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5 Heat, Work, and the First Law of Thermodynamics

5 Heat, Work, and the First Law of Thermodynamics

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268 Chapter 7 | Energy and Chemical Change

Figure 7.7 | Expansion of a gas. (a) A gas is

confined under pressure in a cylinder fitted with a

piston that is held in place by a sliding pin. (b) When

the piston is released, the gas inside the cylinder

expands and pushes the piston upward against the

opposing pressure of the atmosphere. As it does so,

the gas does some work on the surroundings.





Work is

done as








Gas under



Gas now at




Determine qv and qp for this reaction.

n Analysis:

The key to solving this problem is realizing that the heat capacity and temperature change can be used to calculate the heat absorbed by the calorimeter, which is

equal in magnitude to the amount of heat lost by the reaction.

n Assembling

the Tools: The tools are Equation 7.5, q = C D t, which makes it

possible to measure the heat gained by the calorimeter and Equation 7.8, qcalorimeter =

-qreaction, which relates the heat lost by the reaction to the heat gained by the calorimeter.

Run 1 will give the heat at constant volume, qv, since the piston cannot move. Run 2 will

give the heat at constant pressure, qp, because with the piston unlocked the entire reaction

will run under atmospheric pressure.

n Solution:

For Run 1, Equation 7.5 gives the heat absorbed by the calorimeter as

q = C Dt = (8.101 kJ/°C) × (28.91 °C - 24.00 °C) = 39.8 kJ

Because the calorimeter gains heat, this amount of heat is released by the reaction, so q for

the reaction must be negative. Therefore, qv = -39.8 kJ.

For Run 2,

q = C Dt = (8.101 kJ/°C) × (31.54 °C - 27.32 °C) = 34.2 kJ

so qp = -34.2 kJ.

n Is

the Answer Reasonable? The arithmetic is straightforward, but in any calculation

involving heat, always check to see that the signs of the heats in the problem make sense.

The calorimeter absorbs heat, so its heats are positive. The reaction releases this heat, so

its heats must be negative.

Why are qv and qp different for reactions that involve a significant volume change? The

system in this case is the reacting mixture. If the system expands against atmospheric pressure, it is doing work. Some of the energy that would otherwise appear as heat is used up

when the system pushes back the atmosphere. In Example 7.4, the work done to expand

the system against atmospheric pressure is equal to the amount of “missing” heat in the

constant pressure case:

work = (-39.8 kJ) - (-34.2 kJ) = -5.6 kJ

The minus sign indicates that energy is leaving the system. This is called expansion work (or

more precisely, pressure–volume work). A common example of pressure–volume work is the

work done by the expanding gases in a cylinder of a car engine as it moves a piston.

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7.5 | Heat, Work, and the First Law of Thermodynamics


Another example is the work done by expanding gases to lift a rocket from the ground.

The amount of expansion work w done can be computed from atmospheric pressure and

the volume change that the system undergoes:3

w = - P DV


P is the opposing pressure against which the piston pushes, and DV is the change in the

volume of the system (the gas) during the expansion, that is, DV = Vfinal - Vinitial. Because

Vfinal is greater than Vinitial, DV must be positive. This makes the expansion work


n The sign of w confirms that the

system loses energy by doing work 

pushing back the atmosphere.

First Law of Thermodynamics

In chemistry, a minus sign on an energy transfer always means that the system loses energy.

Consider what happens whenever work can be done or heat can flow in the system shown

in Figure 7.7b. If work is negative (as in an expansion) the system loses energy and the

surroundings gain it. We say work is done by the system. If heat is negative (as in an

exothermic reaction) the system loses energy and the surroundings gain it. Either event

should cause a drop in internal energy. On the other hand, if work is positive (as in a

compression), the system gains energy and the surroundings lose it. We say that work is

done on the system. If heat is positive (as in an endothermic reaction), again, the system

will gain energy and the surroundings will lose it. Either positive work or positive heat

should cause a positive change in internal energy.

Work and heat are simply alternative ways to transfer energy. Using this sign convention, we can relate the work w and the heat q that go into the system to the internal energy

change (DE ) the system undergoes:

DE = q + w


First law of thermodynamics

In Section 7.2 we pointed out that the internal energy depends only on the current state

of the system; we said that internal energy, E, is a state function. This statement (together

with Equation 7.10, which is a definition of internal energy change) is a statement of the

first law of thermodynamics. The first law implies that we can move energy around in various

ways, but we cannot create energy or destroy it.

DE is independent of how a change takes place; it depends only on the state of the

system at the beginning and the end of the change. The values of q and w depend, however, on what happens between the initial and final states. Thus, neither q nor w is a state

function. Their values depend on the path of the change. For example, consider the discharge of an automobile battery by two different paths (see Figure 7.8). Both paths take

us between the same two states, one being the fully charged state and the other the fully

discharged state. Because E is a state function and because both paths have the same initial

and final states, DE must be the same for both paths. But how about q and w?

P DV must have units of energy if it is referred to as work. Work is accomplished when an opposing

force, F, is pushed through some distance or length, L. The amount of work done is equal to the strength

of the opposing force multiplied by the distance the force is moved:


Work = F × L

Because pressure is force (F ) per unit area, and area is simply length squared, L2, we can write the following

equation for pressure.


P = 2


Volume (or a volume change) has dimensions of length cubed, L3, so pressure times the volume change is

P ∆V =


× L3 = F × L


but, as explained previously, F × L also equals work.

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270 Chapter 7 | Energy and Chemical Change

Figure 7.8 | Energy, heat, and

work. The complete discharge of a

battery along two different paths

yields the same total amount of

energy, DE. However, if the

battery is simply shorted with a

heavy wrench, as shown in path 1,

this energy appears entirely as heat.

Path 2 gives part of the total

energy as heat, but much of the

energy appears as work done by

the motor.


Electricity moving

through the wrench

does no work but

releases heat.


"Dead" battery


Fully charged


Fully discharged



Energy from the

battery is represented

by heat and work,

turning the electrical


In path 1, we simply short the battery by placing a heavy wrench across the terminals.

Sparks fly and the wrench becomes very hot as the battery quickly discharges. Heat is

given off, but the system does no work (w = 0). All of DE appears as heat.

In path 2, we discharge the battery more slowly by using it to operate a motor. Along

this path, much of the energy represented by DE appears as work (running the motor) and

only a relatively small amount appears as heat (from the friction within the motor and the

electrical resistance of the wires).

There are two vital lessons here. The first is that neither q nor w is a state function.

Their values depend entirely on the path between the initial and final states. The second

lesson is that the sum of q and w, DE, is the same regardless of the path taken, as long as

we start and stop at the same places.

7.6 | Heats of Reaction

As we have shown in Section 7.5, the change of energy in a system can be determined by measuring the amount of heat and work that is exchanged in a system. Since the amount of work

that is either done by a system or on a system depends on both pressure and volume, the heat

of a reaction is measured under conditions of either constant volume or constant pressure.

DE, Constant-Volume Calorimetry

The heats of specific reactions are sometimes given labels, for example, the heat produced by

a combustion reaction is called the heat of combustion. Because combustion reactions require

oxygen and produce gaseous products, we have to measure heats of combustion in a sealed

container. Figure 7.9 shows the apparatus that is usually used to determine heats of combustion. The instrument is called a bomb calorimeter because the vessel holding the reaction itself

resembles a small bomb. The “bomb” has rigid walls, so the change in volume, DV, is zero

when the reaction occurs. This means, of course, that P DV must also be zero, and no expansion work is done, so w in Equation 7.10 is zero. Therefore, the heat of reaction measured in

a bomb calorimeter is the heat of reaction at constant volume, q v, and corresponds to DE.

DE = qv

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7.6 | Heats of Reaction

To let gas in under pressure

For electrical ignition





Figure 7.9 | A bomb calorimeter. The water bath

is usually equipped with devices for adding or

removing heat from the water, thus keeping its

temperature constant up to the moment when the

reaction occurs in the bomb. The reaction chamber

is of fixed volume, so P DV must equal zero for

reactions in this apparatus.








Food scientists determine dietary calories in foods and food ingredients by burning them in

a bomb calorimeter. The reactions that break down foods in the body are complex, but they

have the same initial and final states as the combustion reaction for the food.

Example 7.5

Bomb Calorimetry

(a) When 1.000 g of olive oil was completely burned in pure oxygen in a bomb calorimeter like the one shown in Figure 7.9, the temperature of the water bath increased from

22.000 °C to 26.049 °C. How many dietary Calories are in olive oil, per gram? The heat

capacity of the calorimeter is 9.032 kJ/°C. (b) Olive oil is almost pure glyceryl trioleate,

C57H104O6. The equation for its combustion is

C57H104O6 (l ) + 80O2( g ) → 57CO2( g ) + 52H2O(l )

What is the change in internal energy, DE, for the combustion of one mole of glyceryl

trioleate? Assume the olive oil burned in part (a) was pure glyceryl trioleate.

n Analysis:

Bomb calorimetry measures qv, which is equal to the internal energy change

for the reaction. In part (a), the question is asking for the amount of heat released in the

combustion reaction of olive oil. All of the heat released by the reaction is absorbed by

the calorimeter. By calculating the heat absorbed by the calorimeter we will be able to

then calculate the amount of heat released by the reaction. Also, since only one gram was

burned, the amount of heat will be per gram.

For part (b), the question is asking for the change in energy released when one mole of

glyceryl trioleate is burned. We can use the information from part (a) in which we calculated the change in energy for 1 g of the substance and the molar mass of the substance to

calculate the DE for one mole of glyceryl trioleate.

n Assembling the Tools: In part (a), Equation 7.5, qcalorimeter = C D t, is the tool to compute the heat absorbed by the calorimeter from its temperature change and heat capacity.

Then we use the tool for heat transfer, qcalorimeter = -qreaction, and we place a minus sign in

front of the result to get the heat released by the combustion reaction.

Since the heat capacity is in kJ, the heat released will be in kJ, too. We’ll need to convert kJ

to dietary Calories, which are actually kilocalories (kcal). We can use the relationship

1 kcal ⇔ 4.184 kJ

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272 Chapter 7 | Energy and Chemical Change

Part (b) asks for the change in internal energy, DE, per mole. The heat calculated in

part (a) is equal to DE for the combustion of 1.000 g of glyceryl trioleate. The molar mass

of glyceryl trioleate is the tool to convert DE per gram to DE per mole.

n Solution: First, we compute the heat absorbed by the calorimeter when 1.000 g of olive

oil is burned, using Equation 7.5:

qcalorimeter = C D t = (9.032 kJ/°C) × (26.049 °C - 22.000 °C) = 36.57 kJ

Changing the algebraic sign gives the heat of combustion of 1.000 g of olive oil,

qv = -36.57 kJ. Part (a) asks for the heat in dietary Calories. We convert to kilocalories

(kcal), which are equivalent to dietary Calories (Cal):

−36.57 kJ

1 kcal

1 Cal



= −8.740 Cal/g oil

1.000 g oil 4.184 kJ 1 kcal

or we can say “8.740 dietary Calories are released when one gram of olive oil is burned.”

For part (b), we’ll convert the heat produced per gram to the heat produced per mole,

using the molar mass of C57H104O6, 885.4 g/mol:

885.4 g C 57 H104O6

−36.57 kJ


= −3.238 × 104 kJ/mol

1 mol C 57 H104O6

1.000 g C 57 H104O6

Since this is heat at constant volume, we have DE = qv = -3.238 × 104 kJ for combustion of 1 mol of C57H104O6.

n Is

the Answer Reasonable? Always check the signs of calculated heats first in calorimetry calculations since the signs follow the heat flow. If heat is given off, the sign

for q is negative, and in combustion reactions, heat is released, so the sign of q must be

negative. A tablespoon of olive oil has about 100 dietary Calories (with a tablespoon

being 15 mL). Even though olive oil is less dense than water, 15 mL would weigh about

15 g. Therefore, 1 g of olive oil would have about 120 Cal/15 g ≈ 9 Cal/g. So, it would

seem reasonable that a gram of olive oil would have about 9 dietary Calories per gram.

Practice Exercises

7.4 | The heat of combustion of methyl alcohol, CH3OH, is -715 kJ mol-1. When

2.85 g of CH3OH was burned in a bomb calorimeter, the temperature of the calorimeter

changed from 24.05 °C to 29.19 °C. What is the heat capacity of the calorimeter in units

of kJ/°C? (Hint: Which equation relates temperature change, energy, and heat capacity?)

7.5 | A 1.50 g sample of carbon is burned in a bomb calorimeter that has a heat capacity

of 8.930 kJ/°C. The temperature of the water jacket rises from 20.00 °C to 25.51 °C.

What is DE for the combustion of 1 mole of carbon?

DH, Constant-Pressure Calorimetry

n Biochemical reactions also occur at

constant pressure.

n From the Greek en + thalpein,

meaning to heat or to warm.

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Most reactions that are of interest to us do not occur at constant volume. Instead, they run

in open containers such as test tubes, beakers, and flasks, where they experience the constant pressure of the atmosphere, and we can measure the heat of reaction at constant pressure, qp.

When reactions are run under constant pressure, they may transfer energy as heat, qp, and

as expansion work, w, so to calculate DE we need Equation 7.11:


DE = qp + w

This is inconvenient. If we want to calculate the internal energy change for the reaction,

we’ll have to measure its volume change and then use Equation 7.9. To avoid this problem,

scientists have defined a “corrected” internal energy called enthalpy, or H. Enthalpy is

defined by the equation

H = E + PV

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7.6 | Heats of Reaction


At constant pressure,

DH = DE + P DV

               = (qp + w) + P DV

From Equation 7.9, P DV = -w, so

         DH = (qp + w) + (-w)

DH = qp


Like E, H is a state function.

As with internal energy, an enthalpy change, DH , is defined by the equation

DH = Hfinal - Hinitial

For a chemical reaction, this can be rewritten as follows:

DH = Hproducts - Hreactants


Positive and negative values of DH have the same interpretation as positive and negative values of DE.

Significance of the sign of DH:

For an endothermic change, DH is positive.

For an exothermic change, DH is negative.

Algebraic sign of DH

The difference between DH and DE for a reaction equals P DV. This difference can be

fairly large for reactions that produce or consume gases, because these reactions can have

very large volume changes. For reactions that involve only solids and liquids, though, the

values of DV are tiny, so DE and DH for these reactions are nearly identical.

A very simple constant-pressure calorimeter, dubbed

the coffee-cup calorimeter, is made of two nested and

capped cups made of foamed polystyrene, a very good


insulator (Figure 7.10). A reaction occurring between

aqueous solutions in such a calorimeter exchanges very


little heat with the surroundings, particularly if the

reaction is fast. The temperature change is rapid and

easily measured. We can use Equation 7.5, q = C D t,



to find the heat of reaction, if we have determined the

heat capacity of the calorimeter and its contents.

The coffee cup and the thermometer absorb only a

tiny amount of heat, and we can usually ignore them in

our calculations, and just assume that all of the


heat from the reaction is transferred to the reaction

in solution


Two nested



n Research-grade calorimeters have

greater accuracy and precision than

the coffee-cup calorimeter.

Figure 7.10 | A coffee cup

calorimeter used to measure heats

of reaction at constant pressure.

Typically, aqueous solutions of the

reactants, whose temperatures have

been measured, are combined in the

calorimeter and the temperature

change of the mixture is noted as the

reaction proceeds to completion.

The temperature change and the

heat capacities of the solutions

containing the reactants are used to

calculate the heat released or

absorbed in the reaction.

Example 7.6

Constant-Pressure Calorimetry

The reaction of hydrochloric acid and sodium hydroxide is very rapid and exothermic.

The equation is

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O

In one experiment, a student placed 50.0 mL of 1.00 M HCl at 25.5 °C in a

coffee cup calorimeter. To this was added 50.0 mL of 1.00 M NaOH solution also at

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274 Chapter 7 | Energy and Chemical Change

25.5 °C. The mixture was stirred, and the temperature quickly increased to 32.2 °C.

What is DH expressed in kJ per mole of HCl? Because the solutions are relatively

dilute, we can assume that their specific heats are close to that of water, 4.184 J g-1 °C-1.

The density of 1.00 M HCl is 1.02 g mL-1 and that of 1.00 M NaOH is 1.04 g mL-1.

(We will neglect the heat lost to the coffee cup itself, to the thermometer, or to the surrounding air.)

n Analysis:

The reaction is taking place at constant pressure, so the heat we’re calculating

is qp, which equals DH. We’re given the specific heat of the solutions, so to calculate qp we

also need the system’s total mass and the temperature change. The mass here refers to the

total grams of the combined solutions. We’ve been given volumes, however, so we have to

use their densities as a tool to calculate their masses, which you learned to do in Chapter 2.

n Assembling

the Tools: To calculate qp, we will use Equation 7.7, q = ms D t, as the

tool to calculate the heat gained by the solution, and then Equation 7.8, q1 = -q2, which

relates the heat gained by the solution to the heat released by the reaction. In order to

determine the mass of the solution, we will also need to use the volume and the density

tools from Chapter 2. Finally, we will need to use the tool from Chapter 5 that converts

molarity to moles using volume to calculate qp in terms of moles of HCl.

n Solution:

For the HCl solution, the density is 1.02 g mL-1 and we have

1.02 g

× 50.0 mL = 51.0 g

1.00 mL

mass (HCl) =

Similarly, for the NaOH solution, the density is 1.04 g mL-1 and

mass (NaOH) =

1.04 g

× 50.0 mL = 52.0 g

1.00 mL

The mass of the final solution is thus the sum, 103.0 g.

The reaction changes the system’s temperature by (tfinal - tinitial), so

D t = 32.2 °C - 25.5 °C = 6.7 °C

Now we can calculate the heat absorbed by the solution using Equation 7.7.

Heat absorbed by the solution = mass × specific heat × D t

= 103.0 g × 4.184 J g-1 °C-1 × 6.7 °C

= 2.9 × 103 J = 2.9 kJ

According to the first law of thermodynamics, energy cannot be created or destroyed,

so all of the heat absorbed by the solution must come from the reaction, and in which

case we can write:

qreaction = -qsolution

qreaction = -2.9 kJ

However, this is qp specifically for the mixture prepared; the problem calls for kilojoules

per mole of HCl. The tool we use to calculate the number of moles of HCl is the molarity,

which provides a conversion factor connecting volume and moles. In 50.0 mL of HCl

solution (0.0500 L) we have

0 . 0500 L HCl soln ×

1 . 00 mol HCl

= 0 . 0 5 00 mol HCl

1 . 00 L HCl soln

The neutralization of 0.0500 mol of acid has qp = -2.9 kJ. To calculate the heat

released per mole, DH, we simply take the ratio of joules to moles.

∆H per mole of HCl =

−2.9 kJ

= −58 kJ mol −1

0.0500 mol HCl

Thus, DH for neutralizing HCl by NaOH is -58 kJ mol-1 HCl.

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7.7 | Thermochemical Equations


n Is the Answer Reasonable? Let’s first review the logic of the steps we used. Notice

how the logic is driven by definitions, which carry specific units. Working backward, knowing that we want units of kilojoules per mole in the answer, we must calculate separately

the number of moles of acid neutralized and the number of kilojoules that evolved. The

latter will emerge when we multiply the solution’s mass (g) and specific heat (J g-1 °C-1)

by the degrees of temperature increase (°C). A simple change from joules to kilojoules is

also required.

We can check the answer with some simplified arithmetic. Let’s start with the mass of

the solution. Because the densities are close to 1 g mL-1, each solution has a mass of about

50 g, so the mixture weighs about 100 g.

The heat evolved will equal the specific heat times the mass times the temperature

change. The specific heat is around 4 J g-1 °C-1, the mass is about 100 g, and the temperature change is about 7 °C. The heat evolved will be approximately 4 × 700, or 2800 J.

That’s equal to 2.8 × 103 J or 2.8 kJ. Our answer of 2.9 kJ is certainly reasonable. This

much heat is associated with neutralizing 0.05 mol HCl, so the heat per mole is 2.9 kJ

divided by 0.05 mol, or 58 kJ mol-1.

7.6 | For Example 7.6, calculate DH per mole of NaOH. (Hint: How many moles of

NaOH were neutralized in the reaction?)

Practice Exercises

7.7 | When pure sulfuric acid dissolves in water, heat is given off. To measure it, 175 g of

water was placed in a coffee cup calorimeter and chilled to 10.0 °C. Then 4.90 g of

sulfuric acid (H2SO4), also at 10.0 °C, was added, and the mixture was quickly stirred

with a thermometer. The temperature rose rapidly to 14.9 °C. Assume that the value of

the specific heat of the solution is 4.184 J g-1 °C-1, and that the solution absorbs all the

heat evolved. Calculate the heat evolved in kilojoules by the formation of this solution.

(Remember to use the total mass of the solution, the water plus the solute.) Calculate also

the heat evolved per mole of sulfuric acid.

7.8 | Dissolving NH4NO3 in water is an endothermic process. In fact, this reaction is

often used in cold packs sold at the drug store. When 20.0 g of NH4NO3 was rapidly dissolved in 75.0 mL of water in a coffee cup calorimeter, the temperature dropped from

25.7 °C to 10.4 °C. Assuming that the specific heat of the solution is 4.184 J g-1 °C-1 and

that the solution was isolated from the surroundings, calculate the heat absorbed by the

reaction in kJ per mole of NH4NO3.

7.7 | Thermochemical Equations

As we have seen, the amount of heat a reaction produces or absorbs depends on the number

of moles of reactants we combine. It makes sense that if we burn two moles of carbon, we’re

going to get twice as much heat as we would if we had burned one mole. For heats of reaction to have meaning, we must describe the system completely. Our description must

include amounts and concentrations of reactants, amounts and concentrations of products,

temperature, and pressure, because all of these things can influence heats of reaction.

Chemists have agreed to a set of standard states to make it easier to report and compare

heats of reaction. Most thermochemical data are reported for a pressure of 1 bar, or (for

substances in aqueous solution) a concentration of 1 M. A temperature of 25 °C (298 K)

is often specified as well, although temperature is not part of the definition of standard

states in thermochemistry.

n Unless we specify otherwise, 

whenever we write DH we mean DH

for the system, not the surroundings.

n IUPAC recommends the use of

the bar, instead of atmospheres, for

standard states.

DH °, Enthalpy Change for a Reaction at Standard State

The standard heat of reaction is the value of DH for a reaction occurring under standard

conditions and involving the actual numbers of moles specified by the coefficients of the

equation. To show that DH is for standard conditions, a degree sign is added to DH to

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276 Chapter 7 | Energy and Chemical Change

make DH ° (pronounced “delta H naught” or “delta H zero”). The units of DH ° are really


To illustrate clearly what we mean by DH °, let us use the reaction between gaseous

nitrogen and hydrogen that produces gaseous ammonia.

N2( g ) + 3H2( g ) → 2NH3( g )

When 1.000 mol of N2 and 3.000 mol of H2 react to form 2.000 mol of NH3 at 25 °C

and 1 bar, the reaction releases 92.38 kJ. Hence, for the reaction as given by the preceding

chemical equation, DH ° = -92.38 kJ. Often the enthalpy change is given immediately

after the equation; for example,

N2( g ) + 3H2( g ) → 2NH3( g )

Thermochemical equations

DH ° = -92.38 kJ

An equation that also shows the value of DH ° is called a thermochemical equation. It

always gives the physical states of the reactants and products, and its DH ° value is true only

when the coefficients of the reactants and products are taken to mean moles of the corresponding

substances. The above equation, for example, shows a release of 92.38 kJ if two moles of

NH3 form. If we were to make twice as much or 4.000 mol of NH3 (from 2.000 mol of

N2 and 6.000 mol of H2), then twice as much heat (184.8 kJ) would be released. On the

other hand, if only 0.5000 mol of N2 and 1.500 mol of H2 were to react to form 1.000

mole of NH3, then only half as much heat (46.19 kJ) would be released. To describe the

various sizes of the reactions just described, we write the following thermochemical


N2( g ) + 3H2( g ) → 2NH3( g )

2N2( g ) + 6H2( g ) → 4NH3( g )


N ( g ) + 32 H2( g ) → NH3( g )

2 2

DH ° = -92.38 kJ

DH ° = -184.8 kJ

DH ° = -46.19 kJ

Because the coefficients of a thermochemical equation always mean moles, not molecules,

we may use fractional coefficients. (In the kinds of equations you’ve seen up till now, fractional coefficients were not allowed because we cannot have fractions of molecules, but we

can have fractions of moles in a thermochemical equation.)

You must write down physical states for all reactants and products in thermochemical

equations. The combustion of 1 mol of methane, for example, has different values of DH °

if the water produced is in its liquid or its gaseous state.

CH4( g ) + 2O2( g ) → CO2( g ) + 2H2O(l )

CH4( g ) + 2O2( g ) → CO2( g ) + 2H2O( g )

DH ° = -890.5 kJ

DH ° = -802.3 kJ

(The difference in DH ° values for these two reactions is the amount of energy that would

be released by the physical change of 2 mol of water vapor at 25 °C to 2 mol of liquid

water at 25 °C.)

Example 7.7

Writing a Thermochemical Equation

The following thermochemical equation is for the exothermic reaction of hydrogen and

oxygen that produces water.

2H2( g ) + O2( g ) → 2H2O(l )

DH ° = -571.8 kJ

What is the thermochemical equation for this reaction when 1.000 mol of H2O is


n Analysis:

The given equation is for 2.000 mol of H2O, and any changes in the coefficient for water must be made identically to all other coefficients, as well as to the value

of DH °.

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7.8 | Hess’s Law


n Assembling

the Tools: We will use the tools for the definition of a thermochemical

equation, the relationship between coefficients and DH, and the fact that when we change

the amount of reactants, we change the amount of heat for the reaction. Since we are

dividing the number of moles of water in half, we must divide the rest of the coefficients

by 2 and the heat of the reaction by 2.

n Solution:

We divide everything by 2, to obtain

H 2 ( g ) + 12 O2 ( g ) 

→ H 2O(l )

∆H ° = −258 . 9 kJ

n Is

the Answer Reasonable? Compare the equation just found with the initial equation

to see that the coefficients and the value of DH ° are all divided by 2.

7.9 | The combustion of methane can be represented by the following thermochemical


CH4( g ) + 2O2( g ) → CO2( g ) + 2H2O(l )

Practice Exercises

DH ° = -890.5 kJ

Write the thermochemical equation for the combustion of methane when 1/2 mol of

H2O(l ) is formed. [Hint: What do you have to do to the coefficient of H2O(l ) to give

1 H O(l ) ?]



7.10 | What is the thermochemical equation for the formation of 5.000 mol of H2O(l )

from H2( g ) and O2( g )?

7.8 | Hess’s Law

Recall from Section 7.6 that enthalpy is a state function. This important fact permits us to

calculate heats of reaction for reactions that we cannot actually carry out in the laboratory.

You will see soon that to accomplish this, we will combine known thermochemical equations, which usually requires that we manipulate them in some way.

Manipulating Thermochemical Equations

In the last section you learned that if we change the size of a reaction by multiplying or

dividing the coefficients of a thermochemical equation by some factor, the value of DH °

is multiplied by the same factor. Another way to manipulate a thermochemical equation

is to change its direction. For example, the thermochemical equation for the combustion

of carbon in oxygen to give carbon dioxide is

C(s) + O2( g ) → CO2( g )

DH ° = -393.5 kJ

The reverse reaction, which is extremely difficult to carry out, would be the decomposition of carbon dioxide to carbon and oxygen. The law of conservation of energy requires

that its value of DH ° equals +393.5 kJ.

CO2( g ) → C(s) + O2( g )

DH ° = +393.5 kJ

In effect, these two thermochemical equations tell us that the combustion of carbon is

exothermic (as indicated by the negative sign of DH °) and that the reverse reaction is

endothermic. The same amount of energy is involved in both reactions—it is just the

direction of energy flow that is different. The lesson to learn here is that we can reverse any

thermochemical equation as long as we change the sign of its DH °.

jespe_c07_253-302hr.indd 277

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278 Chapter 7 | Energy and Chemical Change

Enthalpy of Reactions

What we’re leading up to is a method for combining known thermochemical equations in

a way that will allow us to calculate an unknown DH ° for some other reaction. Let’s revisit

the combustion of carbon to see how this works.

We can imagine two paths leading from 1 mol each of carbon and oxygen to 1 mol of

carbon dioxide.

One-Step Path. Let C and O2 react to give CO2 directly.

C(s) + O2( g ) → CO2( g )

DH ° = -393.5 kJ

Two-Step Path. Let C and O2 react to give CO, and then let CO react with more O2

to give CO2.

Step 1.

C(s) + 12 O2( g ) → CO( g )

DH ° = -110.5 kJ

Step 2.

CO( g ) + 12 O2( g ) → CO2( g )

DH ° = -283.0 kJ

Overall, the two-step path consumes 1 mol each of C and O2 to make 1 mol of CO2,

just like the one-step path. The initial and final states for the two routes to CO2, in other

words, are identical.

Because DH ° depends only on the initial and final states and is independent of path,

the values of DH ° for both routes should be identical. We can see that this is exactly true

simply by adding the equations for the two-step path and comparing the result with the

equation for the single step.

Step 1.

C(s) + 12 O2( g ) → CO( g )

Step 2. CO( g ) +



O2( g ) → CO2( g )

CO( g ) + C(s) + O2( g ) → CO2( g ) + CO( g )

DH ° = -110.5 kJ

DH ° = -283.0 kJ

DH ° = -110.5 kJ + (-283.0 kJ)

DH ° = -393.5 kJ

The equation resulting from adding Steps 1 and 2 has CO( g ) appearing identically on

opposite sides of the arrow. We can cancel them to obtain the net equation. Such a cancellation is permitted only when both the formula and the physical state of a species are identical

on opposite sides of the arrow. The net thermochemical equation for the two-step process,

therefore, is

C(s) + O2( g ) → CO2( g )

DH ° = -393.5 kJ

The results, chemically and thermochemically, are thus identical for both routes to CO2.

Enthalpy Diagrams

Enthalpy diagram

jespe_c07_253-302hr.indd 278

The energy relationships among the alternative pathways for the same overall reaction

can be depicted using a graphical construction called an enthalpy diagram. Figure 7.11 is an

enthalpy diagram for the formation of CO2 from C and O2. The vertical axis is an energy

scale for enthalpy. Each horizontal line corresponds to a certain total amount of enthalpy,

which we can’t actually measure, for all the substances listed on the line in the physical

states specified. We can measure differences in enthalpy, however, and that’s what we use

the diagram for. Lines higher up the enthalpy scale represent larger amounts of enthalpy,

so going from a lower line to a higher line corresponds to an increase in enthalpy and a

positive value for DH ° (an endothermic change). The size of DH ° is represented by the

vertical distance between the two lines. Likewise, going from a higher line to a lower one

represents a decrease in enthalpy and a negative value for DH ° (an exothermic change).

11/11/10 5:12 PM

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