An Introduction to Thermodynamics phần 3 docx

It should also be possible to relate the volume of any liquid or solid to its temperature and pressure, or to express such other properties as refractive index, heat capacity at constant

without changing their effect, and therefore decreases the pressure on the walls, requiring the

“supplement” to P to fit the ideal gas equation form; see Phys Teach 34, 248-249 (April, 1996).

The reader with some knowledge of special relativity may recognize that the total 16

energy of any system is measured by its mass, multiplied by the square of the speed of light However, it would be necessary to measure masses about a million times more accurately than is now possible to be able to determine energies to the accuracy required in thermochemistry

It should also be possible to relate the volume of any liquid or solid to its temperature and pressure, or to express such other properties as refractive index, heat capacity at constant volume

or pressure, thermal conductivity, heats of vaporization or fusion, or vapor pressures of solids or liquids, in terms of the temperature and pressure Some of these equations will be encountered in later chapters

Thermochemistry

The application to chemical reactions of the principles developed thus far is called

thermochemistry In particular, the heats of reaction are measured and tabulated and from these

and from measured heat capacities the enthalpy changes are calculated for other reactions or for other experimental conditions

HESS’S LAW. The enthalpy change for a chemical reaction, such as the oxidation of sulfur dioxide

to sulfur trioxide —

2 SO (g) + O 2 2&6 2 SO (liq)3

— can be expressed as the difference between the enthalpies of the initial and final states

∆Hreaction = Hfinal - Hinitial = H(2 SO ) - H(2 SO ) - H(O )3 2 2

There is no way within thermodynamics of measuring an absolute energy, or an absolute16

enthalpy Only energy, and enthalpy, changes can be determined However, knowing that these energy and enthalpy changes depend only on the initial and final states, it is possible to add and subtract chemical reactions and add and subtract the corresponding enthalpy changes That is, we may quite arbitrarily select a reference energy and/or enthalpy level and measure all values from that arbitrary level In particular, it is possible to tabulate “heats of formation”, the enthalpy changes in the reaction of the elements to form each compound, and from these to calculate

enthalpies of other reactions This principle is known as Hess’s law.

The reactions for the formation of the gases SO and SO from the elements are2 3

S + O - SO2 2

S + 3/2 O - SO (liq)2 3

The measured enthalpy changes for these reactions at 25 C and 1 atm pressure are -296.90 kJ/mol and -437.94 kJ/mol Subtraction of the first reaction from the second gives

SO + ½ O 2 2&6 SO (liq)3 and subtraction of the enthalpy changes gives -141.04 kJ/mol, which is the heat of reaction for the oxidation of SO to SO (liq).2 3

Exactly the same elements, in the same quantities, always appear on both sides of a chemical equation (which is why reactions as written are called “equations”) Subtraction of the elements from both sides of an equation will yield, on each side, product minus reactants for the reactions

of formation of each of the substances appearing in the original equation In the example above, the original equation was SO + ½ O 2 2&6 SO Subtract 1 mol of S and 3/2 mol of O from each3 2

side The equation can then be written as the formation of each compound (i.e., of SO , O , and2 2

SO ) from the elements.3

(SO - S - O ) + (½ O - ½ O ) 2 2 2 2 &6 (SO - S - 3 3/2 O )2 and therefore

∆Hreaction = ∆Hform(SO ) - ∆H3 form(SO ) - ∆H2 form(½ O )2

= -437.94 kJ/mol -296.90 kJ/mol - 0

= -141.04 kJ/mol(SO liq)3 (Notice that the heat of formation of any element, in its standard state, is necessarily zero.)

An entirely equivalent way of obtaining the same numbers is to consider the enthalpy of each compound on a scale taken with reference to the elements Such enthalpy values are called

standard enthalpies of the compounds; they are identical with the standard enthalpies of

formation.

Hess’s law can often be applied to find heats of reaction that could not be directly measured experimentally For example, the reaction of two molecules of ethylene, C H , to form2 4

cyclobutane, C H , would not readily occur quantitatively under conditions conducive to4 8

measurement of the heat of reaction But both ethylene and cyclobutane can be burned in

oxygen, and subtraction of these reactions gives the reaction equation desired

2 C H + 8 O 2 4 2&6 4 CO + 4 H O2 2

C H + 8 O 4 8 2&6 4 CO + 4 H O2 2 Subtraction of the second from the first gives

2 C H 2 4&6 C H4 8

and, therefore, subtraction of the ∆H for the second combustion from the ∆H for the first

combustion gives ∆H for the condensation reaction Heats of combustion (equal to - ∆Hreaction ) are comparatively easy to measure and are often tabulated

KIRCHHOFF’S LAW The heat of reaction at a temperature other than that given in a table can be

( ) ( )

=

2

2 1

products 1

reactants

T

T

T P

C H

+

∆

=

1

) reactants ( products

1 2

T

H H

found by calculating enthalpy changes along an arbitrary path The total enthalpy change is

independent of this choice of path The method is known as Kirchhoff’s law.

Assume that ∆H is known for a reaction at a temperature T and the ∆H at another temperature,1

T , is to be found Starting with the hot reactants at T (Figure 4), the reaction could be carried2 2

out isothermally to obtain products at the same temperature An alternative path would be to

cool the reactants to the temperature T , carry out the reaction isothermally at T , and warm the1 1 products to T The heat of reaction at T is already known and if the heat capacities at constant2 1

pressure are known, the enthalpy changes can be calculated for the processes of cooling reactants

and warming products This path must give the same ∆H as the isothermal reaction at T 2

or,

because interchanging limits of an integral will change the sign,

If the difference in heat capacities is independent of temperature, this may be rewritten in the form

∆H = ∆H + [C2 1 p (products) - C P (reactants)](T - T ) (18) 2 1

For example, given that the heat of reaction for rhombic sulfur burning in oxygen to yield

sulfur dioxide gas is - 296.9 kJ/mol at 25 C (298 K), find ∆H at 95 C (368 K) The heato o

capacities are given in Table 2 Insertion of the numerical values into equation 18 gives

Average values (in J/mol-K) for temperature ranges indicated

Compound C P Temperature, C

o

*For rough calculations it is sufficient to set

C (steam) = C (ice) = ½ C (liq H O).P P P 2

This property associated with state functions has sometimes been confused with a

17

conservation principle Enthalpy is not conserved

∆H368 = - 296,900 J/mol + (41.9- 29.4 - 23.7) x 70 J/mol

= - 296.1 J/mol

Sometimes there will be a phase transition during the warming or cooling process Sulfur has

a phase change at 95 C, at which point rhombic sulfur goes to monoclinic sulfur; the monoclinico sulfur melts at 119 C The enthalpy changes are 11.78 and 39.24 kJ/mol The heat of reactiono for liquid sulfur burning in oxygen to form SO at 119 C (392 K) can be calculated as follows (see2 o Figure 5)

∆H = - ∆H392 fusion - C (m) (119 - 95) - ∆H - C (r) (95 - 25) - C (O ) (119 - 25) P tr P P 2

+ ∆H + C (SO ) (119 - 25)298 P 2

∆H = - 39,240 - 25.9 x 24 - 11,780 - 23.7 x 70 - 29.4 x 94 - 296,900 + 41.9 x 94 J/mol392

= - 349.0 kJ/mol

Note that temperature differences can be found without conversion to the Kelvin scale.

Both Hess’s law and Kirchhoff’s law are simply applications of the principle that changes in

a state function, such as the enthalpy, are completely determined by the initial and final states 17

This principle is combined with the equation arising from the first law that shows that i f the

pressure is constant, the enthalpy change will be equal to the heat absorbed by the system Thus

the “heat of reaction,” by which we mean ∆Hreaction (at a particular temperature, pressure, and

concentrations of reactants and products), is only equal to the heat absorbed i f the reaction

proceeds at constant pressure (and at the specified temperature and concentrations) It is

sometimes more convenient to carry out a reaction at constant volume Then the heat absorbed is

not equal to the “heat of reaction” (that is, to ∆Hreaction), but it is still determinate because heat absorbed equals the change in energy when the system follows a constant-volume path and

because ∆Ereaction is fixed by the initial and final states

The experimental determination of a heat of reaction is called calorimetry A typical

calorimeter (Figure 6) consists of a reaction chamber, surrounded by a layer of water, enclosed

by sufficient insulation to prevent heat loss to the surrounding laboratory The reactants, at room temperature, are placed in the reaction chamber, the calorimeter is closed, and the reaction is initiated by an electrically heated wire or other controlled energy source The reaction will normally be exothermic and the reaction chamber will therefore become quite hot, but the heat is conducted into the surrounding water layer so that the products, and the water, reach a final

temperature only slightly above the initial temperature The ∆Ereaction is the same as if the entire process had occurred at the initial temperature even though the materials may have become quite hot during the course of the reaction The heat given off by the reaction is calculated by

observing the temperature rise of the water, using the condition that all heat given off by the reaction must have been absorbed by the water Small corrections are required for the change of temperature, from the initial room temperature, of the products, and for the small amount of energy added by the hot wire or other initiation method Some systems may also require a correction for changes of concentrations during the reaction

1 At room temperature the heat capacity (C ) of most solid elements (except the very light ones) V

is about 3R Specific heat is the heat capacity per unit mass, or the value in J/g·K (For solids the

difference between constant volume and constant pressure conditions is small and may be

neglected here.)

a Find the specific heat of Pb (molar mass = 207.19)

b Find the specific heat of Cu (molar mass = 63.546)

c Find ∆E for 25 g of Cu when it is warmed from 20 C to 35 C.o o

2 Estimate the final temperature when 10 cm of iron (density 7.86 g/cm ) at 80 C is added to 303 3 o

ml of water at 20 C.o

3 Find the work done when 3 mol of N gas at 4 atm pressure expands slowly at 25 C against a2 o constant pressure of 1 atm

4 Calculate the work done when 0.25 mol of SO at 27 C and 1 atm expands reversibly and2 o

isothermally to a final pressure of 0.20 atm Find Q for the gas during this process.

5 Calculate the final temperature if an ice cube (25 g) at - 5 C is added to a cup of coffee (150o

ml, or “2/3 cup”) at 90 C, neglecting heat loss to the surroundings.o

6 The heat of vaporization of benzene, C H , is 30.72 kJ/mol and the normal boiling point is6 6 80.1 C.o

a What is ∆E if the vapor is an ideal gas?vap

b Part of the thermal energy absorbed by a liquid as it vaporizes to a gas is required to perform work on the atmosphere as the substance expands to a vapor What fraction of the total

Q goes into work against the atmosphere when benzene (density about 0.88 g/cm ) vaporizes at3 its normal boiling point?

7 Find ∆E at 25 C for the oxidation of S (rhombic) to give SO (gas) and for the oxidation of So 2

to give SO (liq) The heats of reaction are - 296.81 and - 441.0 kJ/mol Assume gases are ideal.3

8 Calculate ∆H and ∆E for the reaction, at 25 C,o

SO (g) + ½ O 2 2 6 SO (g)3

The heat of vaporization of SO at 25 C is 43.14 kJ/mol.3 o

9 Manganese can be prepared by a thermite process,

3 Mn O + 8 Al -3 4 6 9 Mn + 4 Al O2 3

The standard enthalpy of formation of Mn O is - 1387.8 kJ/mol and for Al O , - 1657.7 kJ/mol 3 4 2 3 Find the amount of thermal energy given off by the reaction as written above, starting with the reactants at room temperature and ending with the products at room temperature

10 The interconversion of graphite and diamond does not occur at room temperature Explain

how you could determine ∆H for this transition by measurements in the laboratory.

11 Calculate the heat of vaporization of water at 50 C.o

12 Find ∆H for the hydrogenation of ethylene to produce ethane,

C H + H -2 4 2 6 C H2 6

The torr is a unit of pressure equal to 1/760 atm It is numerically equivalent to the unit

18

“mm of Hg” but is less awkward, especially when mercury is one of several vapors present The torr is named after Evangelista Torricelli, who invented the barometer in the 17 century.th

at 150 C The standard heats of formation, at 25 C, of ethylene and ethane are 52.4 and - 84.0

kJ/mol, respectively Heat capacities (C ) of ethylene, hydrogen, and ethane within this P

temperature interval are approximately 42.9, 28.8, and 52.5 J/mol·K All three compounds are gases at room temperature and above

13 Benzene, C H , is reduced commercially with hydrogen to give cyclohexane, C H 6 6 6 12

C H + 3 H -6 6 2 6 C H6 12

At room temperature (25 C) the benzene and cyclohexane are liquids and the heat of reaction is o

- 205.5 kJ/mol Benzene boils at 80.1 C with a heat of vaporization of 47.8 kJ/mol; cyclohexaneo boils at 80.7 C with a heat of vaporization of 33.0 kJ/mol The average heat capacity of liquido benzene is 136.0 J/mol·K, of liquid cyclohexane 154.9 J/mol·K, of benzene vapor 82.4 J/mol·K,

of cyclohexane vapor 150 J/mol·K, and of hydrogen 28.8 J/mol·K Find ∆H for the reduction of

benzene with hydrogen at 150 C.o

[0.44 cal/g·K; 0.41; 26.7 35.8 cal/mol·K; 6.94 cal/mol·K]Check numbers

14 Air is approximately 20% O and 80% N , by volume Find the effective molar mass of air 2 2 Calculate the density of air at 25 C and 1 atm in kg/m , assuming it acts as an ideal gas.o 3

15 A gaseous compound, containing only sulfur and fluorine, is 62.7% (by weight) sulfur At

27 C and 750 torr the density of the gas is 4.09 g/L What is the molecular formula of theo 18 compound? Assume the gas is ideal

16 Find the pressure exerted by 32 g of methane in a 2 L steel bomb at 127 C, assuming the gaso

is ideal

17 Find the pressure exerted by 32 g of methane in a 2 L steel bomb at 127 C assuming the gaso

obeys van der Waals’ equation The constants may be taken as a = 2.25 L atm/mol and b =2 2 0.0428 L/mol

18 Find the volume occupied by 32 g of methane at a pressure of 5 atm at 27 C if the gas iso

ideal Calculate, with this value, the n a/V correction term in the van der Waals equation and2 2 find, from this, an approximate value for the volume occupied by the methane if it obeys van der Waals’ equation Recalculate the correction term, with this better volume, and recalculate the volume if necessary

19 A good vacuum obtained with a mechanical pump and a mercury diffusion pump will

measure 10 torr, or better, of “noncondensable” gases (such as air), but there is 10 torr of-5 -3 mercury vapor present unless this has been trapped out Very good vacuum systems can give 10 torr

9

a How many molecules/cm are there at a pressure of 10 torr?3 -3

b How many molecules/cm are there at a pressure of 10 torr?3 -9

c What pressure would be required to achieve 1 molecule/cm ?3

The first law prohibits certain perpetual-motion machines, called “perpetual-motion 1

machines of the first kind.” The more interesting attempts, called “perpetual-motion machines of the second kind,” do satisfy the first law

2 The Second Law of Thermodynamics

During the 19 century several men, of quite different backgrounds and interests, struggledth with the basic problems of thermodynamics Brilliant flashes of understanding were followed by years of doubting, testing, and interpreting Among the fundamental difficulties was a confusion between two concepts One of these was the concept of energy The other, which is related to the general concept of equilibrium and the direction of changes with respect to equilibrium, was stated first, but the language was such that it was long misunderstood and therefore rejected It is now accepted as the second law of thermodynamics

Very few people today who have any acquaintance with modern science would doubt the following generalizations:

1 Perpetual-motion machines don’t work

2 Bodies in equilibrium have the same temperature When two bodies in contact have

different temperatures, energy flows, as heat (Q), from the warmer to the cooler body.

3 Bodies in equilibrium have the same pressure When two bodies in contact, at the same temperature, have different pressures, the body at the higher pressure tends to expand and

compress the body at the lower pressure

None of these statements is required by the first law, which requires energy balance in any1 process but does not say whether a process will actually occur For example, both exothermic and endothermic reactions are known that do proceed without external forcing There is, despite the variety in these generalizations, a similarity of pattern and intent — saying, for example, whether a specific process will or will not occur — that suggests a common basis The basis is

found in the postulate known as the second law of thermodynamics.

The Spread Function

We have seen that thermodynamics largely ignores the atomic structure of matter It is quite

sufficient, for almost all purposes, to measure the macroscopic (large scale) variables such as

temperature, pressure, volume, and energy The difficulty with this approach is that it effectively

hides from us one key property that may be called the “spread function,” S.

Consider the “thought experiment” of adding 10 grains of yellow rice to a large bag of white rice, which will subsequently be mixed and poured and just thoroughly scrambled Initially the yellow grains were carefully arranged, on top in one corner Where will they be a little later after mixing?

There is no energy advantage for the yellow grains to move to the bottom, or back to the top, or to any other specific location We know that, once mixed, there is close to zero

(1)

T

Q

∆

If the ten yellow grains were regarded as indistinguishable, we should remove 10! = 3.6 x 2

10 possibilities that only interchange yellow grains among themselves That would still leave6 more than 10 possibilities, scarcely different qualitatively from 10 The problem of overlapping53 60 positions (two yellow grains being assigned to the same location) may safely be ignored at this level of probability

probability that they will spontaneously reassemble Even if the bag of rice is only 100 grains

deep, 100 grains wide, and 100 grains in the third direction, each yellow grain has 100 x 100 x

100 = 10 possible locations, which means that two grains have 10 x 10 = 10 possible locations6 6 6 12 and the 10 grains have (10 ) = 10 possible locations The probability they will all collect in6 10 60 one location is not much more than 1 in 10 , which is awfully close to zero for most purposes.60 2 Now imagine a very small number of molecules (say 10 ) to which we add 10 molecules that6 are “excited” — perhaps they are each vibrating, or rotating, or otherwise have some extra energy that can be transferred to a neighboring molecule After a very short time, the extra energy will

be spread throughout the 10 molecules, so there is no significant chance the energy will ever6 again (in the lifetime of the universe) be collected in the original 10 molecules With a more typical number of molecules (such as 6 x 10 ) and many ways for each to store energy, it is not at23 all unreasonable to say the probability of “unspreading” the energy goes quickly to zero

If we have a small steel ball (which will typically have 10 or more atoms), one way of23 storing energy in the ball is to start it rolling Another very large group of ways of storing energy

in the ball is to let some of the atoms begin to vibrate The mechanical energy of rolling may be converted by rolling friction to thermal energy of internal motions There are so many possible

internal motion states that we don’t expect the energy to ever spontaneously reappear as

mechanical energy of rolling

Entropy

Fortunately, there is an easy way to keep track of S, the spread function, without even having

to count molecules Any energy passing into an object by thermal energy transfer, Q, is “gone” in

terms of having been spread around It shows up afterward only as an increase (perhaps slight) in

the temperature of the system So the first step is to keep track of Q, and in particular, we want

Q for a reversible path, or one that is as close as possible to equilibrium

Then we want to know whether this amount of energy is important to the system, or whether there is already so much energy spread around that this addition is relatively unimportant That

average energy we will find is nicely measured by the (absolute) temperature of the system So

we measure the increase in average energy as Q and divide by the temperature, T rev

Definition of Entropy Change

The name was chosen by Clausius from the Greek verb entrepo (εντρεπω), meaning “to 3

turn”, implying “change”, and because it somewhat resembles the word energy, one of the

important quantities involved in finding the number of states Entropy, S, is proportional to (or for suitable units, equal to) the logarithm of the relative probability, i.e., to the number of states accessible to the system Unfortunately, it is also vaguely suggestive of enthalpy, H, which has

different dimensions, units, and meaning

We also assume here that the process satisfies the first-law equation, ∆E = Q + W

4

However, equations 1 and 5, and equation 6 (obtained below), are not subject to this restriction against other forms of work

The “official” name of the spread function, S, is entropy By the methods of statistical mechanics we can show that S is a measure of the actual number of individual states over which the energy of the system is spread The greater the number of states (hence S), the greater the

probability associated with the overall state

It is, of course, possible to transfer energy to our system without increasing the spread of

energy For example, we can transfer just enough energy, in the proper form, to make the steel

ball start rolling In a reversible process, this energy transfer is measured as work As we saw for ideal gas expansions and compressions, the work done on the system is a minimum (and Q is a

maximum) when the process is reversible When the process is purely one of transferring energy

to a mechanical mode, such as rolling or moving in a straight line, the amount of thermal energy

transfer, Q = Q max, is equal to zero There is no increase in entropy

An alternative way of approaching the second law is to recognize that ∆E is independent of path, so we can write, for any process at constant temperature, between fixed end points4

(T) ∆E = Q + W = Q + W = Q rev rev max + W min (2)

There can be only one value of Q max and one value of W , so we can represent these, also, as min values “independent of path”, which for the moment we will label as ∆A (= W ) and as ∆B (= rev

Q ) and regard A and B as state functions rev

(T) ∆E = ∆A + ∆B

or

E = A + B (3)

In any isothermal process in which the energy of the system changes by ∆E, some part of this energy change, equal to ∆A, must appear as work done on the system The second energy term, called ∆B for the moment, cannot be lost, because of the first law (conservation of energy);

however, it may be regarded as “spilled energy”, which is lost for useful purposes during the transfer operation because it has become spread through the system

From equation 3, two important functions are obtained One, temporarily given the symbol

A, is called the Helmholtz free energy, or sometimes the work function, and is characterized, as

above, by the constant temperature equation