954. Check Your Learning -
A 92.9-g piece of a silver/gray metal is heated to $178.0^{\circ} \mathrm{C}$, and then quickly transferred into 75.0 mL of water initially at $24.0^{\circ} \mathrm{C}$. After 5 minutes, both the metal and the water have reached the same temperature: $29.7^{\circ} \mathrm{C}$. Determine the specific heat and the identity of the metal. (Note: You should find that the specific heat is close to that of two different metals. Explain how you can confidently determine the identity of the metal).


955. Answer: -
$c_{\text {metal }}=0.13 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$
This specific heat is close to that of either gold or lead. It would be difficult to determine which metal this was based solely on the numerical values. However, the observation that the metal is silver/gray in addition to the value for the specific heat indicates that the metal is lead.

When we use calorimetry to determine the heat involved in a chemical reaction, the same principles we have
been discussing apply. The amount of heat absorbed by the calorimeter is often small enough that we can neglect it (though not for highly accurate measurements, as discussed later), and the calorimeter minimizes energy exchange with the outside environment. Because energy is neither created nor destroyed during a chemical reaction, the heat produced or consumed in the reaction (the "system"), $q_{\text {reaction }}$, plus the heat absorbed or lost by the solution (the "surroundings"), $q_{\text {solution }}$, must add up to zero:

$$
q_{\text {reaction }}+q_{\text {solution }}=0
$$

This means that the amount of heat produced or consumed in the reaction equals the amount of heat absorbed or lost by the solution:

$$
q_{\text {reaction }}=-q_{\text {solution }}
$$

This concept lies at the heart of all calorimetry problems and calculations.


957. Heat Produced by an Exothermic Reaction -
When 50.0 mL of $1.00 \mathrm{M} \mathrm{HCl}(\mathrm{aq})$ and 50.0 mL of $1.00 \mathrm{M} \mathrm{NaOH}(\mathrm{aq})$, both at $22.0^{\circ} \mathrm{C}$, are added to a coffee cup calorimeter, the temperature of the mixture reaches a maximum of $28.9^{\circ} \mathrm{C}$. What is the approximate amount of heat produced by this reaction?

$$
\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)
$$


958. Solution -
To visualize what is going on, imagine that you could combine the two solutions so quickly that no reaction took place while they mixed; then after mixing, the reaction took place. At the instant of mixing, you have 100.0 mL of a mixture of HCl and NaOH at $22.0^{\circ} \mathrm{C}$. The HCl and NaOH then react until the solution temperature reaches $28.9^{\circ} \mathrm{C}$.

The heat given off by the reaction is equal to that taken in by the solution. Therefore:

$$
q_{\text {reaction }}=-q_{\text {solution }}
$$

(It is important to remember that this relationship only holds if the calorimeter does not absorb any heat from the reaction, and there is no heat exchange between the calorimeter and the outside environment.)

Next, we know that the heat absorbed by the solution depends on its specific heat, mass, and temperature change:

$$
q_{\text {solution }}=(c \times m \times \Delta T)_{\text {solution }}
$$

To proceed with this calculation, we need to make a few more reasonable assumptions or approximations. Since the solution is aqueous, we can proceed as if it were water in terms of its specific heat and mass values. The density of water is approximately $1.0 \mathrm{~g} / \mathrm{mL}$, so 100.0 mL has a mass of about $1.0 \times 10^{2} \mathrm{~g}$ (two significant figures). The specific heat of water is approximately $4.184 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}$, so we use that for the specific heat of the solution. Substituting these values gives:

$$
q_{\text {solution }}=\left(4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\right)\left(1.0 \times 10^{2} \mathrm{~g}\right)\left(28.9^{\circ} \mathrm{C}-22.0^{\circ} \mathrm{C}\right)=2.9 \times 10^{3} \mathrm{~J}
$$


958. Solution -
Finally, since we are trying to find the heat of the reaction, we have:

$$
q_{\text {reaction }}=-q_{\text {solution }}=-2.9 \times 10^{3} \mathrm{~J}
$$

The negative sign indicates that the reaction is exothermic. It produces 2.9 kJ of heat.


959. Check Your Learning -
When 100 mL of $0.200 \mathrm{M} \mathrm{NaCl}(\mathrm{aq})$ and 100 mL of $0.200 \mathrm{M} \mathrm{AgNO}_{3}(\mathrm{aq})$, both at $21.9^{\circ} \mathrm{C}$, are mixed in a coffee cup calorimeter, the temperature increases to $23.5^{\circ} \mathrm{C}$ as solid AgCl forms. How much heat is produced by this precipitation reaction? What assumptions did you make to determine your value?


960. Answer: -
$1.34 \times 1.3 \mathrm{~kJ}$; assume no heat is absorbed by the calorimeter, no heat is exchanged between the calorimeter and its surroundings, and that the specific heat and mass of the solution are the same as those for water


962. Thermochemistry of Hand Warmers -
When working or playing outdoors on a cold day, you might use a hand warmer to warm your hands (Figure 9.15). A common reusable hand warmer contains a supersaturated solution of $\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ (sodium acetate) and a metal disc. Bending the disk creates nucleation sites around which the metastable $\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ quickly crystallizes (a later chapter on solutions will investigate saturation and supersaturation in more detail).

The process $\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(s)$ is exothermic, and the heat produced by this process is absorbed by your hands, thereby warming them (at least for a while). If the hand warmer is reheated, the $\mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ redissolves and can be reused.

FIGURE 9.15 Chemical hand warmers produce heat that warms your hand on a cold day. In this one, you can see the metal disc that initiates the exothermic precipitation reaction. (credit: modification of work by Science Buddies TV/YouTube)

Another common hand warmer produces heat when it is ripped open, exposing iron and water in the hand warmer to oxygen in the air. One simplified version of this exothermic reaction is
$2 \mathrm{Fe}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$. Salt in the hand warmer catalyzes the reaction, so it produces heat more rapidly; cellulose, vermiculite, and activated carbon help distribute the heat evenly. Other types of hand warmers use lighter fluid (a platinum catalyst helps lighter fluid oxidize exothermically), charcoal (charcoal oxidizes in a special case), or electrical units that produce heat by passing an electrical current from a battery through resistive wires.


963. LINK TO LEARNING -
This link (http://openstax.org/l/16Handwarmer) shows the precipitation reaction that occurs when the disk in a chemical hand warmer is flexed.


965. Heat Flow in an Instant Ice Pack -
When solid ammonium nitrate dissolves in water, the solution becomes cold. This is the basis for an "instant ice pack" (Figure 9.16). When 3.21 g of solid $\mathrm{NH}_{4} \mathrm{NO}_{3}$ dissolves in 50.0 g of water at $24.9^{\circ} \mathrm{C}$ in a calorimeter, the temperature decreases to $20.3^{\circ} \mathrm{C}$.

Calculate the value of $q$ for this reaction and explain the meaning of its arithmetic sign. State any assumptions that you made.

FIGURE 9.16 An instant cold pack consists of a bag containing solid ammonium nitrate and a second bag of water. When the bag of water is broken, the pack becomes cold because the dissolution of ammonium nitrate is an endothermic process that removes thermal energy from the water. The cold pack then removes thermal energy from your body.


966. Solution -
We assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself), in which case:

$$
q_{\mathrm{rxn}}=-q_{\mathrm{soln}}
$$

with "rxn" and "soln" used as shorthand for "reaction" and "solution," respectively.
Assuming also that the specific heat of the solution is the same as that for water, we have:

$$
\begin{aligned}
& q_{\mathrm{rxn}}=-q_{\mathrm{soln}}=-(c \times m \times \Delta T)_{\text {soln }} \\
& =-\left[\left(4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\right) \times(53.2 \mathrm{~g}) \times\left(20.3^{\circ} \mathrm{C}-24.9^{\circ} \mathrm{C}\right)\right] \\
& =-\left[\left(4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\right) \times(53.2 \mathrm{~g}) \times\left(-4.6^{\circ} \mathrm{C}\right)\right] \\
& +1.0 \times 10^{3} \mathrm{~J}=+1.0 \mathrm{~kJ}
\end{aligned}
$$

The positive sign for $q$ indicates that the dissolution is an endothermic process.


967. Check Your Learning -
When a $3.00-\mathrm{g}$ sample of KCl was added to $3.00 \times 10^{2} \mathrm{~g}$ of water in a coffee cup calorimeter, the temperature decreased by $1.05^{\circ} \mathrm{C}$. How much heat is involved in the dissolution of the KCl ? What assumptions did you make?


968. Answer: -
1.33 kJ ; assume that the calorimeter prevents heat transfer between the solution and its external environment (including the calorimeter itself) and that the specific heat of the solution is the same as that for water

If the amount of heat absorbed by a calorimeter is too large to neglect or if we require more accurate results, then we must take into account the heat absorbed both by the solution and by the calorimeter.

The calorimeters described are designed to operate at constant (atmospheric) pressure and are convenient to measure heat flow accompanying processes that occur in solution. A different type of calorimeter that operates at constant volume, colloquially known as a bomb calorimeter, is used to measure the energy produced by reactions that yield large amounts of heat and gaseous products, such as combustion reactions. (The term "bomb" comes from the observation that these reactions can be vigorous enough to resemble explosions that would damage other calorimeters.) This type of calorimeter consists of a robust steel container (the "bomb") that contains the reactants and is itself submerged in water (Figure 9.17). The sample is placed in the bomb, which is then filled with oxygen at high pressure. A small electrical spark is used to ignite the sample. The energy produced by the reaction is absorbed by the steel bomb and the surrounding water. The temperature increase is measured and, along with the known heat capacity of the calorimeter, is used to calculate the energy produced by the reaction. Bomb calorimeters require calibration to determine the heat capacity of the calorimeter and ensure accurate results. The calibration is accomplished using a reaction with a known $q$, such as a measured quantity of benzoic acid ignited by a spark from a nickel fuse wire that is weighed before and after the reaction. The temperature change produced by the known reaction is used to determine the heat capacity of the calorimeter. The calibration is generally performed each time before the calorimeter is used to gather research data.

FIGURE 9.17 (a) A bomb calorimeter is used to measure heat produced by reactions involving gaseous reactants or products, such as combustion. (b) The reactants are contained in the gas-tight "bomb," which is submerged in water and surrounded by insulating materials. (credit a: modification of work by "Harbor1"/Wikimedia commons)


969. LINK TO LEARNING -
Click on this link (http://openstax.org/l/16BombCal) to view how a bomb calorimeter is prepared for action. This site (http://openstax.org/l/16Calorcalcs) shows calorimetric calculations using sample data.


971. Bomb Calorimetry -
When 3.12 g of glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$, is burned in a bomb calorimeter, the temperature of the calorimeter increases from $23.8^{\circ} \mathrm{C}$ to $35.6^{\circ} \mathrm{C}$. The calorimeter contains 775 g of water, and the bomb itself has a heat
capacity of $893 \mathrm{~J} /{ }^{\circ} \mathrm{C}$. How much heat was produced by the combustion of the glucose sample?


972. Solution -
The combustion produces heat that is primarily absorbed by the water and the bomb. (The amounts of heat absorbed by the reaction products and the unreacted excess oxygen are relatively small and dealing with them is beyond the scope of this text. We will neglect them in our calculations.)

The heat produced by the reaction is absorbed by the water and the bomb:

$$
\begin{aligned}
& q_{\mathrm{rxn}}=-\left(q_{\text {water }}+q_{\text {bomb }}\right) \\
& =-\left[\left(4.184 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}\right) \times(775 \mathrm{~g}) \times\left(35.6^{\circ} \mathrm{C}-23.8{ }^{\circ} \mathrm{C}\right)+893 \mathrm{~J} /{ }^{\circ} \mathrm{C} \times\left(35.6^{\circ} \mathrm{C}-23.8{ }^{\circ} \mathrm{C}\right)\right] \\
& =-(38,300 \mathrm{~J}+10,500 \mathrm{~J}) \\
& =-48,800 \mathrm{~J}=-48.8 \mathrm{~kJ}
\end{aligned}
$$

This reaction released 48.7 kJ of heat when 3.12 g of glucose was burned.


973. Check Your Learning -
When 0.963 g of benzene, $\mathrm{C}_{6} \mathrm{H}_{6}$, is burned in a bomb calorimeter, the temperature of the calorimeter increases by $8.39^{\circ} \mathrm{C}$. The bomb has a heat capacity of $784 \mathrm{~J} /{ }^{\circ} \mathrm{C}$ and is submerged in 925 mL of water. How much heat was produced by the combustion of the benzene sample?


974. Answer: -
$\mathrm{q}_{\mathrm{rx}}=-39.0 \mathrm{~kJ}$ (the reaction produced 39.0 kJ of heat)

Since the first one was constructed in 1899, 35 calorimeters have been built to measure the heat produced by a living person. ${ }^{2}$ These whole-body calorimeters of various designs are large enough to hold an individual human being. More recently, whole-room calorimeters allow for relatively normal activities to be performed, and these calorimeters generate data that more closely reflect the real world. These calorimeters are used to measure the metabolism of individuals under different environmental conditions, different dietary regimes, and with different health conditions, such as diabetes.

For example Carla Prado's team at University of Alberta undertook whole-body calorimetry to understand the energy expenditures of women who had recently given birth. Studies like this help develop better recommendations and regimens for nutrition, exercise, and general wellbeing during this period of significant physiological change. In humans, metabolism is typically measured in Calories per day. A nutritional calorie (Calorie) is the energy unit used to quantify the amount of energy derived from the metabolism of foods; one Calorie is equal to 1000 calories ( 1 kcal ), the amount of energy needed to heat 1 kg of water by $1^{\circ} \mathrm{C}$.


976. Measuring Nutritional Calories -
In your day-to-day life, you may be more familiar with energy being given in Calories, or nutritional calories, which are used to quantify the amount of energy in foods. One calorie (cal) = exactly 4.184 joules, and one Calorie (note the capitalization) $=1000 \mathrm{cal}$, or 1 kcal . (This is approximately the amount of energy needed to heat 1 kg of water by $1^{\circ} \mathrm{C}$.)

The macronutrients in food are proteins, carbohydrates, and fats or oils. Proteins provide about 4 Calories per gram, carbohydrates also provide about 4 Calories per gram, and fats and oils provide about 9 Calories/ g. Nutritional labels on food packages show the caloric content of one serving of the food, as well as the breakdown into Calories from each of the three macronutrients (Figure 9.18).

[^7]

FIGURE 9.18 (a) Macaroni and cheese contain energy in the form of the macronutrients in the food. (b) The food's nutritional information is shown on the package label. In the US, the energy content is given in Calories (per serving); the rest of the world usually uses kilojoules. (credit a: modification of work by "Rex Roof"/Flickr)

For the example shown in (b), the total energy per 228-g portion is calculated by:
$(5 \mathrm{~g}$ protein $\times 4$ Calories $/ \mathrm{g})+(31 \mathrm{~g}$ carb $\times 4$ Calories $/ \mathrm{g})+(12 \mathrm{~g}$ fat $\times 9$ Calories $/ \mathrm{g})=252$ Calories
So, you can use food labels to count your Calories. But where do the values come from? And how accurate are they? The caloric content of foods can be determined by using bomb calorimetry; that is, by burning the food and measuring the energy it contains. A sample of food is weighed, mixed in a blender, freezedried, ground into powder, and formed into a pellet. The pellet is burned inside a bomb calorimeter, and the measured temperature change is converted into energy per gram of food.


976. Measuring Nutritional Calories -
Today, the caloric content on food labels is derived using a method called the Atwater system that uses the average caloric content of the different chemical constituents of food, protein, carbohydrate, and fats. The average amounts are those given in the equation and are derived from the various results given by bomb calorimetry of whole foods. The carbohydrate amount is discounted a certain amount for the fiber content, which is indigestible carbohydrate. To determine the energy content of a food, the quantities of carbohydrate, protein, and fat are each multiplied by the average Calories per gram for each and the products summed to obtain the total energy.


977. LINK TO LEARNING -
Click on this link (http://openstax.org/l/16USDA) to access the US Department of Agriculture (USDA) National Nutrient Database, containing nutritional information on over 8000 foods.


978. 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 ( $\boldsymbol{U}$ ), 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:

$$
\Delta U=q+w
$$


978. LEARNING OBJECTIVES -
The relationship between internal energy, heat, and work can be represented by the equation:

$$
\Delta U=q+w
$$

as shown in Figure 9.19. 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.

FIGURE 9.19 The internal energy, $U$, of a system can be changed by heat flow and work. If heat flows into the system, $q_{i n}$, or work is done on the system, $w_{\text {on }}$, its internal energy increases, $\Delta U>0$. If heat flows out of the system, $q_{\text {out }}$, or work is done by the system, $w_{\text {by }}$, its internal energy decreases, $\Delta U<0$.
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.


979. LINK TO LEARNING -
This view of an internal combustion engine (http://openstax.org/l/16combustion) 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 $\Delta U=q+w$. Internal energy is an example of 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 9.20). The distances traveled would differ (distance is not a state function) but the elevation reached would be the same (altitude is a state function).

FIGURE 9.20 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 ( $V$ ):

$$
H=U+P V
$$


979. LINK TO LEARNING -
$$
H=U+P V
$$

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 ( $\boldsymbol{\Delta H} \boldsymbol{H})$ is:

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

The mathematical product $P \Delta V$ represents work ( $w$ ), namely, expansion or pressure-volume work as noted. By their definitions, the arithmetic signs of $\Delta V$ 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:

$$
\begin{aligned}
& \Delta H=\Delta U+P \Delta V \\
& =q_{\mathrm{p}}+w-w \\
& =q_{\mathrm{p}}
\end{aligned}
$$

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 $\left(q_{p}\right)$ and enthalpy change $(\Delta H)$ 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 (Figure 9.17) is not equal to $\Delta H$ 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 $q=$ $\Delta H$, which makes enthalpy the most convenient choice for determining heat changes for chemical reactions.

The following conventions apply when using $\Delta H$ :


979. LINK TO LEARNING -
The following conventions apply when using $\Delta H$ :

- A negative value of an enthalpy change, $\Delta H<0$, indicates an exothermic reaction; a positive value, $\Delta H>0$, indicates an endothermic reaction. If the direction of a chemical equation is reversed, the arithmetic sign of its $\Delta H$ 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 $\Delta H$ value following the equation for the reaction. This $\Delta H$ 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:

$$
\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-286 \mathrm{~kJ}
$$

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 ( $\Delta H$ is an extensive property):
(two-fold increase in amounts)


979. LINK TO LEARNING -
$$
\begin{array}{lr}
2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H=2 \times(-286 \mathrm{~kJ})=-572 \mathrm{~kJ} \\
\text { (two-fold decrease in amounts) } & \\
\frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{4} \mathrm{O}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{H}_{2} \mathrm{O}(l) & \Delta H=\frac{1}{2} \times(-286 \mathrm{~kJ})=-143 \mathrm{~kJ}
\end{array}
$$

- 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.

$$
\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H=-242 \mathrm{~kJ}
$$


981. Writing Thermochemical Equations -
When 0.0500 mol of $\mathrm{HCl}(\mathrm{aq})$ reacts with 0.0500 mol of $\mathrm{NaOH}(\mathrm{aq})$ to form 0.0500 mol of $\mathrm{NaCl}(\mathrm{aq}), 2.9 \mathrm{~kJ}$ of heat are produced. Write a balanced thermochemical equation for the reaction of one mole of HCl .

$$
\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)
$$


982. Solution -
For the reaction of 0.0500 mol acid $(\mathrm{HCl}), q=-2.9 \mathrm{~kJ}$. The reactants are provided in stoichiometric amounts (same molar ratio as in the balanced equation), and so the amount of acid may be used to calculate a molar enthalpy change. Since $\Delta H$ is an extensive property, it is proportional to the amount of acid neutralized:

$$
\Delta H=1 \mathrm{molHCt} \times \frac{-2.9 \mathrm{~kJ}}{0.0500 \mathrm{molHCt}}=-58 \mathrm{~kJ}
$$

The thermochemical equation is then

$$
\mathrm{HCl}(a q)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H=-58 \mathrm{~kJ}
$$


983. Check Your Learning -
When $1.34 \mathrm{~g} \mathrm{Zn}(s)$ reacts with 60.0 mL of $0.750 \mathrm{M} \mathrm{HCl}(\mathrm{aq}), 3.14 \mathrm{~kJ}$ of heat are produced. Determine the enthalpy change per mole of zinc reacting for the reaction:

$$
\mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g)
$$


984. Answer: -
$\Delta H=-153 \mathrm{~kJ}$

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


986. Writing Thermochemical Equations -
A gummy bear contains 2.67 g sucrose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$. When it reacts with 7.19 g potassium chlorate, $\mathrm{KClO}_{3}, 43.7$ kJ of heat are produced. Write a thermochemical equation for the reaction of one mole of sucrose:

$$
\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+8 \mathrm{KClO}_{3}(a q) \longrightarrow 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l)+8 \mathrm{KCl}(a q) .
$$


987. Solution -
Unlike the previous example exercise, this one does not involve the reaction of stoichiometric amounts of reactants, and so the limiting reactant must be identified (it limits the yield of the reaction and the amount of thermal energy produced or consumed).

The provided amounts of the two reactants are

$$
\begin{aligned}
& (2.67 \mathrm{~g})(1 \mathrm{~mol} / 342.3 \mathrm{~g})=0.00780 \mathrm{~mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11} \\
& (7.19 \mathrm{~g})(1 \mathrm{~mol} / 122.5 \mathrm{~g})=0.0587 \mathrm{~mol} \mathrm{KCIO}
\end{aligned}
$$

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

$$
0.0587 \mathrm{~mol} \mathrm{KCIO}_{3} / 0.00780 \mathrm{~mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}=7.52
$$

The balanced equation indicates $8 \mathrm{~mol}_{\mathrm{KClO}}^{3}$ are required for reaction with $1 \mathrm{~mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$. Since the provided amount of $\mathrm{KClO}_{3}$ is less than the stoichiometric amount, it is the limiting reactant and may be used to compute the enthalpy change:

$$
\triangle \mathrm{H}=-43.7 \mathrm{~kJ} / 0.0587 \mathrm{~mol} \mathrm{KCIO}_{3}=744 \mathrm{~kJ} / \mathrm{mol} \mathrm{KCIO}_{3}
$$

Because the equation, as written, represents the reaction of $8 \mathrm{~mol}_{\mathrm{KClO}}^{3}$, the enthalpy change is


987. Solution -
Because the equation, as written, represents the reaction of $8 \mathrm{~mol}_{\mathrm{KClO}}^{3}$, the enthalpy change is

$$
(744 \mathrm{~kJ} / \mathrm{mol} \mathrm{KCIO} 3)\left(8 \mathrm{~mol} \mathrm{KCIO}_{3}\right)=5960 \mathrm{~kJ}
$$

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

$$
\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+8 \mathrm{KClO}_{3} \longrightarrow 12 \mathrm{CO}_{2}+11 \mathrm{H}_{2} \mathrm{O}+8 \mathrm{KCl} \quad \Delta H=-5960 \mathrm{~kJ}
$$


988. Check Your Learning -
When 1.42 g of iron reacts with 1.80 g of chlorine, 3.22 g of $\mathrm{FeCl}_{2}(s)$ and 8.60 kJ of heat is produced. What is the enthalpy change for the reaction when 1 mole of $\mathrm{FeCl}_{2}(s)$ is produced?


989. Answer: -
$\Delta H=-338 \mathrm{~kJ}$

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


990. Standard Enthalpy of Combustion -
Standard enthalpy of combustion ( $\Delta \boldsymbol{H}_{C}^{\circ}$ ) 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, $-1366.8 \mathrm{~kJ} / \mathrm{mol}$, is the amount of heat produced when one mole of ethanol undergoes complete combustion at $25^{\circ} \mathrm{C}$ and 1 atmosphere pressure, yielding products also at $25^{\circ} \mathrm{C}$ and 1 atm .

$$
\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}+3 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta H^{\circ}=-1366.8 \mathrm{~kJ}
$$

Enthalpies of combustion for many substances have been measured; a few of these are listed in Table 9.2. 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.


990. Standard Enthalpy of Combustion -
| Substance | Combustion Reaction | Enthalpy of Combustion, $\Delta H_{c}^{\circ}$
 $\left(\frac{\mathrm{kJ}}{\mathrm{mol}}\right.$ at $\left.25^{\circ} \mathrm{C}\right)$ |
| :--- | :--- | :--- |
| carbon | $\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)$ | -393.5 |
| hydrogen | $\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ | -285.8 |
| magnesium | $\mathrm{Mg}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{MgO}(s)$ | -601.6 |
| sulfur | $\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)$ | -296.8 |
| carbon
 monoxide | $\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)$ | -283.0 |
| methane | $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$ | -890.8 |
| acetylene | $\mathrm{C}_{2} \mathrm{H}_{2}(g)+\frac{5}{2} \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)$ | -1301.1 |
| ethanol | $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)$ | -1366.8 |


990. Standard Enthalpy of Combustion -
| methanol | $\mathrm{CH}_{3} \mathrm{OH}(l)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$ | -726.1 |
| isooctane | $\mathrm{C}_{8} \mathrm{H}_{18}(l)+\frac{25}{2} \mathrm{O}_{2}(g) \longrightarrow 8 \mathrm{CO}_{2}(g)+9 \mathrm{H}_{2} \mathrm{O}(l)$ | -5461 |


990. Standard Enthalpy of Combustion -
TABLE 9.2


992. Using Enthalpy of Combustion -
As Figure 9.21 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 $\mathrm{g} / \mathrm{mL}$.

FIGURE 9.21 The combustion of gasoline is very exothermic. (credit: modification of work by "AlexEagle"/Flickr)


993. 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 9.2 gives this value as -5460 kJ per 1 mole of isooctane $\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)$.
Using these data,

The combustion of 1.00 L of isooctane produces $33,100 \mathrm{~kJ}$ 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:

$$
\begin{aligned}
& 1.00 \mathrm{LC}_{8} \mathrm{H}_{18} \longrightarrow 1.00 \times 10^{3} \mathrm{~mL} \mathrm{C}_{8} \mathrm{H}_{18} \\
& 1.00 \times 10^{3} \mathrm{mLC}_{8} \mathrm{H}_{18} \longrightarrow 692 \mathrm{~g} \mathrm{C}_{8} \mathrm{H}_{18} \\
& 692 \mathrm{~g} \mathrm{C}_{8} \mathrm{H}_{18} \longrightarrow 6.07 \mathrm{~mol} \mathrm{C}_{8} \mathrm{H}_{18} \\
& 6.07 \mathrm{~mol} \mathrm{C}_{8} \mathrm{H}_{18} \longrightarrow-3.31 \times 10^{4} \mathrm{~kJ}
\end{aligned}
$$


994. Check Your Learning -
How much heat is produced by the combustion of 125 g of acetylene?


995. Answer: -
$6.25 \times 10^{3} \mathrm{~kJ}$


997. 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 9.22). 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.

FIGURE 9.22 (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 $\frac{1}{7}$ 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. ${ }^{\frac{3}{3}}$ The process used to produce algal fuel is as follows: grow the algae (which use sunlight as their energy source and $\mathrm{CO}_{2}$ 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 9.23).

FIGURE 9.23 Algae convert sunlight and carbon dioxide into oil that is harvested, extracted, purified, and transformed into a variety of renewable fuels.


998. LINK TO LEARNING -
Click here (http://openstax.org/l/16biofuel) to learn more about the process of creating algae biofuel.


999. Standard Enthalpy of Formation -
A standard enthalpy of formation $\Delta H_{\mathrm{f}}^{\circ}$ 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 $\mathrm{CO}_{2}(\mathrm{~g})$ is $-393.5 \mathrm{~kJ} / \mathrm{mol}$. This is the enthalpy change for the exothermic reaction:

$$
\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H_{\mathrm{f}}^{\circ}=\Delta H^{\circ}=-393.5 \mathrm{~kJ}
$$

starting with the reactants at a pressure of 1 atm and $25^{\circ} \mathrm{C}$ (with the carbon present as graphite, the most stable form of carbon under these conditions) and ending with one mole of $\mathrm{CO}_{2}$, also at 1 atm and $25^{\circ} \mathrm{C}$. For nitrogen dioxide, $\mathrm{NO}_{2}(\mathrm{~g}), \Delta \boldsymbol{H}_{\mathrm{f}}^{\circ}$ is $33.2 \mathrm{~kJ} / \mathrm{mol}$. This is the enthalpy change for the reaction:

$$
\frac{1}{2} \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \quad \Delta H_{\mathrm{f}}^{\circ}=\Delta H^{\circ}=+33.2 \mathrm{~kJ}
$$


999. Standard Enthalpy of Formation -
A reaction equation with $\frac{1}{2}$ mole of $\mathrm{N}_{2}$ and 1 mole of $\mathrm{O}_{2}$ is correct in this case because the standard enthalpy of formation always refers to 1 mole of product, $\mathrm{NO}_{2}(g)$.

You will find a table of standard enthalpies of formation of many common substances in Appendix G. These values indicate that formation reactions range from highly exothermic (such as $-2984 \mathrm{~kJ} / \mathrm{mol}$ for the formation of $\mathrm{P}_{4} \mathrm{O}_{10}$ ) to strongly endothermic (such as $+226.7 \mathrm{~kJ} / \mathrm{mol}$ for the formation of acetylene, $\mathrm{C}_{2} \mathrm{H}_{2}$ ). By definition, the standard enthalpy of formation of an element in its most stable form is equal to zero under

3 For more on algal fuel, see http://www.theguardian.com/environment/2010/feb/13/algae-solve-pentagon-fuel-problem.
standard conditions, which is 1 atm for gases and 1 M for solutions.


1001. Evaluating an Enthalpy of Formation -
Ozone, $\mathrm{O}_{3}(\mathrm{~g})$, forms from oxygen, $\mathrm{O}_{2}(\mathrm{~g})$, 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, $\Delta \boldsymbol{H}_{\mathrm{f}}^{\circ}$ of ozone from the following information:

$$
3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}_{3}(g) \quad \Delta H^{\circ}=+286 \mathrm{~kJ}
$$


1002. Solution -
$\Delta H_{\mathrm{f}}^{\circ}$ 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, $\Delta H_{\mathrm{f}}^{\circ}$ for $\mathrm{O}_{3}(g)$ is the enthalpy change for the reaction:

$$
\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{3}(g)
$$

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

$$
\Delta H^{\circ} \text { for } 1 \text { mole of } \mathrm{O}_{3}(g)=1 \mathrm{mel}_{3} \times \frac{286 \mathrm{~kJ}}{2 \mathrm{~mol} \Theta_{3}}=143 \mathrm{~kJ}
$$

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


1003. Check Your Learning -
Hydrogen gas, $\mathrm{H}_{2}$, reacts explosively with gaseous chlorine, $\mathrm{Cl}_{2}$, to form hydrogen chloride, $\mathrm{HCl}(g)$. What is the enthalpy change for the reaction of 1 mole of $\mathrm{H}_{2}(\mathrm{~g})$ with 1 mole of $\mathrm{Cl}_{2}(\mathrm{~g})$ if both the reactants and products are at standard state conditions? The standard enthalpy of formation of $\mathrm{HCl}(\mathrm{g})$ is $-92.3 \mathrm{~kJ} / \mathrm{mol}$.


1004. Answer: -
For the reaction $\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g) \quad \Delta H^{\circ}=-184.6 \mathrm{~kJ}$


1006. Writing Reaction Equations for $\Delta H_{\mathrm{f}}^{\circ}$ -
Write the heat of formation reaction equations for:
(a) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{I})$
(b) $\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)$


1007. Solution -
Remembering that $\Delta H_{\mathrm{f}}^{\circ}$ reaction equations are for forming 1 mole of the compound from its constituent elements under standard conditions, we have:
(a) $2 \mathrm{C}(s$, graphite $)+3 \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)$
(b) $3 \mathrm{Ca}(s)+\frac{1}{2} \mathrm{P}_{4}(s)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)$

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


1008. Check Your Learning -
Write the heat of formation reaction equations for:
(a) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}($ I $)$
(b) $\mathrm{Na}_{2} \mathrm{CO}_{3}(s)$


1009. Answer: -
(a) $4 \mathrm{C}\left(s\right.$, graphite) $+5 \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5}(l)$; (b)
$2 \mathrm{Na}(s)+\mathrm{C}(s$, graphite $)+\frac{3}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)$


1010. 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:

$$
\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-394 \mathrm{~kJ}
$$

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

$$
\mathrm{C}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) \quad \Delta H^{\circ}=-111 \mathrm{~kJ}
$$

Then, carbon monoxide reacts further to form carbon dioxide:

$$
\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta H^{\circ}=-283 \mathrm{~kJ}
$$

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


1010. Hess's Law -
The equation describing the overall reaction is the sum of these two chemical changes:

$$
\begin{aligned}
& \text { Step 1: } \mathrm{C}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) \\
& \text { Step 2: } \mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \\
& \text { Sum: } \mathrm{C}(s)+\frac{1}{2} \mathrm{O}_{2}(g)+\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g)+\mathrm{CO}_{2}(g)
\end{aligned}
$$

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

$$
\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)
$$

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

$$
\begin{array}{ll}
\mathrm{C}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}(g) & \Delta H^{\circ}=-111 \mathrm{~kJ} \\
\frac{\mathrm{CO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)}{\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)} & \frac{\Delta H^{\circ}=-283 \mathrm{~kJ}}{\Delta H^{\circ}=-394 \mathrm{~kJ}}
\end{array}
$$


1010. Hess's Law -
The result is shown in Figure 9.24. We see that $\Delta H$ of the overall reaction is the same whether it occurs in one step or two. This finding (overall $\Delta H$ for the reaction = sum of $\Delta H$ values for reaction "steps" in the overall reaction) is true in general for chemical and physical processes.

FIGURE 9.24 The formation of $\mathrm{CO}_{2}(g)$ 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 $\Delta H$.

1. $\Delta H$ is directly proportional to the quantities of reactants or products. For example, the enthalpy change for the reaction forming 1 mole of $\mathrm{NO}_{2}(g)$ is +33.2 kJ :

$$
\frac{1}{2} \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) \quad \Delta H=+33.2 \mathrm{~kJ}
$$

When 2 moles of $\mathrm{NO}_{2}$ (twice as much) are formed, the $\Delta H$ will be twice as large:

$$
\mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \quad \Delta H=+66.4 \mathrm{~kJ}
$$

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. $\Delta H$ for a reaction in one direction is equal in magnitude and opposite in sign to $\Delta H$ for the reaction in the reverse direction. For example, given that:

$$
\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g) \quad \Delta H=-184.6 \mathrm{~kJ}
$$


1010. Hess's Law -
$$
\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g) \quad \Delta H=-184.6 \mathrm{~kJ}
$$

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

$$
2 \mathrm{HCl}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \quad \Delta H=+184.6 \mathrm{~kJ}
$$


1012. Stepwise Calculation of $\Delta H_{\mathrm{f}}^{\circ}$ Using Hess's Law -
Determine the enthalpy of formation, $\Delta H_{f}^{\circ}$, of $\mathrm{FeCl}_{3}(s)$ from the enthalpy changes of the following two-step process that occurs under standard state conditions:

$$
\begin{array}{cr}
\mathrm{Fe}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{FeCl}_{2}(s) & \Delta H^{\circ}=-341.8 \mathrm{~kJ} \\
\mathrm{FeCl}_{2}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{FeCl}_{3}(s) & \Delta H^{\circ}=-57.7 \mathrm{~kJ}
\end{array}
$$


1013. Solution -
We are trying to find the standard enthalpy of formation of $\mathrm{FeCl}_{3}(s)$, which is equal to $\Delta H^{\circ}$ for the reaction:

$$
\mathrm{Fe}(s)+\frac{3}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{FeCl}_{3}(s) \quad \Delta H_{\mathrm{f}}^{\circ}=?
$$

Looking at the reactions, we see that the reaction for which we want to find $\Delta H^{\circ}$ is the sum of the two reactions with known $\Delta H$ values, so we must sum their $\Delta H \mathrm{~s}$ :

$$
\begin{array}{ll}
\mathrm{Fe}(s)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{FeCl}_{2}(s) & \Delta H^{\circ}=-341.8 \mathrm{~kJ} \\
\frac{\mathrm{FeCl}_{2}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{FeCl}_{3}(s)}{\mathrm{Fe}(s)+\frac{3}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{FeCl}_{3}(s)} & \frac{\Delta H^{\circ}=-57.7 \mathrm{~kJ}}{\Delta H^{\circ}=-399.5 \mathrm{~kJ}}
\end{array}
$$

The enthalpy of formation, $\Delta \boldsymbol{H}_{\mathrm{f}}^{\circ}$, of $\mathrm{FeCl}_{3}(s)$ is $-399.5 \mathrm{~kJ} / \mathrm{mol}$.


1014. Check Your Learning -
Calculate $\Delta H$ for the process:

$$
\mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)
$$

from the following information:

$$
\begin{aligned}
\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) & \Delta H=180.5 \mathrm{~kJ} \\
\mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g) & \Delta H=-57.06 \mathrm{~kJ}
\end{aligned}
$$


1015. Answer: -
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 $\Delta H$ ) if they are difficult to determine experimentally.


1017. A More Challenging Problem Using Hess's Law -
Chlorine monofluoride can react with fluorine to form chlorine trifluoride:
(i) $\mathrm{ClF}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{ClF}_{3}(g)$
$\Delta H^{\circ}=$ ?

Use the reactions here to determine the $\Delta H^{\circ}$ for reaction (i):

$$
\text { (ii) } 2 \mathrm{OF}_{2}(g) \longrightarrow \mathrm{O}_{2}(g)+2 \mathrm{~F}_{2}(g) \quad \Delta H_{(i i)}^{\circ}=-49.4 \mathrm{~kJ}
$$

$$
\text { (iii) } 2 \mathrm{ClF}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Cl}_{2} \mathrm{O}(g)+\mathrm{OF}_{2}(g) \quad \Delta H_{(i i i)}^{\circ}=+214.0 \mathrm{~kJ}
$$

$$
\text { (iv) } \mathrm{ClF}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{Cl}_{2} \mathrm{O}(g)+\frac{3}{2} \mathrm{OF}_{2}(g) \quad \Delta H_{(i v)}^{\circ}=+236.2 \mathrm{~kJ}
$$


1018. Solution -
Our goal is to manipulate and combine reactions (ii), (iii), and (iv) such that they add up to reaction (i). Going from left to right in (i), we first see that $\operatorname{ClF}(g)$ is needed as a reactant. This can be obtained by multiplying reaction (iii) by $\frac{1}{2}$, which means that the $\Delta H^{\circ}$ change is also multiplied by $\frac{1}{2}$ :

$$
\mathrm{ClF}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{Cl}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{OF}_{2}(g) \quad \Delta H^{\circ}=\frac{1}{2}(214.0)=+107.0 \mathrm{~kJ}
$$

Next, we see that $\mathrm{F}_{2}$ is also needed as a reactant. To get this, reverse and halve reaction (ii), which means that the $\Delta H^{\circ}$ changes sign and is halved:

$$
\frac{1}{2} \mathrm{O}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{OF}_{2}(g) \quad \Delta H^{\circ}=+24.7 \mathrm{~kJ}
$$

To get $\mathrm{ClF}_{3}$ as a product, reverse (iv), changing the sign of $\Delta H^{\circ}$ :

$$
\frac{1}{2} \mathrm{Cl}_{2} \mathrm{O}(g)+\frac{3}{2} \mathrm{OF}_{2}(g) \longrightarrow \mathrm{ClF}_{3}(g)+\mathrm{O}_{2}(g) \quad \Delta H^{\circ}=-236.2 \mathrm{~kJ}
$$

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


1018. Solution -
Now check to make sure that these reactions add up to the reaction we want:

$$
\begin{array}{ll}
\mathrm{ClF}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{Cl}_{2} \mathrm{O}(g)+\frac{1}{2} \mathrm{OF}_{2}(g) & \Delta H^{\circ}=+107.0 \mathrm{~kJ} \\
\frac{1}{2} \mathrm{O}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{OF}_{2}(g) & \Delta H^{\circ}=+24.7 \mathrm{~kJ} \\
\frac{1}{2} \mathrm{Cl}_{2} \mathrm{O}(g)+\frac{3}{2} \mathrm{OF}_{2}(g) \longrightarrow \mathrm{ClF}_{3}(g)+\mathrm{O}_{2}(g) & \frac{\Delta H^{\circ}=-236.2 \mathrm{~kJ}}{\Delta H^{\circ}=-104.5 \mathrm{~kJ}}
\end{array}
$$

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


1018. Solution -
$$
\Delta H^{\circ}=(+107.0 \mathrm{~kJ})+(24.7 \mathrm{~kJ})+(-236.2 \mathrm{~kJ})=-104.5 \mathrm{~kJ}
$$


1019. Check Your Learning -
Aluminum chloride can be formed from its elements:
(i) $2 \mathrm{Al}(s)+3 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{AlCl}_{3}(s) \quad \Delta H^{\circ}=$ ?

Use the reactions here to determine the $\Delta H^{\circ}$ for reaction (i):
(ii) $\mathrm{HCl}(g) \longrightarrow \mathrm{HCl}(a q) \quad \Delta H_{(i i)}^{\circ}=-74.8 \mathrm{~kJ}$
(iii) $\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g) \quad \Delta H_{(i i i)}^{\circ}=-185 \mathrm{~kJ}$
(iv) $\mathrm{AlCl}_{3}(a q) \longrightarrow \mathrm{AlCl}_{3}(s) \quad \Delta H_{\text {(iv) }}^{\circ}=+323 \mathrm{~kJ} / \mathrm{mol}$
(v) $2 \mathrm{Al}(s)+6 \mathrm{HCl}(a q) \longrightarrow 2 \mathrm{AlCl}_{3}(a q)+3 \mathrm{H}_{2}(g) \quad \Delta H_{(v)}^{\circ}=-1049 \mathrm{~kJ}$


1020. Answer: -
-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 $\sum$ representing "the sum of" and $n$ standing for the stoichiometric coefficients:

$$
\Delta H_{\text {reaction }}^{\circ}=\sum n \times \Delta H_{\mathrm{f}}^{\circ}(\text { products })-\sum n \times \Delta H_{\mathrm{f}}^{\circ}(\text { reactants })
$$

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.

EXAMPLE 9.15


1021. Using Hess's Law -
What is the standard enthalpy change for the reaction:

$$
3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \quad \Delta H^{\circ}=?
$$


1022. Solution: Using the Equation -
Use the special form of Hess's law given previously, and values from Appendix G:

$$
\left.\left.\begin{array}{rl}
& \Delta H_{\text {reaction }}^{\circ}=\sum n \times \Delta H_{\mathrm{f}}^{\circ}(\text { products })-\sum n \times \Delta H_{\mathrm{f}}^{\circ}(\text { reactants }) \\
= & {[2 \mathrm{molHNO}(\mathrm{qq})}
\end{array} \times \frac{-207.4 \mathrm{~kJ}}{m+1 \mathrm{mNO}(\mathrm{~mol})}+1 \mathrm{NO}(\mathrm{~g})-\times \frac{+90.2 \mathrm{~kJ}}{\mathrm{molNO}(\mathrm{~g})}\right]\right] .
$$


1023. Solution: Supporting Why the General Equation Is Valid -
Alternatively, we can write this reaction as the sum of the decompositions of $3 \mathrm{NO}_{2}(g)$ and $1 \mathrm{H}_{2} \mathrm{O}(I)$ into their constituent elements, and the formation of $2 \mathrm{HNO}_{3}(\mathrm{aq})$ and $1 \mathrm{NO}(g)$ from their constituent elements. Writing out these reactions, and noting their relationships to the $\Delta H_{\mathrm{f}}^{\circ}$ values for these compounds (from Appendix G), we have:

$$
\begin{gathered}
3 \mathrm{NO}_{2}(g) \longrightarrow 3 / 2 \mathrm{~N}_{2}(g)+3 \mathrm{O}_{2}(g) \quad \Delta H_{1}^{\circ}=-99.6 \mathrm{~kJ} \\
\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \quad \Delta H_{2}^{\circ}=+285.8 \mathrm{~kJ}\left[-1 \times \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right)\right] \\
\mathrm{H}_{2}(g)+\mathrm{N}_{2}(g)+3 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HNO}_{3}(a q) \quad \Delta H_{3}^{\circ}=-414.8 \mathrm{~kJ}\left[2 \times \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{HNO}_{3}\right)\right] \\
\frac{1}{2} \mathrm{~N}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{NO}(\mathrm{~g})
\end{gathered} \Delta H_{4}^{\circ}=+90.2 \mathrm{~kJ}[1 \times(\mathrm{NO})] \mathrm{l}
$$

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


1023. Solution: Supporting Why the General Equation Is Valid -
Summing these reaction equations gives the reaction we are interested in:

$$
3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g)
$$

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

$$
\begin{aligned}
\Delta H_{\mathrm{rxn}}^{\circ} & =\Delta H_{1}^{\circ}+\Delta H_{2}^{\circ}+\Delta H_{3}^{\circ}+\Delta H_{4}^{\circ}=(-99.6 \mathrm{~kJ})+(+285.8 \mathrm{~kJ})+(-414.8 \mathrm{~kJ})+(+90.2 \mathrm{~kJ}) \\
& =-138.4 \mathrm{~kJ}
\end{aligned}
$$

So the standard enthalpy change for this reaction is $\Delta H^{\circ}=-138.4 \mathrm{~kJ}$.
Note that this result was obtained by (1) multiplying the $\Delta H_{\mathrm{f}}^{\circ}$ of each product by its stoichiometric coefficient and summing those values, (2) multiplying the $\Delta H_{\mathrm{f}}^{\circ}$ 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.


1024. Check Your Learning -
Calculate the heat of combustion of 1 mole of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(I)$, when $\mathrm{H}_{2} \mathrm{O}(I)$ and $\mathrm{CO}_{2}(g)$ are formed. Use the following enthalpies of formation: $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{I}),-278 \mathrm{~kJ} / \mathrm{mol} ; \mathrm{H}_{2} \mathrm{O}(\mathrm{I}),-286 \mathrm{~kJ} / \mathrm{mol}$; and $\mathrm{CO}_{2}(\mathrm{~g}),-394 \mathrm{~kJ} / \mathrm{mol}$.


1025. Answer: -
-1368 kJ/mol


1026. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the energetics of covalent and ionic bond formation and breakage
- Use the Born-Haber cycle to compute lattice energies for ionic compounds
- Use average covalent bond energies to estimate enthalpies of reaction

A bond's strength describes how strongly each atom is joined to another atom, and therefore how much energy is required to break the bond between the two atoms.

It is essential to remember that energy must be added to break chemical bonds (an endothermic process), whereas forming chemical bonds releases energy (an exothermic process). In the case of $\mathrm{H}_{2}$, the covalent bond is very strong; a large amount of energy, 436 kJ , must be added to break the bonds in one mole of hydrogen molecules and cause the atoms to separate:

$$
\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{H}(\mathrm{~g}) \quad \text { bond energy }=436 \mathrm{~kJ}
$$

Conversely, the same amount of energy is released when one mole of $\mathrm{H}_{2}$ molecules forms from two moles of H atoms:

$$
2 \mathrm{H}(g) \longrightarrow \mathrm{H}_{2}(g) \quad \text { bond energy }=-436 \mathrm{~kJ}
$$


1027. Bond Strength: Covalent Bonds -
Stable molecules exist because covalent bonds hold the atoms together. We measure the strength of a covalent bond by the energy required to break it, that is, the energy necessary to separate the bonded atoms. Separating any pair of bonded atoms requires energy (see Figure 4.4). The stronger a bond, the greater the energy required to break it.

The energy required to break a specific covalent bond in one mole of gaseous molecules is called the bond energy or the bond dissociation energy. The bond energy for a diatomic molecule, $\mathrm{D}_{\mathrm{X}-\mathrm{Y}}$, is defined as the standard enthalpy change for the endothermic reaction:

$$
\mathrm{XY}(g) \longrightarrow \mathrm{X}(g)+\mathrm{Y}(g) \quad \mathrm{D}_{\mathrm{X}-\mathrm{Y}}=\Delta H^{\circ}
$$

For example, the bond energy of the pure covalent $\mathrm{H}-\mathrm{H}$ bond, $\mathrm{D}_{\mathrm{H}-\mathrm{H}}$, is 436 kJ per mole of $\mathrm{H}-\mathrm{H}$ bonds broken:

$$
\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g) \quad \mathrm{D}_{\mathrm{H}-\mathrm{H}}=\Delta H^{\circ}=436 \mathrm{~kJ}
$$

Molecules with three or more atoms have two or more bonds. The sum of all bond energies in such a molecule is equal to the standard enthalpy change for the endothermic reaction that breaks all the bonds in the molecule. For example, the sum of the four $\mathrm{C}-\mathrm{H}$ bond energies in $\mathrm{CH}_{4}, 1660 \mathrm{~kJ}$, is equal to the standard enthalpy change of the reaction:


1027. Bond Strength: Covalent Bonds -
The average $\mathrm{C}-\mathrm{H}$ bond energy, $\mathrm{D}_{\mathrm{C}-\mathrm{H}}$, is $1660 / 4=415 \mathrm{~kJ} / \mathrm{mol}$ because there are four moles of $\mathrm{C}-\mathrm{H}$ bonds broken per mole of the reaction. Although the four $\mathrm{C}-\mathrm{H}$ bonds are equivalent in the original molecule, they do not each require the same energy to break; once the first bond is broken (which requires $439 \mathrm{~kJ} / \mathrm{mol}$ ), the remaining bonds are easier to break. The $415 \mathrm{~kJ} / \mathrm{mol}$ value is the average, not the exact value required to break any one bond.

The strength of a bond between two atoms increases as the number of electron pairs in the bond increases. Generally, as the bond strength increases, the bond length decreases. Thus, we find that triple bonds are stronger and shorter than double bonds between the same two atoms; likewise, double bonds are stronger and shorter than single bonds between the same two atoms. Average bond energies for some common bonds appear in Table 9.3, and a comparison of bond lengths and bond strengths for some common bonds appears in

Table 9.4. When one atom bonds to various atoms in a group, the bond strength typically decreases as we move down the group. For example, $\mathrm{C}-\mathrm{F}$ is $439 \mathrm{~kJ} / \mathrm{mol}, \mathrm{C}-\mathrm{Cl}$ is $330 \mathrm{~kJ} / \mathrm{mol}$, and $\mathrm{C}-\mathrm{Br}$ is $275 \mathrm{~kJ} / \mathrm{mol}$.

Bond Energies (kJ/mol)


1027. Bond Strength: Covalent Bonds -
| Bond | Bond Energy | Bond | Bond Energy | Bond | Bond Energy |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $\mathrm{H}-\mathrm{H}$ | 436 | C-S | 260 | $\mathrm{F}-\mathrm{Cl}$ | 255 |
| H-C | 415 | $\mathrm{C}-\mathrm{Cl}$ | 330 | $\mathrm{F}-\mathrm{Br}$ | 235 |
| $\mathrm{H}-\mathrm{N}$ | 390 | $\mathrm{C}-\mathrm{Br}$ | 275 | Si-Si | 230 |
| H-O | 464 | C-I | 240 | Si-P | 215 |
| H-F | 569 | $\mathrm{N}-\mathrm{N}$ | 160 | Si-S | 225 |
| H-Si | 395 | $\mathrm{N}=\mathrm{N}$ | 418 | $\mathrm{Si}-\mathrm{Cl}$ | 359 |
| $\mathrm{H}-\mathrm{P}$ | 320 | $\mathrm{N} \equiv \mathrm{N}$ | 946 | $\mathrm{Si}-\mathrm{Br}$ | 290 |
| H-S | 340 | $\mathrm{N}-\mathrm{O}$ | 200 | Si-I | 215 |
| $\mathrm{H}-\mathrm{Cl}$ | 432 | N-F | 270 | P-P | 215 |
| $\mathrm{H}-\mathrm{Br}$ | 370 | N-P | 210 | P-S | 230 |
| H-I | 295 | $\mathrm{N}-\mathrm{Cl}$ | 200 | $\mathrm{P}-\mathrm{Cl}$ | 330 |
| C-C | 345 | $\mathrm{N}-\mathrm{Br}$ | 245 | $\mathrm{P}-\mathrm{Br}$ | 270 |
| $\mathrm{C}=\mathrm{C}$ | 611 | O-O | 140 | P-I | 215 |


1027. Bond Strength: Covalent Bonds -
| $\mathrm{C}=\mathrm{C}$ | 611 | O-O | 140 | P-I | 215 |
| $\mathrm{C} \equiv \mathrm{C}$ | 837 | $\mathrm{O}=\mathrm{O}$ | 498 | S-S | 215 |
| $\mathrm{C}-\mathrm{N}$ | 290 | O-F | 160 | $\mathrm{S}-\mathrm{Cl}$ | 250 |
| $\mathrm{C}=\mathrm{N}$ | 615 | O-Si | 370 | $\mathrm{S}-\mathrm{Br}$ | 215 |
| $\mathrm{C} \equiv \mathrm{N}$ | 891 | O-P | 350 | $\mathrm{Cl}-\mathrm{Cl}$ | 243 |
| $\mathrm{C}-\mathrm{O}$ | 350 | $\mathrm{O}-\mathrm{Cl}$ | 205 | $\mathrm{Cl}-\mathrm{Br}$ | 220 |
| $\mathrm{C}=\mathrm{O}$ | 741 | O-I | 200 | Cl-I | 210 |
| $\mathrm{C} \equiv \mathrm{O}$ | 1080 | F-F | 160 | $\mathrm{Br}-\mathrm{Br}$ | 190 |
| C-F | 439 | F-Si | 540 | $\mathrm{Br}-\mathrm{I}$ | 180 |
| $\mathrm{C}-\mathrm{Si}$ | 360 | F-P | 489 | I-I | 150 |


1027. Bond Strength: Covalent Bonds -
TABLE 9.3

| Bond | Bond Energy | Bond | Bond Energy | Bond | Bond Energy |
| :---: | :---: | :---: | :---: | :---: | :---: |
| C-P | 265 | F-S | 285 |  |  |

TABLE 9.3

Average Bond Lengths and Bond Energies for Some Common Bonds

| Bond | Bond Length (Å) | Bond Energy (kJ/mol) |
| :--- | :--- | :--- |
| $\mathrm{C}-\mathrm{C}$ | 1.54 | 345 |
| $\mathrm{C}=\mathrm{C}$ | 1.34 | 611 |
| $\mathrm{C} \equiv \mathrm{C}$ | 1.20 | 837 |
| $\mathrm{C}-\mathrm{N}$ | 1.43 | 290 |
| $\mathrm{C}=\mathrm{N}$ | 1.38 | 615 |
| $\mathrm{C} \equiv \mathrm{N}$ | 1.16 | 891 |
| $\mathrm{C}-\mathrm{O}$ | 1.43 | 741 |
| $\mathrm{C}=\mathrm{O}$ | 1.23 | 1.13 |

TABLE 9.4


1027. Bond Strength: Covalent Bonds -
TABLE 9.4

The bond energy is the difference between the energy minimum (which occurs at the bond distance) and the energy of the two separated atoms. This is the quantity of energy released when the bond is formed. Conversely, the same amount of energy is required to break the bond. For the $\mathrm{H}_{2}$ molecule shown in Figure 5.2, at the bond distance of 74 pm the system is $7.24 \times 10^{-19} \mathrm{~J}$ lower in energy than the two separated hydrogen atoms. This may seem like a small number. However, as we will learn in more detail later, bond energies are often discussed on a per-mole basis. For example, it requires $7.24 \times 10^{-19} \mathrm{~J}$ to break one $\mathrm{H}-\mathrm{H}$ bond, but it takes $4.36 \times 10^{5} \mathrm{~J}$ to break 1 mole of $\mathrm{H}-\mathrm{H}$ bonds. A comparison of some bond lengths and energies is shown in Figure 5.2 and Table 9.3. We can find many of these bonds in a variety of molecules, and this table provides average values. For example, breaking the first $\mathrm{C}-\mathrm{H}$ bond in $\mathrm{CH}_{4}$ requires $439.3 \mathrm{~kJ} / \mathrm{mol}$, while breaking the first $\mathrm{C}-\mathrm{H}$ bond in $\mathrm{H}-\mathrm{CH}_{2} \mathrm{C}_{6} \mathrm{H}_{5}$ (a common paint thinner) requires $375.5 \mathrm{~kJ} / \mathrm{mol}$.


1027. Bond Strength: Covalent Bonds -
As seen in Table 9.3 and Table 9.4, an average carbon-carbon single bond is $347 \mathrm{~kJ} / \mathrm{mol}$, while in a carboncarbon double bond, the $\pi$ bond increases the bond strength by $267 \mathrm{~kJ} / \mathrm{mol}$. Adding an additional $\pi$ bond causes a further increase of $225 \mathrm{~kJ} / \mathrm{mol}$. We can see a similar pattern when we compare other $\sigma$ and $\pi$ bonds. Thus, each individual $\pi$ bond is generally weaker than a corresponding $\sigma$ bond between the same two atoms. In a $\sigma$ bond, there is a greater degree of orbital overlap than in a $\pi$ bond.

We can use bond energies to calculate approximate enthalpy changes for reactions where enthalpies of formation are not available. Calculations of this type will also tell us whether a reaction is exothermic or endothermic. An exothermic reaction ( $\Delta H$ negative, heat produced) results when the bonds in the products are stronger than the bonds in the reactants. An endothermic reaction ( $\Delta H$ positive, heat absorbed) results when the bonds in the products are weaker than those in the reactants.

The enthalpy change, $\Delta H$, for a chemical reaction is approximately equal to the sum of the energy required to break all bonds in the reactants (energy "in", positive sign) plus the energy released when all bonds are formed in the products (energy "out," negative sign). This can be expressed mathematically in the following way:

$$
\Delta H=\Sigma \mathrm{D}_{\text {bonds broken }}-\Sigma \mathrm{D}_{\text {bonds formed }}
$$


1027. Bond Strength: Covalent Bonds -
$$
\Delta H=\Sigma \mathrm{D}_{\text {bonds broken }}-\Sigma \mathrm{D}_{\text {bonds formed }}
$$

In this expression, the symbol $\Sigma$ means "the sum of" and D represents the bond energy in kilojoules per mole, which is always a positive number. The bond energy is obtained from a table (like Table 9.4) and will depend on whether the particular bond is a single, double, or triple bond. Thus, in calculating enthalpies in this manner, it is important that we consider the bonding in all reactants and products. Because D values are typically averages for one type of bond in many different molecules, this calculation provides a rough estimate, not an exact value, for the enthalpy of reaction.

Consider the following reaction:

$$
\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{HCl}(g)
$$

or

$$
\mathrm{H}-\mathrm{H}(g)+\mathrm{Cl}-\mathrm{Cl}(g) \longrightarrow 2 \mathrm{H}-\mathrm{Cl}(g)
$$

To form two moles of HCl , one mole of $\mathrm{H}-\mathrm{H}$ bonds and one mole of $\mathrm{Cl}-\mathrm{Cl}$ bonds must be broken. The energy required to break these bonds is the sum of the bond energy of the $\mathrm{H}-\mathrm{H}$ bond $(436 \mathrm{~kJ} / \mathrm{mol})$ and the $\mathrm{Cl}-\mathrm{Cl}$ bond ( $243 \mathrm{~kJ} / \mathrm{mol}$ ). During the reaction, two moles of $\mathrm{H}-\mathrm{Cl}$ bonds are formed (bond energy $=432 \mathrm{~kJ} / \mathrm{mol}$ ), releasing 2 $\times 432 \mathrm{~kJ}$; or 864 kJ . Because the bonds in the products are stronger than those in the reactants, the reaction releases more energy than it consumes:


1027. Bond Strength: Covalent Bonds -
$$
\begin{aligned}
\Delta H & =\sum \mathrm{D}_{\text {bonds broken }}-\sum \mathrm{D}_{\text {bonds formed }} \\
\Delta H & =\left[\mathrm{D}_{\mathrm{H}-\mathrm{H}}+\mathrm{D}_{\mathrm{Cl}-\mathrm{Cl}}\right]-2 \mathrm{D}_{\mathrm{H}-\mathrm{Cl}} \\
& =[436+243]-2(432)=-185 \mathrm{~kJ}
\end{aligned}
$$

This excess energy is released as heat, so the reaction is exothermic. Appendix $G$ gives a value for the standard molar enthalpy of formation of $\mathrm{HCl}(\mathrm{g}), \Delta H_{\mathrm{f}}^{\circ}$, of $-92.307 \mathrm{~kJ} / \mathrm{mol}$. Twice that value is -184.6 kJ , which agrees well with the answer obtained earlier for the formation of two moles of HCl .


1029. Using Bond Energies to Calculate Approximate Enthalpy Changes -
Methanol, $\mathrm{CH}_{3} \mathrm{OH}$, may be an excellent alternative fuel. The high-temperature reaction of steam and carbon produces a mixture of the gases carbon monoxide, CO , and hydrogen, $\mathrm{H}_{2}$, from which methanol can be produced. Using the bond energies in Table 9.4, calculate the approximate enthalpy change, $\Delta H$, for the reaction here:

$$
\mathrm{CO}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{~g})
$$


1030. Solution -
First, we need to write the Lewis structures of the reactants and the products:

From this, we see that $\Delta H$ for this reaction involves the energy required to break a $\mathrm{C}-\mathrm{O}$ triple bond and two $\mathrm{H}-\mathrm{H}$ single bonds, as well as the energy produced by the formation of three $\mathrm{C}-\mathrm{H}$ single bonds, a $\mathrm{C}-\mathrm{O}$ single bond, and an $\mathrm{O}-\mathrm{H}$ single bond. We can express this as follows:

$$
\begin{aligned}
\Delta H & =\Sigma \mathrm{D}_{\text {bonds broken }}-\Sigma \mathrm{D}_{\text {bonds formed }} \\
\Delta H & =\left[\mathrm{D}_{\mathrm{C} \equiv \mathrm{O}}+2\left(\mathrm{D}_{\mathrm{H}-\mathrm{H}}\right)\right]-\left[3\left(\mathrm{D}_{\mathrm{C}-\mathrm{H}}\right)+\mathrm{D}_{\mathrm{C}-\mathrm{O}}+\mathrm{D}_{\mathrm{O}-\mathrm{H}}\right]
\end{aligned}
$$

Using the bond energy values in Table 9.4, we obtain:

$$
\begin{aligned}
\Delta H & =[1080+2(436)]-[3(415)+350+464] \\
& =-107 \mathrm{~kJ}
\end{aligned}
$$

We can compare this value to the value calculated based on $\Delta H_{\mathrm{f}}^{\circ}$ data from Appendix G :

$$
\begin{aligned}
\Delta H & =\left[\Delta H_{\mathrm{f}}^{\circ} \mathrm{CH}_{3} \mathrm{OH}(g)\right]-\left[\Delta H_{\mathrm{f}}^{\circ} \mathrm{CO}(g)+2 \times \Delta H_{\mathrm{f}}^{\circ} \mathrm{H}_{2}\right] \\
& =[-201.0]-[-110.52+2 \times 0] \\
& =-90.5 \mathrm{~kJ}
\end{aligned}
$$


1030. Solution -
Note that there is a fairly significant gap between the values calculated using the two different methods. This occurs because D values are the average of different bond strengths; therefore, they often give only rough agreement with other data.


1031. Check Your Learning -
Ethyl alcohol, $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$, was one of the first organic chemicals deliberately synthesized by humans. It has many uses in industry, and it is the alcohol contained in alcoholic beverages. It can be obtained by the fermentation of sugar or synthesized by the hydration of ethylene in the following reaction:

Using the bond energies in Table 9.4, calculate an approximate enthalpy change, $\Delta H$, for this reaction.


1032. Answer: -
-35 kJ


1033. Ionic Bond Strength and Lattice Energy -
An ionic compound is stable because of the electrostatic attraction between its positive and negative ions. The lattice energy of a compound is a measure of the strength of this attraction. The lattice energy ( $\boldsymbol{\Delta} \boldsymbol{H}_{\text {lattice }}$ ) of an ionic compound is defined as the energy required to separate one mole of the solid into its component gaseous ions. For the ionic solid MX, the lattice energy is the enthalpy change of the process:

$$
\mathrm{MX}(s) \longrightarrow \mathrm{M}^{n+}(g)+\mathrm{X}^{n-}(g) \quad \Delta H_{\text {lattice }}
$$

Note that we are using the convention where the ionic solid is separated into ions, so our lattice energies will be endothermic (positive values). Some texts use the equivalent but opposite convention, defining lattice energy as the energy released when separate ions combine to form a lattice and giving negative (exothermic) values. Thus, if you are looking up lattice energies in another reference, be certain to check which definition is being used. In both cases, a larger magnitude for lattice energy indicates a more stable ionic compound. For sodium chloride, $\Delta H_{\text {lattice }}=769 \mathrm{~kJ}$. Thus, it requires 769 kJ to separate one mole of solid NaCl into gaseous $\mathrm{Na}^{+}$ and $\mathrm{Cl}^{-}$ions. When one mole each of gaseous $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$ions form solid $\mathrm{NaCl}, 769 \mathrm{~kJ}$ of heat is released.

The lattice energy $\Delta H_{\text {lattice }}$ of an ionic crystal can be expressed by the following equation (derived from Coulomb's law, governing the forces between electric charges):

$$
\Delta H_{\text {lattice }}=\frac{\mathrm{C}\left(\mathrm{Z}^{+}\right)\left(\mathrm{Z}^{-}\right)}{\mathrm{R}_{\mathrm{o}}}
$$


1033. Ionic Bond Strength and Lattice Energy -
in which C is a constant that depends on the type of crystal structure; $\mathrm{Z}^{+}$and $\mathrm{Z}^{-}$are the charges on the ions; and $R_{0}$ is the interionic distance (the sum of the radii of the positive and negative ions). Thus, the lattice energy
of an ionic crystal increases rapidly as the charges of the ions increase and the sizes of the ions decrease. When all other parameters are kept constant, doubling the charge of both the cation and anion quadruples the lattice energy. For example, the lattice energy of $\mathrm{LiF}\left(\mathrm{Z}^{+}\right.$and $\mathrm{Z}^{-}=1$ ) is $1023 \mathrm{~kJ} / \mathrm{mol}$, whereas that of $\mathrm{MgO}\left(\mathrm{Z}^{+}\right.$and $\mathrm{Z}^{-}=2$ ) is $3900 \mathrm{~kJ} / \mathrm{mol}\left(\mathrm{R}_{\mathrm{o}}\right.$ is nearly the same-about 200 pm for both compounds).

Different interatomic distances produce different lattice energies. For example, we can compare the lattice energy of $\mathrm{MgF}_{2}(2957 \mathrm{~kJ} / \mathrm{mol})$ to that of $\mathrm{MgI}_{2}(2327 \mathrm{~kJ} / \mathrm{mol})$ to observe the effect on lattice energy of the smaller ionic size of $\mathrm{F}^{-}$as compared to $\mathrm{I}^{-}$.


1035. Lattice Energy Comparisons -
The precious gem ruby is aluminum oxide, $\mathrm{Al}_{2} \mathrm{O}_{3}$, containing traces of $\mathrm{Cr}^{3+}$. The compound $\mathrm{Al}_{2} \mathrm{Se}_{3}$ is used in the fabrication of some semiconductor devices. Which has the larger lattice energy, $\mathrm{Al}_{2} \mathrm{O}_{3}$ or $\mathrm{Al}_{2} \mathrm{Se}_{3}$ ?


1036. Solution -
In these two ionic compounds, the charges $\mathrm{Z}^{+}$and $\mathrm{Z}^{-}$are the same, so the difference in lattice energy will depend upon $\mathrm{R}_{0}$. The $\mathrm{O}^{2-}$ ion is smaller than the $\mathrm{Se}^{2-}$ ion. Thus, $\mathrm{Al}_{2} \mathrm{O}_{3}$ would have a shorter interionic distance than $\mathrm{Al}_{2} \mathrm{Se}_{3}$, and $\mathrm{Al}_{2} \mathrm{O}_{3}$ would have the larger lattice energy.


1037. Check Your Learning -
Zinc oxide, ZnO , is a very effective sunscreen. How would the lattice energy of ZnO compare to that of NaCl ?


1038. Answer: -
ZnO would have the larger lattice energy because the Z values of both the cation and the anion in ZnO are greater, and the interionic distance of ZnO is smaller than that of NaCl .


1039. The Born-Haber Cycle -
It is not possible to measure lattice energies directly. However, the lattice energy can be calculated using the equation given in the previous section or by using a thermochemical cycle. The Born-Haber cycle is an application of Hess's law that breaks down the formation of an ionic solid into a series of individual steps:

- $\Delta H_{\mathrm{f}}^{\circ}$, the standard enthalpy of formation of the compound
- IE, the ionization energy of the metal
- $E A$, the electron affinity of the nonmetal
- $\Delta H_{s}^{\circ}$, the enthalpy of sublimation of the metal
- $D$, the bond dissociation energy of the nonmetal
- $\Delta H_{\text {lattice }}$, the lattice energy of the compound

Figure 9.25 diagrams the Born-Haber cycle for the formation of solid cesium fluoride.

FIGURE 9.25 The Born-Haber cycle shows the relative energies of each step involved in the formation of an ionic solid from the necessary elements in their reference states.

We begin with the elements in their most common states, $\operatorname{Cs}(s)$ and $\mathrm{F}_{2}(g)$. The $\Delta H_{s}^{\circ}$ represents the conversion of solid cesium into a gas, and then the ionization energy converts the gaseous cesium atoms into cations. In the next step, we account for the energy required to break the $F-F$ bond to produce fluorine atoms. Converting one mole of fluorine atoms into fluoride ions is an exothermic process, so this step gives off energy (the electron affinity) and is shown as decreasing along the $y$-axis. We now have one mole of Cs cations and one mole of F anions. These ions combine to produce solid cesium fluoride. The enthalpy change in this step is the negative of the lattice energy, so it is also an exothermic quantity. The total energy involved in this conversion is equal to the experimentally determined enthalpy of formation, $\Delta \boldsymbol{H}_{\mathrm{f}}^{\circ}$, of the compound from its elements. In this case, the overall change is exothermic.


1039. The Born-Haber Cycle -
Hess's law can also be used to show the relationship between the enthalpies of the individual steps and the enthalpy of formation. Table 9.5 shows this for cesium fluoride, CsF.


1039. The Born-Haber Cycle -
| Enthalpy of
 sublimation of
 $\mathrm{Cs}(s)$ | $\mathrm{Cs}(s) \longrightarrow \mathrm{Cs}(g)$ | $\Delta H=\Delta H_{s}^{\circ}=76.5 \mathrm{~kJ} / \mathrm{mol}$ |
| :--- | :--- | :--- |
| One-half of the
 bond energy of
 $\mathrm{F}_{2}$ | $\frac{1}{2} \mathrm{~F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{F}(\mathrm{g})$ | $\Delta H=\frac{1}{2} D=79.4 \mathrm{~kJ} / \mathrm{mol}$ |
| Ionization
 energy of $\mathrm{Cs}(g)$ | $\mathrm{Cs}(g) \longrightarrow \mathrm{Cs}^{+}(g)+\mathrm{e}^{-}$ | $\Delta H=I E=375.7 \mathrm{~kJ} / \mathrm{mol}$ |
| Electron
 affinity of F | $\mathrm{F}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{F}^{-}(g)$ | $\Delta H=E A=-328.2 \mathrm{~kJ} / \mathrm{mol}$ |
| Negative of the
 lattice energy
 of CsF $(s)$ | $\mathrm{Cs}{ }^{+}(g)+\mathrm{F}^{-}(g) \longrightarrow \mathrm{CsF}(s)$ | $\Delta H=-\Delta H_{\text {lattice }}=?$ |
| Enthalpy of
 formation of
 CsF $(s)$, add
 steps $1-5$ | $\Delta H=\Delta H_{f}^{\circ}=\Delta H_{s}^{\circ}+\frac{1}{2} D+I E+(E A)+\left(-\Delta H_{\mathrm{lattice}}\right)$ | $\Delta H=-553.5 \mathrm{~kJ} / \mathrm{mol}$ |


1040. TABLE 9.5 -
Thus, the lattice energy can be calculated from other values. For cesium fluoride, using this data, the lattice energy is:

$$
\Delta H_{\text {lattice }}=76.5+79.4+375.7+(-328.2)-(-553.5)=756.9 \mathrm{~kJ} / \mathrm{mol}
$$

The Born-Haber cycle may also be used to calculate any one of the other quantities in the equation for lattice energy, provided that the remainder is known. For example, if the relevant enthalpy of sublimation $\Delta H_{s}^{\circ}$, ionization energy (IE), bond dissociation enthalpy (D), lattice energy $\Delta H_{\text {lattice, }}$, and standard enthalpy of formation $\Delta H_{\mathrm{f}}^{\circ}$ are known, the Born-Haber cycle can be used to determine the electron affinity of an atom.

Lattice energies calculated for ionic compounds are typically much higher than bond dissociation energies measured for covalent bonds. Whereas lattice energies typically fall in the range of 600-4000 kJ/mol (some even higher), covalent bond dissociation energies are typically between $150-400 \mathrm{~kJ} / \mathrm{mol}$ for single bonds. Keep in mind, however, that these are not directly comparable values. For ionic compounds, lattice energies are associated with many interactions, as cations and anions pack together in an extended lattice. For covalent bonds, the bond dissociation energy is associated with the interaction of just two atoms.


1041. Key Terms -
bomb calorimeter device designed to measure the energy change for processes occurring under conditions of constant volume; commonly used for reactions involving solid and gaseous reactants or products
bond energy (also, bond dissociation energy) energy required to break a covalent bond in a gaseous substance
Born-Haber cycle thermochemical cycle relating the various energetic steps involved in the formation of an ionic solid from the relevant elements
calorie (cal) unit of heat or other energy; the amount of energy required to raise 1 gram of water by 1 degree Celsius; 1 cal is defined as 4.184 J
calorimeter device used to measure the amount of heat absorbed or released in a chemical or physical process
calorimetry process of measuring the amount of heat involved in a chemical or physical process
chemical thermodynamics area of science that deals with the relationships between heat, work, and all forms of energy associated with chemical and physical processes
endothermic process chemical reaction or physical change that absorbs heat
energy capacity to supply heat or do work
enthalpy ( $\boldsymbol{H}$ ) sum of a system's internal energy and the mathematical product of its pressure and volume
enthalpy change ( $\mathbf{\Delta H}$ ) heat released or absorbed by a system under constant pressure during a chemical or physical process
exothermic process chemical reaction or physical change that releases heat
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
heat (q) transfer of thermal energy between two bodies
heat capacity (C) extensive property of a body of matter that represents the quantity of heat required to increase its temperature by 1 degree Celsius (or 1 kelvin)
Hess's law if a process can be represented as the sum of several steps, the enthalpy change of the process equals the sum of the enthalpy changes of the steps
hydrocarbon compound composed only of hydrogen and carbon; the major component of fossil fuels
internal energy ( $\boldsymbol{U}$ ) total of all possible kinds of energy present in a substance or substances


1041. Key Terms -
hydrocarbon compound composed only of hydrogen and carbon; the major component of fossil fuels
internal energy ( $\boldsymbol{U}$ ) total of all possible kinds of energy present in a substance or substances
joule (J) SI unit of energy; 1 joule is the kinetic energy of an object with a mass of 2 kilograms moving with a velocity of 1 meter per second, 1 J $=1 \mathrm{~kg} \mathrm{~m}^{2} / \mathrm{s}$ and $4.184 \mathrm{~J}=1 \mathrm{cal}$
kinetic energy energy of a moving body, in joules, equal to $\frac{1}{2} m v^{2}$ (where $m=$ mass and $v=$ velocity)
lattice energy ( $\boldsymbol{\Delta} \boldsymbol{H}_{\text {lattice }}$ ) energy required to separate one mole of an ionic solid into its component gaseous ions
nutritional calorie (Calorie) unit used for quantifying energy provided by digestion of foods, defined as 1000 cal or 1 kcal
potential energy energy of a particle or system of particles derived from relative position, composition, or condition
specific heat capacity (c) intensive property of a substance that represents the quantity of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius (or 1 kelvin)
standard enthalpy of combustion $\left(\Delta H_{c}^{\circ}\right)$ heat released when one mole of a compound undergoes complete combustion under standard conditions
standard enthalpy of formation $\left(\Delta H_{\mathrm{f}}^{\circ}\right)$ 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
surroundings all matter other than the system being studied
system portion of matter undergoing a chemical or physical change being studied


1041. Key Terms -
state function property depending only on the state of a system, and not the path taken to reach that state
surroundings all matter other than the system being studied
system portion of matter undergoing a chemical or physical change being studied
temperature intensive property of matter that is a quantitative measure of "hotness" and "coldness"
thermal energy kinetic energy associated with the random motion of atoms and molecules
thermochemistry study of measuring the amount
of heat absorbed or released during a chemical reaction or a physical change
work (w) energy transfer due to changes in
external, macroscopic variables such as pressure and volume; or causing matter to move against an opposing force


1042. Key Equations -
$q=c \times m \times \Delta T=c \times m \times\left(T_{\text {final }}-T_{\text {initial }}\right)$
$\Delta U=q+w$
$\Delta H_{\text {reaction }}^{\circ}=\sum n \times \Delta H_{\mathrm{f}}^{\circ}($ products $)-\sum n \times \Delta H_{\mathrm{f}}^{\circ}($ reactants $)$
Bond energy for a diatomic molecule: XY $(g) \longrightarrow \mathrm{X}(g)+\mathrm{Y}(g)$

$$
\mathrm{D}_{\mathrm{X}-\mathrm{Y}}=\Delta H^{\circ}
$$

Enthalpy change: $\Delta H=\Sigma \mathrm{D}_{\text {bonds broken }}-\Sigma \mathrm{D}_{\text {bonds formed }}$
Lattice energy for a solid MX: MX $(s) \longrightarrow \mathrm{M}^{n+}(g)+\mathrm{X}^{n-}(g) \quad \Delta H_{\text {lattice }}$
Lattice energy for an ionic crystal: $\Delta H_{\text {lattice }}=\frac{\mathrm{C}\left(\mathrm{Z}^{+}\right)\left(\mathrm{Z}^{-}\right)}{\mathrm{R}_{\mathrm{o}}}$


1043. Summary - 1043.1. Energy Basics
Energy is the capacity to supply heat or do work (applying a force to move matter). Kinetic energy (KE) is the energy of motion; potential energy is energy due to relative position, composition, or condition. When energy is converted from one form into another, energy is neither created nor destroyed (law of conservation of energy or first law of thermodynamics).

The thermal energy of matter is due to the kinetic energies of its constituent atoms or molecules. Temperature is an intensive property of matter reflecting hotness or coldness that increases as the average kinetic energy increases. Heat is the transfer of thermal energy between objects at different temperatures. Chemical and physical processes can absorb heat (endothermic) or release heat (exothermic). The SI unit of energy, heat, and work is the joule ( J ).

Specific heat and heat capacity are measures of the energy needed to change the temperature of a substance or object. The amount of heat absorbed or released by a substance depends directly on the type of substance, its mass, and the temperature change it undergoes.


1043. Summary - 1043.2. Calorimetry
Calorimetry is used to measure the amount of thermal energy transferred in a chemical or physical process. This requires careful measurement of the temperature change that occurs during the process and the masses of the system and surroundings. These measured quantities are then used to compute the amount of heat produced or consumed
in the process using known mathematical relations.
Calorimeters are designed to minimize energy exchange between their contents and the external environment. They range from simple coffee cup calorimeters used by introductory chemistry students to sophisticated bomb calorimeters used to determine the energy content of food.


1043. Summary - 1043.3. Enthalpy
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 $\Delta H$, or $\Delta H^{\circ}$ for reactions occurring under standard state conditions at 298 K . The value of $\Delta H$ for a reaction in one direction is equal in magnitude, but opposite in sign, to $\Delta H$ for the reaction in the opposite direction, and $\Delta H$ is directly proportional to the quantity of reactants and products. The standard enthalpy of formation, $\Delta \boldsymbol{H}_{\mathrm{f}}^{\circ}$, is the enthalpy change accompanying the formation of 1 mole of a substance from the elements in their most stable states at 1 bar 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.


1043. Summary - 1043.4. Strengths of Ionic and Covalent Bonds
The strength of a covalent bond is measured by its bond dissociation energy, that is, the amount of energy required to break that particular bond in a mole of molecules. Multiple bonds are stronger than
single bonds between the same atoms. The enthalpy of a reaction can be estimated based on the energy input required to break bonds and the energy released when new bonds are formed. For ionic bonds, the lattice energy is the energy required to separate one mole of a compound into its gas phase
ions. Lattice energy increases for ions with higher charges and shorter distances between ions. Lattice energies are often calculated using the Born-Haber cycle, a thermochemical cycle including all of the energetic steps involved in converting elements into an ionic compound.


1044. Exercises - 1044.1. Energy Basics
1. A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not?
2. Prepare a table identifying several energy transitions that take place during the typical operation of an automobile.
3. Explain the difference between heat capacity and specific heat of a substance.
4. Calculate the heat capacity, in joules and in calories per degree, of the following:
(a) 28.4 g of water
(b) 1.00 oz of lead
5. Calculate the heat capacity, in joules and in calories per degree, of the following:
(a) 45.8 g of nitrogen gas
(b) 1.00 pound of aluminum metal
6. How much heat, in joules and in calories, must be added to a $75.0-\mathrm{g}$ iron block with a specific heat of $0.449 \mathrm{~J} / \mathrm{g}{ }^{\circ} \mathrm{C}$ to increase its temperature from $25^{\circ} \mathrm{C}$ to its melting temperature of $1535{ }^{\circ} \mathrm{C}$ ?
7. How much heat, in joules and in calories, is required to heat a $28.4-\mathrm{g}(1-\mathrm{oz})$ ice cube from $-23.0^{\circ} \mathrm{C}$ to -1.0 ${ }^{\circ} \mathrm{C}$ ?
8. How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added?
9. If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase?
10. A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from $23.2^{\circ} \mathrm{C}$ to $89.6^{\circ} \mathrm{C}$.
(a) What is the specific heat of the substance?
(b) If it is one of the substances found in Table 9.1, what is its likely identity?