concise physical chemistry (rogers/concise physical chemistry) || thermochemistry

P1: OTA/XYZ P2: ABCc04 JWBS043-Rogers September 13, 2010 11:24 Printer Name: Yet to Come

4THERMOCHEMISTRY

Einstein once said, “Some things are simple but not easy.” Although he did nothave thermochemistry in mind, his comment applies to this time-honored branchof physical chemistry. Thermochemistry is simple: Run a chemical reaction andmeasure the temperature change (if any). But it is not easy. Anyone trying to dothis job at state-of-the-art precision will soon be enmeshed in technical problemsthat try the patience of Job. Entire institutes of experimental science exist just forprecise measurement of the heat of chemical reactions. If governments are willing tospend millions of dollars to support acquisition of thermochemical data, there mustbe some significant advantage to be gained from them. That, in part, will be discussedin this chapter. Today, computers play a large role in this field. The most significantadvance in thermochemistry in the last decade is the calculation of thermochemicalquantities from quantum mechanical first principles. That also will be introduced inthis chapter.

4.1 CALORIMETRY

A calorimeter is a device intended to measure heat given out or taken up whena chemical reaction or a physical change, such as a change of state, takes place.A Styrofoam coffee cup with a thermometer is a crude calorimeter. The word hashistorical significance; it was once supposed to measure the amount of “caloric”flowing into or out of a system. We have long since discarded the caloric theory but

Concise Physical Chemistry, by Donald W. RogersCopyright C© 2011 John Wiley & Sons, Inc.

56


ENERGIES AND ENTHALPIES OF FORMATION 57

we still measure the amount of heat in calories or kilocalories, related through theconversion factor 4.184 to the number of joules or kilojoules.

4.2 ENERGIES AND ENTHALPIES OF FORMATION

Many compounds can be formed by direct combination of their elements. An exampleis carbon dioxide (CO2). When a small measured amount of carbon in the commonform of graphite, C(gr), is burned in O2 in a closed steel container called a bomb,the process causes a small temperature rise �T in the immediate surroundings calledthe bath. The heat capacity of the bath having been previously determined by anelectrical calibration, one can find qV , the amount of heat given off by the combustionat constant volume:

C(gr) + O2(g) → CO2(g) + qV

Straightforward proportional calculation gives the amount of heat that would havebeen given off if the measured amount of C(gr) had been one gram (qV =32.76 kJ g−1) or if it had been one mole (qV = �U 298 = −393.51 kJ mol−1). Thelatter qV gives the molar energy change �U 298 of the system, which is negative be-cause the system gives off heat to the surroundings. The heat is given off at constantvolume of the closed bomb; hence it is an energy change. In this reaction, the numberof moles of gas used up is the same as the number produced �ngas = 0, so the energychange is the same as the enthalpy change:

�H = �U + �(pV ) = �U + �ngas RT

We can write � f U 298(CO2) = � f H 298(CO2) = −393.5 kJ mol−1 to indicate theenergy or the enthalpy of formation of CO2 for an experiment carried out at 298 K.

The heat of combustion of a gas—for example, hydrogen—can be found using aflame calorimetric apparatus in which a known amount of gas is burned and the heatgiven off is measured by measuring the temperature rise of a suitably positioned bath.The apparatus is a fancy Bunsen burner heating up a container of water equipped witha thermometer. It functions at constant pressure, so the heat given out is the enthalpydecrease of the system:

H2(g) + 12 O2(g) → H2O(l) + qp

which leads to the molar enthalpy of formation of liquid water:

qp = � f H 298(H2O(l)) = −285.6 kJ mol−1

In the formation reaction, 32 mol of gas are consumed to produce 1.0 mol of liquid

water, which has a negligible volume compared to the gas burned. Energy and enthalpyare related by H = U + pV ; hence �H = �U + p�V at constant pressure and


58 THERMOCHEMISTRY

�H = �U + �n RT under the ideal gas assumption. For the formation of one moleof water, n = − 3

2

� f U 298 = � f H 298 − �n RT = � f H 298 + 32 RT

= −285,600 J mol−1 + 32 [8.31(298)] ∼= −281.9 kJ mol−1

For comparison with theoretical calculations, one often needs to know thethermodynamic properties of molecules in the gaseous state—for example,� f H 298(H2O(g)). This is handled by adding the heat of vaporization of water to� f H 298(H2O(l)) to obtain

� f H 298(H2O(g)) = −285.6 + 44.0 = −241.6 kJ mol−1

However, water vapor is not in its standard state.

4.3 STANDARD STATES

One can burn a diamond, C(dia), in an oxygen bomb calorimeter. When this is done,the measured enthalpy of formation of CO2(g) is about 2 kJ mol−1 more negativethan the � f H 298(CO2(g)) found when C(gr), carbon in the standard state, is burned.The difference is not in the CO2(g) produced, but in the crystalline form of diamond,which is not the standard state for carbon. Since the path from C to CO2(g) is about2 kJ mol−1 longer in the diamond combustion, the starting point C(dia) must havebeen about 2 kJ mol−1 higher in enthalpy than C(gr). Differences like this lead usto define the standard state of all elements as the stable form at 1.000 atm pressure.By the nature of enthalpy and energy (thermodynamic properties), we can set anyarbitrary point to zero as a reference point. Hence we define the enthalpy of formationof any element in its standard state as zero at all temperatures. This definitionworks because elements are not converted from one to another in ordinary chemicalreactions.

Our previous observation that � f H 298(C(dia)) �= 0 but is about 2 kJ mol−1 sug-gests that, given this small enthalpy change, it might be possible to convert commongraphite into the nonstandard state of diamond. Indeed it is. Production of smallindustrial diamonds for cutting tools is commercially feasible.

4.4 MOLECULAR ENTHALPIES OF FORMATION

According to the first law of thermodynamics, the enthalpy of formation of a moleculecan be determined even if the formation reaction from its elements cannot be carriedout in an actual laboratory experiment. This is illustrated by the simple example ofcarbon monoxide CO(g). We know from painful examples that this poisonous gas isproduced by incomplete combustion of hydrocarbon fuels, but the simple controlled


MOLECULAR ENTHALPIES OF FORMATION 59

-393.5

-283.0

FIGURE 4.1 Combustion of C(gr) and CO(g).

experiment C(gr) + 12 O2(g) in a limited supply of O2(g) gives a mixture of products.

It is not a “clean” reaction. The reaction

CO(g) + 12 O2(g) → CO2(g)

is clean, however, and gives well-defined qp and qV . We find that the enthalpy changeis �r H 298 = −283.0 kJ mol−1 for this reaction (Fig. 4.1). Thus we have two pathsconnecting the same initial and final thermodynamic states. By a thermochemicalprinciple equivalent to the first law of thermodynamics known as Hess’s law, theenthalpy change over both paths must be the same. The third leg of the trianglemust be

� f H 298(CO(g)) = −393.5 − (−283.0) = −110.5 kJ mol−1

This method of indirect determination of � f H 298 is capable of wide extension.For example, knowing � f H 298(H2O(l)) and � f H 298(CO2(g)), one can determine� f H 298 of a hydrocarbon like methane by burning it in a flame calorimeter andsetting up a state diagram that is a little more complicated than the triangle justdescribed but that works on the same principle. The combustion reaction

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) �r H 298 = −890 kJ

has the same final state as

2H2(g) + C(gr) + 2O2(g) → CO2(g) + 2H2O(l) �r H 298 = −966 kJ

In Fig. 4.2, the thermodynamic state of the products of combustion of methane,CO2(g) + 2H2O(l) is reproduced by burning 2 mol of hydrogen and 1 mol of C(gr).The heats of combustion are arranged in an enthalpy diagram so as to make everythingcome out even except for one missing leg of the quadrangle, that of methane. Thethermodynamic state of methane is found by difference; it is 76 kJ mol−1 belowthat of the elements. Formation of methane from its elements would give off 76 kJper mole of methane produced; therefore � f H 298(methane) = −76 kJ mol−1. (Thetabulated NIST value is −74.9 kJ mol−1.)


60 THERMOCHEMISTRY

2(286)=572

394

-890

-76

methane

elements

CO2 + 2H2O

FIGURE 4.2 A Thermochemical cycle for determining � f H 298(methane). Not to scale.

The indirect method shown for methane has been extended to very many hydro-carbons and other organic compounds. Although many inorganic substances do notburn, they do react. Inorganic reaction cycles similar to Fig. 4.2 can often be set upto obtain thermochemical data. A free thermochemical database is maintained bythe National Institutes of Standards and Technology (NIST) of the US government(.gov). Go to webbook.nist.gov.

4.5 ENTHALPIES OF REACTION

Suppose we know the enthalpies of formation � f H 298 of acetylene CH CH, etheneCH2 CH2, and ethane CH3CH3 (226.7, 52.5, and −84.7 kJ mol−1) by the combustionmethod shown in Fig. 4.1. The enthalpy change �r H 298 of the reaction

CH CH(g) + 2H2(g) → CH3CH3(g)

can be found by comparing the level given by � f H 298(CH CH(g)) with that of� f H 298(CH3CH3(g)). (Remember that � f H 298 is zero for elemental hydrogen.)

�r H 298 = � f H 298(CH3CH3(g)) − � f H 298(CH CH(g))

= −84.7 − 226.7 = −311.4 kJ mol−1

This reaction has been carried out with the result �r H 298 = −312.1 ± 0.6kJ mol−1.

A similar reaction, the partial hydrogenation,

CH CH(g) + H2(g) → CH2 CH2(g)

cannot be carried out in the laboratory. Hydrogenation doesn’t stop at CH2 CH2(g),but goes on to the fully hydrogenated product CH3CH3(g) or gives a mixed product


ENTHALPIES OF REACTION 61

under all known experimental conditions. That doesn’t matter because of Hess’s law.The enthalpies of formation yield

�r H 298 = � f H 298(CH2 CH2(g)) − � f H 298(CH CH(g))

= 52.5 − 226.7 = −174.4 kJ mol−1

An equivalent way of finding �r H 298 of partial hydrogenation is by measuringthe enthalpy change −136.3 kJ mol−1 for the reaction

CH2 CH2(g) + H2 → CH3CH3(g)

When this is done, one has a first law triangle like that of Fig. 4.1, with only �r H 298

of partial hydrogenation yet to be determined. It is

�r H 298(partial) = −312.1 − (−136.3) = −175.8 kJ mol−1

as compared to the previous value of −174.4 kJ mol−1.One can also find the enthalpy of hydrogenation of ethene:

�r H 298 = � f H 298(CH3CH3(g)) − � f H 298(CH2 CH2(g))

= −84.7 − 52.5 = −137.2 kJ mol−1

This is a complete reaction to a well-defined final state. It has been carried out in thelaboratory. The experimental value is −136.3 ± 0.2 kJ mol−1.

Although, for simplicity, we shall use hydrocarbons to illustrate the principlesof thermochemistry, there is no such restriction in practice. For example, Pitzeret al. (1961) give the standard enthalpies of formation � f H ◦(KCl) = −435.9 and� f H ◦(KClO3) = −391.2 kJ mol−1. (The superscript ◦ indicates the standard state inthis notation; please do not confuse it with a superscripted zero of temperature 0.)These values lead to the enthalpy of reaction �r H ◦ necessary to convert solid KCl(s)to solid KClO3(s):

KCl(s) + 32 O2(g) → KClO3(s)

�r H ◦ = −391.2 − (−435.9) = 44.7 kJ mol−1

Including the notation (s) to indicate the solid state is a precaution rather than anecessity because in the standard states, both KCl and KClO3 are solids.

Even from these few simple examples, it should be clear that with a sufficientdatabase of enthalpies of formation, it is possible to calculate the enthalpies of analmost limitless array of chemical reactions of industrial, medical, and pharmaceuticalimportance. This is the reason why so much effort has gone into populating the


62 THERMOCHEMISTRY

database with a variety of entries (webbook.nist.gov) and why the most meticulouscare is exercised to be sure that the data entered are accurate.

All of the previous examples are consistent with the equations

�rU 298 =∑

� f U 298(products) −∑

� f U 298(reactants)

and

�r H 298 =∑

� f H 298(products) −∑

� f H 298(reactants)

where it is to be understood that products and reactants are multiplied by theirstoichiometric coefficients and that they are in their standard states. We shall soonsee that all thermodynamic properties follow analogous equations, which is whythermodynamics is so useful.

Although combustion thermochemistry on C(gr), CH4 (methane), and so on, is theprime source of � f H 298 data, other reactions can contribute as well. For example,suppose we know that � f H 298(ethane) = −83.8 ± 0.4 kJ mol−1 but we don’t know� f H 298(ethene). Measurement of the enthalpy of the hydrogenation reaction ethene→ ethane gives −136.3 ± 0.2 kJ mol−1; hence � f H 298(ethene) must be higher inenthalpy than � f H 298(ethane) by just that amount. By this reasoning,

� f H 298(ethene) = −83.8 ± 0.4 + 136.3 ± 0.2 = 52.5 ± 0.4 kJ mol−1

The tabulated value is 52.5 ± 0.4 kJ mol−1.Molar enthalpies of physical processes like phase changes (vaporization and melt-

ing), solution of a solute in a solvent, mixing of miscible solvents, and dilution aretreated by slight modifications of the methods shown, always in accord with the firstlaw of thermodynamics, which has never been violated in a controlled, reproducibleexperiment.

4.6 GROUP ADDITIVITY

Upon scanning a series of heats of combustion �c H 298, one notices a regular increasewith molecular weight. For example, for the alkanes in their standard states (g), wehave the following:

Methane Ethane Propane n-Butane

�c H 298 (g) −890.8 −1560.7 −2219.2 −2877.6Difference −669.9 −658.5 −658.4


GROUP ADDITIVITY 63

If we were asked to predict �c H 298(n-pentane(g)), a reasonable answer would be toadd another negative change just like the last two:

�c H 298(n-pentane(g)) = −2877.6 − 658.4 = −3536 kj mol−1.

The experimental value is 3535.4 ± 1.0.Even simpler, we can note that, given the reference point �c H 298(ethane(g)) =

−1560.7 kJ mol−1, the heats of combustion of n-alkanes (except for methane) obey alinear function of the number of additional hydrogen atoms over those in the referencecompound with slope equal to –658.5/2 = −329.2 kJ mol−1. These CH2 hydrogenatoms are called secondary hydrogens as distinct from CH3, hydrogens, which areprimary, and isolated C H atoms, which are tertiary. Now any �c H 298(n-alkane(g))can be found by counting secondary hydrogen atoms and adding the count times−329.2 to the base value of −1560.7.

Because the enthalpies of formation � f H 298 of n-alkanes(g) are proportionalto �c H 298, one can estimate � f H 298 values in the same way that we used in thecombustion case. Simply count hydrogen atoms and add to a base value for ethane:

Ethane Propane n-Butane

� f H 298 (g) −84.0 −104.7 −125.6Difference −20.7 −20.9

Counting secondary hydrogens for n-octane and multiplying by −20.8/2 = −10.4gives 12(−10.4) = −124.8. Adding the base value, −84.0 gives −208.8 kJ mol−1.The experimental value is −208.4 ± 0.8 kJ mol−1.

Zavitsas et al. (2008) have extended this method to cover primary, secondary, andtertiary hydrogens by the equation

� f H ◦ = −14.0n p + (−10.4ns) + (−6.65nt ) =∑

ci ni

where n p, ns, and nt are the numbers of primary, secondary, and tertiary hydrogenatoms in any alkane or cycloalkane. By this system, the enthalpy of formation of4-methylheptane, an isomer of n-octane, is

H|CH3CH2CH2CCH2CH2CH3|

CH3

� f H ◦ = −14.0n p + (−10.4ns) + (−6.65nt )

� f H 298(4−methylheptane(g)) = −14.0(9) + (−10.4(8)) + −6.65(1)

= −215.8 kJ mol−1

The experimental value is −212.1.


64 THERMOCHEMISTRY

4.7 �fH298(g) FROM CLASSICAL MECHANICS

One of the drawbacks of the hydrogen-atom counting method is that account isnot taken of classical mechanical properties of molecules such as mechanical strainenergy induced in distorted or crowded molecules. Other mechanical features notaccounted for are those due to the quantum mechanical influences on electronprobability densities which, in turn, influence both molecular energy and molecularstructure.

The influence of molecular strain can be seen in the discrepancy between thehydrogen-atom counting estimate of � f H 298(cyclopentane(g)) = −24.9 kJ mol−1

as contrasted to the experimental value of −16.3 kJ mol−1. The experimental valueis higher (less negative) than the estimate. Evidently, crowding 10 hydrogen atomsinto the small space afforded by a 5-membered ring increases the interatomic in-terference energy that contributes to � f H 298. N. L. Allinger at the University ofGeorgia has developed a series of molecular mechanics programs called MM1 toMM4 that incorporate energies due to bond bending, bond stretching, ring distortion,and so on, into the calculation of molecular structure and � f H 298. The method asimplemented on a computer is extremely fast and is therefore well-suited to largemolecules.

4.8 THE SCHRODINGER EQUATION

In 1926 Erwin Schrodinger published an equation that gives correct solutions for theenergy levels of the hydrogen atom. Shortly afterward, Heitler and London showedthat the Schrodinger equation, as it has come to be called, predicts the existence of achemical bond between H and H and that it gives an approximate strength of the H Hbond. A powerful refinement and extension of this molecular orbital calculation calledGAMESS is available as a site license at no cost for your microcomputer (academicor similar affiliation must be specified):

http://www.msg.ameslab.gov/GAMESS/GAMESS.html

We shall use the GAMESS program to determine the bond energies of a number ofmolecules starting with the simplest case, that of the hydrogen molecule when it isformed from two hydrogen atoms:

2H· → H H

The GAMESS output for calculation of the total energy of H· is E = −0.4998 Eh ,where Eh is the hartree, a unit of energy. A GAMESS output for the molecule H H is

E = −1.1630349978Eh


VARIATION OF �H WITH T 65

Hence the bond energy is the energy of the final state (the molecule) minus theenergy of the initial state (the two atoms):

E(H H) − 2E(H·)= −1.1630 − (2 ×− 0.4998) = − 0.1630Eh = −102.3 kcal mol−1=−428 kJ mol−1

as compared to the experimental value of −431 kJ mol−1. The bond energy isnegative because the molecule is more stable than the isolated atoms (energygoes downhill). Thus our first molecular orbital calculation comes to within about1% of the experimental value. The conversions 1.0 Eh = 627.51 kcal mol−1 and1.0 kcal mol−1 ≡ 4.184 kJ mol−1 have been used. This calculation can be carried outon much more complicated molecules, ions, and free radicals, some of which will bedescribed in later chapters The advent of these powerful computer programs and thehardware to run them has made it possible to study reactions that are not accessibleby experimental means. Computational thermochemistry is an active research area atpresent.

4.9 VARIATION OF �H WITH T

The definition of the heat capacity at constant pressure C p = (∂ H/∂T )P leads to theinfinitesimal enthalpy change with temperature of a pure substance d H = CP dT .Over a reasonably short temperature interval �T , the equation �H = CP�T isapproximately true. The heat capacity of a mixture is the sum of the molar heatcapacities of its components multiplied by the number of moles of each componentpresent. When a chemical reaction takes place, the number of moles of the reactantsdecreases and the number of moles of products appears with different heat capacities.The difference in heat capacities between the reactant state and the product state is

�CP =∑

CP (products) −∑

CP (reactants)

Applying the definition of C p to all of the component species of a chemical reaction,we get

�C p =(

∂�H

∂T

)

P

Selecting the hydrogenation of ethene to ethane for simplicity, experimen-tal values (webbook.nist.gov) are C p = 42.90 J K−1mol−1 for ethene and C p =52.49 J K−1mol−1 for ethane; and taking an experimental value (Atkins, 1994) ofC p = 28.87 J K−1mol−1 for hydrogen, we obtain

�C p = 52.49 − 42.90 − 28.87 = −19.28 J K−1 mol−1


66 THERMOCHEMISTRY

The change in �r H for relatively small changes in T is �CP�T . We can test thisresult against experimental values for the simple hydrogenation of ethene (Kisti-akowsky et al., 1935). Using the computed result for �CP�T over the temperaturerange 355–298, the range between the temperature at which his experiments werecarried out (355 K) and standard temperature, we get

��hyd H 355 = −19.28 [− (355 − 298)] = 1.099 kJ mol−1

for the change in �hyd H 355, from 355 K, the temperature at which the hydrogenationwas actually carried out to room temperature 298 K. This small temperature correctiongives

�hyd H 298 = −137.44 + 1.10 = −136.34 kJ mol−1

for the enthalpy of hydrogenation of ethene at 298 K. This value is to be compared tothe result of a series of experiments carried out at 298 K by a different experimentalmethod which led to �hyd H 298 = −136.29 ± 0.21 kJ mol−1. The computed heatcapacities are probably reliable over a temperature range of ±50 K or so. Furthermore,the enthalpy of hydrogenation itself is insensitive to temperature, so we may takeexperimental determinations of �hyd H carried out under normal laboratory conditionsas essentially the same as the standard state �hyd H ◦

298.When calculations are carried out over larger temperature ranges, as they often are

in industrial applications, a polynomial approximation to the heat capacity is used.

C p = α + βT + γ T 2 + · · ·

This enables one to determine C p for the reactants and products of a chemical reactionat some new temperature other than 298 K, thereby enabling one to determine thenew change in heat capacity �C p for the reaction:

�CP =∑

CP (products) −∑

CP (reactants)

Variation in the�H of physical and chemical processes with variation in pressurecan be calculated from equations of state or by acquisition of experimental data andcurve fitting. Many reactions are less sensitive to pressure change than to temperaturechange over comparable ranges. Metiu (2006) has treated both temperature andpressure variation in the industrial production of ammonia from its elements.

4.10 DIFFERENTIAL SCANNING CALORIMETRY

When moderate amounts of heat are supplied to a solution of simple salts in water,we expect a smooth heating curve between, say 290 and 320 K, like the lower curvein Fig. 4.3. The heat capacity of water is nearly constant over this temperature range,


DIFFERENTIAL SCANNING CALORIMETRY 67

and it will be little affected by small amounts of dissolved salts. Electrical circuitryexists that permits one to supply heat to a dilute solution in an adiabatic (insulated)calorimeter in very small pulses which may be regarded as infinitesimals dq. We nor-mally carry out the experiment at constant pressure, so the definition of heat capacityat constant pressure C p = dqp/dT is satisfied. The gradual temperature rise overmany small pulses can be followed by means of a thermistor circuit or its equivalent.

If, instead of a dilute solution of simple salts, the calorimeter contains a solutethat is capable of undergoing a thermal reaction, which is a reaction brought aboutby heat, the heating curve is more complicated. Thermal reactions are importantin many areas, especially in biochemistry. Proteins undergo heat denaturation. Heatdenaturation involves unfolding of the native protein and requires breaking of someor many of the bonds holding it in its native structure. Heat denaturation may bequite specific as to the temperature at which it occurs, and it may bring about subtlechanges in the protein, like changes in physiological activity, or it may bring aboutgross changes in the form of the protein as in the cooking of an egg.

Because heat denaturation involves breaking of internal bonds in the protein, itrequires an enthalpy input at constant pressure. The reaction is endoenthalpic. Adilute solution of salt and protein takes more heat to bring about a small temperaturechange than would the solution without the protein. The difference is observed onlyat or near the temperature of denaturation. Thus we have a normal temperature riseuntil denaturation begins, after which the heat capacity of the solution is abnormallylarge until we achieve complete thermal denaturation whereupon the temperature risedrops back to the normal baseline of a salt solution. Plotting C p as a function of T ,we see a peak at the denaturation temperature. This is the upper line in Fig. 4.3. Itis a simple matter to interface a computer to the scanning calorimeter output and tointegrate under the experimental curve:

�den H =∫ T f

Ti

C p dT

Cp

T

FIGURE 4.3 Schematic diagram of the thermal denaturation of a water-soluble protein. Thestraight line is the baseline of salt solution without protein. The peak is due to endoenthalpicdenaturation of the protein.


68 THERMOCHEMISTRY

The heat capacity curve of the simple salt solution (the baseline) is subtracted fromthe experimental result. There may be multiple peaks if there is more than one proteinin the test solution or if the protein is capable of unfolding in sequential steps.

PROBLEMS AND EXAMPLE

Example 4.1 Oxygen Bomb Calorimetry

Exactly 0.5000 g of benzoic acid C6H5COOH were burned under oxygen. Thecombustion produced a temperature rise of 1.236 K. The same calorimetric setup wasused to burn 0.3000 g of naphthalene (C10H8) and the resulting temperature rise was1.128 K. The heat of combustion of benzoic acid is qV = �cU 298 = –3227 kJ mol−1

(exothermic). What is the heat of combustion qV = �cU 298 of naphthalene?

Solution 4.1 The corresponding molar masses are: benzoic acid, 122.12 g mol−1;and naphthalene, 128.19 g mol−1. The temperature rise for each combustion was: ben-zoic acid, 1.236/0.5000 = 2.472 K g−1; and naphthalene, 1.128/0.3000 = 3.760 K g−1.Multiplying by the molar masses in each case, one obtains 301.9 K mol−1 for benzoicacid and 482.0 K mol−1 for naphthalene. This gives us the ratio

301.9

482.0= −3227

x

x = −3227482.0

301.9= −5152 kJ mol−1

qV = �cU 298 = −5152 kJ mol−1

Notice that the units cancel on the left; thus x has the units of kJ mol−1, not K mol−1.The handbook value is �cU = −5156 kJ mol−1.

Problem 4.1

A resistor of precisely 1 ohm is immersed in a liter (1 dm3) of water in a perfectlyinsulated container. Suppose that precisely 1 ampere flows through the resistor forprecisely 1 second. What is the temperature rise of the water?

Problem 4.2

Exactly one gram of a solid substance is burned in a bomb calorimeter. The bombabsorbs as much heat as 300 g of water would absorb. (Its water equivalent is 300 g.)The bomb was immersed in 1700 g of water in an insulated can. During combustionof the sample, the temperature went from 24.0◦C to 26.35◦C. What is the heat ofcombustion per gram of the sample? What is the molar enthalpy of combustion if themolar mass of the substance is 60.0 g mol−1 and 2 mol of gas are formed in excessof the O2 burned?


PROBLEMS AND EXAMPLE 69

Problem 4.3

The enthalpy of formation of liquid acetic acid CH3COOH(l) is � f H ◦ =−484.5 kJ mol−1. What is �c H?

Problem 4.4

The enthalpy of combustion of solid α-d-glucose, C6H12O6(s) is −2808 kJ mol−1.What is its enthalpy of formation?

Problem 4.5

Estimate �c H 298(n-octane(g)) of n-octane by the hydrogen-atom counting methodfor alkanes.

Problem 4.6

Find � f H 298(2,4-dimethylpentane(g)) by the hydrogen atom counting method. Whatis the enthalpy of isomerization of n-heptane(g) to 2,4-dimethylpentane(g) accordingto this method? Compare your answer with the experimental result of −14.6 ±1.7 kJ mol−1.

Problem 4.7

The input file for a Gaussian C© quantum mechanical calculation will be discussedin later chapters. Briefly, it consists of a few lines of instructions to the computerconcerning memory requirements, the number of processors to be used, and theGaussian procedure to be used, followed by an approximate geometry of the molecule.In simple cases, the input geometry can be merely a guess based on what we learnedin general chemistry. The machine specifications will vary from one installation toanother. Our input file for the water molecule is

%mem=1800Mw%nproc=1

# g3

water

0 1H -1.012237 0.210253 0.097259O -0.260862 0.786229 0.119544H 0.489699 0.209212 0.142294

Adapt this input file for your system and run the water molecule. What is the optimizedgeometry in the form of Cartesian coordinates (like the input file)? What is the O Hbond length? What is the H O H bond angle? What energy is given for water?


70 THERMOCHEMISTRY

Problem 4.8

Plot the heat capacity of ethylene from the following data set:

Temperature (K) Heat capacity, Cp (J K−1 mol−1)

300.0000 43.1000400.0000 53.0000500.0000 62.5000600.0000 70.7000700.0000 77.7000800.0000 83.9000900.0000 89.20001000.0000 93.9000

Problem 4.9

Suppose that the heat capacities C p for N2, H2, and NH3 in the standard state areconstant with temperature change (they aren’t) at 29.1, 28.8, and 35.6 J K−1 mol−1.Suppose further that �r H ◦ for the reaction

N2 + 3H2 → 2NH3

is −92.2 kJ mol−1 (of N2 consumed) at 298 K. What is �C p for the reaction? Whatis �r H ◦ at 398 K? What is the (hypothetical) �r H ◦ at 0 K?

concise physical chemistry (rogers/concise physical chemistry) || thermochemistry

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