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.
Atmospheric
pressure
Atmospheric
pressure
Work is
done as
piston
moves.
Pin
prevents
piston
from
moving.
Gas under
pressure
(a)
Gas now at
atmospheric
pressure
(b)
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
269
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
(7.9)
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
negative.
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
(7.10)
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:
3
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.
F
P = 2
L
Volume (or a volume change) has dimensions of length cubed, L3, so pressure times the volume change is
P ∆V =
F
× L3 = F × L
L2
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.
Heat
Electricity moving
through the wrench
does no work but
releases heat.
1
"Dead" battery
Motor
Fully charged
Work
Fully discharged
Heat
2
Energy from the
battery is represented
by heat and work,
turning the electrical
motor.
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
Thermometer
Stirrer
Valve
271
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.
Lid
The
“bomb”
Heavily
insulated
vat
Water
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:
(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
273
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
(7.12)
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
(7.13)
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
Thermometer
insulator (Figure 7.10). A reaction occurring between
aqueous solutions in such a calorimeter exchanges very
Stirrer
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,
Insulated
cover
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
Reactants
heat from the reaction is transferred to the reaction
in solution
mixture.
Two nested
Styrofoam
cups
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
275
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
kilojoules/reaction.
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
equations:
N2( g ) + 3H2( g ) → 2NH3( g )
2N2( g ) + 6H2( g ) → 4NH3( g )
1
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
produced?
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
277
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
equation:
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 ) ?]
2
2
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 °.
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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 ) +
1
2
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
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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).
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