Thermodynamics : Entropy & Second law of thermodynamics

 

Spontaneous and Non-spontaneous processes

(1) Definition : A process which can take place by itself under the given set of conditions once it has been initiated if necessary, is said to be a spontaneous process. In other words, a spontaneous process is a process that can occur without work being done on it. The spontaneous processes are also called feasible or probable processes.

On the other hand, the processes which are forbidden and are made to take place only by supplying energy continuously from outside the system are called non-spontaneous processes. In other words, non spontaneous processes can be brought about by doing work.

Note :  A spontaneous process is unidirectional and irreversible.

 

(2) Examples of Spontaneous and Non-spontaneous processes

(i) A river flows from a mountain towards the sea is spontaneous process.

(ii) Heat flows from a conductor at high temperature to another at low temperature is spontaneous process. 

(iii) A ball rolls down the hill is spontaneous process.

(iv) The diffusion of the solute from a concentrated solution to a dilute solution occurs when these are brought into contact is spontaneous process.

(v) Mixing of different gases is spontaneous process.

(vi) Heat flows from a hot reservoir to a cold reservoir is spontaneous process.

(vii) Electricity flows from high potential to low potential is spontaneous process.

(viii) Expansion of an ideal gas into vacuum through a pinhole is spontaneous process.

All the above spontaneous processes becomes non-spontaneous when we reverse them by doing work.

 

(3) Spontaneous process and Enthalpy change : A spontaneous process is accompanied by decrease in internal energy or enthalpy, i.e., work can be obtained by the spontaneous process. It indicates that only exothermic reactions are spontaneous. But the melting of ice and evaporation of water are endothermic processes which also proceeds spontaneously. It means, there is some other factor in addition to enthalpy change (ΔH) which explains the spontaneous nature of the system. This factor is entropy.

 

Second law of thermodynamics

All the limitations of the first law of thermodynamics can be remove by the second law of thermodynamics. This law is generalisation of certain experiences about heat engines and refrigerators. It has been stated in a number of ways, but all the statements are logically equivalent to one another.

 

(1) Statements of the law

(i) Kelvin statement : “It is impossible to derive a continuous supply of work by cooling a body to a temperature lower than that of the coldest of its surroundings.”

(ii) Clausius statement : “It is impossible for a self acting machine, unaided by any external agency, to convert heat from one body to another at a higher temperature or Heat cannot itself pass from a colder body to a hotter body, but tends invariably towards a lower thermal level.”

(iii) Ostwald statement : “It is impossible to construct a machine functioning in cycle which can convert heat completely into equivalent amount of work without producing changes elsewhere, i.e., perpetual motions are not allowed.”

(iv) Carnot statement: “It is impossible to take heat from a hot reservoir and convert it completely into work by a cyclic process without transferring a part of it to a cold reservoir.”

 

(2) Proof of the law : No rigorous proof is available for the second law. The formulation of the second law is based upon the observations and has yet to be disproved. No deviations of this law have so far been reported. However, the law is applicable to cyclic processes only.

 

Conversion of heat into work: The Carnot cycle

Carnot, a French engineer, in 1824 employed merely theoretical and an imaginary reversible cycle known as carnot cycle to demonstrate the maximum convertibility of heat into work.

The system consists of one mole of an ideal gas enclosed in a cylinder fitted with a piston, which is subjected to a series of four successive operations. The four operations are

(1) Isothermal reversible expansion,

q2 = – w1 = RT2 log    \frac { { V }_{ 2 } }{ { V }_{ 1 } }                          ….(i) 

(2) Adiabatic reversible expansion,

ΔE = – w2 = – CV(T2 – T1)                         ….(ii)    

(3) Isothermal reversible compression,

q1 = – w3 = RT2 log    \frac { { V }_{ 4 } }{ { V }_{ 3 } }                          ….(iii)

(4) Adiabatic reversible compression,

w4 = CV(T2 – T1)                                       ..…(iv) 

The net heat absorbed, q, by the ideal gas in the cycle is given by

q = q2 + (-q1) = RT2 loge    \frac { { V }_{ 2 } }{ { V }_{ 1 } }  + RT1 log  \frac { { V }_{ 4 } }{ { V }_{ 3 } }   = RT2 loge    \frac { { V }_{ 2 } }{ { V }_{ 1 } }   – RT loge    \frac { { V }_{ 3 } }{ { V }_{ 4 } }                          ….(V)

According to the expression governing adiabatic changes,

          \frac { T_{ 2 } }{ T_{ 1 } } =\left( \frac { { V }_{ 3 } }{ { V }_{ 2 } } \right) ^{ \gamma -1 }        (For adiabatic expansion)

         \frac { T_{ 1 } }{ T_{ 2 } } =\left( \frac { { V }_{ 1 } }{ { V }_{ 4 } } \right) ^{ \gamma -1 }    (For adiabatic compression)

          or      \frac { { V }_{ 3 } }{ { V }_{ 2 } } =\frac { { V }_{ 4 } }{ { V }_{ 1 } }      or    \frac { { V }_{ 2 } }{ { V }_{ 1 } } =\frac { { V }_{ 3 } }{ { V }_{ 4 } }

Substituting the value of    \frac { { V }_{ 3 } }{ { V }_{ 4 } }   in eq. (v),

q = q2 – q1 =RT2 loge   \frac { { V }_{ 2 } }{ { V }_{ 1 } }    q2 = R(T2-T1) loge  \frac { { V }_{ 2 } }{ { V }_{ 1 } }                                  …(vi)

Similarly, net work done by the gas is given by

w = -w1 – w2 + w3 + w4         

          So,   w = R(T2-T1) loge  \frac { { V }_{ 2 } }{ { V }_{ 1 } }                                                                  ..(vii)

Thus, q = w. For cyclic process, the essential condition is that net work done is equal to heat absorbed. This condition is satisfied in a carnot cycle. Dividing Equation (vii) by (vi) we get,

                   \frac { w }{ { q }_{ 2 } } =\frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 } } =  Thermodynamic efficiency

Thus, the larger the temperature difference between high and low temperature reservoirs, the more the heat converted into work by the heat engine. 

Since   \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 } }   < 1, it follows that w < q2. This means that only a part of heat absorbed by the system at the higher temperature is transformed into work. The rest of the heat is given out to surroundings. The efficiency of the heat engine is always less then 1. This has led to the following enunciation of the second law of thermodynamics.

It is impossible to convert heat into work without compensation.

 

Example ­1 :      An engine operating between 150°C and 25°C takes 500 J heat from a higher temperature reservoir if there are no frictional losses then work done by engine is

(a) 147.7J           (b) 157.75 J                  (c) 165.85J                   (d) 169.95J

Solution (a) :    T2 = 150 + 273 = 423K

                              T1 = 25 + 273 = 298K

                             q2 = 500J

                             \frac { w }{ { q }_{ 2 } } =\frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 } }

                             w = 500    \left[ \frac { 423-298 }{ 423 } \right] = 147.7J

 

Example ­2 :      Find out the working capacity of an engine which is working between 110° to 25°C, if temperature of boyler is increased up to 140°C and sink temperature equal to its.

(a) I – 20.2%, II – 18%                             (b) I – 22.2%, II – 27.8%     

(c) I – 28%, II – 30%                                (d) None of these

Solution (b) :    First condition

                   T2 = 110oC = 110 + 273K = 383K         ή = ?

                   T1 = 25oC = 25 + 273K =298K

                    \eta =\frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 } } =\frac { 383-298 }{ 383 }   = 0.222   or   22.2%

                   Second condition

                   T2 = 140oC = 140 + 273K = 413K         ή = ?

                   T1 = 25oC = 25C = 25 + 273K = 29K

                  \eta =\frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 2 } } =\frac { 413-298 }{ 413 }   = 0.278 or 27.8%

 

Entropy and Entropy change

(1) Definition : Entropy is a thermodynamic state quantity which is a measure of randomness or disorder of the molecules of the system.

Entropy is represented by the symbol “S”. It is difficult to define the actual entropy of a system. It is more convenient to define the change of entropy during a change of state.

The entropy change of a system may be defined as the integral of all the terms involving heat exchanged (q) divided by the absolute temperature (T) during each infinitesimally small change of the process carried out reversibly at constant temperature.

ΔS = Sfinal – Sinitial =   \frac { { q }_{ rev } }{ T }                            

If heat is absorbed, then ΔS = + ve and if heat is evolved, then ΔS = – ve

 

(2) Units of entropy : Since entropy change is expressed by a heat term divided by temperature, it is expressed in terms of calorie per degree, i.e., cal deg-1 . In SI units, the entropy is expressed in terms of joule per degree Kelvin, i.e., JK–1.

 

(3) Characteristics of entropy : The important characteristics of entropy are summed up below

(i) Entropy is an extensive property. Its value depends upon the amount of the substance present in the system.

(ii) Entropy of a system is a state function. It depends upon the state variables (T, p, V, n).

(iii) The change in entropy in going from one state to another is independent of the path.

(iv) The change in entropy for a cyclic process is always zero.

(v) The total entropy change of an isolated system is equal to the entropy change of system and entropy change of the surroundings. The sum is called entropy change of universe.

ΔSuniverse = – ΔSsys + ΔSsurr

(a) In a reversible process, ΔSuniverse = 0 and, therefore

ΔSsys = – ΔSsurr

(b) In an irreversible process, ΔSuniverse > 0. This means that there is increase in entropy of universe is spontaneous changes.

(vi) Entropy is a measure of unavailable energy for useful work.

                   Unavailable energy = Entropy × Temperature

(vii) Entropy, S  is related to thermodynamic probability (W) by the relation,

S = k loge W and S = 2.303k log10 W ;  where, k is Boltzmann’s constant

 

(4) Entropy changes in system & surroundings and total entropy change for Exothermic and Endothermic reactions : Heat increases the thermal motion of the atoms or molecules and increases their disorder and hence their entropy. In case of an exothermic process, the heat escapes into the surroundings and therefore, entropy of the surroundings increases on the other hand in case of endothermic process, the heat enters the system from the surroundings and therefore. The entropy of the surroundings decreases.

In general, there will be an overall increase of the total entropy (or disorder) whenever the disorder of the surroundings is greater than the decrease in disorder of the system. The process will be spontaneous only when the total entropy increases.

 

(5) Entropy change during phase transition : The change of matter from one state (solid, liquid or gas) to another is called phase transition. Such changes occur at definite temperature such as melting point (solid to liquid). boiling point (liquid to vapours) etc, and are accompanied by absorption or evolution of heat.

When a solid changes into a liquid at its fusion heat). Let ΔHf be the molar heat of fusion. The entropy change will be

ΔSf =  \frac { { \Delta H }_{ f } }{ T_{ f } }

Similarly, if the latent heat of vaporisation and sublimation are denoted by ΔHvap and ΔHsub respectively, the entropy of vaporisation and sublimation are given by

ΔSvap = \frac { { \Delta H }_{ vap } }{ T_{ b } }   and  ΔSsub =  \frac { { \Delta H }_{ sub } }{ T_{ s } }

Since ΔHf, ΔHvap and ΔHsub are all positive, these processes are accompanied by increase of entropy.

The reverse processes are accompanied by decrease in entropy.

Note     Entropy increases not only in phase transition but also when the number of moles of products is greater than the number of moles of reactants. (nproduct > nreactant  i.e., Δn = + ve)

 

(6) Entropy change for an ideal gas : In going from initial to final state, the entropy change, for an ideal gas is given by the following relations,

(i) When T and V are two variables,

ΔS = nCv  In   \frac { { T }_{ 2 } }{ T_{ 1 } }   +  nR  In   \frac { { V }_{ 2 } }{ V_{ 1 } } . Assuming CV is constant

(ii) When T and p are two variables,

ΔS = nCp In  \frac { { T }_{ 2 } }{ T_{ 1 } }   –  nR  In   \frac { { p }_{ 2 } }{ p_{ 1 } } . Assuming Cp is constant

(a) Thus, for an isothermal process (T constant),

ΔS = nR  In   \frac { { V }_{ 2 } }{ V_{ 1 } }     or = – nR  In   \frac { { p }_{ 2 } }{ p_{ 1 } }  

(b) For isobaric process (p constant), ΔS = n  Cp  In   \frac { { T }_{ 2 } }{ T_{ 1 } }  

(c) For isochoric process (V constant), ΔS = n  Cv  In   \frac { { T }_{ 2 } }{ T_{ 1 } }

(d) Entropy change during adiabatic expansion : In such process q=0  at all stages. Hence ΔS = 0. Thus, reversible adiabatic processes are called isoentropic process.

 

Example ­3 :      The enthalpy of vapourisation of a liquid is 30 kJ mol–1 and entropy of vapourisation is 75mol–1 K. The boiling point of the liquid at 1 atm is

(a) 250 K                      (b)  400 K            (c) 450 K              (d) 600 K

Solution (b) :         ΔSv = \frac { \Delta { H }_{ v } }{ { T }_{ b } }    30 kJ = 30000 J

                    ΔHV = 30000 J; ΔSV = 75 mole–1KTb =  \frac { 3000 }{ 75 }   = 400K

 

Example ­4 :      If 900 J/g of heat is exchanged at boiling point of water, then what is increase in entropy

(a) 43.4 J/mole             (b) 87.2 J/mole   (c) 900 J/mole     (d) Zero

Solution (a):     Boiling point (Tb) = 100°C; ΔHV = 9000J/g

                            ΔSvap = \frac { \Delta { H }_{ v } }{ { T } }  ;  Molecular weight of water = 18      

                            ΔSvap = \frac { 900\times 18 }{ 375 }   = 43.4JK-1 mol-1

 

Example 5 :      The standard entropies of CO2(g), C(s) and O2(g) are 213.5, 5.690 and 205 JK–1 respectively. The standard entropy of formation of CO2(g) is

(a)1.86 JK–1                  (b)1.96 JK–1        (c) 2.81 JK–1       (d) 2.86 JK–1

Solution (c) :     Formation of CO2 is,  C(s) + O2(g) → CO2(g)

ΔSo = ΔS0(product) – ΔS0(reactants) = 213.5 – [5.690 + 205] = 2.81 JK–1

 

Free energy and Free energy change

Gibb’s free energy (G) is a state function and is a measure of maximum work done or useful work done from a reversible reaction at constant temperature and pressure.

 

(1) Characteristics of free energy

(i) The free energy of a system is the enthalpy of the system minus the product of absolute temperature and entropy i.e., G = H – TS

(ii) Like other state functions E, H and S, it is also expressed as ΔG. Also ΔG = ΔH – TΔSsystem where ΔS is entropy change for system only. This is Gibb’s Helmholtz equation.

(iii) At equilibrium ΔG = 0

(iv) For a spontaneous process decrease in free energy is noticed i.e., ΔG = – ve.

(v) At absolute zero, TΔS is zero. Therefore if ΔG cis – ve, should be – ve or only exothermic reactions proceed spontaneously at absolute zero.

(vi) ΔGsystem = TΔSuniverse

(vii) The standard free energy change, ΔGo = – 2.303RT log10 where K is equilibrium constant.

(a) Thus if K > 1, then ΔGo = – ve thus reactions with equilibrium constant K>1 are thermodynamically spontaneous.

(b) If K<1, then ΔGo = + ve and thus reactions with equilibrium constant K<1 are thermodynamically spontaneous in reverse direction.

 

(2) Criteria for spontaneity of reaction : For a spontaneous change ΔGo = – ve and therefore use of ΔG = ΔH – TΔS, provides the following conditions for a change to be spontaneous.

ΔG Reaction characteristics Example
Always negative Reaction is spontaneous at all temperatures 2O3(g) → 3O2(g)
Always positive Reaction is non spontaneous at all temperatures 3O2(g) →2O3(g)
Negative at low temperature but positive at high temperature Reaction is spontaneous at low temperature but becomes non spontaneous at high temperature CaO(s) + CO2(g) → CaCO3(g)
Positive at low temperature but negative at high temperature Reaction is non spontaneous at low temperature but becomes spontaneous at high temperature CaCO3(g) → CaO(s) + CO2(g)

 

Example 6 :      For a reaction at 25°C enthalpy change and entropy change are – 11.7×103 J mol–1 and –105J mol–1K–1 respectively, what is the Gibb’s free energy.       

(a) 5.05 kJ                     (b) 19.59 kJ        (c) 2.55 kJ           (d) 22.55 kJ

Solution (b) :    ΔG = ΔH – TΔS, T = 25 + 273 = 298 K

                                ΔG = – 11.7 × 103 – 298 × (–105) = 19590J = 19.59 kJ

 

Example­ 7 :      For reaction Ag2O(s) → 2Ag(s) + \frac { 1 }{ 2 }  O2(g) the value of ΔH = 30.56kJ mol–1 and ΔS = 0.066kJ mol–1 K–1 Temperature at which free energy change for reaction will be zero is

(a) 373 K                       (b) 413 K             (c) 463 K              (d) 493 K

Solution (c) :     ΔG = ΔH – TΔS, ΔH = 30.56kJ mol–1; ΔS = 0.066 kJ mol–1 K–1; ΔG = 0 at equilibrium;  T = ?

ΔH = TΔS or 30.56 = T × 0.066 , T = 463 K

 

Example 8 :      The free energy change for the following reactions are given below

                   C2H2(g) +  \frac { 5 }{ 2 }   O2(g) → 2CO2(g) + H2O(l) ; ΔGo = – 1234 kJ

                   C(s) + O2(g) → CO2(g); ΔGo = – 394 kJ

                   H2(g) +  \frac { 1 }{ 2 }  O2(g) → H2O(l)   ; ΔGo = – 273 kJ

What is the standard free energy change for the reaction

H2(g) + 2C(s) → C2H2(g)     

(a) – 209 kJ                  (b) – 2259 kJ      (c) +2259 kJ       (d) 209 kJ

Solution(a) :     ΔG = ΔG0(product) – ΔG0(Reactants)      

                             = (–1234) – [–237 + (–394)] = –1234 + 1025 = – 209 kJ

 

Example 9 :      Calculate the free energy per mole when liquid water boils against 1 atm pressure (ΔHvap = 2.0723 kJ/g)                          

(a) 0                               (b) 0.2                  (c) 0.3                   (d) 0.4

Solution (a) :  ΔH = 2.0723 × 18 kJ/mole ; T = 373 K ; ΔGvap = ?

                         ΔS =  \frac { \Delta { H }_{ vap } }{ { T } }  \frac { 37.3kJ }{ 373 } = 0.1kJK-1 mol-1

                    ΔGvap = ΔHvap – TΔSvap = 37.3 – 373 × 0.1 = 0

 

Example 10 :    ΔGo for the reaction x + y = z is – 4.606 kcal. The value of equilibrium constant of the reaction at 227oC is (R = 2.0 kcal mol-1 K-1

(a) 100                           (b) 10                   (c) 2                     (d) 0.01

Solution (a) :    R = 2.0 kcal mol–1K–1; T = 500 K ; ΔG0 = – 4.606 kcal;

                   ΔG0 = – 2.303 RT log K

                         – 4.606 = – 2.303 × 0.002 × 500 log K

log K = 2, K = 100

                  

Third law of thermodynamics

This law was first formulated by German chemist Walther Nernst in 1906. According to this law,

“The entropy of all perfectly crystalline solids is zero at the absolute zero temperature. Since entropy is a measure of disorder, it can be interpreted that at absolute zero, a perfectly crystalline solid has a perfect order of its constituent particles.”

The most important application of the third law of thermodynamics is that it helps in the calculation of absolute entropies of the substance at any temperature T.

                             S = 2.303 CP log T

Where Cp is the heat capacity of the substance at constant pressure and is supposed to remain constant in the range of 0 to T.  

Limitations of the law

(1) Glassy solids even at 0K has entropy greater than zero.

(2) Solids having mixtures of isotopes do not have zero entropy at 0K. For example, entropy of solid chlorine is not zero at 0K.

(3) Crystals of Co, N2O, NO, H2O, etc. do not have perfect order even at 0K thus their entropy is not equal to zero.