thermochemistry and calorimetry · pdf fileheat and enthalpy changes is known as...

2/24/2015 Thermochemistry and calorimetry

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Thermochemistry and calorimetry

Chemical Energetics and thermodynamics - 4The heat that flows across the boundaries of a systemundergoing a change is a fundamental property thatcharacterizes the process. It is easily measured, and ifthe process is a chemical reaction carried out atconstant pressure, it can also be predicted from thedifference between the enthalpies of the products andreactants. The quantitative study and measurement ofheat and enthalpy changes is known as thermochemistry.

1 Thermochemical equations and standard statesIn order to define the thermochemical properties of a process, it is first necessary to write a thermochemicalequation that defines the actual change taking place, both in terms of the formulas of the substances involvedand their physical states (temperature, pressure, and whether solid, liquid, or gaseous.

To take a very simple example, here is the completethermochemical equation for the vaporization of water at its normal boiling point:

H2O(l, 373 K, 1 atm) → H2O(g, 373 K, 1 atm) ΔH = 40.7 kJ mol–1

The quantity 40.7 is known as the enthalpy of vaporization (often referred to as “heat of vaporization”) ofliquid water.

The following points should be kept in mind when writing thermochemical equations:

• Any thermodynamic quantity such as ΔH that is associated with athermochemical equation always refers to the number of moles of substancesexplicitly shown in the equation.

Thus for the synthesis of water we can write

2 H2(g) + O2(g)→ 2 H2O(l) ΔH = –572 kJ

or

H2(g) + ½ O2(g)→ H2O(l) ΔH = –286 kJ

• Thermochemical equations for reactions taking place in solution must alsospecify the concentrations of the dissolved species.



The standard enthalpy of formation isa fundamental property of anycompound. Most chemistry textbookscontain tables of Hf ° values (usuallyalongside values of otherthermodynamic properties) in theirappendices. There are also manyWeb-based sources, such as the oneat chemistry.about.com.

For example, the enthalpy of neutralization of a strong acid by a strong base is given by

H+(aq, 1M, 298 K, 1 atm) + OH–(aq, 1M, 298 K, 1 atm) →

H2O(l, 373 K, 1 atm) ΔH = —56.9 kJ mol–1

in which the abbreviation aq refers to the hydrated ions as they exist in aqueous solution. Since mostthermochemical equations are written for the standard conditions of 298 K and 1 atm pressure, we can leavethese quantities out if these conditions apply both before and after the reaction. If, under these same conditions,the substance is in its preferred (most stable) physical state, then the substance is said to be in its standard state.

Thus the standard state of water at 1 atm is the solid below 0°C, and the gas above 100°C. A thermochemicalquantity such as ΔH that refers to reactants and products in their standard states is denoted by ΔH°.

• In the case of dissolved substances, the standard state of a solute is that inwhich the “effective concentration”, known as the activity, is unity.

For non-ionic solutes the activity and molarity are usually about the same for concentrations up to about 1M,but for an ionic solute this approximation is generally valid only for solutions more dilute than 0.001-0.01M,depending on electric charge and size of the particular ion.

2 Standard enthalpy of formationThe enthalpy change for a chemical reaction is the difference

ΔH = Hproducts – Hreactants

If the reaction in question represents the formation of one mole of the compound from its elements in theirstandard states, as in

H2(g) + ½ O2(g)→ H2O(l) ΔH = –286 kJ

then we can arbitrarily set the enthalpy of the elements to zero and write

Hf ° = ΣHf °products – ΣHf °reactants= –286 kJ – 0 = –268 kJ mol–1

which defines the standard enthalpy of formation of water at 298K.

The value Hf ° = –268 kJ tells us that when hydrogen and oxygen, each at a pressure of 1 atm and at 298 K (25° C)react to form 1 mole of liquid water also at 25°C and 1 atm pressure, 268 kJ will have passed from the sytstem (thereaction mixture) into the surroundings. The negative sign indicates that the reaction is exothermic: the enthalpy ofthe product is smaller than that of the reactants.

The standard enthalpy of formationof a compound is defined as the heat

associated with the formation ofone mole of the compound from itselements in their standard states.

http://chemistry.about.com/od/thermodynamics/a/Heats-Of-Formation.htm



In general, the standard enthalpy change for a reaction is given by the expression

important ⇒ Δ ΣHf °products – ΣHf °reactants (2-1)

in which the ΣHf ° terms indicate the sums of the standard enthalpies of formations of all products and reactants.The above definition is one of the most important in chemistry because it allows us to predict the enthalpychange of any reaction without knowing any more than the standard enthalpies of formation of the products andreactants, which are widely available in tables.

The following examples illustrate some important aspects of the standard enthalpy of formation of substances.

• The thermochemical equation defining Hf ° is always written in terms of one mole of the substance inquestion:

½ N2(g) + 3/2 H2(g)→ NH3(g) ΔH° = –46.1 kJ (per mole of NH3)

•The standard heat of formation of a compound is always taken in reference to the forms of the elementsthat are most stable at 25°C and 1 atm pressure.

A number of elements, of which sulfur and carbon are common examples, can exist in more then one solidcrystalline form.

In the case of carbon, the graphite modification is the more stable form.

C(graphite) + O2(g) → CO2(g) ΔH° ≡ Hf ° = –393.5 kJ mol–1

C(diamond) + O2(g) → CO2(g) ΔH° = –395.8 kJ mol–1

• The physical state of the product of the formation reaction must be indicated explicitly if it is not themost stable one at 25°C and 1 atm pressure:

H2(g) + ½ O2(g) → H2O(aq) ΔH° ≡ Hf ° = –285.8 kJ mol–1

H2(g) + ½ O2(g) → H2O(g) ΔH° = –241.8 kJ mol–1

Notice that the difference between these two ΔH° values is just the heat of vaporization of water.

• Although the formation of most molecules from their elements is an exothermic process, the formationof some compounds is mildly endothermic:

½ N2(g) + O2(g)→ NO2(g) ΔH° ≡ Hf ° = +33.2 kJ mol–1

A positive heat of formation is frequently associated with instability— the tendency of a molecule to decomposeinto its elements, although it is not in itself a sufficient cause. In many cases, however, the rate of thisdecomposition is essentially zero, so it is still possible for the substance to exist. In this connection, it is worthnoting that all molecules will become unstable at higher temperatures.

• The thermochemical reactions that define the heats of formation of most compounds cannot actuallytake place.

For example, the direct synthesis of methane from its elements

C(graphite) + 2 H2(g) → CH4(g)



cannot be observed directly owing to the large number of other possible reactions between these two elements.However, the standard enthalpy change for such a reaction be found indirectly from other data, as explained inthe next section.

• The standard enthalpy of formation of gaseous atoms from the element is known as the heat ofatomization.

Heats of atomization are always positive, and are important in the calculation of bond energies.

Fe(s) → Fe(g) ΔH° = 417 kJ mol–1

• The standard enthalpy of formation of an ion dissolved in water is expressed on a separate scale in

which that of H+(aq) is defined as zero.

The standard heat of formation of a dissolved ion such as Cl– (that is, formation of the ion from the element)cannot be measured because it is impossible to have a solution containing a single kind of ion. For this reason,ionic enthalpies are expressed on a separate scale on which Hf° of the hydrogen ion at unit activity (1 Meffective concentration) is defined as zero:

½H2(g) → H+(aq) ΔH° ≡ Hf ° = 0 kJ mol

–1

Other ionic enthalpies (as they are commonly known) are found by combining appropriate thermochemicalequations (as explained in Section 3 below). For example, Hf° of HCl(aq) is found from the enthalpies of

formation and solution of HCl(g), yielding

½H2(g) + ½Cl2(g) → HCl(aq) ΔH° ≡ Hf ° = –167 kJ mol

Because Hf° for H+(aq) is zero, this value establishes the standard enthalpy of the chloride ion.

The standard enthalpy of formation of Ca2+(aq), given by

Ca(s) + Cl2(g) → CaCl2(aq)

could then be calculated by combining other measurable quantities such as the enthalpies of formation andsolution of CaCl2(s) to find Hf° for CaCl2(aq), from which Hf° of Ca2+(aq) is found by difference from that ofCl–(aq).

Tables of the resulting ionic enthalpies are widely available (see here) and are often printed in general chemistrytextbooks.

3 Hess’ law and thermochemical calculationsYou probably know that two or more chemical equations can be combined algebraically to give a new equation.Even before the science of thermodynamics developed in the late nineteenth century, it was observed that theheats associated with chemical reactions can be combined in the same way to yield the heat of another reaction.For example, the standard enthalpy changes for the oxidation of graphite and diamond can be combined toobtain ΔH° for the transformation between these two forms of solid carbon, a reaction that cannot be studiedexperimentally.

C(graphite) + O2(g)→ CO2(g) ΔH° = –393.51 kJ mol–1

http://chemistry.about.com/od/chartstables/a/heatoformions.htm



(16 min)(6 min);(10 min)

C(diamond) + O2(g)→ CO2(g) ΔH° = –395.40 kJ mol–1

Subtraction of the second reaction from the first (i.e., writing thesecond equation in reverse and adding it to the first one) yields

C(graphite) → C(diamond) ΔH° = 1.89 kJ mol–1

This principle, known as Hess’ law of independent heatsummation is a direct consequence of the enthalpy being a statefunction. Hess’ law is one of the most powerful tools of chemistry,for it allows the change in the enthalpy (and in other

thermodynamic functions) of huge numbers of chemical reactions to be predicted from a relatively small base ofexperimental data.



Germain Henri Hess (1802-1850) was a Swiss-bornprofessor of chemistry at St. Petersburg, Russia. Heformulated his famous law, which he discoveredempirically, in 1840. Very little appears to be knownabout his other work in chemistry.

Standard enthalpies of combustion

Because most substances cannot be prepared directly from their elements, heats of formation of compounds areseldom determined by direct measurement. Instead, Hess’ law is employed to calculate enthalpies of formationfrom more accessible data. The most important of these are the standard enthalpies of combustion. Mostelements and compounds combine with oxygen, and many of these oxidations are highly exothermic, makingthe measurement of their heats relatively easy.

For example, by combining the heats of combustion of carbon, hydrogen, and methane, we can obtain thestandard enthalpy of formation of methane, which as we noted above, cannot be determined directly.

Problem Example 1Use the following heat of formation/combustion information to estimate the standard heat of formation ofmethane CH4.

C(graphite) + O2(g) → CO2(g) ΔH° = –394 kJ mol–1 (P1-1)

H2(g) + ½O2(g) → H2O(g) ΔH° = –242 kJ mol–1 (P1-2)

CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) ΔH° = –890 kJ mol–1 (P1-3)

Solution: The standard heat of formation of methane is defined by the reaction

C(graphite) + 2H2(g) → CH4(g) ΔH° = ??? (P1-4)

Our task is thus to combine the top three equations in such a way that they add up to (4).

1) Begin by noting that (3), the combustion of methane, is the only equation that contains the CH4 term, so weneed to write it in reverse (not forgetting to reverse the sign of ΔH°!) so that CH4 appears as the product.

CO2(g) + 2H2O(g) → CH4(g) + 2O2(g) ΔH° = +393 kJ mol–1 (P1-3Rev)

2) Since H2O should not appear in (4), add two times (2) to cancel both out. Notice that this also cancels oneof the oxygens in (3Rev):


2 H2(g) + O2(g) → 2H2O(g) ΔH° = –484 kJ mol–1 (P1-2)

3) Finally, get rid of the remaining O2 and CO2 by adding (1); this also adds a needed C:


2 H2(g) + O2(g) → 2H2O(g) ΔH° = –484 kJ mol–1 (P1-2)

C(graphite) + O2(g) → CO2(g) ΔH° = –76.4 kJ mol–1 (P1-1)

4) So our creative cancelling has eliminated all except the substances that appear in (4). Just add up theenthalpy changes and we are done:

C(graphite) + 2H2(g) → CH4(g) ΔH° = ??? (P1-4)



More on calorimetry: ChemEd digitallibrary

4 Calorimetry: measuring ΔH in the laboratoryHow are enthalpy changes determined experimentally?First, you must understand that the only thermal quantitythat can be observed directly is the heat q that flows into or

out of a reaction vessel, and that q is numerically equal to ΔH° only under the special condition ofconstant pressure. Moreover, q is equal to the standard enthalpy change only when the reactants andproducts are both at the same temperature, normally 25°C.

The measurement of q is generally known as calorimetry.

The most common types of calorimeters contain a known quantity of water which absorbs the heatreleased by the reaction. Because the specific heat capacity of water (4.184 J g–1 K–1) is known to highprecision, a measurement of its temperature rise due to the reaction enables one to calculate the quantityof heat released.

The calorimeter constant

In all but the very simplest calorimeters, some of the heat released by the reaction is absorbed by thecomponents of the calorimeter itself. It is therefore necessary to "calibrate" the calorimeter by measuringthe temperature change that results from the introduction of a known quantity of heat. The resultingcalorimeter constant, expressed in J K–1, can be regarded as the “heat capacity of the calorimeter”. Theknown source of heat is usually produced by passing a known quantity of electric current through aresistor within the calorimeter, but it can be measured by other means as described in the followingproblem example.

Problem Example 2In determining the heat capacity of a calorimeter, a student mixes 100.0 g of water at 57.0 °C with 100.0 g of water, already in the calorimeter, at 24.2°C. (The specific

heat of water is 4.184 J g–1 K–1)

After mixing and thermal equilibration with the calorimeter, the temperature of the water stabilizes at38.7°C. Calculate the heat capacity of the calorimeter in J/K.

Solution:

The hot water loses heat, the cold water gains heat, and the calorimter itself gains heat, so this isessentially a thermal balance problem. Conservation of energy requires that

qhot + qcold + qcal = 0

We can evaluate the first two terms from the observed temperature changes:

qhot = (100 g) (38.7 – 57.0) K (4.184 J g–1 K–1) = –7657 J

qcold = (100 g) (38.7 – 24.2) K (4.184 J g–1 K–1) = 6067 J

So qcal = (7657 – 6067) J = 1590 J

The calorimeter constant is (1590 J) / (38.7 – 24.2) K = 110 J K–1

Note: Strictly speaking, there is a fourth thermal balance term that must be considered in a highlyaccurate calculation: the water in the calorimeter expands as it is heated, performing work on theatmosphere.

http://chempaths.chemeddl.org/services/chempaths/?q=book/General%20Chemistry%20Textbook/Thermodynamics%3A%20Atoms,%20Molecules%20and%20Energy/1487/measuring-enthalpy-



For reactions that can be initiated by combining two solutions, the temperature rise of the solution itselfcan provide an approximate value of the reaction enthalpy if we assume that the heat capacity of thesolution is close to that of the pure water — which will be nearly true if the solutions are dilute.

For example, a very simple calorimetric determination of the standard enthalpy of the reaction H+(aq) +

OH–(aq) → H2O(l) could be carried out by combining equal volumes of 0.1M solutions of HCl and ofNaOH initially at 25°C. Since this reaction is exothermic, a quantity of heat q will be released into thesolution. From the tmperature rise and the specific heat of water, we obtain the number of joules of heatreleased into each gram of the solution, and q can then be calculated from the mass of the solution. Sincethe entire process is carried out at constant pressure, we have ΔH° = q.

For reactions that cannot be carried out in dilute aqueous solution, the reaction vessel is commonlyplaced within a larger insulated container of water. During the reaction, heat passes between the innerand outer containers until their temperatures become identical. Again, the temperature change of thewater is observed, but in this case we need to know the value of the calorimeter constant describedabove.

The bomb calorimeter

Most serious calorimetry carried out in research laboratories involves the determination of heats ofcombustion, since these are essential to the determination of standard enthalpies of formation of thethousands of new compounds that are prepared and characterized each month.

In order to ensure complete combustion, the experiment is carried out in the presence of oxygen aboveatmospheric pressure. This requires that the combustion be confined to a fixed volume.

Since the process takes place at constantvolume, the reaction vessel must beconstructed to withstand the high pressureresulting from the combustion process, whichamounts to a confined explosion. The vesselis usually called a “bomb”, and the techniqueis known as bomb calorimetry. The reactionis initiated by discharging a capacitor througha thin wire which ignites the mixture.

Another consequence of the constant-volumecondition is that the heat released correspondsto qv , and thus to the internal energy changeΔU rather than to ΔH. The enthalpy change iscalculated according to the formjula

ΔH = qv + ΔngRT

in which Δng  is the change in the number of moles of gases in the reaction.

(See the second lesson in this set for the derivation of this equation).

http://www.chem1.com/acad/webtext/energetics/CE-2.html#S4A



Problem Example 3

A sample of biphenyl (C6H5)2 weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C, producinga temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid C6H5COOHweighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K. For

benzoic acid, the heat of combustion at constant pressure is known to be 3226 kJ mol–1 (that is, ΔU° = –3226kJ mol–1.) Use this information to determine the standard enthalpy of combustion of biphenyl.

Solution. Begin by working out the calorimeter constant:

Moles of benzoic acid: (0.825 g) / (122.1 g mol–1) = .00676 mol

Heat relesed to calorimeter: (.00676 mol) x (3226 kJ mol–1) = 21.80 kJ

Calorimeter constant: (21.80 kJ) / (1.94 K) = 11.24 kJ K–1

Now determine ΔUcombustion of the biphenyl ("BP"):

moles of biphenyl: (.526 g) / (154.12 g mol–1) = .00341 mol

heat relesed to calorimeter: (1.91 K) x (11.24 kJ K–1) = 21.46 kJ

heat released per mole of biphenyl: (21.46 kJ) / (.00341 mol) = 6293 kJ mol–1

ΔUcombustion (BP) = –6293 kJ mol–1

(This is the heat change at constant volume, qV ; the negative sign indicates that the reaction is exothermic, asall combustions are.)

From the reaction equation

(C6H5)2(s) + 19/2 O2(g) → 12 CO2(g) + 5 H2O(l)

we have Δng = 12 – (19/2) = –5/2. Thus the volume of the system decreases when the reaction takes place.Converting to ΔH, we substitute intoΔH = qV + ΔngRT = ΔU + ΔngRT

ΔH° = ΔU° – (5/2)(8.314 J mol–1K–1)(298 K) = Note that the use of standard temperature, 298 K, defines ΔH°, as opposed to plain ΔH.

ΔH° = (–6293 kJ mol–1) – (6194 J mol–1) = (–6293 – 6.2) kJ mol–1 = –6299 kJ mol–1A common mistake here is to forget that the subtracted term is in J, not kJ. Note that the additional 6.2 kJ in ΔH° compared to ΔU° reflects the work that the surroundings do on the system as the voume of gases decreasesaccording to the reaction equation.

Calorimeters

Although calorimetry is simple in principle, its practice is a highly exacting art, especially when appliedto processes that take place slowly or involve very small heat changes, such as the germination of seeds.

Calorimeters can be assimple as a foam plastic coffee cup, which is often used in studentlaboratories.

Research-grade calorimeters, able to detect minute temperature changes,are more likely to occupy table tops, or even entire rooms:



The ice calorimeter is an important tool formeasuring the heat capacities of liquids and solids,as well as the heats of certain reactions. Thissimple yet ingenious apparatus is essentially adevice for measuring the change in volume due tomelting of ice. To measure a heat capacity, a warmsample is placed in the inner compartment, whichis surrounded by a mixture of ice and water.

The heat withdrawn from the sample as it coolscauses some of the ice to melt. Since ice is lessdense than water, the volume of water in theinsulated chamber decreases. This causes anequivalent volume of mercury to be sucked into

the inner reservoir from the outside container. The loss in weight of this container gives the decrease involume of the water, and thus the mass of ice melted. This, combined with the heat of fusion of ice, givesthe quantity of heat lost by the sample as it cools to 0°C.

What you should be able to doMake sure you thoroughly understand the following essential concepts that have been presented above.

Write a balanced thermochemical equation that expresses the enthalpy change for a given chemical reaction. Besure to correctly specificy the physical state (and, if necessary, the concentration) of each component.

Write an equation that defines the standard enthalpy of formation of a given chemical species.

Define the standard enthalpy change of a chemical reaction. Use a table of standard enthalpies of formation toevaluate ΔHf°.

State the principle on which Hess' law depends, and explain whythis law is so useful.

Describe a simple calorimeter and explain how it is employed and how its heat capacity is determined.

Concept Map



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2/24/2015 The First Law: Applications


Chem1 General Chemistry Virtual Textbook → Energetics → Applications

Some applications of enthalpy and the FirstLaw

Chemical energetics and thermodynamics -5Virtually all chemical processes involve the absorption or release of heat, and thus changes inthe internal energy of the system. In this section, we survey some of the more commonchemistry-related applications of enthalpy and the First Law. While the first two sectionsrelate mainly to chemistry, the remaining ones impact the everyday lives of everyone.

1 Enthalpy diagrams and their usesComparison and interpretation of enthalpy changes is materially aided by a graphical construction in which therelative enthalpies of various substances are represented by horizontal lines on a vertical energy scale. The zeroof the scale can be placed anywhere, since energies are always arbitrary; it is generally most useful to locate theelements at zero energy, which reflects the convention that their standard enthlapies of formation are zero.

This very simple enthalpy diagram for carbon andoxygen and its two stable oxides shows the changes inenthalpy associated with the various reactions thissystem can undergo. Notice how Hess’ law is implicitin this diagram; we can calculate the enthalpy changefor the combustion of carbon monoxide to carbondioxide, for example, by subtraction of the appropriatearrow lengths without writing out the thermochemicalequations in a formal way.

The zero-enthalpy reference states refer to graphite,the most stable form of carbon, and gaseous oxygen.All temperatures are 298 K.



This enthalpy diagram for the hydrogen-oxygensystem shows the known stable configurations ofthese two elements. Reaction of gaseous H2 and O2to yield one mole of liquid water releases 285 kJ ofheat .

If the H2O is formed in the gaseous state, theenergy release will be smaller.

Notice also that...

The heat of vaporization of water (an endothermicprocess) is clearly found from the diagram.

Hydrogen peroxide H2O2, which spontaneously decomposes into O2 and H2O,releases some heat in this process. Ordinarily this reaction is so slow that theheat is not noticed. But the use of an appropriate catalyst can make the reaction sofast that it has been used to fuel a racing car.

Why you can't run your car on water

You may have heard the venerable urban legend, probably by now over 80years old, that some obscure inventor discovered a process to do this, but theinvention was secretly bought up by the oil companies in order to preservetheir monopoly. The enthalpy diagram for the hydrogen-oxygen system showswhy this cannot be true — there is simply no known compound of H and Othat resides at a lower enthalpy level.

Enthalpy diagrams are especially useful for comparing groups of substances having some common feature. Thisone shows the molar enthalpies of species relating to two hydrogen halides, with respect to those of theelements. From this diagram we can see at a glance that the formation of HF from the elements is considerablymore exothermic than the corresponding formation of HCl. The upper part of this diagram shows the gaseousatoms at positive enthalpies with respect to the elements. The endothermic processes in which the H2 and thedihalogen are dissociated into atoms can be imagined as taking place in two stages, also shown. From theenthalpy change associated with the dissociation of H2 (218 kJ mol

–1), the dissociation enthalpies of F2 and Cl2can be calculated and placed on the diagram.

2 Bond enthalpies and bond energiesThe enthalpy change associated with the reaction

HI(g) → H(g)+ I(g)

http://www.ecofriend.org/entry/eco-cars-hydrogen-peroxide-powered-drag-racer-is-green-and-fast/



is the enthalpy of dissociation of the HImolecule; it is also the bond energy of thehydrogen-iodine bond in this molecule.Under the usual standard conditions, itwould be expressed either as the bondenthalpy H°(HI,298K) or internal energy U°(HI,298); in this case the two quantitiesdiffer from each other by ΔPV = RT. Sincethis reaction cannot be studied directly, theH–I bond enthalpy is calculated from theappropriate standard enthalpies offormation:

½ H2(g)→ H(g) + 218 kJ

½ I2(g)→ I(g) +107 kJ

½ H2(g) + 1/2 I2(g)→ HI(g) –36 kJ

HI(g) → H(g) + I(g) +299 kJ

Bond energies and enthalpies are important properties of chemical bonds, and it is very important to be able toestimate their values from other thermochemical data. The total bond enthalpy of a more complex moleculesuch as ethane can be found from the following combination of reactions:

C2H6(g)→ 2C(graphite) + 3 H2(g) 84.7 kJ

3 H2(g)→ 6 H(g) 1308 kJ

2 C(graphite) → 2 C(g) 1430 kJ

C2H6(g)→ 2 C(g) + 6H(g) 2823 kJ

When a molecule in its ordinary state is broken upinto gaseous atoms, the process is known asatomization.

The standard enthalpy of atomization refers to thetransformation of an element into gaseous atoms:

C(graphite) → C(g) ΔH° = 716.7 kJ

... which is always, of course, an endothermicprocess. Heats of atomization are most commonlyused for calculating bond energies. They areusually measured spectroscopically.



Pauling’s rule and average bond energy

The total bond energy of a molecule can be thought of as the sum of the energies of the individual bonds. Thisprinciple, known as Pauling’s Rule, is only an approximation, because the energy of a given type of bond is notreally a constant, but depends somewhat on the particular chemical environment of the two atoms. In otherwords, all we can really talk about is the average energy of a particular kind of bond, such as C–O, for example,the average being taken over a representative sample of compounds containing this type of bond, such as CO,CO2, COCl2, (CH3)2CO, CH3COOH, etc.

Despite the lack of strict additivity of bond energies, Pauling’s Rule is extremelyuseful because it allows one to estimate the heats of formation of compoundsthat have not been studied, or have not even been prepared. Thus in theforegoing example, if we know the enthalpies of the C–C and C–H bonds fromother data, we could estimate the total bond enthalpy of ethane, and then workback to get some other quantity of interest, such as ethane’s enthalpy offormation.

By assembling a large amount of experimental information of this kind, a consistent set of average bondenergies can be obtained. The energies of double bonds are greater than those of single bonds, and those oftriple bonds are higher still.

Average energies of some single bonds (KJ/mol)

H C N O F Cl Br I Si

H 436 415 390 464 569 432 370 295 395

C 345 290 350 439 330 275 240 360

N 160 200 270 200 270

O 140 185 205 185 200 370

F 160 255 160 280 540

Cl 243 220 210 359

Br 190 180 290

I 150 210

Si 230



3 Energy content of fuelsA fuel is any substance capable of providing useful amounts of energy through a process that can be carried outin a controlled manner at economical cost. For most practical fuels, the process is combustion in air (in whichthe oxidizing agent O2 is available at zero cost.) The enthalpy of combustion is obviously an important criterionfor a substance’s suitability as a fuel, but it is not the only one; a useful fuel must also be easily ignited, and inthe case of a fuel intended for self-powered vehicles, its energy density in terms of both mass (kJ kg–1) andvolume (kJ m–3) must be reasonably large. Thus substances such as methane and propane which are gases at 1atm must be stored as pressurized liquids for transportation and portable applications.

Energy densities of some common fuels

fuel MJ kg–1

wood (dry) 15

coal (poor) 15

coal (premium) 27

ethanola 30

petroleum-derived products 45

methane, liquified natural gas 54

hydrogenb 140

Notes on the above table

a Ethanol is being strongly promoted as a motor fuel by the U.S. agricultural industry. Note,however, that according to some estimates, it takes 46 MJ of energy to produce 1 kg of ethanolfrom corn. Some other analyses, which take into account optimal farming practices and the useof by-products arrive at different conclusions; see, for example, this summary with links toseveral reports.

b Owing to its low molar mass and high heat of combustion, hydrogen possesses anextraordinarily high energy density, and would be an ideal fuel if its critical temperature (33 K,the temperature above which it cannot exist as a liquid) were not so low. The potential benefitsof using hydrogen as a fuel have motivated a great deal of research into other methods ofgetting a large amount of H2 into a small volume of space. Simply compressing the gas to a very

high pressure is not practical because the weight of the heavy-walled steel vessel required towithstand the pressure would increase the effective weight of the fuel to an unacceptably largevalue. One scheme that has shown some promise exploits the ability of H2 to “dissolve” in

certain transition metals. The hydrogen can be recovered from the resulting solid solution(actually a loosely-bound compound) by heating.

4 Energy content of foods

http://journeytoforever.org/ethanol_energy.html



What, exactly, is meant by the statement that a particular food “contains 1200 calories” per serving? Thissimply refers to the standard enthalpy of combustion of the foodstuff, as measured in a bomb calorimeter. Note,however, that in nutritional usage, the calorie is really a kilocalorie (sometimes called “large calorie”), that is,4184 J. Although this unit is still employed in the popular literature, the SI unit is now commonly used in thescientific and clinical literature, in which energy contents of foods are usually quoted in kJ per unit of weight.

Although the mechanisms of oxidation of a carbohydrate such as glucose to carbon dioxide and water in a bombcalorimeter and in the body are complex and entirely different, the net reaction involves the same initial andfinal states, and must be the same for any possible pathway:

C6H12O6 + 6 O2 → 6 CO2 + 6 H2O ΔH° – 20.8 kJ mol–1

Glucose is a sugar, a breakdown product of starch, and is the most important energy source at the cellular level;fats, proteins, and other sugars are readily converted to glucose. By writing balanced equations for thecombustion of sugars, fats, and proteins, a comparison of their relative energy contents can be made. Thestoichiometry of each reaction gives the amounts of oxygen taken up and released when a given amount of eachkind of food is oxidized; these gas volumes are often taken as indirect measures of energy consumption andmetabolic activity; a commonly-accepted value that seems to apply to a variety of food sources is 20.1 J (4.8kcal) per liter of O2 consumed.

For some components of food, particularly proteins, oxidation may not always be complete in the body, so theenergy that is actually available will be smaller than that given by the heat of combustion. Mammals, forexample, are unable to break down cellulose (a polymer of sugar) at all; animals that derive a major part of theirnutrition from grass and leaves must rely on the action of symbiotic bacteria which colonize their digestivetracts.

The amount of energy available from a food can be found by measuring the heat of combustion of the wasteproducts excreted by an organism that has been restricted to a controlled diet, and subtracting this from the heat ofcombustion of the food.

Energy content and availability of the major food components

type of food foodΔH°,

kJ g–1percent

availability

Protein meat 22.4 92

egg 23.4

Fat butter 38.2

animal fat 39.2 95

Carbohydrate starch 17.2

glucose(sugar) 15.5 99

ethanol 29.7 100

The amount of energy an animal requires depends on the age, sex, surface area of the body, and of course on the



amount of physical activity. The rate at which energy is expended is expressed in watts: 1 W = 1 J sec–1. Forhumans, this value varies from about 200-800 W. This translates into daily food intakes having energyequivalents of about 10-15 MJ for most working adults. In order to just maintain weight in the absence of anyphysical activity, about 6 MJ per day is required.

Metabolic rates of some animals

animal kJ hr–1 kJ kg–1hr–1

mouse 82 17

cat 34 6.8

dog 78 3.3

sheep 193 2.2

human 300 2.1

horse 1430 1.1

elephant 5380 0.7

The above table is instructive in that although larger animalsconsume more energy, the energy consumption per unit of body weight decreases with size. This reflects thefact the rate of heat loss to the environment depends largely on the surface area of an animal, which increaseswith mass at a greater rate than does an animal's volume ("size").



What is an adiabatic process? See here

5 Thermodynamics and the weather

Hydrogen bonds at work

It is common knowledge that large bodies of water have a “moderating” effect on the local weather, reducingthe extremes of temperature that occur in other areas. Water temperatures change much more slowly than dothose of soil, rock, and vegetation, and this effect tends to affect nearby land masses. This is largely due to thehigh heat capacity of water in relation to that of land surfaces— and thus ultimately to the effects of hydrogenbonding. The lower efficiency of water as an absorber and radiator of infrared energy also plays a role.

Adiabatic heating and cooling of the atmosphere

The specific heat capacity of water is about fourtimes greater than that of soil. This has a directconsequence to anyone who lives near the oceanand is familiar with the daily variations in thedirection of the winds between the land and thewater. Even large lakes can exert a moderatinginfluence on the local weather due to water'srelative insensitivity to temperature change.

During the daytime the land and sea receiveapproximately equal amounts of heat from theSun, but the much smaller heat capacity of theland causes its temperature to rise more rapidly.This causes the air above the land to heat,reducing its density and causing it to rise. Cooleroceanic air is drawn in to vill the void, thusgiving rise to the daytime sea breeze.

In the evening, both land and ocean lose heat byradiation to the sky, but the temperature of thewater drops less than that of the land, continuing

to supply heat to the oceanic air and causing it to rise, thus reversing the direction of air flow and producing theevening land breeze.

The above images are from the NDBC site. For a more detailed explanation ofland/sea breezes and links to some interesting meteorological images, see thisU. Wisconsin page. An animated movie that correlates time and temperaturedata with wind directions can be seen at the Earth Science textbook site.

http://cimss.ssec.wisc.edu/wxwise/seabrz.html

http://seaboard.ndbc.noaa.gov/educate/seabreeze_ans.shtml

http://www.chem1.com/acad/webtext/energetics/CE-2.html#S3A

http://earthsci.terc.edu/content/visualizations/es1903/es1903page01.cfm?chapter_no=visualization



Why it gets colder as you go higher: the adiabatic lapse rate

The First Law at work

The air receives its heat by absorbing far-infrared radiation from the earth, which of course receives its heatfrom the sun. The amount of heat radiated to the air immediately above the surface varies with what's on it(forest, fields, water, buildings) and of course on the time and season. When a parcel of air above a particularlocation happens to be warmed more than the air immediately surrounding it, this air expands and becomes lessdense. It therefore rises up through the surrounding air and undergoes further expansion as it encounters lowerpressures at greater altitudes.

Whenever a gas expands against an opposing pressure, it doeswork on the surroundings. According to the First LawΔU = q + w, if this work is not accompanied by a compensatingflow of heat into the system, its internal energy will fall, and so,therefore, will its temperature. It turns out that heat flow andmixing are rather slow processes in the atmosphere in comparisonto the convective motion we are describing, so the First Law canbe written as ΔU = w (recall that w is negative when a gasexpands.) Thus as air rises above the surface of the earth itundergoes adiabatic expansion and cools. The actual rate oftemperature decrease with altitude depends on the composition ofthe air (the main variable being its moisture content) and on itsheat capacity. For dry air, this results in an adiabatic lapse rate of9.8 C° per km of altitude.

Santa Anas and chinooks: those warm, wild winds

Just the opposite happens when winds developin high-altitude areas and head downhill. As theair descends, it undergoes compression from thepressure of the air above it. The surroundingsare now doing work on the system, and becausethe process occurs too rapidly for the increasedinternal energy to be removed as heat, thecompression is approximately adiabatic.

[image]

The resulting winds are warm (and thereforedry) and are often very irritating to mucousmembranes. These are known generically as

Föhn winds (which is the name given to those that originate in the Alps.) In North America they are oftencalled chinooks (or, in winter, "snow melters") when they originate along the Rocky Mountains.

Among the most notorious are the Santa Ana winds of Southern California which pick up extra heat (and dust)

http://www.stuffintheair.com/Blowin_in_the_Wind-ValleyandMountainWinds.html

http://en.wikipedia.org/wiki/Foehn_wind

http://en.wikipedia.org/wiki/Santa_Ana_winds

http://en.wikipedia.org/wiki/Chinook_wind



as they pass over the Mohave Desert before plunging downinto the Los Angeles basin. Their dryness and high velocitiesfeed many of the disasterous wildfires that afflict the region.

[image]

[image]

http://fantabuloushimbo.blogspot.com/2008/07/legend-of-santa-ana-winds.html

http://fireground.com/?p=1819l



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© 2005, 2012 by Stephen Lower - last modified 2012-08-31

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