37

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 Δ𝐻° for reactions occurring under standard state conditions at 298 K. 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. The standard enthalpy of formation, Δ𝐻°_{f}, is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar and 298.15 K. If the enthalpies of formation are available for the reactants and products of a reaction, the enthalpy change can be calculated using Hess’s law: 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.

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

- 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

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*):

$$H=U+PV$$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 (Δ*H*) is:

$$\text{\Delta}H=\text{\Delta}U+P\text{\Delta}V$$

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

$$P\text{\Delta}V=\text{\u2212}w$$

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

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

where *q _{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 (*q _{p}*) and enthalpy change (Δ

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 Δ*H* because the closed, constant-volume metal container prevents the pressure from remaining constant (it may increase or decrease if the reaction yields increased or decreased amounts of gaseous species). Chemists usually perform experiments under normal atmospheric conditions, at constant external pressure with *q* = Δ*H*, which makes enthalpy the most convenient choice for determining heat changes for chemical reactions.

The following conventions apply when using Δ*H*:

A negative value of an enthalpy change, Δ

*H*< 0, indicates an exothermic reaction; a positive value, Δ*H*> 0, indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its Δ*H*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 Δ

*H*value following the equation for the reaction. This Δ*H*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:$${\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(l)\phantom{\rule{2em}{0ex}}\text{\Delta}\text{H}=\mathrm{-286}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$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 (Δ

*H*is an extensive property):$$\begin{array}{l}\text{(two-fold increase in amounts)}\\ 2{\text{H}}_{2}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{H}}_{2}\text{O}(l)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}(\mathrm{-286}\phantom{\rule{0.2em}{0ex}}\text{kJ})=\mathrm{-572}\phantom{\rule{0.2em}{0ex}}\text{kJ}\\ \left(\text{two-fold decrease in amounts}\right)\\ \frac{1}{2}{\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{4}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{H}}_{2}\text{O}(l)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}(\mathrm{-286}\phantom{\rule{0.2em}{0ex}}\text{kJ})=\mathrm{-143}\phantom{\rule{0.2em}{0ex}}\text{kJ}\end{array}$$The enthalpy change of a reaction depends on the physical states of the reactants and products, 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.

$${\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\mathrm{-242}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

$$\text{HCl}(aq)+\text{NaOH}(aq)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{NaCl}(aq)+{\text{H}}_{2}\text{O}(l)$$

$$\text{\Delta}\text{H}=1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol HCl}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\mathrm{-2.9}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{0.0500\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol HCl}}}\phantom{\rule{0.2em}{0ex}}=\mathrm{-58}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

The thermochemical equation is then

$$\text{HCl}(aq)+\text{NaOH}(aq)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{NaCl}(aq)+{\text{H}}_{2}\text{O}(l)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\mathrm{-58}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

$$\text{Zn}(s)+2\text{HCl}(aq)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{ZnCl}}_{2}(aq)+{\text{H}}_{2}(g)$$

Δ*H* = −153 kJ

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

$${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}(aq)+8{\text{KClO}}_{3}(aq)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}12{\text{CO}}_{2}(g)+11{\text{H}}_{2}\text{O}(l)+8\text{KCl}(aq).$$

The provided amounts of the two reactants are

$$\begin{array}{l}\left(2.67\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(1\phantom{\rule{0.2em}{0ex}}\text{mol}\text{/}342.3\phantom{\rule{0.2em}{0ex}}\text{g}\right)=0.00780\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{12}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{22}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{11}\\ \left(7.19\phantom{\rule{0.2em}{0ex}}\text{g}\right)\left(1\phantom{\rule{0.2em}{0ex}}\text{mol}\text{/}122.5\phantom{\rule{0.2em}{0ex}}\text{g}\right)=0.0587\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}\text{KCI}{\text{O}}_{3}\end{array}$$

The provided molar ratio of perchlorate-to-sucrose is then

$$0.0587\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}\text{KCI}{\text{O}}_{3}\text{/}0.00780\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}=7.52$$

The balanced equation indicates 8 mol KClO_{3} are required for reaction with 1 mol C_{12}H_{22}O_{11}. Since the provided amount of KClO_{3} is less than the stoichiometric amount, it is the limiting reactant and may be used to compute the enthalpy change:

$$\u25b3\text{H}=-43.7\text{}\phantom{\rule{0.2em}{0ex}}\text{kJ}\text{/}0.0587\phantom{\rule{0.2em}{0ex}}\text{mol KCI}{\text{O}}_{3}=744\phantom{\rule{0.2em}{0ex}}\text{kJ}\text{/}\text{mol}\phantom{\rule{0.2em}{0ex}}\text{KCI}{\text{O}}_{3}$$

Because the equation, as written, represents the reaction of 8 mol KClO_{3}, the enthalpy change is

$$\left(744\phantom{\rule{0.2em}{0ex}}\text{kJ}\text{/}\text{mol}\phantom{\rule{0.2em}{0ex}}\text{KCI}{\text{O}}_{3}\right)\left(8\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}\text{KCI}{\text{O}}_{3}\right)=5960\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

The enthalpy change for this reaction is −5960 kJ, and the thermochemical equation is:

$${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}+8{\text{KClO}}_{3}\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}12{\text{CO}}_{2}+11{\text{H}}_{2}\text{O}+8\text{KCl}\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\mathrm{-5960}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

Δ*H* = −338 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 Δ*H* of a reaction changes very little with such small changes in pressure (1 bar = 0.987 atm), Δ*H* 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 $(\text{\Delta}H\text{\xb0})$ is used to indicate an enthalpy change for a process occurring under these conditions. (The symbol Δ*H* 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 Δ*H* for specific amounts of reactants). However, we often find it more useful to divide one extensive property (Δ*H*) by another (amount of substance), and report a per-amount *intensive* value of Δ*H*, 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 $(\text{\Delta}{H}_{C}^{\text{\xb0}})$ 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.

$${\text{C}}_{2}{\text{H}}_{5}\text{OH}(l)+3{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{CO}}_{2}+3{\text{H}}_{2}\text{O}(l)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\text{\u22121366.8 kJ}$$

Enthalpies of combustion for many substances have been measured; a few of these are listed in Table 37.1. 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.

Standard Molar Enthalpies of Combustion

Substance | Combustion Reaction | Enthalpy of Combustion, $\text{\Delta}{H}_{c}^{\xb0}$$(\frac{\text{kJ}}{\text{mol}}\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}25\phantom{\rule{0.2em}{0ex}}\text{\xb0C})$ |
---|---|---|

carbon | $\text{C}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)$ | −393.5 |

hydrogen | ${\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(l)$ | −285.8 |

magnesium | $\text{Mg}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{MgO}(s)$ | −601.6 |

sulfur | $\text{S}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{SO}}_{2}(g)$ | −296.8 |

carbon monoxide | $\text{CO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)$ | −283.0 |

methane | ${\text{CH}}_{4}(g)+2{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)+2{\text{H}}_{2}\text{O}(l)$ | −890.8 |

acetylene | ${\text{C}}_{2}{\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{5}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{CO}}_{2}(g)+{\text{H}}_{2}\text{O}(l)$ | −1301.1 |

ethanol | ${\text{C}}_{2}{\text{H}}_{5}\text{OH}(l)+3{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{2CO}}_{2}(g)+3{\text{H}}_{2}\text{O}(l)$ | −1366.8 |

methanol | ${\text{CH}}_{3}\text{OH}(l)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)+2{\text{H}}_{2}\text{O}(l)$ | −726.1 |

isooctane | ${\text{C}}_{8}{\text{H}}_{18}(l)+\phantom{\rule{0.1em}{0ex}}\frac{25}{2}\phantom{\rule{0.1em}{0ex}}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}8{\text{CO}}_{2}(g)+9{\text{H}}_{2}\text{O}(l)$ | −5461 |

Using these data,

$$1.00\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1000\phantom{\rule{0.2em}{0ex}}\overline{)\text{mL}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{L}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{0.692\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mL}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}{114\phantom{\rule{0.2em}{0ex}}\overline{)\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{-5460\phantom{\rule{0.2em}{0ex}}\text{kJ}}{1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}}}\phantom{\rule{0.2em}{0ex}}=\mathrm{-3.31}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\text{kJ}$$

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:

$$\begin{array}{l}\\ 1.00\phantom{\rule{0.2em}{0ex}}\text{L}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}1.00\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{mL}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\\ 1.00\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{mL}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}692\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\\ 692\phantom{\rule{0.2em}{0ex}}\text{g}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}6.07\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\\ 6.07\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{C}}_{8}{\text{H}}_{18}\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\mathrm{-3.31}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\text{kJ}\end{array}$$

6.25 $\times $ 10^{3} kJ

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 37.2). 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.^{3} The process used to produce algal fuel is as follows: grow the algae (which use sunlight as their energy source and CO_{2} 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 37.3).

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

A standard enthalpy of formation $\text{\Delta}{H}_{\text{f}}^{\xb0}$ 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 CO_{2}(*g*) is −393.5 kJ/mol. This is the enthalpy change for the exothermic reaction:

$$\text{C}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}=\text{\Delta}H\text{\xb0}=\mathrm{-393.5}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

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 CO_{2}, also at 1 atm and 25 °C. For nitrogen dioxide, NO_{2}(*g*), $\text{\Delta}{H}_{\text{f}}^{\xb0}$ is 33.2 kJ/mol. This is the enthalpy change for the reaction:

$$\frac{1}{2}{\text{N}}_{2}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}=\text{\Delta}H\text{\xb0}=\text{+33.2 kJ}$$

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

These values indicate that formation reactions range from highly exothermic (such as −2984 kJ/mol for the formation of P_{4}O_{10}) to strongly endothermic (such as +226.7 kJ/mol for the formation of acetylene, C_{2}H_{2}). 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.

$$3{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{O}}_{3}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\text{+286 kJ}$$

$$\frac{3}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}(g)$$

For the formation of 2 mol of O_{3}(*g*), $\text{\Delta}H\text{\xb0}=\text{+286 kJ.}$ This ratio, $\left(\frac{286\phantom{\rule{0.2em}{0ex}}\text{kJ}}{2\phantom{\rule{0.2em}{0ex}}\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}\right),$ can be used as a conversion factor to find the heat produced when 1 mole of O_{3}(*g*) is formed, which is the enthalpy of formation for O_{3}(*g*):

$$\text{\Delta}\text{H}\text{\xb0 for}\phantom{\rule{0.2em}{0ex}}1\phantom{\rule{0.2em}{0ex}}\text{mole of}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}(g)=1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{286\phantom{\rule{0.2em}{0ex}}\text{kJ}}{2\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{O}}_{3}}}\phantom{\rule{0.2em}{0ex}}=143\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

Therefore, $\text{\Delta}{H}_{\text{f}}^{\xb0}[{\text{O}}_{3}(g)]=\text{+143 kJ/mol}.$

For the reaction ${\text{H}}_{2}(g)+{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2\text{HCl}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\mathrm{-184.6}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

(a) C_{2}H_{5}OH(*l*)

(b) Ca_{3}(PO_{4})_{2}(*s*)

(a) $2\text{C}(s,\phantom{\rule{0.2em}{0ex}}\text{graphite})+3{\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}\text{OH}(l)$

(b) $3\text{Ca}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{P}}_{4}(s)+4{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{Ca}}_{3}({\text{PO}}_{4}{)}_{2}(s)$

Note: The standard state of carbon is graphite, and phosphorus exists as P_{4}.

(a) C_{2}H_{5}OC_{2}H_{5}(*l*)

(b) Na_{2}CO_{3}(*s*)

(a) $4\text{C}(s,\phantom{\rule{0.2em}{0ex}}\text{graphite})+5{\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{C}}_{2}{\text{H}}_{5}{\text{OC}}_{2}{\text{H}}_{5}(l);$ (b) $2\text{Na}(s)+\text{C}(s,\phantom{\rule{0.2em}{0ex}}\text{graphite})+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{Na}}_{2}{\text{CO}}_{3}(s)$

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:

$$\text{C}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\mathrm{-394}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

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

$$\text{C}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{CO}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\mathrm{-111}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

Then, carbon monoxide reacts further to form carbon dioxide:

$$\text{CO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=\mathrm{-283}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

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

$$\begin{array}{}\\ \text{Step 1: C}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{CO}(g)\\ \underset{\xaf}{\text{Step 2: CO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)}\\ \text{Sum: C}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)+\text{CO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{CO}(g)+{\text{CO}}_{2}(g)\end{array}$$

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

$$\text{C}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)$$

According to Hess’s law, the enthalpy change of the reaction will equal the sum of the enthalpy changes of the steps.

$$\begin{array}{ll}\text{C}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{CO}(g)\hfill & \text{\Delta}H\text{\xb0}=\mathrm{-111}\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \\ \frac{\text{CO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)}{\text{C}(s)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{CO}}_{2}(g)\phantom{\rule{1em}{0ex}}}\hfill & \frac{\text{\Delta}H\text{\xb0}=\mathrm{-283}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{\Delta}H\text{\xb0}=\mathrm{-394}\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}$$

The result is shown in Figure 37.4. We see that Δ*H* of the overall reaction is the same whether it occurs in one step or two. This finding (overall Δ*H* for the reaction = sum of Δ*H* values for reaction “steps” in the overall reaction) is true in general for chemical and physical processes.

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

Δ

*H*is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of NO_{2}(*g*) is +33.2 kJ:$$\frac{1}{2}{\text{N}}_{2}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\text{+33.2 kJ}$$When 2 moles of NO

_{2}(twice as much) are formed, the Δ*H*will be twice as large:$${\text{N}}_{2}(g)+2{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{NO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\text{+66.4 kJ}$$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.

Δ

*H*for a reaction in one direction is equal in magnitude and opposite in sign to Δ*H*for the reaction in the reverse direction. For example, given that:$${\text{H}}_{2}(g)+{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2\text{HCl}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\mathrm{-184.6}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$Then, for the “reverse” reaction, the enthalpy change is also “reversed”:

$$2\text{HCl}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}(g)+{\text{Cl}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\text{+184.6 kJ}$$

$$\text{Fe}(s)+{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{2}(s)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=\mathrm{-341.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

$${\text{FeCl}}_{2}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}(s)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=\mathrm{-57.7}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

$$\text{Fe}(s)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}(s)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}=?$$

Looking at the reactions, we see that the reaction for which we want to find Δ*H*° is the sum of the two reactions with known Δ*H* values, so we must sum their Δ*H*s:

$$\begin{array}{lll}\text{Fe}(s)+{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{2}(s)\hfill & \hfill & \text{\Delta}H\text{\xb0}=\mathrm{-341.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \\ \frac{{\text{FeCl}}_{2}(s)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}(s)}{\text{Fe}(s)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{FeCl}}_{3}(s)\phantom{\rule{1em}{0ex}}}\hfill & \hfill & \frac{\text{\Delta}H\text{\xb0}=\mathrm{-57.7}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{\Delta}H\text{\xb0}=\mathrm{-399.5}\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}$$

The enthalpy of formation, $\text{\Delta}{H}_{\text{f}}^{\xb0},$ of FeCl_{3}(*s*) is −399.5 kJ/mol.

$${\text{N}}_{2}(g)+2{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{NO}}_{2}(g)$$

from the following information:

$${\text{N}}_{2}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2\text{NO}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=180.5\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

$$\text{NO}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}=\mathrm{-57.06}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

66.4 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 Δ*H*) if they are difficult to determine experimentally.

*(i)* $\text{ClF}(g)+{\text{F}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=?$

Use the reactions here to determine the Δ*H*° for reaction *(i)*:

*(ii)* $2{\text{OF}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{O}}_{2}(g)+2{\text{F}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(ii)}^{\xb0}=\mathrm{-49.4}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

*(iii)* $2\text{ClF}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{Cl}}_{2}\text{O}(g)+{\text{OF}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(iii)}^{\xb0}=\text{+214.0 kJ}$

*(iv)* ${\text{ClF}}_{3}(g)+{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\text{O}(g)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{OF}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(iv)}^{\xb0}=\text{+236.2 kJ}$

$$\text{ClF}(g)+\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\text{O}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{OF}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=\phantom{\rule{0.2em}{0ex}}\frac{1}{2}(214.0)=\text{+107.0 kJ}$$

Next, we see that F_{2} is also needed as a reactant. To get this, reverse and halve reaction *(ii)*, which means that the Δ*H*° changes sign and is halved:

$$\frac{1}{2}{\text{O}}_{2}(g)+{\text{F}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{OF}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=\text{+24.7 kJ}$$

To get ClF_{3} as a product, reverse *(iv)*, changing the sign of Δ*H*°:

$$\frac{1}{2}{\text{Cl}}_{2}\text{O}(g)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{OF}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}(g)+{\text{O}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=\text{\u2212236.2 kJ}$$

Now check to make sure that these reactions add up to the reaction we want:

$$\begin{array}{lll}\text{ClF}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\frac{1}{2}{\text{Cl}}_{2}\text{O}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{OF}}_{2}(g)\hfill & \hfill & \text{\Delta}H\text{\xb0}=\text{+107.0 kJ}\hfill \\ \frac{1}{2}{\text{O}}_{2}(g)+{\text{F}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{OF}}_{2}(g)\hfill & \hfill & \text{\Delta}H\text{\xb0}=\text{+24.7 kJ}\hfill \\ \frac{\frac{1}{2}{\text{Cl}}_{2}\text{O}(g)+\phantom{\rule{0.1em}{0ex}}\frac{3}{2}{\text{OF}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}(g)+{\text{O}}_{2}(g)}{\text{ClF}(g)+{\text{F}}_{2}\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{ClF}}_{3}(g)\phantom{\rule{8em}{0ex}}}\hfill & \hfill & \frac{\text{\Delta}H\text{\xb0}=\mathrm{-236.2}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\text{\Delta}H\text{\xb0}=\mathrm{-104.5}\phantom{\rule{0.2em}{0ex}}\text{kJ}}\hfill \end{array}$$

Reactants $\frac{1}{2}{\text{O}}_{2}$ and $\frac{1}{2}{\text{O}}_{2}$ cancel out product O_{2}; product $\frac{1}{2}{\text{Cl}}_{2}\text{O}$ cancels reactant $\frac{1}{2}{\text{Cl}}_{2}\text{O;}$ and reactant $\frac{3}{2}{\text{OF}}_{2}$ is cancelled by products $\frac{1}{2}{\text{OF}}_{2}$ and OF_{2}. This leaves only reactants ClF(*g*) and F_{2}(*g*) and product ClF_{3}(*g*), which are what we want. Since summing these three modified reactions yields the reaction of interest, summing the three modified Δ*H*° values will give the desired Δ*H*°:

$$\text{\Delta}H\text{\xb0}=(\mathrm{+107.0}\phantom{\rule{0.2em}{0ex}}\text{kJ})+(24.7\phantom{\rule{0.2em}{0ex}}\text{kJ})+(\mathrm{-236.2}\phantom{\rule{0.2em}{0ex}}\text{kJ})=\mathrm{-104.5}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

*(i)* $2\text{Al}(s)+3{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{AlCl}}_{3}(s)\phantom{\rule{3em}{0ex}}\text{\Delta}H\text{\xb0}=?$

Use the reactions here to determine the Δ*H*° for reaction *(i)*:

*(ii)* $\text{HCl}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{HCl}(aq)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(ii)}^{\xb0}=\mathrm{-74.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

*(iii)* ${\text{H}}_{2}(g)+{\text{Cl}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2\text{HCl}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(iii)}^{\xb0}=\mathrm{-185}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

*(iv)* ${\text{AlCl}}_{3}(aq)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{AlCl}}_{3}(s)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(iv)}^{\xb0}=\mathrm{+323}\phantom{\rule{0.2em}{0ex}}\text{kJ/mol}$

*(v)* $\text{2Al}(s)+6\text{HCl}(aq)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{AlCl}}_{3}(aq)+3{\text{H}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{(v)}^{\xb0}=\mathrm{-1049}\phantom{\rule{0.2em}{0ex}}\text{kJ}$

−1407 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:

$$\text{\Delta}{H}_{\text{reaction}}^{\xb0}={\displaystyle \sum n}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}(\text{products})-{\displaystyle \sum n}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}(\text{reactants})$$

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.

$$3{\text{NO}}_{2}(g)+{\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}(aq)+\text{NO}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}\text{H}\text{\xb0}=?$$

$$\text{\Delta}{H}_{\text{reaction}}^{\xb0}={\displaystyle \sum n\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}\text{(products)}}-{\displaystyle \sum n}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}(\text{reactants})$$

$$\begin{array}{l}\\ \\ \\ =\left[2\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}(aq)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\mathrm{-207.4}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{3}(aq)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol NO}(g)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\text{+90.2 kJ}}{\overline{)\text{mol NO}(g)}}\right]\\ -\left[3\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\text{+33.2 kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{NO}}_{2}(g)}}\phantom{\rule{0.1em}{0ex}}+1\phantom{\rule{0.2em}{0ex}}\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(l)}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\frac{\mathrm{-285.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}}{\overline{)\text{mol}\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}(l)}}\right]\\ =[2\times \left(\mathrm{-206.64}\right)+90.25]-[3\times 33.2+-\left(\mathrm{-285.83}\right)]\\ =\mathrm{\u2013323.03}+186.23\\ =\mathrm{-136.80}\phantom{\rule{0.2em}{0ex}}\text{kJ}\end{array}$$

$$3{\text{NO}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{3/2N}}_{2}(g)+{\text{3O}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{1}^{\xb0}=\mathrm{-99.6}\phantom{\rule{0.2em}{0ex}}\text{kJ}$$

$${\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{2}^{\xb0}=\text{+285.8 kJ}\phantom{\rule{0.2em}{0ex}}\mathrm{[-1}\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}({\text{H}}_{2}\text{O})]$$

$${\text{H}}_{2}(g)+{\text{N}}_{2}(g)+3{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}(aq)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{3}^{\xb0}=\mathrm{-414.8}\phantom{\rule{0.2em}{0ex}}\text{kJ}\phantom{\rule{0.2em}{0ex}}[2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}\text{\Delta}{H}_{\text{f}}^{\xb0}({\text{HNO}}_{3})]$$

$$\frac{1}{2}{\text{N}}_{2}(g)+\phantom{\rule{0.1em}{0ex}}\frac{1}{2}{\text{O}}_{2}(g)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}\text{NO}(g)\phantom{\rule{3em}{0ex}}\text{\Delta}{H}_{4}^{\xb0}=\text{+90.2 kJ}\phantom{\rule{0.2em}{0ex}}[1\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}(\text{NO})]$$

Summing these reaction equations gives the reaction we are interested in:

$${\text{3NO}}_{2}(g)+{\text{H}}_{2}\text{O}(l)\phantom{\rule{0.2em}{0ex}}\u27f6\phantom{\rule{0.2em}{0ex}}2{\text{HNO}}_{3}(aq)+\text{NO}(g)$$

Summing their enthalpy changes gives the value we want to determine:

$$\begin{array}{cc}\hfill \text{\Delta}{H}_{\text{rxn}}^{\xb0}& =\text{\Delta}{H}_{1}^{\xb0}+\text{\Delta}{H}_{2}^{\xb0}+\text{\Delta}{H}_{3}^{\xb0}+\text{\Delta}{H}_{4}^{\xb0}=(\mathrm{-99.6}\phantom{\rule{0.2em}{0ex}}\text{kJ})+(\text{+285.8 kJ})+(\mathrm{-414.8}\phantom{\rule{0.2em}{0ex}}\text{kJ})+(\text{+90.2 kJ})\hfill \\ & =\mathrm{-138.4}\phantom{\rule{0.2em}{0ex}}\text{kJ}\hfill \end{array}$$

So the standard enthalpy change for this reaction is Δ*H*° = −138.4 kJ.

Note that this result was obtained by (1) multiplying the $\text{\Delta}{H}_{\text{f}}^{\xb0}$ of each product by its stoichiometric coefficient and summing those values, (2) multiplying the $\text{\Delta}{H}_{\text{f}}^{\xb0}$ of each reactant by its stoichiometric coefficient and summing those values, and then (3) subtracting the result found in (2) from the result found in (1). This is also the procedure in using the general equation, as shown.

−1368 kJ/mol

Supplemental exercises are available if you would like more practice with these concepts.

- For more on algal fuel, see http://www.theguardian.com/environment/2010/feb/13/algae-solve-pentagon-fuel-problem.

Previous Citation(s)

Flowers, P., Neth, E. J., Robinson, W. R., Theopold, K., & Langley, R. (2019). Chemistry in Context. In Chemistry: Atoms First 2e. OpenStax. https://openstax.org/books/chemistry-atoms-first-2e/pages/9-3-enthalpy

This content is provided to you freely by BYU-I Books.

Access it online or download it at https://books.byui.edu/general_college_chemistry/enthalpy_part_1.