Thermodynamics [03] Section [54 - 91]

LOCUSLOCUSLOCUSLOCUSLOCUS 54

Chemistry : Thermodynamics

In the introduction to this chapter we pointed out that the study of thermodynamics is divided into two mainaspects. The first is thermochemistry which is concerned with how we observe, measure and predict energychanges for chemical reactions. This is what we have studied so far in this chapter. We now take up the study ofthe second part that we said is concerned with learning to use energy changes to tell whether or not a givenprocess can occur under specified conditions and how to make a process more(or less) favourable.

However, before we enter this realm of thermodynamics, let us see if we can answer to any extent, the question ofwhether or not a process can occur spontaneously using our knowledge of thermochemistry.

If we hold a brick in the air and let it go, we know what will happen : It will fall as the force of gravity pulls it towardearth. A falling brick is an example of a process that the laws of thermodynamics tell us is favoured to happen. Werefer to such a thermodynamically favoured process as a spontaneous process (we will talk about this in detail,later in this chapter.)

Chemical processes can be spontaneous as well. The enthalpy of a reaction gives one indication so as to whetherit is likely to be spontaneous. For example, the combustion of 2 ( )H g and 2 ( )O g is a highly exothermic process:

2 2 21( ) ( ) ( ) ; 2422

H g O g H O g H kJ+ → ∆ = −

During this reaction, large amounts of heat are transferred from the system to the surroundings. Generally, it isfavourable for the enthalpy of the system to decrease by transferring the heat to the surroundings, as by doing thatthe products, (and hence, the system) acquire a lower energy state and hence are generally more stable.

So, we see that a large value of negative H∆ has helped us predict the spontaneity of a reaction. So, from this canwe conclude that by measuring H∆ of a reaction, we can predict the spontaneity of any reaction ? The answer isa big NO! Let us see so as to why. Consider the melting of ice :

2 2( ) ( ) ; 6.01H O s H O l H kJ→ ∆ = +

This is an endothermic process. Even though it is endothermic, it is spontaneous at temperatures above the freezingpoint of water (0°C). (This we know even by common observation in day to day life that when block of ice is keptat room temperature (>0°C), it melts spontaneously.) So, we have a counter example to our assertion that onlyexothermic reactions are spontaneous.

However, the validity of our earlier argument that a negative H∆ should favour spontaneity is still very much inplace. This might seem a little confusing to you. To clarify matters, lets deviate a little from thermodynamics tocricket ! (I am sure most of you would not mind it ! )

Suppose we want to study the possibility of a boundary (either fours or sixes) being hit by none other than M. S.Dhoni. Although there can be various factors (like type of bowling, bowler, size of the field, etc.) for a successfulboundary being hit, we will consider only two major factors i.e. the strength with which the ball is hit and the qualityof the fielder. We can see that the probability of the boundary being hit increases with the first factor i.e. thestrength with which the ball is hit by Mr. Dhoni. However, the possibility of the boundary suffers if the quality

Section - 3 SECOND LAW OF THERMODYNAMICS AND RELATED TOPICS



of the fielder is good. So, we see that the two factors are opposing to each other in a certain way as far as thesuccessful hitting of the boundary is concerned. However, the two factors can also support each other in the hittingof a boundary. For example, even if the strength with which the ball is hit is not too great, the low quality of thefielder can still favour the boundary being hit. And even if the strength with which the ball is hit is high ( a likely caseif Mr. Dhohi is the batsman), a Jonty Rhodes (although he is retired now) standing at the silly point can complicatematters for Mr. Dhoni’s intentions by his superb fielding!

The crux of the matter is that a higher strength of hitting the ball only favours the boundary being hit but is not thesole factor in determining the success of the boundary as it depends on the other factor of the quality of fielder aswell. And the other factor may increase as well as decrease the probability of boundary being hit.

Lets come back now to thermodynamics from cricket (no matter how painful it is!). We can compare spontaneityof a reaction to the successful boundary being hit, the strength of hitting the ball to ,H∆ and the quality of fielderto S∆ (a term we will study, shortly).

So, just as the strength of hitting the ball favours the successful boundary, but does not quite guarantee it completely,a high value of negative H∆ favours the spontaneity of a reaction but does not quite guarantee it completely aswell. However, just as complete knowledge of both the strength of hitting and quality of fielder may help usascertain (roughly) if the boundary will be hit or not, we will see that complete knowledge of H∆ and S∆ (to bediscussed, shortly) will help us ascertain if the reaction will be spontaneous or not. But first, we must characterisespontaneous process clearly for ourselves and study two new terms (reversible and irreversible processes)essentialto understanding the key ideas of this section. We will need the last two terms specially when we discuss a part ofthis section in the unit on equilibrium.

SPONTANEOUS PROCESSES

A spontaneous process is simply a physical or chemical change that occurs by itself. It requires no continuingoutside agency to make it happen. For example, a round rock at the top of a hill rolls down, heat flows from a hotobject to a cold one, an iron object rusts in moist air. These processes occur spontaneously, or naturally, withoutrequiring an outside force or agency. They continue until a certain equilibrium is reached.

If these processes were to go in the opposite direction, they would be non-spontaneous i.e. they will not happenby themselves and will need a continuing outside agency to make them happen. For example, the rolling of a roundrock uphill by itself is not a natural process; it is non spontaneous. The rock could be moved to the top of the hillbut work would have to be expended. Similarly, heat can be made to flow from cold to a hot object, but a heatpump or refrigerator is needed. Rust can be converted to iron, but the process requires chemical reactions whichare not spontaneous.

Any spontaneous change has a natural direction. For example in the rusting of iron, the direction is towards therusting. The direction in which a process is spontaneous can be very dependent on the temperature of the system.For example, let’s reconsider the endothermic process of melting of ice under atmospheric pressure. When 0T C> °ice melts spontaneously; the reverse process, liquid water turning into ice, is not spontaneous at these temperatures(refer the figure below). However, when 0 ,T C< ° the opposite is true. Liquid water converts into ice spontaneously,and the conversion of ice into water is not spontaneous.



Spontaneous for T > °C0

Spontaneous for T < °C0

The spontaneity of a process can depend on the temperature. At T > °C, ice melts spontaneously to liquid water. At T < °C, the reverse process, water freezing to ice, is spontaneous. At T = °C, the two states are in equilibrium and neither conversion occurs spontaneously.

00 0

iceliquidwater

fig.12

But what happens at 0 ,T C= ° the normal melting point of ice ? At the normal melting point of a substance, thesolid and liquid phases are in equilibrium. At this particular temperature the two phases are inter converting intoeach other at the same rate, and there is no preferred direction for the process : Both the forward and reverseprocesses occur with equal preference, and the process is not spontaneously favoured in one direction over theother.

The special situation that occurs at equilibrium can be related to the particular way in which the system changesas it goes from one state to another. Let us understand this with an example. Imagine melting 1 mole of ice at 0°C,1 atm to form 1 mole of liquid water at 0°C, 1 atm. We can achieve this change by adding a certain amount ofheat, fusion .q H= ∆ If we want to return the system to its original state (ice at 0°C), we simply reverse the procedure.We can remove the same amount of heat that we originally added. The melting and freezing of water at 0°C is anexample of a reversible process, one in which we can go back and forth between the states along exactly thesame path. In contrast, melting 1 mole of ice by placing it in a laboratory at room temperature is an irreversibleprocess. We can’t simply follow the same path back to the original state, because we must lower the watertemperature to 0°C to form ice again. Whenever a chemical system is in equilibrium, we can go reversibly betweenthe reactants and products. We will see in this chapter that in any spontaneous process the path betweenreactants and products is irreversible.

Before we move ahead, one more important point to understand is that if a process is spontaneous, this does notmean that it will occur at an observable rate. A spontaneous reaction can be very fast, as in the case of acid-baseneutralisation, or very slow, as in the case of rusting of iron. Thermodynamics can tell us the direction and extentof a reaction, but it tells nothing about its speed. This aspect is covered by chemical kinetics.



DISORDER AND ENTROPY

Consider these three spontaneous processes. So as to why they are spontaneous is not explained conclusively bythe theories and laws studied till now in this chapter :

Illustration (1): We have already studied above that when ice melts above its melting point, the process isspontaneous even though it is endothermic. So, exothermicity alone cannot explain thespontaneity of a reaction as discussed.

Illustration (2): When ( ),KCl s a crystal salt dissolves in water, the heat of solution, solution 0.H∆ > Yet wefind that the dissolution is spontaneous. Why ?

Illustration (3): Consider the following figure:

( )b

( )( )

( ),( ) ( )

a A sample of gas in which all molecules of one gas are in one bulb and all moleculesof the other gas are in the other bulb. b A sample of gas that contains the same number

of each kind of molecule as in a but with the two kinds randomly mixed in the two bulbs. Sample b has greater disorder higher entropy , and is thus more probable.

spontaneous

non-spontaneous( )a

fig. 13

Here in fig - 13 (a), consider the system to be isolated. Both the bulbs contain different kind of molecules. As thestop cock is opened the two gases will spontaneously merge into each other and the system will become moredisordered. Here as the system is isolated, net heat transfer, 0.q = Besides, as no work is done against anyexternal pressure, 0.w = So, by first law of thermodynamics, 0.U q w∆ = + = Now imagine the reverse processof this process i.e. the molecules in the fig-13 (b). acquiring the configuration of fig - 13(a) on their own accord!Seems unlikely; isn’t it ? But even for this process according to the first law of thermodynamics, 0.U∆ = So, wesee that the first law of thermodynamics is insufficient in predicting the spontaneity of a process. Clearly somefactor other than heat or work is important in making the expansion of the gas spontaneous.

The “missing factor” in the explanation of the spontaneity of the above processes is entropy. Let us study it in detailand see if it can help explain the spontaneity of the above mentioned processes. We will focus on developing anintuitive idea of entropy before we quantify it.

ENTROPY

Entropy, S is a thermodynamic quantity that is a measure of the randomness or disorder of a system. It is astate function. The greater the disorder of a system, the higher is its entropy. For any substance, the particles aremore highly ordered in the solid state than in the liquid state. Thus, the entropy of any substance increases as thesubstance goes from solid to liquid to gas (see figure:)



Increasing disorderIncreasing entropy, S

As a sample changes from solid to liquid to gas,its particles become increasingly less ordered (more disordered), so its entropy increases.

Solid liquid

Gas

fig. 14

Now, let us go back to the three motivating examples at the beginning of this section and make some observations:

Illustration (1): We see that when ice melts, the solid state changes into liquid state. Hence, the entropy of thesystem increases.

Illustration(2): When ( )KCl s dissolves, its highly ordered crystal structure is broken and it turns into randomly

distributed ( )K aq+ ions and ( )Cl aq− ions i.e. the entropy of the system increases.

Illustration(3): When the gaseous molecules spread over a wider volume, the randomness/disorder of thesystem increases. Here, also, the entropy of the system increases.

Do these observations suggest that the increase in entropy of a system might have something to do with explanationabout spontaneity of the above processes?

We will answer this question after the studying the second law of thermodynamics :

Second law of thermodynamics states that in spontaneous changes the universe tends toward a state of greater disorder.

This can be restated equivalently as :

The total entropy of a system and its surroundings always increases for a spontaneous process

The second law of thermodynamics is based on our experiences. Some examples illustrate this law in themacroscopic world. When a mirror is dropped, it can shatter. When a drop of colour is added to a glass of water,it diffuses until a homogeneously coloured solution results. These are examples of spontaneous processes. And wesee that the entropy ( or disorder) of the universe is increasing in each of these processes.



The reverse of any spontaneous change is non-spontaneous, because if it did occur, the universe would tendtoward a state of greater order. This is contrary to our experience We would be very surprised if we droppedsome pieces of silvered glass on the floor and a mirror spontaneously assembled.

Now, in order to further study the second law, we must give it a mathematical form. Mathematically the secondlaw can be written as :

( )universe system surroundings 0 for spontaneous processS S S∆ = ∆ + ∆ >

Compare this with the two boxed statements of second law above. Convince yourself that they are all equivalent.

The most important observation about the second law interpreted in the mathematical form above is that if theentropy of a system increases during a process, the spontaneity of the process is only favoured but not required.This is because if the entropy of the system increases it tends to make the universe 0S∆ > but does not quite

guarantee it as the sign of universeS∆ is dependent upon surroundingsS∆ as well. (We will later derive a criterion forspontaneity from the second law itself using which we shall be able to predict the spontaneity of a process using theproperties of the system alone.) So, the second law of thermodynamics says that the entropy of the universe (notthe system) increases during a spontaneous process.

Having stated and studied the second law let us see if we can now account for the spontaneity of the threeprocesses we discussed earlier in three illustrations.

Illustration (1): When ice melts above its melting point, solid phase changes into liquid phase and hence theentropy of the system increases and

systemS∆ is > 0. But the system absorbs heat from

surroundings in the process (as the process in endothermic.) This must have decreased themotion of the molecules of the surroundings. And hence surroundings 0.S∆ < But for the processto be spontaneous according to the second law of thermodynamics, the sum of these twomust be greater than zero i.e. sysS∆ must be greater in magnitude than surroundings .S∆Experimentally, this is found to be true. And that explains the spontaneity of this process.

Question: Account for the spontaneous conversion of pure liquid water into ice at temperature below0°C on the same lines as in illustration (1).

Illustration (2): When ( )KCl s dissolves into water, it dissociates into K + and Cl− ions as follows:

2( ) ( ) ( ).H OKCl s K aq Cl aq+ − → +

solution 0H∆ > for this process i.e. heat is absorbed from the surroundings and as discussed in illustration (1)above, the entropy of surroundings must have decreased because of this i.e. surr 0.S∆ < But we see that the highly

ordered crystal lattice of KCl is converted into the randomly arranged hydrated K + and Cl − ions i.e. entropy ofsystem has increased. So, sys 0.S∆ > I leave it upon you to argue and account for so as to why this process isspontaneous.



An anomaly : In illustration (2) above, although the entropy of the system due to dissolution of KCl increases, wemust note that the water molecules which were free to more anywhere have been localised due to hydration of K +

and Cl − ions. So, the entropy of the system due to localisation of water molecules must have decreased. Yet weclaim that sys 0.S∆ > This is because the magnitude of entropy change due to dissolution of ions is much greaterthan that of the entropy change due to localisation of the water molecules in this particular case. However, forsmall, highly charged ions such as Al3+ and Fe3+, the decrease in entropy due to hydration can outweigh theincrease in entropy due to mixing and dissociation so that the entropy change for the overall process can actuallybe negative.

Illustration (3): This is the simplest of the cases. As the system is isolated, there is no change in entropy of thesurroundings. Therefore, surr 0.S∆ =

universe systemS S∴ ∆ = ∆

in this case. When we open the stopcock between the two bulbs, we expect the gases to mixspontaneously, with an increase in disorder of the system i.e. systemS∆ is positive. Therefore,

universe system 0.S S∆ = ∆ >

So, according to second law this process should be spontaneous which is in perfect conformity with what observein real life !

Please attempt the following problems in order to crystallize the concepts learned so far. (see the solutions onlyafter having attempted the problems yourself.)

• Which of the following are spontaneous processes?

(a) A cube of sugar dissolves in a cup of hot tea.

(b) A rusty iron piece turns shiny.

(c) Butane from a lighter burns in air.

(d) A clock pendulum, initially stopped, begins swinging

(e) Hydrogen and oxygen gases bubble out from a glass of pure water.

Solution: (a) Spontaneous. Sugar dissolves spontaneously in hot water.

(b) Nonspontaneous. Rust does not spontaneously change to iron; rather iron spontaneously rustsin air.

(c) Spontaneous. The burning of butane in air is a spontaneous reaction.

(d) Nonspontaneous. A pendulum once stopped will not spontaneously begin moving again.

(e) Nonspontaneous. Water will not spontaneously decompose into its elements.



• Predict the sign of the entropy change for each of the following processes.

(a) A drop of food colouring diffuses throughout a glass of water.

(b) A lake freezes over in the winter.

(c) Rainwater on the pavement evaporates.

Solution: (a) Entropy increases; S∆ is positive; there is an increase of disorder when the food colouring dispersesthroughout the water.

(b) Entropy decreases; S∆ is negative; as a liquid changes to solid, there is an increase of order anda decrease of entropy.

(c) Entropy increases; S∆ is positive; as a liquid changes to vapour, there is a decrease of order andthe entropy increases.

• For each of the following statements, indicate whether it is true or false.

(a) A spontaneous reaction always releases heat.

(b) A spontaneous reaction is always a fast reaction.

(c) The entropy of a system always increases for a spontaneous change.

(d) The entropy of a system and its surroundings always increases for a spontaneous change.

(e) The energy of a system always increases for a spontaneous change.

Solution: (a) False. The enthalpy change (heat of reaction) has no direct relation to spontaneity.

(b) False. The rate of a reaction has nothing to do with the spontaneity (thermodynamics) of a reaction.

(c) False. The entropy may increase or decrease during a spontaneous reaction.

(d) True. The entropy of the system plus surroundings always increases during a spontaneous change.

(e) False. The energy may increase or decrease during a spontaneous reaction.

• In illustration 1, we saw that the melting of ice is spontaneous for temperatures above its melting point. Weattributed this to the fact that univS∆ is > 0 for this process. Can you account for the entropy changes thatoccurs at 0°C for this process ?

Solution: At the melting point, surroundingsS∆ is equal in magnitude and opposite in sign to system.S∆ Therefore,

universeS∆ is zero both for melting and freezing. Here, the system is at equilibrium.

We have seen that second law of thermodynamics is sufficient in predicting the spontaneity of a process. Howeverwe have to measure the change in the entropy of the universe (system as well as surroundings) in order to use thesecond law to predict spontaneity of a process. This is generally difficult to do. However, as we will see, S∆ systemis easy to calculate.



So is it possible to have a criterion for spontaneity which is described only in terms of the properties of a system?The answer is YES! But before we take it up we must study so as to how entropy and entropy changes arequantified.

The third law of thermodynamics establishes the zero of the entropy scale.

The third law of thermodynamics states that the entropy of a pure, perfect, crystalline substance (perfectly ordered) is zero at absolute zero (0 Kelvin)

As the temperature of a substance increases, the constituent particles of the substance vibrate more vigorously.

So, the entropy increases.(see the figure:)

heating

fig. 15

+ – + – + – + – + –+ – + – + – + – + –+ – + – + – + – + –+ – + – + – + – + –+ – + – + – + – + –

(a)

+ – + – + – + – + –+ – + – + – + – + –+ – + – + – + – + –+ – + – + – + – + –+ – + – + – + – + –

(b)A simplified representation of a side view of a "perfect" crystal of a polar substance at 0 K. Note the perfect alignment of the dipoles in all molecules in a perfect crystal. This causes its entropy to be zero at 0 K. However, there are no perfect crystals, because even the purest substances that scientists have prepared are contaminated by traces of impurities that occupy a few of the positions in the crystal structure.

A simplified representation of the same "perfect" crystal at a temperature above 0 K. Vibrations of the individual molecules within the crystal cause some dipoles to be oriented in directions other than those in a perfect arrangement. The entropy of such a crystalline solid is greater than zero, because there is disorder in the crystal.

As we go on supplying heat to the substance, either the temperature increases (which means still higher entropy orphase transitions (melting, boiling or sublimation) take place (which again means higher entropy).

The entropy of a substance at any condition is its absolute entropy, also sometimes called as standard molarentropy or simply, standard entropy. The unit of entropy as we will show later is J/K.

Now, consider the absolute entropies at 298 kelvin of the substances listed in following table.

Substance S° (J/K.mole) H2O (l) H2O (g) Br2 (l) Br2 (g)

C (diamond) C (graphite)

69.91 188.7 152.33 245.4 2.38 5.74

At 298 K, any substance is more disordered than if it were in a perfect crystalline state at absolute zero. So, wesee that 0

298S values of all the above substances are positive. In fact 0298S of any substance is found to be positive.



Important Note:0298S of an element unlike its 0

fH∆ , is not equal to zero. For example, in the table above, 0298S of C(graphite) is

not zero. This is because, the reference state for absolute entropy is specified by the third law of thermodynamics.

It is different from the reference state for 0fH∆ .

Now, when a reaction occurs, the entropy of the products and reactants is generally different. This entropy change

(also called standard entropy change S∆ ° if reaction is carried out under standard conditions) of a reaction can be

determined from the absolute entropies of reactants and products, as follows:

0 0 0reaction products reactantsS n S m S∆ = Σ ⋅ − Σ ⋅

(Here, n and m denote the coefficients of the reactants and products in the balanced equation.)

So, we have seen how we can calculate the accompanying entropy changes in a system as it undergoes a process.But when we defined entropy we said that it is a measure of the randomness or disorder of a system. We cannotquantify entropy changes based on this definition. But can we at least predict if entropy has increased or decreasedusing this definition? Yes, we can as we illustrate it in various cases below:

(a) Temperature changes: For example, warming a gas from 100°C to 150°C. As any sample is warmed,the molecules undergo more (random) motion. Hence, we can predict that entropyincreases (i.e. 0sysS∆ > ) as temperature increases. Similarly, as we raise thetemperature of a solid, the particles vibrate more vigorously about their positionsin the crystal, so that at any instant there is a large average displacement from theirmean positions. This results in an increase in entropy.

(b) Volume changes: When the volume of sample of gas increases, the molecules can occupy morepositions and hence, are more randomly arranged than when they are closertogether in a smaller volume. Hence, an expansion is accompanied by an increasein entropy. Conversely, as a sample is compressed, the molecules are more restrictedin their locations and a situation of greater order (lower entropy) results.

(c) Changes in no. of moles: Processes that result in an increase in the number of moles of gaseous substances (of gaseous substances) have 0sysS∆ > . Consider the following reaction:

2 4 2 2 2( ) 2 ( ) ( ) 4 ( )N H l H O l N g H O g+ → +

We see that there are no gaseous reactants, but the products comprise of 5 molesof gases. We can predict that the entropy change is positive. In fact, experimentalresults show that S∆ ° for the above reaction is +606 J/mol. K which is in perfectagreement with our prediction. Conversely the process 2 22 ( ) ( )H g O g+

22 ( )H O g→ has a negative S∆ ° value. Can you see so as to why?



(d) Mixing of substances: Mixing of substances even without chemical reaction may lead to a state of greaterentropy. Situations in which the molecules are more ‘‘mixed up” are moredisordered, and hence are at higher entropy. We have already encountered asimple example of this type in illustration (3) on page - 57. We pointed out thatthe mixed gases (fig.-13(b), page - 57) were more disordered than the separatedgases (fig.-13(a), page - 57) and that the former was a situation of higher entropy.We see, then, that mixing of gases by diffusion is a process for which 0sysS∆ > .We know from experience that it is always spontaneous.

Another example we could consider is that of dissolution of solids into liquids.The increase in disorder (entropy increase) that accompanies this dissolution (ormixing) often provides the driving force for solubility of one substance in another.For example, when one mole of NaCl dissolves in water, i.e.

( ) ( )NaCl s NaCl aq→ , the entropy increases from 72 J/mol.K to 116 J/mol.K i.e. 43.1 / .S J mol K∆ ° = + .

(e) Phase change: When melting occurs, the molecules or ions are taken from their ordered crytallinearrangement to a more disordered one in which they are able to move past oneanother in the liquid. Thus a melting process is always accompanied by an entropyincrease. For further study of how entropy is affected by phase changes refer tothe figure below:

Solid Liquid Gas

Boiling

1 2 3

Solid-state phase change

Melting

Temperature (K)0

0

Entro

py,S

fig. 16

Solids can exist in different forms (Example: Carbon exists as diamond and graphitein the solid state) and generally one of them has more entropy than the other. Asa solid is heated, it might change its form of existence to another form. This solid-state phase change is depicted in the part 1 of the figure above. For example, Tinis a substance that undergoes such a solid-state phase change. Below 13°C, asolid form called gray tin is stable. When heated, gray tin converts to anothersolid form called white tin, which has a higher entropy.The figure further depicts how entropy varies with temperature for a sample. Thechange in entropy with temperature (we have already studied this in part (a) above.)is gradual upto the solid-state phase change. There is a sharp increase in entropyat that temperature. The entropy then continues to increase (gradually again) withincreasing temperature up to the melting point of the solid.



When the solid melts, the atoms or molecules are free to more about the entirevolume of the substance. This added freedom of motion for the individual moleculesgreatly increases the entropy of the substance. At the melting point, we thereforesee a large increase in entropy. After all the solid has melted, the temperatureagain increases, and with it the entropy.

At the boiling point of the liquid another big increase in entropy occurs. Thisincreased entropy results largely from the increased volume in which the moleculesmay be found. The increase in volume means an increase in randomness. As thegas is heated thereafter, the entropy increases gradually again as more energy isstored in the gas molecules.

So, what we have essentially seen is that though the increase in temperature increasesthe entropy of a substance gradually in a particular phase, the phase change bringsabout a “jump” in the entropy of the substance.

And quite evidently for the reverse processes of freezing, condensation, anddeposition, entropy decreases because order increases.

_______________________________________________________________________________________________________________

SECOND LAW OF THERMODYNAMICS; REVISITED

Recall that the second law of thermodynamics states that:

0universe system surroundingsS S S∆ = ∆ + ∆ > (for spontaneous process).

Our aim now is to state the above in terms of the properties of the system alone. systemS∆ is already in terms of aproperty (entropy) of the system. So, we only have to try and express surroundingsS∆ in terms of the properties ofthe system.

∆ surroundingsS in terms of the properties of system

When a system undergoes a process, either heat is released or absorbed. If heat is released (exothermic), the heattransferred to the surroundings enhances the motion of the molecules in the surroundings. This leads to an increasein the entropy of the surroundings. Conversely, in an endothermic process, the system absorbs heat from thesurroundings and so, decreases the entropy of the surroundings because molecular motion decreases. So, we canconclude that,

surroundings systemS Hα∆ − ∆ ...(1)

for a constant pressure process. The minus sign is used because if the process is endothermic, systemH∆ is positiveand surroundingsS∆ is a negative quantity, indicating a decrease in entropy. On the other hand, for an exothermicprocess, systemH∆ is negative and the negative sign in front of it makes sure that the entropy of the surroundingsincreases.



The change in entropy of a system for a certain amount of heat absorbed is also dependent upon the temperatureat which the process occurs. In fact this relation between the sysS∆ and temperature is an inversely proportionalone. Let us see so as to why by an analogy: Suppose a classmate of yours has cough (may god bless him!) and hecoughs in a crowdy and noisy market place. Can he disturb many people by the noise he creates in the alreadynoisy market place? Not much; isn’t it?

And now consider him visiting your school library where there is a pin drop silence. If he coughs here too, howmany people will he disturb? Nearly all!

Using this analogy we can understand how the change in entropy and the temperature at which the processoccurs are inversely proportional to each other. If the temperature of the surroundings is high, the molecules arealready quite energetic. Therefore, the absorption of heat from an exothermic process in the system will haverelatively little impact on molecular motion and the resulting increase in entropy of the surroundings will be small.This is analogous to the “coughing in market place” situation discussed above.

Now, if the temperature of the surroundings is low, then the addition of the same amount of heat will cause a muchsharper increase in molecular motion of the surroundings and hence a larger increase in entropy. This is analogousto the “coughing in library” situation discussed above.

So, we may conclude:

1surroundingsS

Tα∆ ...(2)

Using equations (1) and (2) above we can write the complete relationship between ,surr sysS H∆ ∆ and absolute

temperature, T as

syssurr

HS

T−∆

∆ = . ...(3)

So, we have achieved what we had set out for. i.e. expressing surroundingsS∆ in terms of the properties of thesystem. Let us see next how we can use this information to arrive at a criterion for spontaneity explained in termsof the properties of the system alone.

We know that for a spontaneous process

0univ sys surrS S S∆ = ∆ + ∆ > ...(4)

We substitute (3) in (4) to obtain:

0sysuniv sys

HS S

T∆

∆ = ∆ − > ...(5)



Multiplying both sides of (3) by T (which is positive as it is in kelvins) we get,

0univ sys sysT S T S H∆ = ∆ − ∆ >

0univ sys sysT S H T S⇒ − ∆ = ∆ − ∆ < ...(6)

We see in (6) that if we know the changes in the properties of the system (occurring at constant pressure and

temperature) alone (namely, andsys sysH S∆ ∆ ) and find ( )sys sysH T S∆ − ∆ to be less than zero, then we canconclude that ( )univT S− ∆ is less than zero as well. As T is positive, univS∆ must be positive too for ( )univT S− ∆ tobe negative. And if univS∆ is positive, the process is spontaneous! So, we see that we have concluded that aprocess is spontaneous using the properties of the system alone. We can thus make out that eqn. (6) above mustbe a very important relationship and to express what it says more directly, we introduce another thermodynamicfunction called Gibbs free energy!

GIBBS FREE ENERGYGibbs free energy, G, is defined as:

G H TS= −

where H, T, S are enthalpy, absolute temperature and entropy of a system. Gibbs free energy (or simply freeenergy) is a state function.

We can also see that G has units of energy (as both H and TS are in energy units).Now, let us see how this new state function can express what is expressed by eqn. (6) above, more conveniently.The change in free energy ( )G∆ of a system for a constant temperature process is

G H T S∆ = ∆ − ∆ ...(7)

If we compare the above equation with equation (6) we see that univG T S∆ = − ∆ . Now, according to the secondlaw of thermodynamics, a reaction is spontaneous if univS∆ is positive, non spontaneous if univS∆ is negative, andat equilibrium if univS∆ is zero. Now, univG T S∆ = − ∆ and since T is positive (absolute scale) always, G∆ and

univS∆ have opposite signs. We can sum this up as:

Process S ign of ∆S univ � T∆S univ = ∆G Spontaneous > 0 < 0

N on-spontaneous < 0 > 0 Equilibrium = 0 = 0

( this is for a process occurring at constant tem perature and pressure)

So, we may conclude from the above that, in any spontaneous process at constant temperature and pressure, thefree energy of the system always decreases.At this point, when we have achieved our goal of having a criterion for spontaneity in terms of the properties of thesystem alone and defined a new thermodynamic function in the process, let us pause and study the newthermodynamic function, G, in a little depth.



Important points

(1) Consider the combustion of petrol to move vehicles. This combustion reaction is exothermic; its ∆H mustbe negative. This reaction is also spontaneous and hence its ∆G also must be negative. It can be shownthat the maximum useful work (moving the vehicle in this case) for the above reaction is not given by ∆Hbut by ∆G i.e.

wmaximum = ∆G.

The term free energy comes from this result. ∆G is the portion of the energy change of a spontaneousreaction that is free to do useful work. The remainder of the energy (i.e ∆H – ∆G) enters the environmentas heat. Hence, the free energy change is the maximum energy available, or free, to do useful work.(However, please note that although in principle the complete ∆G is available to do useful work, inpractice, it is impossible to convert all of this available energy into useful work.). So, we can generalise thisand say that, the spontaneous reactions can be used to obtain useful work and the maximum obtainableuseful work (theoretically) is given by ∆G accompanying that reaction.

(2) When ∆G = 0, there is no net transfer of free energy; both the forward and reverse processes are equallyfavourable. Thus, G∆ = 0 describes a system at equilibrium. (For those of you who have not studiedequilibrium yet, it is sufficient to understand for the time being that a reaction is said to be at equilibriumwhen besides the reactants combining to form the products, the process of products combining to formthe reactants also occurs simultaneously and the rate of both the reactions is same. Therefore, there is nonet change when a system is at equilibrium.)

(3) For processes that are not spontaneous ( ),G∆ > 0 the free energy change is a measure of the minimumamount of work that must be done to cause the process to occur. (However, in actual cases we alwaysneed to do more than this theoretical minimum amount because of the insufficiencies in the way the changesoccur.)

Before we move further, please attempt the following problems:

• Here is a simple experiment. Take a rubber band and stretch it. (Is this a spontaneous process? How doesthe Gibbs free energy change?) Place the rubber band against your lips: note how warm the rubber bandhas become. (How does the enthalpy change?) According to polymer chemists, the rubber band consistsof long, coiled molecules. On stretching the rubber band, these long molecules uncoil and align themselvesin a more ordered state. Show how the experiment given here is in accord with this molecular view of therubber band.

Solution: The process is not spontaneous; you have to stretch the rubber band (the opposite process, a stretchedrubber band snapping to its normal shorter shape, is spontaneous). Thus, G∆ is positive. The factthat the stretched rubber band feels warm means that H∆ is negative (exothermic). Note that

; so ( ) /G H T S S G H T∆ = ∆ − ∆ ∆ = − ∆ − ∆ . This implies that S∆ is negative which is consistentwith an increase in order.



• Hypothetical elements A(g) and B(g) are introduced into a container and allowed to react according to thereaction 2( ) 2 ( ) ( )A g B g AB g+ → . The container depicts the reaction mixture after equilibrium has beenattained.

= A

= B

fig . 17

a. Is the value of S∆ for the reaction positive, negative, or zero?

b. Is the value of H∆ for the reaction positive, negative, or zero?

c. Prior to equilibrium, is the value of G∆ for the reaction positive, negative, or zero?

d. At equilibrium, is the value of G∆ for the reaction positive, negative, or zero?

Solution: a. Since this is a gas phase reaction, calculate the change in the number of moles of gas molecules,1 3 2n∆ = − = − . This corresponds to a decrease in entropy, so S∆ is negative for this reaction.

b. Since this is an equilibrium state, 0, so 0G H T S∆ = ∆ − ∆ = also. This gives H T S∆ = ∆ and,since S∆ is positive, so is H∆ .

c. The reaction spontaneously reestablishes equilibrium, so G∆ is negative prior to equilibrium.

d. At equilibrium, 0G∆ = .

Let us now consider the equation G H T S∆ = ∆ − ∆ again and analyse its structure. We see that the more exothermica reaction is, more are the chances of G∆ to be negative and consequently the reaction to be spontaneous.However no matter what is the value of H∆ , a larger positive value of ( )T S− ∆ will make the reactionnon-spontaneous. So a large magnitude of negative H∆ can only favour the spontaneity of a reaction and cannotguarantee it (as we had pointed out at the beginning of section - 3). We can similarly observe that a positive S∆can only favour the spontaneity of a reaction but cannot guarantee it. However, besides H∆ and S∆ we see thetemperature (T) term also in the equation G H T S∆ = ∆ − ∆ . So, is the spontaneity of a reaction also dependentupon the temperature at which it occurs? Quite obviously, the answer to that is yes! We next take up so as to how:

The temperature T acts as weighing factor that determines the relative importance of H∆ and S∆ contributionsto G∆ in the free energy equation G H T S∆ = ∆ − ∆ . If H∆ and S∆ are either both negative or both positive, thesign of G∆ (and therefore the spontaneity of the reaction) depends on the temperature. If H∆ and S∆ are bothnegative, the reaction will be spontaneous only if the absolute value of H∆ is larger than the absolute valueof T S∆ .



This is most likely (and not guaranteed) at low temperatures, where the weighing factor T in T S∆ is small andhence, tends to make the whole term small in magnitude. If H∆ and S∆ are both positive, the reaction will bespontaneous only if T S∆ is larger than H∆ which is most likely (and again, not guaranteed) at high temperatures.The other two cases are that of “ H∆ being negative and S∆ being positive” and “ H∆ being positive and S∆being negative”. Convince yourself that the former case will be spontaneous at all temperatures and the latter casewill be non-spontaneous at all temperatures. This discussion is summarised in the following table:

Case

Sign of change

∆H ∆S ∆G

Possibility (impossibility) of spontaneous proceeding of reaction

I – + – Possible at any temperature

II – – + Possible at sufficiently low temperature

III + + + Possible at sufficiently high temperature

IV + – + Impossible at any temperature

Directions of reactions with different signs of H∆ and S∆

So, we see that in case - II above, the reaction will be spontaneous at sufficiently low temperatures. But exactlyhow low? Let us figure it out. Suppose we are given 10H kJ∆ = − and 200 /S J K∆ = − . Then we can evaluatevarious G∆ for the given H∆ and S∆ at various temperatures (assume that H∆ and S∆ do not vary appreciablywith temperature). We tabulate the various G∆ for certain temperatures in the following table:

∆H ∆S T ∆G = ∆H � T∆S Sign of ∆G

– 10000 J – 200 J/K 100 K ∆G = – 10000 – (100 × – 200) = + 10000 J. +

-do- -do- 80 K ∆G = – 10000 – (80 × – 200) = + 6000 J +

-do- -do- 60 K ∆G = – 10000 – (60 × – 200) = + 2000 J +

-do- -do- 40 K ∆G = – 10000 – (40 × – 200) = – 2000 J

–

-do- -do- 20 K ∆G = – 10000 – (20 × – 200) = – 6000 J –



We observe that as we lower the temperature the value of G∆ (although positive in first three cases) movestowards turning negative. And once it is negative, it turns more negative as we lower the temperature. The questionnow is where exactly (at what temperature) does this “transition” from positive G∆ to negative G∆ come about?It must be the temperature at which 0G∆ = . Let us try to find it out for the above considered case:

G H T S∆ = ∆ − ∆

Substituting appropriate values we get:

0 = – 10,000 J – T. (– 200 J/K).

0 10,000 200J T⇒ = − +

200 10,000T J⇒ =

50T K⇒ =

i.e. above 50 K the reaction would be non-spontaneous and below 50K the reaction would be spontaneous. Wesee that the set of data tabulated above conforms to this result.

Question: If 10,000H J∆ = + and 200S∆ = + J/K, then find out the temperature where 0G∆ = . What is thetemperature range of spontanteity for this reaction?

We will discuss the exact reasons for the above discussion in the unit on chemical equilibrium. However, for thetime being we can summarise the above discussion as:

We can estimate the temperature range over which a chemical reaction is spontaneous by evaluating H∆ and S∆for the reaction. The temperature at which 0G∆ = (for the reaction) is the temperature limit of spontaneity. Thesign of S∆ tells us whether the reaction is spontaneous below or above this limit. This is figuratively explained in thefollowing figure.:



∆H = +, S = - (

Case-IV

)

∆

∆H = –, S = -

∆

(Case-II)

∆H = –, S = + (Case-I)

∆

∆H = +, S = +∆(Case-III)

T

–

( 0 )∆H <

( 0)∆H >

SPONTANEOUS

NOT SPONTANEOUS

+

∆G

Fig. 18 : A graphical representation of the dependence of G∆ and spontaneity

on temperature for each of the four cases of reactions.

Please note that for case I and IV if we equated G∆ to zero we would get a negative value of temperature(on kelvin scale) which is impossible! This indicates that reaction falling under case - IV and case - I arenon-spontaneous and spontaneous respectively at all temperatures.

(PS: For discussing further topics of this section, we need to study chemical equilibrium. So, the remaining part ofthis section (section -3) will be taken up in the unit on equilibrium as we had pointed out in the introduction tothis chapter.)

At a certain temperature T, the endothermic reaction A → B proceeds virtually to the end. Determine:(a) the sign of ∆S of the reaction;(b) the sign of ∆G for the reaction B → A at the temperature T;(c) the possibility of the reaction B → A proceeding at low temperatures.

Critical thinking

We have been given an endothermic reaction. Saying that it proceeds virtually to the end means that thereaction is spontaneous. This information is enough to answer the question. Can you see how?

Example – 19



Solution: (a) Spontaneous proceeding of the reaction A B→ indicates that 0.G∆ < Since 0,H∆ > it followsfrom the equation G H T S∆ = ∆ − ∆ that 0.S∆ > For the reverse reaction ,B A→ we have

0.S∆ <

(b) For the reaction ,A B→ we have 0.G∆ < Hence, for the reverse reaction at the same temperature,0.G∆ >

(c) The reaction ,B A→ which is the reverse reaction of ,A B→ is exothermic ( 0).H∆ < Themagnitude of the term T S∆ is small for low temperatures, so that the sign of G∆ is determined bythat of .H∆ Hence at sufficiently low temperatures, the reaction B A→ can proceed.

Estimate the temperature range for which the following reaction is spontaneous.

2 2 4( ) 2 ( ) 2 ( ) ( ) 2 ( )SiO s C graphite Cl g SiCl g CO g+ + → +

Given :

4( , ) 657 /fH SiCl g kJ mol°∆ = − 4( , ) 330.6 /S SiCl g J mol K° = ⋅

( , ) 110.5 /fH CO g kJ mol°∆ = − ( , ) 197.6 /S CO g J mol K° = ⋅

2( , ) 910.9 /fH SiO s kJ mol°∆ = − 2( , ) 41.84 /S SiO s J mol K° = ⋅

( , graphite) 5.740 /S C J mol K° = ⋅

2( , ) 223.0 /S Cl g J mol K° = ⋅

Critical thinking

When we proceed we find that reactionS °∆ is favourable to spontaneity, whereas reactionH °∆ is unfavourable.

Thus, we know that the reaction becomes spontaneous above some temperature. We can set G∆ ° equal tozero in the equation G H T S∆ ° = ∆ ° − ∆ ° and solve for the temperature at which the system is at equilibrium.This will represent the temperature above which the reaction would be spontaneous.

Solution:

reaction 4 2 2( , ) 2 ( , ) ( , ) 2 ( , graphite) 2 ( , )f f f f fH H SiCl g H CO g H SiO S H C H Cl g° ° ° ° ° ° ∆ = ∆ + ⋅∆ − ∆ + ∆ + ∆

[ ] [ ]( 657) 2 ( 110.5) ( 910.9) 2(0) 2(0)= − + ⋅ − − − + +

32.9 /kJ mol= +

Example – 20



4 2 2reaction ( ) ( ) ( ) ( ) ( )2 2 2SiCl g CO g SiO s C gr Cl gS S S S S S° ° ° ° ° ° ∆ = + − + +

[ ] [ ](330.6) 2(197.6) (41.84) 2(5.740) 2(223.0)= + − + +

226.5 / .J mol K=

= 0.2265 kJ / mol.K

When 0,G∆ ° = neither the forward reaction nor the reverse reaction is spontaneous. Let’s find thetemperature at which 0G∆ ° = and the system is at equilibrium.

0G H T S∆ ° = ∆ ° − ∆ ° =

H T S⇒ ∆ ° = ∆ °

HTS

∆ °⇒ =∆ °

32.9 /0.2265 / .

kJ molTkJ mol K

+⇒ =+

145 .T K⇒ =

At temperature above 145 K, the T S∆ ° would be greater (i.e. T S− ∆ ° would be more negative)than the H∆ ° term, which would make G∆ ° negative. So, the reaction would be spontaneous above145 K.

Show that the reaction 2 2( ) (1/ 2) ( ) ( )CO g O g CO g+ → at 300 K is spontaneous and exothermic, when thestandard entropy change is –0.094 kJ mol–1 K–1. The standard Gibbs free energies of formation for CO2 and COare –394.4 and –137.2 kJ mol–1, respectively.

Critical thinking

To show spontaneity, we need to show G∆ ° to be negative and to show exothermicity, we need to showH∆ ° to be negative. Use the following result to calculate the ºG∆ of reaction (it is analogous to calculating

enthalpy of a reaction using Hess’s law):

reaction products reactantsf fG m G n G° ° °∆ = ⋅∆ − ⋅∆∑ ∑(this equation will be discussed in the unit on equilibrium).

But, we don’t have enough data to calculate H∆ ° by using ( ).H U PV∆ = ∆ + ∆ How will wecalculate ?H∆ °

Example – 21



Solution: The given reaction is 2 21( ) ( ) ( )2CO g O g CO g+ →

Applying the equation given in the box above, we can calculate its reactionG°∆ as follows:

reaction 2 21( , ) ( , ) ( , )2f f fG G CO g G O g G CO g° ° ° °∆ = ∆ − ∆ − ∆

[ ]( 394.4) (0) ( 137.2) /kJ mol= − − − −

257.2 /kJ mol= −

reaction, 0,G°∆ <∵ the reaction is spontaneous.

To calculate the ,H∆ let us use the equation .G H T S∆ = ∆ − ∆

G H T S∆ ° = ∆ ° − ∆ °

H G T S⇒ ∆ ° = ∆ °+ ∆ °

Substituting the given values of T and S∆ ° and obtained value of ,G∆ ° we get,

( 257.2 300( 0.094)) /H kJ mol∆ ° = − + −

285.4 /kJ mol= −

So, 0H∆ ° < as well, and hence, the reaction is exothermic.

Use the thermodynamic data given to estimate the normal boiling point of bromine, Br2 . (Assume andH S∆ ∆ donot change with temperature)

Given : ( )02 , 30.91 /fH Br g kJ mol∆ = ( )0

2 , 245.4 / .S Br g J mol K=

( )02 , 0 /fH Br l kJ mol∆ = ( )0

2 , 152.2 / .S Br l J mol K=

Critical thinking

The process we must consider is

( ) ( )2 2Br l Br g→

By definition, the normal boiling point of a liquid is the temperature at which pure liquid and pure gascoexist in equilibrium at 1 atm. And we have studied that at equilibrium, 0G∆ = (We will cover thisconcept in much more detail in the unit on equilibrium). We can calculate andH S∆ ∆ for the aboveprocess and substitute in G H T S∆ = ∆ − ∆ to get the required boiling point.

Example – 22



Solution: ( ) ( )0 0 0reaction 2 2, ,f fH H Br g H Br l∆ = ∆ − ∆

= (30.91 – 0) kJ / mol = 30.91 kJ / mol

( ) ( )0 0 0reaction 2 2, ,S S Br g S Br l∆ = −

( )245.4 152.2= −

= 93.2 J/mol.K = 0.0932 kJ / mol.K

We can now solve for the temperature at which the system is in equilibrium, i.e., the boiling pointof Br2.

reaction reaction reactionG H T S∆ = ∆ − ∆

reaction reaction0 H T S⇒ = ∆ − ∆

reaction

reaction

HTS

∆⇒ =∆

30.91 /

0.0932 / .kJ mol

kJ mol K=

= 332 K (or 59º C )This is the temperature at which the system is in equilibrium, i.e. the boiling point of Br2.



Q. 1 Indicate which of the following statements are correct(a) endothermic reactions cannot proceed spontaneously(b) endothermic reactions can proceed at sufficiently low temperatures(c) endothermic reactions can proceed at sufficiently high temperatures if the change in the entropy of the

reaction is positive.

Q. 2 Without performing calculations, indicate for which of the processes listed below the change in the entropyis positive:

(a) 2 2( ) ( ) ( ) ( )MgO s H g Mg s H O l+ → +

(b) 2( ) ( ) 2 ( )C graphite CO g CO g+ →

(c) 3 3( ) ( ) ( )CH COOH aq CH COO aq H aq− +→ +

(d) 2 2 24 ( ) ( ) 2 ( ) 2 ( )HCl g O g Cl g H O g+ → +

(e) 4 3 2 2( ) ( ) 2 ( )NH NO s N O g H O g→ +

Q. 3 In which of the following cases is a reaction possible at any temperature?

(a) 0, 0H S∆ ° < ∆ ° >

(b) 0, 0H S∆ ° < ∆ ° <

(c) 0, 0H S∆ ° > ∆ ° >

Q. 4 In which of the following cases is a reaction impossible at any temperature?

(a) 0, 0H S∆ ° > ∆ ° >

(b) 0, 0H S∆ ° > ∆ ° <

(c) 0, 0H S∆ ° < ∆ ° <

Q. 5 If 0H∆ ° < and 0,S∆ ° < in which of the following cases can a reaction proceed spontaneously ?

(a) H T S∆ ° > ∆ °

(b) H T S∆ ° < ∆ °

TRY YOURSELF - III



Q. 6 What is the sign of G∆ for the process of ice melting at 263 K ?(a) 0G∆ >(b) 0G∆ =(c) 0G∆ <

Q. 7 Taking into account that 2 ( )NO g is coloured, and 2 4N O is colourless, and proceeding from the sign of

the change in the entropy in the reaction 2 2 42 ( ) ( ),NO g N O g" predict how the colour will change in

the system 2 2 4NO N O− with elevation of the temperature.

(a) It will increase(b) It will fade out.

Q. 8 For the reaction at 298 K,

2A B C+ →

100H kcal∆ = and 0.050 / .S kcal K∆ = Assuming H∆ and S∆ to be constant over the temperaturerange, at what temperature will the reaction become spontaneous?

Q. 9 A reaction has a value of 40H kcal∆ = − at 400 K. Above 400 K, the reaction is spontaneous; belowthat temperature, it is not. Calculate G∆ and S∆ at 400 K.

Q. 10 For the reaction 22 ( ) ( ),Cl g Cl g→ what are the signs of H∆ and ?S∆



MISCELLANEOUS EXAMPLES

Using the data given below, calculate the bond enthalpy of C — C and C — H bonds.1 1(ethane) 1556.5 ; (propane) 2117.5c cH kJ mol H kJ mol− −∆ ° = − ∆ ° = −

1(graphite) ( ); 719.7C C g H kJ mol−→ ∆ =

Bond enthalpy of H — H = 435.1 kJ mol–1.1 1

2 2( , ) 284.5 ; ( , ) 393.3f fH H O l kJ mol H CO g kJ mol− −∆ ° = − ∆ ° = −

Solution: The plan

This is a slightly involved problem and requires planning before one solves it successfully. Lets try tosolve this problem in steps as that would help us develop our thinking process. This is our plan: Thereare 3 Critical thinking boxes for this problem. It is suggested that you study one critical thinking box ata time and then try to solve the problem. If you cannot then study the next box and try to attempt theproblem again!

Critical thinking Box 1Lets set our target first. We have to calculate the bond enthalpy of C — C and C — H bonds. Till now thepolyatomic molecules which were used to calculate the bond enthalpies contained only one type of bond.For example, CH4 can be used to obtain bond enthalpy for C — H bonds. But the molecules given in thisproblem are C2H6 and C3H8. Thus both have at least one C — C bond besides C — H bonds. So, theycannot be individually used to get bond enthalpy value of any of these bonds. However, enthalpy changeaccompanying dissociation of any of these molecules into free atoms will be a sum of bond enthalpies ofC — C and C — H bonds multiplied by appropriate constants.

Solution: (part - 1)

When C2H6(g) is dissociated into free atoms, one C — C bond and six C — H bonds are broken toform 2 carbon atoms and 6 hydrogen atoms as shown:

H — C — C — H

H H

H H

( )C H2 6

2 6( free atoms)2 ( ) 6 ( ); C HC g H g H → → + ∆ ... (1)

Example – 23



Similarly, the dissociation of C3H8(g) into free atoms will be as follows:

H — C — C — C — H

H H

H H

( )C H2 8

H

H

3 8( free atoms)3 ( ) 8 ( ); C HC g H g H → → + ∆ ...(2)

In this case 2 C — C bonds and 8 C — H bonds have been broken.

If we let x and y represent the bond enthalpy of C — C and C — H bonds respectively then H∆ ofequations (1) and (2) above can be written as:

2 6( free atoms)6 C Hx y H →+ = ∆ ...(3)

3 8( free atoms)2 8 C Hx y H →+ = ∆ ...(4)

(You are requested to try and solve the whole problem yourself now and only after the attempt see thenext critical thinking box).

Critical thinking Box 2

We have to now find H∆ for equations (1) and (2) above from the given set of data. We see we have beengiven enthalpies of combustion for ethane and propane. Using these we can calculate the enthalpy offormation of both ethane and propane. Although we don’t get the required H∆ of (1) and (2) from this, itseems like a reasonable line to move on. Can you see how then the fH∆ values so obtained can becombined with other thermodynamical data to get the required 'H s∆ ?

Solution : (part - 2)The combustion reaction for ethane can be written as:

2 6 2 2 27( ) ( ) 2 ( ) 3 ( ); 1556.5 /2 cC H g O g CO g H O l H kJ mole+ → + ∆ = −

From this we can write,

2 2 2 2 673 ( , ) 2. ( , ) ( , ) ( , )2C f f f fH H H O l H CO g H O g H C H g∆ = ∆ + ∆ − ⋅∆ − ∆

2 671556.5 3 284.5 2.( 393.3) (0) ( , )2 fH C H g⇒ − = × − + − − − ∆

2 6( , ) 83.6 /fH C H g kJ mol⇒ ∆ = −

We can write the thermochemical equation for this as:



2 2 62 (graphite) 3 ( ) ( ); 83.6 /fC H g C H g H kJ mol+ → ∆ = − ...(5)

Similarly, the combustion reaction for propane can be written as:

3 8 2 2 2( ) 5 ( ) 3 ( ) 4 ( )C H g O g CO g H O l+ → +

From this we can write,

2 2 2 3 84 ( , ) 3 ( , ) 5. ( , ) ( , )C f f f fH H H O l H CO g H O g H C H g∆ = ∆ + ⋅∆ − ∆ − ∆

3 82117.5 4 ( 284.5) 3 ( 393.3) 5(0) ( , )fH C H g⇒ − = × − + × − − − ∆

3 8( , ) 101 /fH C H g kJ mol⇒ ∆ = −

We can write the thermochemical equation for this as:

2 3 83 (graphite) 4 ( ) ( ); 101 /fC H g C H g H kJ mol+ → ∆ = − ...(6)

Critical thinking Box 3

If we reverse equations (5) and (6) we can obtain the H∆ for conversion of ethane and propane into itsconstituent elements. But the H∆ that we are seeking in equations (1) and (2) are for conversion of ethaneand propane into their constituent atoms (and not elements!). However, can we use equations (5) and (6)with the given thermodynamical data to obtain H∆ for (1) and (2) using Hess’s law?

Solution: (part - 3)It is given that,

(graphite) ( ); 719.7 /C C g H kJ mol→ ∆ = ...(7)

2 ( ) 2 ( ); 435.1 /H g H g H kJ mol→ ∆ = ...(8)

Using equations (5), (7) and (8) we may write,

2 C ( )graphite 3 H+ ( ) ( )( )

2 2 6 ; 83.6 /

2 2

g C H g H kJ mol

C g C

→ ∆ = −

→ ( )( )

graphite ; 2 719.7 /

6 3

H kJ mol

H g H

∆ = − ×

→ ( )( ) ( ) ( )

2

2 6

; 3 435.1 /

2 6 ; 2828.3 /

g H kJ mol

C g H g C H g H kJ mol

∆ = − ×

+ → ∆ = − ... (9)

↓This is obtained applying Hess’s law.



Now, using equations (6), (7), (8) we may write,

3C ( )graphite 4 H+ ( ) ( )( )

2 3 8 ; 101 /

3 3

g C H g H kJ mol

C g C

→ ∆ = −

→ ( )( )

graphite ; 3 719.7 /

8 4

H kJ mol

H g H

∆ = − ×

→ ( )( ) ( ) ( )

2

3 6

; 4 435.1 /

3 8 ; 4000.5 /

g H kJ mol

C g H g C H g H kJ mol

∆ = − ×

+ → ∆ = − ... (10)

↓

This is obtained applying Hess’s law.

It is the reverse of (9) and (10) that we set out to find. We thus have,

( ) ( ) ( ) ( )2 62 6 free atoms2 6 ; 2828.3 /C HC H g C g H g H kJ mol→→ + ∆ =

( ) ( ) ( ) ( )3 83 8 free atoms3 8 ; 4000.5 /C HC H g C g H g H kJ mol→→ + ∆ =

We can use equations(3) and (4) to rewrite the above as: x + 6y = 2828.3 2x + 8y = 4000.5Solving these two equations simultaneously, we get, Bond enthalpy of C — H bond, y = 414 kJ / mol Bond enthalpy of C — C bond, x = 344.3 kJ / molThis is what we set out to find. It is recommended that you reread the whole problem again and try tointernalise the approach taken here to solve this problem.

Compute the enthalpy of formation of liquid methyl alcohol in kilojoules per mole, using the following data. Enthalpyof vaporization of liquid methyl alcohol = 38 kJ/mol. Enthalpy of gaseous atoms from the elements in their standardstates; H, 218 kJ/mol; C, 715 kJ/mol; O, 249 kJ/mol.

Average bond enthalpy: C — H, 415 kJ/mol; C — O, 356 kJ/mol, and O — H, 463 kJ/mol

Critical thinking box 1Our target equation is:

( ) ( ) ( ) ( )2 2 31graphite 2 ; ?2 fC H g O g CH OH l H+ + → ∆ =

If we can obtain this equation from the summation of other equations, then using Hess’s law we cancalculate the .fH∆

Example – 24



Solution: (part - 1)The given data can be enlisted as:

(1) ( ) ( )3 3 1 38 / .CH OH l CH OH g H kJ mol→ ∆ =

(2) ( ) ( )2 21 218 / .2 H g H g H kJ mol→ ∆ =

(3) 3(graphite) ( ) 715 /C C g H kJ mol→ ∆ =

(4) 2 41 ( ) ( ) 249 /2O g O g H kJ mol→ ∆ =

(5) 5( ) ( ) ( ) 415 /C O g C g H g H kJ mol− → + ∆ =

(6) 6( ) ( ) ( ) 356 /C O g C g O g H kJ mol− → + ∆ =

(7) 7( ) ( ) ( ) 463 /O H g O g H g H kJ mol− → + ∆ =

Our target equation is:

2 2 31(graphite) 2 ( ) ( ) ( ); ?2 fC H g O g CH OH l H+ + → ∆ =

If you try to apply Hess’s law now you will find an anomaly in the data. Generally, when we applyHess’s law to a given set of equations to obtain a target equation, all substances which are not presentin the target equation, are present twice in the given set of equations and we arrange them on oppositesides so that they cancel. But here we see that many substances which are not present in the targetequation are present only at one place in the given set of equations. ( 3 ( )CH OH g and the bonds

— , —C H C O and —O H .) How do we proceed further?

Critical thinking box 2To get our target equation from the given set of equations we need one or more themochemical equationsthat relate the substances which are not present in target equations and are present only once in the givenset of equations. How can we do that here ?

Solution: (part - 2)We see that 3 ( ), ( ), ( ) and ( )CH OH g C H g C O g O H g− − − are not present in the target equationand they can be related together by the following equation:

(8) ... 3 8( ) ( ) 4 ( ) ( ); ?CH OH g C g H g O g H→ + + ∆ =

This equation involves breaking of all the bonds in CH3OH and the accompanying 8H∆ can be calculatedfrom equations (5), (6) and (7) as follows:

8 3. C H C O C HH − − −∆ = ∈ + ∈ + ∈

= (3 × 415 + 356 + 463) kJ/mol = 2064 kJ/mol



Critical thinking box 3If we analyse now, then the target equation can be obtained from the given set of equations (1) - (7) and thenew equation (8)

Solution: (part - 3)

(3) 1: (graphite)C C× → 3( ) 1 715 /g H kJ mol×∆ =

(2) 24 : 2 ( ) 4H g H× → 2( ) 4 4 218 /g H kJ mol× ∆ = ×

(4) 211: ( )2 O g O× → 4( ) 1 249 /g H kJ mol× ∆ =

(8) 1: C× − ( ) 4g H+ ( )g O+ 3( )g CH→ 8( ) 1 2064 /OH g H kJ mol− × ∆ = −

(1) 31: CH× − 3 1( ) ( ) 1 38 /OH g CH OH l H kJ mol→ − × ∆ = −

______________________________________________________________________________________

Target equation: 2 2 31(graphite) 2 ( ) ( ) ( ) 266 /2C H g O g CH OH l H kJ mol+ + → ∆ = −

Hence enthalpy of formation of 3 ( )CH OH l is – 266 kJ/mol

The standard molar enthalpies of formation of cyclohexane (1) and benzene (1) at 25°C are – 156 and + 49 kJmol–1, respectively. The standard enthalpy of hydrogenation of cyclohexene(1) at 25°C is – 119 kJ mol–1. Usethese data to estimate the magnitude of the resonance energy of benzene.

Critical thinking

Resonance energy of benzene is the H∆ ° accompanying the following hypothetical reaction:

(1) ; H∆ ° = resonance energy.

Now, we can obtain the reactionH∆ for the following:

(2)

(using 0fH∆ values given). Can this reaction be written in another sequence of steps such that and

are still the initial and final compounds respectively but involve the last step as ?

Example – 25



Solution: The reaction corresponding to hydrogenation of cyclohexene is:

2( ) ( )l H g+ 119 /H kJ mol∆ ° = −

This means that the creation of one double bond in cyclohexane requires + 119 kJ/mol. And furthercreation of double bonds in the product molecule can also be assumed to be + 119 kJ/mol only.

This motivates us to write the equation (2) in critical thinking box above alternatively as:

(3) ∆H°= + 119 /kJ mol

∆H°= + 119 /kJ mol

. ∆H°= + 119 /kJ mol

resonanceenergy

and we already know that:

(2) ∆H°r

Here, 0 0 0(benzene, l) (cyclohexane, l)r f fH H H∆ = ∆ − ∆

= 49 kJ/mol – (– 156) kJ/mol

= 205 kJ/mol.

By applying Hess’s law we can say that,

0 3.rH H∆ = ∆ ° + resonance energy.

0resonance energy 3rH H⇒ = ∆ − ∆ °

= (205 – 3. 119) kJ/mol

= – 152 kJ/mol.



Fundamentals and first law of thermodynamics

Q. 1 Indicate the correct relationship of the standard changes in the enthalpy for the following reactions:

2 2( ) ( ) ( )H g O g H O g+ → ...(1)

2 2 21( ) ( ) ( )2

H g O g H O g+ → ...(2)

22 ( ) ( ) ( )H g O g H O g+ → ...(3)

(a) 2 1 3;H H H° ° °∆ < ∆ < ∆

(b) 2 1 3.H H H° ° °∆ > ∆ > ∆

Q. 2 If 1500 cal of heat is added to a system while the system does work equivalent to 2500 cal by expandingagainst the surrounding atmosphere, what is the value of U∆ for the system?

Q. 3 Starting with the definition ,H U PV= + prove that the enthalpy change for a process with no work otherthan expansion is equal to the heat added at constant pressure.

Q. 4 Read the following statement and explanation and answer as per the options given below:STATEMENT(S) : The endothermic reactions are favoured at lower temperature and the exothermicreactions are favoured at higher temperature.EXPLANATION(E): When a system in equilibrium is disturbed by changing the temperature, it will tendto adjust itself so as to overcome the effect of change.(a) Both S and E are true, and E is the correct explanation of S.(b) Both S and E are true, and E is not the correct explanation of S.(c) S is true but E is false.(d) S is false but E is true.

Q. 5 For which change :H U∆ ≠ ∆

(a) 2 2 2H I HI+ → (b) HCl NaOH NaCl+ →

(c) ( ) 2 2( ) ( )s g gC O CO+ → (d) 2 2 33 2N H NH+ →

Q. 6 Which one of the following statements is false?(a) Work is a state function(b) Temperature is a state function(c) Change in the state is completely defined when the initial and final states are specified(d) Work appears at the boundary of the system.

EXERCISE



Q. 7 Identify the intensive quantities from the following:(a) Enthalpy (b) Temperature(c) Volume (d) Refractive Index

Q. 8 If 1.000 kcal of heat is added to 1.200 L of oxygen in a cylinder at constant pressure of 1.00 atm, thevolume increases to 1.500 L. Calculate U∆ for the process.

Q. 9 The reaction of cyanamide, 2 ( ),NH CN s with oxygen was run in a bomb calorimeter, and U∆ was found

to be –742.7 kJ/mol of 2 ( )NH CN s at 298 K. Calculate the 298H∆ for the reaction.

2 2 2 2 23( ) ( )2

NH CN s O N CO H O l+ → + +

Q. 10 A sample of 0.20 mol of a gas at 44°C and 1.5 atm pressure is cooled to 27°C and compressed to 3.0atm. Calculate .V∆ Suppose the original sample of gas were heated at constant volume until its pressurewas 3.0 atm and then cooled at constant pressure to 27°C. What would V∆ have been? Is volume astate function? Is your answer limited to this sample of gas?

Hess’s law and its applications

Q. 11 Equal volumes of hydrogen and acetylene (C2H2) taken in identical conditions are burned with the formationof 2 ( ).H O g In which case is more heat liberated, and how many times more?

Q. 12 Determine 298H °∆ for the reaction 2 2 6 63 ( ) ( )C H g C H l→ if 298H °∆ for the reaction of combustion of

acetylene with the formation of 2( )CO g and 2 ( )H O l is –1300 kJ/mol, and 298H °∆ of formation ofbenzene (l) is 82.9 kJ/mol ?

Q. 13 The polymerisation of ethylene to linear polyethylene is represented by the reaction

2 2 2 2(— — —)nn CH CH CH CH= →

where n has a large integral value. Given that the average enthalpies of bond dissociation for C — C— andC — C at 298 K are +590 and +331 kJ mol–1 respectively. Calculate the enthalpy of polymerisation permole of ethylene at 298 K.

Q. 14 The standard molar enthalpies of formation of ethane, carbon dioxide and liquid water are –88.3, –393.7and –285.8 kJ mol–1, respectively. Calculate the standard molar enthalpy of combustion of ethane.

Q. 15 The standard enthalpy of combustion at 25°C of hydrogen, cyclohexene 6 10( )C H and cyclohexane

6 12( )C H are –241, –3800 and –3920 kJ mol–1, respectively. Calculate the standard enthalpy ofhydrogenation of cyclohexene.



Q. 16 An athlete is given 100 g of glucose 6 12 6( )C H O of energy equivalent to 1560 kJ. He utilizes 50% of thisgained energy in the event. In order to avoid storage of energy in the body, calculate the mass of water hewould need to perspire. The enthalpy of evaporation of water is 44 kJ mol–1.

Q. 17 Determine the enthalpy of the reaction

3 8 2 2 6 4( ) ( ) ( ) ( )C H g H g C H g CH g+ → +

at 25°C using the given enthalpy of combustion values under standard conditions:

Compound: 2 ( )H g 4 ( )CH g 2 6 ( )C H g C(graphite)

1( ) :H kJ mol−∆ –285.8 –890.0 –1560.0 –393.5

The standard enthalpy of formation of 3 8 ( )C H g is –103.8 kJ mol–1.

Q. 18 The enthalpy change involved in the oxidation of glucose is –2880 kJ mol–1. Twenty-five per cent of thisenergy is available for muscular work. If 100 kJ of muscular work is needed to walk one kilometer, whatis the maximum distance that a person will be able to walk after eating 120 g of glucose.

Q. 19 Estimate the average S — F bond enthalpy of SF6. The value of standard enthalpy of formation of

6 ( ), ( )SF g S g and ( )F g are : –1100, 275 and 80 kJ mol–1 respectively.

Q. 20 Calculate the bond energy of C C≡ in 2 2C H from the following data:

(i) 2 2 2 2 25( ) ( ) 2 ( ) ; 3102

C H g O g CO g H O H kcal+ → + ∆ = −

(ii) 2 2( ) ( ) ( ); 94C s O g CO g H kcal+ → ∆ = −

(iii) 2 2 21( ) ( ) ; 682

H g O g H O H kcal+ → ∆ = −

Bond energy of C — H bonds = 99 kcalHeat of atomisation of C = 171 kcalHeat of atomisation of H = 52 kcal

Second Law of Thermodynamics and related topics

Q. 21 The enthalpy change for a certain reaction at 298 K is –15.0 kcal/mol. The entropy change under theseconditions is 7.2 / .cal mol K− ⋅ Calculate the free energy change for the reaction, and predict whetherthe reaction may occur spontaneously.

Q. 22 For the reaction

2 ( ) ( ) 2 ( )A g B g D g+ →

298 2.50U kcal°∆ = − and 298 10.5 / .S cal K°∆ = − Calculate 298G°∆ for the reaction, and predict whetherthe reaction may occur spontaneously.



Q. 23 At 25°C, the enthalpy of reaction under standard state conditions for the combustion of CO,

2 21( ) ( ) ( )2

CO g O g CO g+ →

is –282.98 kJ/mol, and G∆ ° for the reaction is –257.19 kJ/mol. At what temperature will this reactionno longer be spontaneous under standard state conditions?

Q. 24 For the reaction at 25°C

2 4 2( ) 2 ( )X O l XO g→

2.1U kcal∆ = and 20 / .S cal K∆ =

(a) Calculate the G∆ for the reaction.(b) Is the reaction spontaneous as written ?

Q. 25 Given the following reactions with their enthalpy changes,

2 2 2 298( ) 2 ( ) 2 ( ) 16.18N g O g NO g H kcal+ → ∆ =

2 2 2 4 298( ) 2 ( ) ( ) 2.31N g O g N O g H kcal+ → ∆ =

calculate the enthalpy of dimerization of NO2. Is N2O4 apt to be stable with respect to NO2 at 25°C?Explain you answer.

Q. 26 Are the following statements true of false? Justify you answer.(a) An endothermic reaction is spontaneous.

(b) If H∆ and S∆ are both positive, then G∆ will decrease when the temperature increases.

(c) A reaction for which sysS∆ is positive is spontaneous

Q. 27 For the reaction

2 2( ) ( ) ( )C graphite O g CO g+ →0 393.51 /H kJ mol∆ = − and 0 2.86 /S J mol K∆ = ⋅ at 25°C. Does this reaction become more or

less favourable as the temperature increases?(b) For the reaction

21( ) ( ) ( )2

C s O g CO g+ →

0 110.52 /H kJ mol∆ = − and 0 89.36 /S J mol K∆ = ⋅ at 25°C. Does this reaction become more orless favorable as the temperature increases ?(c) Compare the temperature dependencies of these reactions.

Q. 28 How does the value of 0G∆ change for the reaction

4 2 3( ) ( ) ( ) ( )CCl l H g HCl g CHCl l+ → +

if the reaction is carried out at 95°C rather than at 25°C? At 025 , 103.75 /C G kJ mol° ∆ = − and0 91.34 /H kJ mol∆ = − for the reaction.



[ ANSWERS KEY ]

TRY YOURSELF - I

Ans. 1 (a) w = 0 (b) – 2.2 kJAns. 2 (a) by the system (b) no work done (c) on the systemAns. 3 (a) ∆U (b) ∆H (c) ∆U (d) ∆HAns. 4 ZeroAns. 5 When q = – w, ∆U = 0 even if the final conditions are not same as initial conditionsAns. 6 + 107 kJAns. 7 The first reaction.Ans. 8 (a) by the system (b) on the system (c) by the systemAns. 9 – 556 kJ

TRY YOURSELF - II

Ans. 2 296.5 litres Ans. 3 –162.1 kJ/molAns. 4 1312 kJ Ans. 5 52.4 kJ/molAns. 6 294 kJ/mol Ans. 7 –173.4 kJ/molAns. 8 –2035 kJ/mol Ans. 9 +60.6 kcal/mole.Ans. 10 2957 calories; –41, 500 calories Ans. 11 –1260 kJ/mol.

TRY YOURSELF - III

Ans. 1 c Ans. 2 b, c, eAns. 3 a Ans. 4 bAns. 5 a Ans. 6 aAns. 7 a Ans. 8 2000 K

Ans. 9 0; 100 /G S cal K∆ = ∆ = − Ans. 10 H∆ = negative; S∆ = negative

TRY YOURSELF



EXERCISE

[ ANSWERS KEY ]

Fundamentals and first law of thermodynamics

Ans. 1 b Ans. 2 –1000 caloriesAns. 4 d Ans. 5 dAns. 6 a Ans. 7 b, dAns. 8 0.993 kcal Ans. 9 –741.5 kJAns. 10 –1.83 litres

Hess’s law and its applications

Ans. 11 In the combustion of 2 2;5.2C H times more.

Ans. 12 –598.7 kJ Ans. 13 –72 kJAns. 14 –1556.5 kJ/mol Ans. 15 –121 kJ/molAns. 16 319.14 grams Ans. 17 –55.7 kJ/molAns. 18 4.80 km Ans. 19 309.2 kJ/molAns. 20 +194 kcal.

Second Law of Thermodynamics and related topics

Ans. 21 –12.9 kcal/mol Ans. 22 cannot be spontaneous.Ans. 23 No such temperature exists i.e. it will be spontaneous at all temperatures.

Ans. 24 2.7G kcal∆ = −

Ans. 25 13.87H kcal∆ = − Ans. 26 (a) False (b) True (c) False

Ans. 27 (a) more favourable (b) more fovourableAns. 28 – 106.66 kJ

Thermodynamics [03] Section [54 - 91]

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Transcript of Thermodynamics [03] Section [54 - 91]