Chapter 6 Thermochemistry

Thermochemistry is the part of

the thermodynamics that deals

with the relationship between

chemical reaction and heat.

Will a chemical change take place under a given set of conditions?

How far a chemical change can go before reaching equilibrium?

If a chemical change occurs, what are the energy changes?

6-1 Getting Started: Some Terminology

6-2 Heat

6-3 Heats of Reaction and Calorimetry

6-4 Work

6-5 The First Law of Thermodynamics

6-6 Heats of Reaction: ΔU and ΔH

6-7 Indirect Determination of ΔH: Hess’s law

6-8 Standard Enthalpies of Formation

Contents

6-1 Getting Started: Some Terminology

Systems and their surroundings

The study of thermochemistry requires the definition of system and surroundings. The system is the particular sample of matter under investigation; the part of the universe that we are going to focus on. And the surroundings include everything else relevant to the change.

Open system

HeatHeat Closed

system

HeatMatter

(water vapor)

Heat

Isolated systema b c

Vacuum flask

a b c

Systems and their Surroundings

A system and its surroundings compose the universe.

a. Open system. The glass cylinder of hot water transfers energy to the surroundings---it loses heat as it cools. Matter

is also transfers in the form of water vapor.

b. Closed system. The glass cylinder with a cover of hot water transfers energy(heat) to the surroundings as it cools. Because the flask is stoppered, no water vapor escapes and no matter is transferred.

c. Isolated system. Hot water in an insulated flask approximates an isolated system. No water vapor escapes, and for a time at least, little heat is transferred to the surroundings.

System

No energy in or out

No matter in or out

Boundary

Isolated System: No interaction across the system boundary.(a system that does not interact with its surroundings)

System is the part of the

universe chosen for study

Isolated System

Energy: The capacityto do work

Kinetic energy: The energy of moving

object

Potential energy: The Energy has the

potential to do work

Work: done when a force acts through a distance

Wok = force × distance = [m(kg) × a(m s-2)] × d(m)

Kinetic energy = ½ × m(kg) × [u(m/s)]2 (6.1)

SI unit of energy: 1 joule(J) = 1 kgm2s-2

1 cal = 4.184J (exactly) (6.2) 1Cal=1000cal=1kcal

6-2 Heat

Heat(q), like work, is energy transferred between a system and its surroundings as a result of a temperature difference.

We will discuss the quantity of heat following three questions:

Heat capacity Specific heatLaw of conservation

of energy

The quantity of heat required to change the temperature of a system by one degree is called the heat capacity of the system. If the system is a mole of substance, we can use the term molar heat

capacity. If it is one gram of substance, we call it specific heat capacity, or more commonly, Specific heat(sp ht). The specific heat of water depends some what on temperature, but, over the range from

0 to 100 ºC, its value is about:

1.00cal/g ºC =1cal g-1 ºC -1 = 4.18J/ g ºC = 4.18 g-1 ºC –1 (6.3)

Example for calculating a quantity of heat

How much heat is required to raise the temperature of 7.35g of water From 21.0 to 98.0 ºC? (Assume the specific heat of water is

4.18J g-1 ºC -1 throughout this temperature range. )

Solution The specific heat is the heat capacity of 1.00g water: 4.18J/g water ºC.

The heat capacity of system(7.35g water) is 7.35 g water×4.18J/ g water ºC = 30.7J/ ºC. The required temperature change in the system is (98.0-21.0) ºC = 77.0 ºC. The heat required

to produce this temperature change = 30.7 7J/ ºC × 77.0 ºC = 2.36 × 103J.

Summarized

quantity of heat = mass of substance × specific heat × temperature change

heat capacity = C (6.4)

q = m × specific heat × ΔT = C × ΔT (6.5)

ΔT = Tf – Ti

Tf is the final temperature;

Ti is the initial temperature.

If the temperature of a system increases: Tf >Ti, ΔT >0

(the heat is absorbed or gained by the system)

If the temperature of a system decreases: Tf <Ti, ΔT <0

(the heat is evolved or lost by the system)

Law of conservation of energy

In interactions between a system and its surroundings, The total the energy remains constant-energy is neither created

nor destroyed. Applied to the exchange of heat,This means that

qsystem + qsurroundings = 0 (6.6)

Heat lost by a system is gained by its surroundings, and vice versa.

qsystem = - qsurroundings (6.7)

Experimental determination of specific heats

a: A 150.0-g sample of lead is heated to the temperature of boiling water

(100 ºC). b: A 50.0-g sample of water is added to a thermally insulted beaker,

and its temperature is found to be 22.0ºC. c: The hot lead is dumped into the cold water, and the temperature of the final lead-water mixture is 28.8 ºC

a b c

22.0ºC 28.8 ºC150.0g Lead

qlead =- qwater

(6.8)

Example for determining a specific heat from experimental data(use data presented in figure)

First, let us use equation(6.5) to calculate qwater

qwater = 50.0g water × 4.18J/ g water ºC × (28.8 - 22.0) ºC = 1.4 × 103JFrom equation (6.8) we can write q lead =- q water = 1.4 × 103JNow, from equation(6.5) again, we obtain qlead = 150.0g lead × specific heat of lead × (28.8 -100.0 ) ºC = -1.4 × 103Jspecific heat of lead = -1.4 × 103J/ 150.0g lead × (28.8 -100.0 ) ºC = -1.4 × 103J/ 150.0g lead × (- 71.2 )ºC = 0.13 J g-1ºC –1

Note: if you know any four of the five quantities-q, m, specific heat, Tf , Ti – you can solve equation (6.5) for the remaining one

Solution

Metals Sp ht Nonmetals Sp ht Metalloids Sp ht

Pb 0.128 Se 0.321 Te 0.202

Cu 0.385 S 0.706 As 0.329

Fe 0.449 P 0.777

Al 0.903

Mg 1.024

Table 6.1 Specific heat of several solid elements(in J g-1ºC -1 )

Significance: The relatively high specific heat of aluminum helps to account for its use as “miracle thaw” products designed to thaw frozen foods rapidly. Compounds tend to have higher specific heats. Water has a specific heat that is more than 30time greater than that of lead.

6-3 Heats of reaction and calorimetry

Thermal energy

Chemical energy

Heat of reactionExothermic and

Endothermic reactions

Calorimetry

40.5

a

CaO(s) + H2O(l) Ca(OH)2(s)

a: An exothermic reaction. The reactants are mixed at room temperature, but the temperature

of the mixture rises to 40. 5ºC.

5.8

b

b: An endothermic reaction. The reactants are mixed at room temperature, but the temperature

of the mixture falls to 5.8ºC.

Heat of reaction, qrxn, is the quantity of heat exchanged between a system and its surroundings when a chemical reaction

occurs within the system,at constant temperature

Exothermic reaction is one that produces a temperature increase

In an isolated system or, in a nonisolated system,

gives off heat to surroundings. (qrxn< 0)

In an endothermicreaction, the correspondingsituation is a

temperature decreasein an isolated system

or a gain of heat from the surroundingsby a nonisolated system.(qrxn> 0)

Heats of reaction are

experimentallydetermined

in a Calorimeter, a device for measuring quantities of heat.

Bomb Calorimeter

Motorized stirrer

Ignition wires

Thermometer

Insulated container

O2 inlet Steel Bomb(reaction chamber)

Sample dish

Water bath

Ignition coil

Constant-Volume Calorimeter

A bomb calorimeter is ideally suited for measuring the heat evolved in a combustion reaction . The system is isolated from its surroundings. When the combustion reaction occurs, chemical energy is converted to thermal energy, and the temperature of the system rises. The heat of reaction is the quantity of heat that the system would have to lose to its surroundings to be restored to its initial temperature. This quantity of heat, in turn, is just the negative of the thermal energy gained by the calorimeter and its contents (qcalorim):

qrxn = － qcalorim(where qcalorim = qwater ＋ qbomb …＋ ) (6.9)

Note : That the temperature of a reaction mixture usually changes during a reaction, so that we must return the mixture to the initial temperature before we assess how much heat is exchanged with surroundings.

We assemble the calorimeter in exactly the same way each time, a heat capacity of the calorimeter is defined. From qalorim, we then establish qrxn, then we can determine the heat of combution of different substances.

qcalorim = heat capacity of calorim × ΔT (6.10)

Example for using bomb calorimeter data to determine a heat of reaction.

The combustion of 1.010 g sucrose, C12H22O11,in a bomb calorimeter causes the temperature torise from 24.92 to 28.33ºC. The heat capacity of

the calorimeter assembly is 4.90kJ/ºC.(a) What is the heat of combustion of sucrose, expressed

in kilojoules per mole of C12H22O11?(b) Verify the claim of sugar producers that one teaspoon

of sugar(about 4.8 g) contains only 19 Calories

Solution

(a) First we can calculate q calorim with equation (7.10).

q calorim = 4.90kJ/ºC × (28.33 –24.92) ºC = (4.90 × 3.14)kJ = 16.7 kJ

Now, using equation (7.9), we get

qrxn = － qcalorim = -16.7 kJ

This is the heat of combustion of the 1.010 g sample.

Per gram C12H22O11

Per mol C12H22O11

qrxn =-16.7kJ

1.010 g C12H22O11= -16.5 kJ/g C12H22O11

qrxn =-16.5kJ

g C12H22O11 1mol C12H22O11

342.3 g C12H22O11

= - 5.65

¡Á

¡Á103 kJ/mol C12H22O11

(b) To determine the caloric content of sucrose, we can use heat of combustion per gram of sucrose determined in part(a), together with a factor to convert from kilojoules to kilocalories. (Because 1 cal = 4.184 J, 1 kcal = 4.184 kJ.)

1 food Calorie (1 Calorie with a capital C) is actually 1000 cal, or 1 kcal. Therefore, 19 kcal = 19 Calories. The claim is justified.

The “Coffee － Cup” Calorimeter

Thermometer

Stirrer Rod

Cork cover

Two Styrofoam cups nested together containing reactants in solution.

Constant-Pressure Calorimeter

Example for using “Coffee － Cup” calorimeter data to determine a heat of reaction.

In the neutralization of a strong acid with a strong base, the essential reaction is the combination of H ＋ (aq) and OH － (aq) to form water. H ＋ (aq) ＋ OH － (aq) H2O(l)two solutions, 100.0mL of 1.00 mol/L HCl(aq) and 100.0 mL of 1.00 mol/L NaOH (aq), both initially at 21.1 ºC, are added to a Styrofoam cup calorimeter and allowed to react. The temperature rises to 27.8 ºC. Determine the heat of the neutralization reaction, expressed per mole H2O formed. Is the reaction endothermic or exothermic?

In addition to assuming that the calorimeter is an isolated system, assume that all there is in the system to absorb heat is 200.0 mL of Water. This assumption ignores the fact that 0.10 mol each of NaCl and H2O are formed in the reaction, that the density of the resulting NaCl (aq) is not quite 1.00 g/mL, And that its specific heat is not quite 4.18 J g － 1ºC –1. Also, ignore the small heat capacity of the Styrofoam cup itself.Solution

Because the reaction is a neutralization reaction, let us call the heat of reaction qneutr. Now, according to equation(6.9), qneutr = － qcalorim, and if we make the assumptions described above,

qcalorim = 200.0 mL × 1.00 g

mL ×4.18 J

g ºC× (27.8 － 21.1 º

C) = 5.6 × 103 J

? Mol HCl H ＋ = 0.1000 L × 1.00 mol HCl

1 L×

qneutr = － qcalorim= － 5.6 × 103 J = － 5.6 kJ

In 100.0 mL of 1.00 mol/L HCl, the amount of H ＋ is

1 mol H ＋

1 mol HCl

= 0.100 mol H ＋

Similarly, in 100.0 mL of 1.00 mol/L NaOH there is 0.100 mol OH － . Thus, the H ＋ and the OH － combine to form 0.100 mol H2O. (The two are in stoichiometric proportions; neither is in excess.)

The amount of heat produced per mole of H2O is

q neutr =－ 5.6kJ

0.100 mol H2O= － 56 kJ/mol H2O

Because q neutr is a negative quantity, the neutralization

is an exothermic reaction.

6-4 Work

Cross-sectionArea=A

Initialstate

Final state

A moving piston does work on the surroundings. The amount of work done is: w = － p Δv

Note: Pressure is a force per unit area (p = F/A), which means that the product of a pressure and an area is a force (p = F × A). The product in the expansion, Δv . It is the “volume” part of pressure-volume work

Now we can use equation (6.1) to calculate the work done. Work (w) = force (F) × distance (h) = p × A × hEquation (6.11) is the expression we will use to represent a quantity of pressure-volume work. w = － pext ΔvTwo significant features to note in equation (6.11) are :(1) The negative sign: it is necessary to conform to sign conventions.When a gas expands, Δv is positive and w is negative, signifying that energy leaves the system as work; When a gas is compressed, Δv is negative and w is positive, signifying that energy (as work) enters the system.(2) The term pext : it is the external pressure-the pressure against whicha system expands or the applied pressure that compresses a system.

6-5 The First Law of Thermodynamics

Nuclear energy

Chemical energy

Reactor

Thermalenergy

Mechanicalenergy

Electricalenergy

Combustion

Batery

Thermoelectricity

Hydroelectricity

Steam turbine Generator

Friction Motor

Schematic diagram of the common energy conversion processes.

Internal energy, U, is the total energy (both kinetic and potential) in a system, including translational kinetic energy of molecules, the energy associated with molecular rotation andvibrations, the energy stored in chemical bonds and intermolecular attractions, and the energy associated with electrons in atom.

• Molecules exhibit several types of motion:– Translational: Movement of the entire molecule from one

place to another.

– Vibrational: Periodic motion of atoms within a molecule.

– Rotational: Rotation of the molecule on about an axis or rotation about bonds.

First law of thermodynamics, the law of conservation of energy, dictates the relationship between heat (q), work (w), and

changes in internal energy ( ΔU ).

ΔU = q + w (6.12)

If we consider that an isolated system is unable to exchange either heat or work with its surroundings, then ΔU isolated system = 0, and we can say The energy of an isolated system is constant.

In using equation (6.12) we must keep these important points in mind.

Any energy entering the system carries a positive sign. Thus, if heat is absorbed by the system, q > 0. If work is done on the system, w > 0

Any energy leaving the system carries a negative sign. Thus, if heat is given off by the system, q < 0. If work is done by the system, w < 0

In general, the internal energy of a system changes as a result of energy entering or leaving the system as heat and/or work. If, on balance, more energy enters the system than leaves, ΔU is positive. If more energy leavesthan enters, ΔU is negative.

Notice: assigning the correct signs to the quantities of heat and work.

Example for relating ΔU , q, and w Through the first law of thermodynamics

A gas, while expanding ( see the figure of 6-4), absorbs 25 J of heat and does 243 J of work. What is ΔU for the gas?

The key to problems of this type lies in assigning the correct signs to the quantities of heat and work. Because heat is absorbed by (enters) the system, q is positive. Because work done by the system represents energy leaving the system, w is negative. You may find it useful to represent the values of q and w, with their correct signs, within parentheses. Then complete the algebra. ΔU = q + w = ( + 25 J) + ( － 243 J) = 25 J － 243 J = － 218 J

Solution

Functions of state: a state function. any property that has a unique value for a specified state of a system.

• A function of state is a property of a system whose change depends only on the initial and final state of the system and not on the path taken between them.

The internal energy of a system is a function of state. Its value can’ t be established. The difference in internal energy betweenthe two states has a unique value, ΔU = U f - U i.

As a further illustration, consider the scheme outlined here and illustrated by the diagram below. Imagine that a system changes from state 1 to state 2 and then back to state 1.

State 1 (U 1) State 2 (U 2) State 1 (U 1) ΔU － ΔU

U 2

U 1

State 2

State 1

Step 1:

ΔU =

U 2 － U

1

Step 2:

－ ΔU =

U 1 － U

2

ΔU 0verall= U 2 － U 1 ＋ U 1 － U 2 = 0

Internal energy

Thus, the overall change in internal energy is

Because U has a unique value in each state, ΔU also has a unique value; it is U 2 － U 1. The change in internal energy when the system is returned From state 2 to state 1 is

－ ΔU = U 1 － U2

ΔU ＋ ( － ΔU )= (U 2 － U 1) ＋ (U 1 － U 2) = 0

This means that the internal energy returns to its initial value of U 1, which it must do, since it is a function of state. It is important to note here that when we reverse the direction of change, we change the sign of ΔU .

•A path function is a property whose change does depend on the path.

Path-Dependent Function: Unlike internal energy and

changes in internal energy, heat (q), and work (w) are not

functions of state. Their values depend on the path

followed when a system undergoes a change.

（ 1 ） 1g of water under a pressure of 101.3KPa ， its

temperature rise from 287.7K to 288.7K, if it gains heat of

2.09J and the surrounding do 2.09J work to it. Calculate the

ΔU. （ 2 ） if the system is an insulated one ， the same

change takes place to 1g of water, how much work should be

done to the water?

Solution: （ 1 ） Q=2.09J ， W=2.09J

∴ ΔU=Q+W=2.09+(2.09)=4.18J

(2) Q=0, ΔU=4.18J

∴ W= ΔU-Q=4.18-0=4.18J

The value of ΔU is the same for two processes because internal energy is a function of state. However, we see that different work is done in the two processes . Work is not a function of state; it is path dependent.

Note: If w differs in the two different processes, q must also differ, and in such a way that q + w the two process are the same. This then makes q + w = ΔU a unique quantity, as required by the first law of thermodynamics.

6-6 Heats of Reaction: ΔU and ΔH

Think of the reactants in a chemical reaction as the initial state of a system and the products as the final state.

reactants

(initial state )

Ui

products

(final state)

Uf

ΔU = Uf － Ui

According to the first law of thermodynamics, we can also say that ΔU = q ＋ w We have previously identified a heat of reaction as qrxn, and so we can write ΔU = qrxn ＋ w

A combustion reaction carried out in a bomb calorimeter. The volume is constant, ΔV = 0. That is, w = -p ΔV = 0. qv is the heat of reaction for a constant-volume reaction, we see that ΔU = q v

ΔU = qrxn ＋ w = qrxn ＋ 0 = qrxn = q v (6.13)

The heat of reaction measured in a bomb calorimeter is equal to ΔU . How does the heat of a reaction measured in a bomb calorimeter compare with the heat of reaction if the reactionis carried out in some other way? Such as in beakers, flasks,and other containers open to the atmosphere and under theatmosphere and under the constant pressure of the atmosphere

Two different paths leading to the same internal energy change in a system.

Internal energy

Initial state Initial state

Final state Final state

qV

w

qp

UiUi

Uf Uf

ΔU = q v ΔU = qp ＋ w(a) (b)

(a) The volume of the system retains constant and no internal energyis converted into work. Think of burning gasoline in a bomb calorimeter

(b) The system does work, so some of the internal energy change is used to do work. Think of burning gasoline in an automobile engine.

From the figure and the first law of thermodynamics, we see that for the same reaction at constant pressure ΔU = qp ＋w , which means ΔU = qV = qp ＋ w . Thus, unless w = 0, qV and qp must be different. So U is a function of state and q and w are not.

We can use the relationship between qV and qp to devise another state function that represents the heat flow for a process at constant pressure. We begin by writing

qV = qp ＋ w

Now, using ΔU = qV , w = -p ΔV and rearranging terms, we obtain ΔU = qp － w

qp = ΔU ＋ p ΔV

Enthalpy – H : a state function, is the sum of the internal energy and the pressure – volume product of a system: H = U ＋ P V .Enthalpy change – ΔH for a process between initial and final state is ΔH = Hf － Hi = (Uf ＋ PfVf) － (Ui ＋ PiVi) ΔH = = (Uf － Ui) ＋ (PfVf － PiVi)

ΔH = ΔU ＋ p ΔV

If Pi = Pf , the work limited to pressure – volume work, the

enthalpy change is

ΔH = ΔU ＋ p ΔV

And the heat flow for the process under these conditions is

ΔH =qp (6.14)

Enthalpy (ΔH ) and internal energy (ΔU ) changes in a chemical reaction

ΔH and ΔU are related by the expression

ΔH = ΔU + p ΔV (6.15)

the work limited to pressure – volume work;

constant pressure

Now let’s consider the combustion of sucrose.

C12H22O11(s) ＋ 12O2 (g) 12 CO2(g) ＋ 11H2O (l)Δ H = － 5.65 × 103 KJ (6.16)

There is no change in volume in the combustion of sucrose:

qp = qV

Standard States and Standard Enthalpy Changes

The standard state of a solid or liquid substance is the pure Element or compound at a pressure of 1 bar (105 Pa)* and a gas behaving as an ideal gas at a pressure of 1 bar. The values given in this text are all for 298.15 K (25 º C) unless otherwise stated.

we say that the standard enthalpy change is the enthalpy change in a reaction in which all the reactants and products are in their standard states. This so-called standard enthalpy of reaction is denoted with a superscript degree symbol, ΔH º.

Enthalpy diagram

En

thalp

y

Enthalpy

Products

Reactants Products

Reactants

ΔH > 0 ΔH < 0

Endothermic reactants

Exothermic reactants

Horizontal lines represent absolute values of enthalpy. The higher a horizontal line, the greater the value of H that it represents. Vertical lines or arrows represent changes in enthalpy.

6-7 Indirect Determination of ΔH: Hess's Law

The following features of enthalpy change (ΔH) make it possible to calculate large numbers of heats of reaction .

1. ΔH is an Extensive Property. Consider the standard enthalpy change in the formation of NO(g) from its elements at 25 ºC.

To express the enthalpy change in terms of one mole of NO(g), we divide all coefficients and the ΔH value by two.

1/2N2(g) ＋ 1/2O2 (g) NO(g) Δ H = ½ ×180.50 KJ = 90.25 KJ

N2(g) ＋ O2 (g) 2 NO(g) Δ H = 180.50 KJ

2. ΔH Changes Sign When a Process Is Reversed. If we reverse a process, the change in a function of state reverses sign. Thus, ΔH for the decomposition of one mole of NO(g) is - ΔH for the formation of one mole of NO(g).

NO(g) 1/2N2(g) ＋ 1/2O2 (g) Δ Hº = － 90.25 KJ

3. Hess's Law of Constant Heat Summation. To describe the standard enthalpy change for the formation of NO2(g) from N2(g) and O2(g).

½ N2(g) ＋ O2 (g) NO2(g) Δ H = ?

We can think of the reaction as proceeding in two steps: First we form NO(g) from N2(g) and O2(g), and then NO2(g) from NO(g) and O2(g). When we add the equations for these two steps, together with their individual and distinctive ΔH values, we get the overall equation and ΔH value that we are seeking.

1/2N2(g) ＋ 1/2O2 (g) NO(g) Δ H = ＋ 90.25 KJ

NO(g) ＋ 1/2O2 (g) NO2(g) Δ H = － 57.07 KJ

1/2N2(g) ＋ O2 (g) NO2(g) Δ H = ＋ 33.18KJ

Note that in summing the two equations, we canceled out NO(g), a species that would have appeared on both sides of the overall equation.

An enthalpy diagram illustrating Hess's law

Whether the reaction occurs through a single step (black arrow) or in two steps (blue arrow), the enthalpy change is Δ Hº = 33.18 kJ for the overall reaction:

1/2N2(g) ＋ O2 (g) NO2(g)

Enthalpy

NO(g) ＋ ½ O2(g)

½ N2(g) ＋ O2(g)

NO2(g)

Δ Hº = ＋ 90.25 KJ

Δ Hº = － 57.07 KJ

Δ Hº = ＋ 33.18 KJ

Example: Calculate the Δ Hº of the following reaction

at 298K:CO(g)(g)OC(graphite) 22

1 （ 1 ）

(g)CO(g)O)C(graphite 221

221

21

1298 197 molKJH

（ 2）

)()()( 2221 gCOgOgCO

1298 283 molKJH

（ 3 ）

We have known that:

1θ molkJ283)(ΔH 3298

? 1H

321 2 HHH

= -2 × 197- （ -283 ） = -111 kJ·mol-1

1298 1972 molKJH )(

∴ (1)=2× (2)-(3)

• In general, if the equation for a chemical

reaction can be obtained by combining the

equations for other chemical reactions, Δ Hº

for the reaction can be calculated by

combining the Δ Hº values for the reactions

in the same way.

6-8 Standard Enthalpies of Formation

Using Hess's law allows us to determine Δ Hº

values for reactions without actually performing them,

as long as we have sufficient data about the steps that

make up the reactions. We do not always have all of the

information necessary to do this. Fortunately, there is

another way to determine the heat of reaction without

actually performing an experiment.

                                                                                                                      

Δ Hfº("standard delta H of formation"; also called

heat of formation) values are tabulated for a large

number of compounds. (Appendix D in your textbook

contains Δ Hfº values.) Δ Hfº is defined as the heat of

reaction for the production of one mole of a substance

from its constituent elements, each in their standard

states. For instance, the Δ Hfº of liquid water is the ΔH for

the following reaction (at 298 K).

(Remember that production of one mole is part of the definition of Δ Hfº )

Relating a standard enthalpy of formation to a chemical equation

The enthalpy of formation of formaldehyde at 298 K is Δ Hfº = － 108.6 KJ/mol HCHO(g).

Solution

The equation must be written for the formation of one mole of gaseous HCHO. The most stable forms of the elements at 298 K and 1 bar are gaseous H2 and O2 and solid carbon in the for of graphite. Note that we need one fractional coefficient in this equation.

H2(g) + C(graphite) + ½ O2(g) HCHO(g) Δ Hfº = － 108.6 KJ

Standard Enthalpy of Reaction

Standard enthalpy of reaction, Δ Hº or Δ Hrxnº, the enthalpychange of a reaction if the reactants and products of a reaction are in their standard states.

Hrxn° = npHf°(product) - nrHf°( reactant)

ΔfHθ

reactantsΔfH

θproducts

ΔHθ

The most stable elementary substance ΔfHθ= 0

（ sea level）

Example 1 Calculating Δ Hº from Tabulated Values of Δ Hfº . Let us apply equation 6.21 to calculate the standard enthalpy of combustion of ethane, C2H6(g), a component of natural gas.

The reaction is

C2H6(g) + 7/2 O2(g) 2 CO2(g) + 3 H2O(l)

The relationship we need is equation (6.21). The data we

substitute into the relation are from Appendix D.

Δ Hº = {2 × Δ Hfº [CO2(g)] + 3 ×Δ Hfº [H2O(l)]}

－ {Δ Hfº [C2H6(g)] ＋ 7/2 × Δ Hfº [O2(g)]}

= 2 × ( － 393.5 ) ＋ 3 × ( － 285.8) － ( － 84.7) － 7/2 × 0

= － 1559.7 KJ/mol

Solution

Example 2 Calculating an unknown Δ Hfº value. Use the data here and in Table 6.2 to calculate Δ Hfº of benzene, C6H6(l).

2 C6H6(l) ＋ 15 O2 12 CO2(g) ＋ 6 H2O(l) Δ Hfº = － 6535 KJSolution

To organize the data needed in the calculation, let’s begin by writing the chemical equation for the reaction, with Δ Hfº data listed under the chemical formulas.

2 C6H6(l) ＋ 15 O2 12 CO2(g) ＋ 6 H2O(l)

Δ Hfº = － 6535 KJ

Δ Hfº , KJ/mol ? 0 － 393.5 － 285.5

Now, we can substitute known data into expression (6.21) and

rearrange the equation to obtain a lone term on the left:

Δ Hfº [C6H6(1)]. The remainder of the problem simply involves

numerical calculations.

Δ Hfº = 12 × ( － 393.5) ＋ 6 × ( － 285.8) － 2 × [ Δ Hfº C2H6 (l)]

= － 6535 KJ

[Δ Hfº C2H6 (l)] =

{ － 4722 KJ － 1715 KJ} ＋ 6535 KJ

2 mol C6H6

= 49 KJ/mol C6H6 (l)