kJ}
    If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation?
  8. A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74 °C to 27.93 °C. What is the heat capacity of the calorimeter and its contents?
  9. Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, 6.00 kcal/mol) was used in household refrigerators. What mass of SO2 must be evaporated to remove as much heat as evaporation of 1.00 kg of CCl2F2 (enthalpy of vaporization is 17.4 kJ/mol)? The vaporization reactions for SO2 and CCl2F2 are
  10. Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 °C, as the heat provided when 100 g of steam is cooled from 110 °C to 100 °C.

Standard Enthalpy of Formation

A standard enthalpy of formation
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 CO2(g) is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:

C(s)+O2(g)CO2(g)ΔHf=ΔH298=393.5kJ\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{CO}}_{2}\left(g\right)\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 CO2, also at 1 atm and 25°C. For nitrogen dioxide, NO2(g),
ΔHf\Delta{H}_{\text{f}}^{\circ }
is 33.2 kJ/mol. This is the enthalpy change for the reaction:

12N2(g)+O2(g)NO2(g)ΔHf=ΔH298=+33.2kJ\frac{1}{2}{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{NO}}_{2}\left(g\right)\Delta{H}_{\text{f}}^{\circ }=\Delta{H}_{\text{298}}^{\circ }=\text{+33.2}\text{kJ}

A reaction equation with
mole of N2 and 1 mole of O2 is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, 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 P4O10) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, 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 4: Evaluating an Enthalpy of Formation

Ozone, O3(g), forms from oxygen, 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,
of ozone from the following information:

3O2(g)2O3(g)ΔH298=+286kJ3{\text{O}}_{2}\left(g\right)\rightarrow 2{\text{O}}_{3}\left(g\right)\Delta{H}_{298}^{\circ }=\text{+286}\text{kJ}

Check Your Learning

Hydrogen gas, H2, reacts explosively with gaseous chlorine, Cl2, to form hydrogen chloride, HCl(g). What is the enthalpy change for the reaction of 1 mole of H2(g) with 1 mole of Cl2(g) if both the reactants and products are at standard state conditions? The standard enthalpy of formation of HCl(g) is -92.3 kJ/mol.

Example 5: Writing Reaction Equations for
ΔHf\Delta{H}_{\text{f}}^{\circ }

Write the heat of formation reaction equations for:

  1. C2H5OH(l)
  2. Ca3(PO4)2(s)

Check Your Learning

Write the heat of formation reaction equations for:

  1. C2H5OC2H5(l)
  2. Na2CO3(s)


  1. Does the standard enthalpy of formation of H2O(g) differ from ΔH° for the reaction
    2H2(g)+O2(g)2H2O(g)?{\text{2H}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow 2{\text{H}}_{2}\text{O}\left(g\right)?
  2. Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO(s) to Hg(l) and O2(g) under standard conditions?
  3. How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn3O4(s) at standard state conditions?
  4. How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Fe2O3(s) at standard state conditions?
  5. The following sequence of reactions occurs in the commercial production of aqueous nitric acid:
    4NH3(g)+5O2(g)4NO(g)+6H2O(l)ΔH=907kJ4{\text{NH}}_{3}\left(g\right)+5{\text{O}}_{2}\left(g\right)\rightarrow 4\text{NO}\left(g\right)+6{\text{H}}_{2}\text{O}\left(l\right)\Delta\text{H}=-907\text{kJ}
    2NO(g)+O2(g)2NO2(g)ΔH=113kJ2\text{NO}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow 2{\text{NO}}_{2}\left(g\right)\Delta\text{H}=-113\text{kJ}
    3NO2+H2O(l)2HNO2(aq)+NO(g)ΔH=139kJ3{\text{NO}}_{2}+{\text{H}}_{2}\text{O}\left(l\right)\rightarrow 2{\text{HNO}}_{2}\left(aq\right)+\text{NO}\left(g\right)\Delta\text{H}=-139\text{kJ}

    Determine the total energy change for the production of one mole of aqueous nitric acid by this process.
  6. Both graphite and diamond burn.
    For the conversion of graphite to diamond:
    C(s,graphite)C(s,diamond)ΔH298=1.90kJ\text{C}\left(s,\text{graphite}\right)\rightarrow\text{C}\left(s,\text{diamond}\right)\Delta{H}_{298}^{\circ }=1.90\text{kJ}
    Which produces more heat, the combustion of graphite or the combustion of diamond?
  7. From the molar heats of formation in Standard Thermodynamic Properties for Selected Substances, determine how much heat is required to evaporate one mole of water:
  8. Which produces more heat?
    Os(s)2O2(g)OsO4(s)\text{Os}\left(s\right)\rightarrow 2{\text{O}}_{2}\left(g\right)\rightarrow{\text{OsO}}_{4}\left(s\right)
    Os(s)2O2(g)OsO4(g)\text{Os}\left(s\right)\rightarrow 2{\text{O}}_{2}\left(g\right)\rightarrow{\text{OsO}}_{4}\left(g\right)
    for the phase change
  9. Calculate
    ΔH298\Delta{H}_{298}^{\circ }
    for the process
    from the following information:
    Sb(s)+32Cl2(g)SbCl3(g)ΔH298=314kJSbCl3(s)+Cl2(g)SbCl5(g)ΔH298=80kJ\begin{array}{l}\text{Sb}\left(s\right)+\frac{3}{2}{\text{Cl}}_{2}\left(g\right)\rightarrow{\text{SbCl}}_{3}\left(g\right)\Delta{H}_{298}^{\circ }=-314\text{kJ}\\ {\text{SbCl}}_{3}\left(s\right)+{\text{Cl}}_{2}\left(g\right)\rightarrow{\text{SbCl}}_{5}\left(g\right)\Delta{H}_{298}^{\circ }=-80\text{kJ}\end{array}
  10. Calculate
    ΔH298\Delta{H}_{298}^{\circ }
    for the process
    from the following information:
    Zn(s)+S(s)ZnS(s)ΔH298=206.0kJZnS(s)+2O2(g)ZnSO4(s)ΔH298=776.8kJ\begin{array}{l}\text{Zn}\left(s\right)+\text{S}\left(s\right)\rightarrow\text{ZnS}\left(s\right)\Delta{H}_{298}^{\circ }=-206.0\text{kJ}\\ \text{ZnS}\left(s\right)+{\text{2O}}_{\text{2}}\left(g\right)\rightarrow{\text{ZnSO}}_{4}\left(s\right)\Delta{H}_{298}^{\circ }=-776.8\text{kJ}\end{array}

Hess’s Law

There are two ways to determine the amount of heat involved in a chemical change: measure it experimentally, or calculate it from other experimentally determined enthalpy changes. Some reactions are difficult, if not impossible, to investigate and make accurate measurements for experimentally. And even when a reaction is not hard to perform or measure, it is convenient to be able to determine the heat involved in a reaction without having to perform an experiment.

This type of calculation usually involves the use of Hess’s law, which states: If a process can be written as the sum of several stepwise processes, the enthalpy change of the total process equals the sum of the enthalpy changes of the various steps. Hess’s law is valid because enthalpy is a state function: Enthalpy changes depend only on where a chemical process starts and ends, but not on the path it takes from start to finish. For example, we can think of the reaction of carbon with oxygen to form carbon dioxide as occurring either directly or by a two-step process. The direct process is written:

C(s)+O2(g)CO2(g)ΔH298=394kJ\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{CO}}_{2}\left(g\right)\Delta{H}_{298}^{\circ }=-394\text{kJ}

In the two-step process, first carbon monoxide is formed:

C(s)+12O2(g)CO(g)ΔH298=111kJ\text{C}\left(s\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow\text{CO}\left(g\right)\Delta{H}_{298}^{\circ }=-111\text{kJ}

Then, carbon monoxide reacts further to form carbon dioxide:

CO(g)+12O2(g)CO(g)ΔH298=283kJ\text{CO}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow\text{CO}\left(g\right)\Delta{H}_{298}^{\circ }=-283\text{kJ}

The equation describing the overall reaction is the sum of these two chemical changes:

Step 1: C(s)+12O2(g)CO(g)Step 2: CO(g)+12O2(g)CO2(g)Sum: C(s)+12O2(g)+CO(g)+12O2(g)CO(g)+CO2(g)\begin{array}{l}\\ \text{Step 1: }\text{C}\left(s\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow\text{CO}\left(g\right)\\ \underline{\text{Step 2: }\text{CO}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow{\text{CO}}_{2}\left(g\right)}\\ \text{Sum: }\text{C}\left(s\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)+\text{CO}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow\text{CO}\left(g\right)+{\text{CO}}_{2}\left(g\right)\end{array}

Because the CO produced in Step 1 is consumed in Step 2, the net change is:


According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps. We can apply the data from the experimental enthalpies of combustion in Table 1 to find the enthalpy change of the entire reaction from its two steps:

C(s)+12O2(g)CO(g)ΔH298=111kJCO(g)+12O2(g)CO2(g)ΔH298=283kJC(s)+O2(g)CO2(g)ΔH298=394kJ\begin{array}{ll}\text{C}\left(s\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow\text{CO}\left(g\right)\qquad & \Delta{H}_{298}^{\circ }=-111\text{kJ}\qquad \\ \underline{\text{CO}\left(g\right)+\frac{1}{2}{\text{O}}_{2}\left(g\right)\rightarrow{\text{CO}}_{2}\left(g\right)}& \underline{\Delta{H}_{298}^{\circ }=-283\text{kJ}}\\{\text{C}\left(s\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{CO}}_{2}\left(g\right)}&{\Delta{H}_{298}^{\circ }=-394\text{kJ}}\qquad \end{array}

The result is shown in Figure 7. We see that ΔH of the overall reaction is the same whether it occurs in one step or two. This finding (overall ΔH for the reaction = sum of ΔH values for reaction “steps” in the overall reaction) is true in general for chemical and physical processes.

A diagram is shown. A long arrow faces upward on the left with the phrase “H increasing.” A horizontal line at the bottom of the diagram is shown with the formula “C O subscript 2 (g)” below it. A horizontal line at the top of the diagram has the formulas “C (s) + O subscript 2 (g)” above it. The top and bottom lines are connected by a downward facing arrow with the value “Δ H = –394 k J” written beside it. Below and to the right of the top horizontal line is a second horizontal line with the equations “C O (g) + one half O subscript 2 (g)” above it. This line and the bottom line are connected by a downward facing arrow with the value “Δ H = –283 k J” written beside it. The same line and the top line are connected by a downward facing arrow with the value “Δ H = –111 k J” written beside it. There are three brackets to the right of the diagram. The first bracket runs from the top horizontal line to the second horizontal line. It is labeled, “Enthalpy of reactants.” The second bracket runs from the second horizontal line to the bottom horizontal line. It is labeled, “Enthalpy of products.” Both of these brackets are included in the third bracket which runs from the top to the bottom of the diagram. It is labeled, “Enthalpy change of exothermic reaction in 1 or 2 steps.” Figure 7. The formation of CO2(g) from its elements can be thought of as occurring in two steps, which sum to the overall reaction, as described by Hess’s law. The horizontal blue lines represent enthalpies. For an exothermic process, the products are at lower enthalpy than are the reactants.

Before we further practice using Hess’s law, let us recall two important features of ΔH.

  1. ΔH is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of NO2(g) is +33.2 kJ:


    When 2 moles of NO2 (twice as much) are formed, the ΔH will be twice as large:

    N2(g)+2O2(g)2NO2(g)ΔH=+66.4kJ{\text{N}}_{2}\left(g\right)+2{\text{O}}_{2}\left(g\right)\rightarrow 2{\text{NO}}_{2}\left(g\right)\Delta\text{H}=\text{+66.4}\text{kJ}

    In general, if we multiply or divide an equation by a number, then the enthalpy change should also be multiplied or divided by the same number.
  2. ΔH for a reaction in one direction is equal in magnitude and opposite in sign to ΔH for the reaction in the reverse direction. For example, given that:

    H2(g)+Cl2(g)2HCl(g)ΔH=184.6kJ{\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\rightarrow 2\text{HCl}\left(g\right)\Delta\text{H}=-184.6\text{kJ}

    Then, for the “reverse” reaction, the enthalpy change is also “reversed:”


Example 6: Stepwise Calculation of
ΔHf\Delta{H}_{\text{f}}^{\circ }
Using Hess’s Law

Determine the enthalpy of formation,
ΔHf\Delta{H}_{\text{f}}^{\circ }
, of FeCl3(s) from the enthalpy changes of the following two-step process that occurs under standard state conditions:

Fe(s)+Cl2(g)FeCl2(s)ΔH=341.8kJ\text{Fe}\left(s\right)+{\text{Cl}}_{2}\left(g\right)\rightarrow{\text{FeCl}}_{2}\left(s\right)\Delta\text{H}^{\circ }=-341.8\text{kJ}

FeCl2(s)+12Cl2(g)FeCl3(s)ΔH=57.7kJ{\text{FeCl}}_{2}\left(s\right)+\frac{1}{2}{\text{Cl}}_{2}\left(g\right)\rightarrow{\text{FeCl}}_{3}\left(s\right)\Delta\text{H}^{\circ }=-57.7\text{kJ}

Check Your Learning

Calculate ΔH for the process:

N2(g)+2O2(g)2NO2(g){\text{N}}_{2}\left(g\right)+2{\text{O}}_{2}\left(g\right)\rightarrow 2{\text{NO}}_{2}\left(g\right)

from the following information:

N2(g)+O2(g)2NO(g)ΔH=180.5kJ{\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow 2\text{NO}\left(g\right)\Delta\text{H}=180.5\text{kJ}


Here is a less straightforward example that illustrates the thought process involved in solving many Hess’s law problems. It shows how we can find many standard enthalpies of formation (and other values of ΔH) if they are difficult to determine experimentally.

Example 7: A More Challenging Problem Using Hess’s Law

Chlorine monofluoride can react with fluorine to form chlorine trifluoride:

ClF(g)+F2(g)ClF3(g)ΔH=?\text{ClF}\left(g\right)+{\text{F}}_{2}\left(g\right)\rightarrow{\text{ClF}}_{3}\left(g\right)\Delta\text{H}^{\circ }=?

Use the reactions here to determine the ΔH° for reaction (i):

2OF2(g)O2(g)+2F2(g)ΔH(ii)=49.4kJ2{\text{OF}}_{2}\left(g\right)\rightarrow{\text{O}}_{2}\left(g\right)+2{\text{F}}_{2}\left(g\right)\Delta{H}_{\left(ii\right)}^{\circ }=-49.4\text{kJ}

2ClF(g)+O2(g)Cl2O(g)+OF2(g)ΔH(iii)=+205.6kJ2\text{ClF}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{Cl}}_{2}\text{O}\left(g\right)+{\text{OF}}_{2}\left(g\right)\Delta{H}_{\left(iii\right)}^{\circ }=\text{+205.6}\text{kJ}

ClF3(g)+O2(g)Cl2O(g)+32OF2(g)ΔH(iv)=+266.7kJ{\text{ClF}}_{3}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{Cl}}_{2}\text{O}\left(g\right)+\frac{3}{2}{\text{OF}}_{2}\left(g\right)\Delta{H}_{\left(iv\right)}^{\circ }=\text{+266.7}\text{kJ}

Check Your Learning

Aluminum chloride can be formed from its elements:

2Al(s)+3Cl2(g)2AlCl3(s)ΔH=?2\text{Al}\left(s\right)+3{\text{Cl}}_{2}\left(g\right)\rightarrow 2{\text{AlCl}}_{3}\left(s\right)\Delta H^{\circ }=?

Use the reactions here to determine the ΔH° for reaction (i):

HCl(g)HCl(aq)ΔH(ii)=74.8kJ\text{HCl}\left(g\right)\rightarrow\text{HCl}\left(aq\right)\Delta{H}_{\left(ii\right)}^{\circ }=-74.8\text{kJ}

H2(g)+Cl2(g)2HCl(g)ΔH(iii)=185kJ{\text{H}}_{2}\left(g\right)+{\text{Cl}}_{2}\left(g\right)\rightarrow 2\text{HCl}\left(g\right)\Delta{H}_{\left(iii\right)}^{\circ }=-185\text{kJ}

AlCl3(aq)AlCl3(s)ΔH(iv)=+323kJ/mol{\text{AlCl}}_{3}\left(aq\right)\rightarrow{\text{AlCl}}_{3}\left(s\right)\Delta{H}_{\left(iv\right)}^{\circ }=+323\text{kJ/mol}

2Al(s)+6HCl(aq)2AlCl3(aq)+3H2(g)ΔH(v)=1049kJ\text{2Al}\left(s\right)+6\text{HCl}\left(aq\right)\rightarrow 2{\text{AlCl}}_{3}\left(aq\right)+3{\text{H}}_{2}\left(g\right)\Delta{H}_{\left(v\right)}^{\circ }=-1049\text{kJ}

We also can use Hess’s law to determine the enthalpy change of any reaction if the corresponding enthalpies of formation of the reactants and products are available. The stepwise reactions we consider are: (i) decompositions of the reactants into their component elements (for which the enthalpy changes are proportional to the negative of the enthalpies of formation of the reactants), followed by (ii) re-combinations of the elements to give the products (with the enthalpy changes proportional to the enthalpies of formation of the products). The standard enthalpy change of the overall reaction is therefore equal to: (ii) the sum of the standard enthalpies of formation of all the products plus (i) the sum of the negatives of the standard enthalpies of formation of the reactants. This is usually rearranged slightly to be written as follows, with ∑ representing “the sum of” and n standing for the stoichiometric coefficients:

ΔHreaction=n×ΔHf(products)n×ΔHf(reactants)\Delta{H}_{\text{reaction}}^{\circ }=\sum n\times \Delta{H}_{\text{f}}^{\circ }\left(\text{products}\right)-\sum n\times \Delta{H}_{\text{f}}^{\circ }\left(\text{reactants}\right)

The following example shows in detail why this equation is valid, and how to use it to calculate the enthalpy change for a reaction of interest.

Example 8: Using Hess’s Law

What is the standard enthalpy change for the reaction:

3NO2(g)+H2O(l)2HNO3(aq)+NO(g)ΔH=?3{\text{NO}}_{2}\left(g\right)+{\text{H}}_{2}\text{O}\left(l\right)\rightarrow 2{\text{HNO}}_{3}\left(aq\right)+\text{NO}\left(g\right)\Delta\text{H}^{\circ }=?

There are two possible solutions for this reaction. The first solution supports why the general equation is valid, the second solution uses the equation.

Check Your Learning

Calculate the heat of combustion of 1 mole of ethanol, C2H5OH(l), when H2O(l) and CO2(g) are formed. Use the following enthalpies of formation: C2H5OH(l), −278 kJ/mol; H2O(l), −286 kJ/mol; and CO2(g), −394 kJ/mol.


  1. Calculate the standard molar enthalpy of formation of NO(g) from the following data:
    N2(g)+2O22NO2(g)ΔH298=66.4kJ2NO(g)+O22NO2(g)ΔH298=114.1kJ\begin{array}{l}{\text{N}}_{2}\left(g\right)+2{\text{O}}_{2}\rightarrow 2{\text{NO}}_{2}\left(g\right)\Delta{H}_{298}^{\circ }=66.4\text{kJ}\\ \text{2NO}\left(g\right)+{\text{O}}_{2}\rightarrow 2{\text{NO}}_{2}\left(g\right)\Delta{H}_{298}^{\circ }=-114.1\text{kJ}\end{array}
  2. Using the data in Standard Thermodynamic Properties for Selected Substances, calculate the standard enthalpy change for each of the following reactions:

    1. N2(g)+O2(g)2NO(g){\text{N}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow 2\text{NO}\left(g\right)
    2. Si(s)+2Cl2(g)SiCl4(g)\text{Si}\left(s\right)+2{\text{Cl}}_{2}\left(g\right)\rightarrow{\text{SiCl}}_{4}\left(g\right)
    3. Fe2O(s)+3H2(g)2Fe(s)+3H2O(l){\text{Fe}}_{2}\text{O}\left(s\right)+3{\text{H}}_{2}\left(g\right)\rightarrow 2\text{Fe}\left(s\right)+3{\text{H}}_{2}\text{O}\left(l\right)
    4. 2LiOH(s)+CO2(g)Li2CO3(s)+H2O(g)2\text{LiOH}\left(s\right)+{\text{CO}}_{2}\left(g\right)\rightarrow{\text{Li}}_{2}{\text{CO}}_{3}\left(s\right)+{\text{H}}_{2}\text{O}\left(g\right)

  3. Using the data in Standard Thermodynamic Properties for Selected Substances, calculate the standard enthalpy change for each of the following reactions:

    1. Si(s)+2F2(g)SiF4(g)\text{Si}\left(s\right)+2{\text{F}}_{2}\left(g\right)\rightarrow{\text{SiF}}_{4}\left(g\right)
    2. 2C(s)+2H2(g)+O2(g)CH3CO2H(l)2\text{C}\left(s\right)+2{\text{H}}_{2}\left(g\right)+{\text{O}}_{2}\left(g\right)\rightarrow{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(l\right)
    3. CH4(g)+N2(g)HCN(g)+NH3(g);{\text{CH}}_{4}\left(g\right)+{\text{N}}_{2}\left(g\right)\rightarrow\text{HCN}\left(g\right){\text{+NH}}_{3}\left(g\right)\text{;}
    4. Cs2(g)+3Cl2(g)CCl4(g)+S2Cl2(g){\text{Cs}}_{2}\left(g\right)+3{\text{Cl}}_{2}\left(g\right)\rightarrow{\text{CCl}}_{4}\left(g\right){\text{+S}}_{2}{\text{Cl}}_{2}\left(g\right)

  4. The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each.

    1. 2Ag2O(s)4Ag(s)+O2(g)2{\text{Ag}}_{2}\text{O}\left(s\right)\rightarrow 4\text{Ag}\left(s\right)+{\text{O}}_{2}\left(g\right)
    2. SnO(s)+CO(g)Sn(s)+CO2(g)\text{SnO}\left(s\right)+\text{CO}\left(g\right)\rightarrow\text{Sn}\left(s\right)+{\text{CO}}_{2}\left(g\right)
    3. Cr2O3(s)+3H2(g)2Cr(s)+3H2O(l){\text{Cr}}_{2}{\text{O}}_{3}\left(s\right)+3{\text{H}}_{2}\left(g\right)\rightarrow 2\text{Cr}\left(s\right)+3{\text{H}}_{2}\text{O}\left(l\right)
    4. 2Al(s)+Fe2O3(s)Al2O3(s)+2Fe(s)2\text{Al}\left(s\right)+{\text{Fe}}_{2}{\text{O}}_{3}\left(s\right)\rightarrow{\text{Al}}_{2}{\text{O}}_{3}\left(s\right)+2\text{Fe}\left(s\right)

  5. The decomposition of hydrogen peroxide, H2O2, has been used to provide thrust in the control jets of various space vehicles. Using the data in Standard Thermodynamic Properties for Selected Substances, determine how much heat is produced by the decomposition of exactly 1 mole of H2O2 under standard conditions.
    2H2O2(l)2H2O(g)+O2(g)2{\text{H}}_{2}{\text{O}}_{2}\left(l\right)\rightarrow 2{\text{H}}_{2}\text{O}\left(g\right)+{\text{O}}_{2}\left(g\right)