COURSE CODE: CHE 212

NAME: MR. OLANREWAJU ALADESUYI

The first law of thermodynamics states that one form of energy can change into another form but the total amount of energy remains the same. But the first law fails to indicate if the process of change we specify, can occur or if so in what direction. The 1st Law of thermodynamics does not give any information about direction of change. Yet experience shows that most natural processes always tend to occur spontaneously in a direction which will lead to equilibrium e.g. flow of heat (ii) electricity (iii) water (iv) gas

SECOND LAW OF THERMODYNAMICS

A process which proceeds of its own accord, without any outside assistance, is termed a spontaneous process.The reverse process which does not proceed on its own, is referred to as a non spontaneous process.In general, the tedency of a process to occur naturally is called the spontaneity. Spontaneous changes proceed until the system reaches an equilibrium state. They are irreversible.Examples: Cliff jumping and mountain climbing. Heat flow from a hot ball to a cold one, never from cold to

hot. When a vessel containing a gas is connected to another

evacuated vesse, the gas spreads throughout spontaneiuoly unless the pressure is the same in both vessels.

What is a Spontaneous process

1. A spontaneous change is one-way or unidirectional.

2. For a spontaneous to occur, time is no factor i.e a spontaneous change may occur slowly or rapidly.

3. If the system is not in equilibrium then a spontaneous is inevitable

4. Once a system is in equilibrium state, it does not undergo any further spontaneous change in state if left undisturbed.

5. A spontaneous change is accompanied by a decrease of internal energy or enthalpy (

Criteria for Spontaneity

This implies that only exothermic reactions will proceed spontaneously due to the fact that exothermic changes results in a decrease in enthalpy. But it has been observed that endothermic changes such as melting of Ice and evaporation of water takes place spontaneously. Clearly there must be some other factor other than enthalpy change which governs spontaneity. Its is the second law which introduces the new factor is entropy factor. . The two factors are related by the equation∆G = ∆H - T∆S (for constant Pressure Process)∆G = Change in Gibbs Free Energy which is the driving force of any chemical or physical change or process.∆S = Change on entropy.

What is a Process?If a change in the value of the properties of a system occurs when the system goes from one state to another, a process is said to have taken place. The process can either be reversible or irreversible.Reversible Process The process is reversible when the driving force is only infinitesimally (minutely) greater than the opposing force, and the process can be reversed by increasing the force by an infinitesimal amount.  For a reversible process p Pex is slightly greater than p the gas is compressed. On the other hand of p is slightly greater than Pex, the gas will undergo expansion.

A reversible process is a change of state such that1. The change occurs in a sequence of steps, each of which is infinitesimally removed from the preceding and succeeding steps.2. The change occurs through a succession of equilibrium steps, each intermediate state being an equilibrium state.A reversible process must be carried out at a very slow rate for condition 2 above to hold. This makes true reversible process in practice unattainable as such a process will require an infinitely long time to go to completion. 

Irreversible Process

Irreversible process is simply defined as one which is not reversible.

Examples of Reversible and Irreversible Process

a) A gas in a cylinder equipped with a movable piston

In both cases A & B, the gas is expanded. In case A the piston is withdrawn at a very slow rate so that time is allowed for the pressure and temperature of the gas to attain uniform value after each change. Hence the expansion goes through a succession of equilibrium states. Therefore the overall process can be considered reversible

Pex Pex

(A) Pex p

(B) Pex >> p

In case B the piston is withdrawn rapidly and no time is allowed for the pressure and temperature of the gas to attain uniform values after each change. The temperature and pressure vary throughout the gas. Hence the expansion does not go through a succession of equilibrium states. Therefore, the overall process can be considered irreversible.

Entropy is a thermodynamic state quantity that is a measure of the randomness or disorder of the molecules of the system. The symbol of entropy is S, while the change accompanying entropy from start to the completion is represented by ∆S. The entropy of a system is a state function and depends only on the initial and final states of the system. The change in entropy, ∆S, for any process is given by the equation,∆S = Sfinal - Sinitial When Sfinal > Sinitial, ∆S is positive. It is however observed that a process accompanied by an increase in entropy always tends to be spontaneous.

Definition of Entropy

Concept of Entropy, S1. It is a state function; unlike work or heat there is no simple physical analogue. However it may be pictured as a measure of disorder or randomness of a system.2. It has the dimension of energy and temperature.For infinitesimal change ds = Jmol-1K-1

or for a finite change ∆S = ; +ve when qr is +ve

-ve when qr is -ve

3. For a given system STotal = Sconfiguration + SThermal

Unlike enthalpy, the absolute entropy of a substance can be calculated (from 3rd Law of Thermodynamics)4. For any process occurring in an isolated system (Universe) ∆S 0

For any reversible process ∆S = 0For any irreversible process ∆S > 0

The second law of thermodynamics states that: whenever a spontaneous process takes place, it is accompanied by an increase in the total energy of the universe.∆Suniv = ∆Ssyst + ∆Ssurr

The second law, as stated above, tells us that when an irreversible spontaneous process occurs, the entropy of the system and the surroundings increases. In other words ∆Suniv > 0. When a reversible process occurs, the entropy of the system remains constant. ∆Suniv =0. since the univers is always experiencing a change the second law can be restated as: the entropy of the system is constantly changing

Physical Interpretation of Entropy – Concept of order and disorder.

Examples

Ordered System Disordered System

1. Pure crystalline solid Liquid, gas

2. Single Substance Mixtures

3.

Magnetized Iron Bar Unmagnetized Iron Bar

The second law of thermodynamics can be stated in different ways, all of which are equivalent to one another. This equivalence is not immediately obvious.1. Clausius StatementIt is impossible for a cyclic process to transfer heat from a body at a lower temperature to one at a higher temperature without simultaneous conversion of work into heat.This implies that when work is used for the continuous transfer of heat from a lower temperature to one at a higher temperature, some work must be wasted by direct conversion into heat

Statement of Second Law

According to Clausius,entropy of a system (not undergoing chemical or physical changes) is a constant quantity when there is no communication of heat. When heat(q) flows into a system, the entropy increases by . Heat flowing out of a system produces a corresponding decrease. Thus entropy could be expresses as ∆S = .If heat is absorbed, then ∆S is positive and there will be increase in entropy. If heat is evolved, ∆S is negative and there is a decrease in entropy.

Kelvin’s StatementIt is impossible for a cyclic process to convert heat into work

without simultaneous transfer of heat from a body at higher temperature to one at lower temperature.

This implies that in the conversion of heat into work, some of the heat in a body at higher temperature must be wasted by direct transfer to a body at a lower temperature.Heat cannot flow spontaneously from a colder to a hotter body.Heat cannot be extracted from a hot source and turned entirely into work.However, the most general law of the second law of thermodynamics states that any process which occurs in Nature is accompanied by an increase in total entropy i.e.

∆STotal = (∆SSystem + ∆SSur) increaseStatistical definition of entropy supposes that it is possible to calculate the degree of disorder in a system using the Boltzmann formula (1896):

S = KlnWWhere k is the Boltzmann’s constant (k = 1.381 x 10-23JK-1) and W = the number of different ways in which the energy of the system can be achieved by rearranging the atoms or molecules among their available states.

Calculation of Entropy Change for an Ideal gasEntropy is a state function and its value depends on two of the three variables T, P and V.a) T and V as variablesFor one mole of an ideal gas,∆S = 2.303Cv log +2.303 R log

For n moles, ∆S = 2.303nCv log +2.303n R log

b. P and T as variablesFor one mole of an ideal gas∆S = 2.303Cp log +2.303 R log

For n moles, ∆S = 2.303nCp log +2.303 nR log

From these equations entropy change can be calculated.

At constant temperature for an isothermal process.

In this case T1 = T2, the equation (1) and (2) reduced to ∆ST = 2.303× n × R log

and ∆ST = 2.303× n × R log

In an isothermal expansion ∆ST is positive, while in isothermal contraction ∆ST is negative At constant pressure ( Isobaric

process) where p1 = p2 the equation (1) reduces to

∆Sp = 2.303× n × CP log

At constant volume for an Isochoric process in this case v1 = v2,

The equation (1) reduces to ∆Sv = 2.303× n × Cv log .

Exercise1. Calculate the change in entropy accompanying the

heating of one mole of helium gas, assumed ideal, from a temperature of to a temperature of 82 at constant pressure given that

2. 2mol of an ideal monoatomic gas expand reversibly from a volume of 2litres and temperature of 289K to a volume of 20litres and temperature of 240K. Assuming , calculate the entropy change for the process

When there is a change of state from solid to liquid or liquid to vapours or solid to vapours, there is always an entropy change. This change may be carried out reversibly at constant temperature as two phases are in equilibrium during the change.Considering the process of melting of 1 mole of the substance reversibly. It absorb molar heat of fusion at temperature equal to its melting point. The entropy change then becomes ∆Sf =

Entropy change Accompanying change of phase

Similarly we have ∆Sv = as the value for entropy change when one mole of liquid is boiled reversibly. and also ∆St = when one mole of a solid changes reversibly from one allotropic form to another at its transition temperature.ExerciseCalculate the entropy increase in the evaporation of 2mol of water at 100 degree centigrade in S. I. units given that the heat of vaporization of water at 100 degree centigrade is 540cal/g.

Entropy Changes in Physical and Chemical Processes

A. Physical Transformation. (Fusion, vapourization, transition) At constant

temperature and Pressure (T & P)

∆ S = Tqr =

TdTC

TH p

1. Consider the reaction.

H2O(s) H2O(l), ∆ H273 = 6.01KJ

∆ St = 113

0.22273

1001.6

kJmolTH

f

f

1. Vapourization at Constant T & P

E.g. H2O(l) H2O(g), ∆ H373 = 40.72

:. ∆ Svap = 113

3731072.40

kJmolTH

b

vap

= 109.1Jmol-1k-1

2. Transition of One Allotropic form to Another at Constant T & P

C(graphite) C(diamond), ∆ H298 = 1.9KJ

:. ∆ St = 113

38.6298

109.1

kJMolTH

t

t

A. Entropy Change in Chemical Reactions.

Consider a reaction:

aA + bB cC + dD + …

∆ S = (c Sc + d Sd + …) – (a Sa + b Sb + …)

OR ∆ Sr = S (Product) - S (Reactant)

For standard entropy change.

Sr = S(Product) - S (Reactant)

The free energy (G) is defined asG = H – TS

where H is the heat content or enthalpy of the system, T is its temperature and S its entropy. It is a single valued function of thermodynamic state of the system and is an extensive property.

Let us consider a system which undergoes a change of state from (1) to (2) at constant temperature.We haveG2 – G1 =(H2 – H1) – T(S2 – S1)

Free Energy (G)

or ∆G = ∆H - T∆SRecall ∆H = ∆E + P∆VSubstituting for ∆H ∆G =∆E + P∆V - T∆SAlso we know that ∆A = ∆E - T∆STherefore, ∆G = ∆A + P∆V P∆V is the work done as a result of expansion against a constant external pressure P. Therefore, it is clear that the decrease in free energy (- ∆G ) accompanying a process taking place at constant temperature and pressure is equal to the maximum work done obtainable from the system other than the work of expansion. This quantity is referred to as the network.Total work/ network = w - P∆V = - ∆G This quantity is of great importance in thermodynamics because the change in free energy is a measure of network which may be electrical, chemical or surface work.

G= H – TSand H = E + PV ( first law of thermodynamics)G = E + PV – TSDifferentiatingdG = dE + PdV + VdP – TdS – SdT……….(i)For an infinitesimal stage of a reversible process. dq = dE + dwRecall dS = dq = dE + PdV ( first law of thermodynamics)Therefore TdS =dE + PdV…………(ii)

How Free Energy Varies with Temperature and Pressure

Substituting eqn. (ii) into (i) we have dG= dE + PdV + VdP – dE - PdV – SdT

= VdP – SdT…………. (iii)At constant pressure eqn (iii) becomesdGp = -SdTp

Or ()p = -S………………………. (iv)At constant temperature eqn (iii) becomesdGT = VdPT

Or ()T = V………………………. (V)Eqn (iv) and (v) shows how free energy varies with temperature and pressure respectively.

Isothermal Change in Free energyFor an isothermal process eqn (iii) ,dG= dE + PdV + VdP – dE - PdV – SdT

= VdP – SdTis reduced to dG = VdP………….. (vi)Integrating, ∆G = G2 – G1 =……. (V)For one mole of a gas PV = RTOr V = ,Substitute for V in eqn (v)dG = RT

= RT In For n moles of the gas, ∆G =n RT In ….. (vi)

From the general gas equation P1V1 = P2V2

Or = Substituting into eqn. (vi) gives∆G = RT In And for n number of moles∆G =n RT In Or ∆G =2.303 n RT log

Conditions for Equilibrium and Criteria for a Spontaneous Changea. In terms of Entropy changeThe entropy of a system remains constant in a reversible change while it increases in an irreversible change.∑dS = 0 for reversible∑dS > 0 for irreversible changeConsidering both the system and its surroundingsdSsystem + dS surrounding =0And dSsystem + dS surrounding > 0………….. (Vii)Combining we havedSsystem + dS surrounding ≥ 0………………..(viii)

From first law of thermodynamicsdS =- = - , dE is the change in internal energy and dw is the wort done by the system.From eqn (viii) we getdSsystem –( )≥ 0…………..(ix)dSsystem - ≥ 0…………..(x)When E and V are constantEqn. (x) reduces to dSE.V ≥ 0dSE.V > 0 for an irreversible process (spontaneous)dSE.V = 0 for a reversible process (equilibrium)

When S and V are constantdS = 0 and dV = 0Eqn (x) becomes reduced to –dE ≥ 0–dE >0 for an irreversible change(spontaneous)–dE = 0 for an reversible change (equilibrium)

b) In terms of enthalpy change H = E + PVDifferentiating,dH = dE + PdV + VdP Or –dE – PdV = -dH + VdP Putting into eqn (x)Tds – dH +VdP ≥ 0Rearranging we get dH -Tds –VdP ≤ 0…………..(xi)

At constant S and PdS = 0 and dP = 0Therefore dH ≤ 0…………..(xii)So dH < 0 for an irreversible change (spontaneous) dH = 0 for a reversible change (equilibrium)

c. In terms of Free Energy ChangeG = H – TSG = E + PV - TS (H = E + PV)Differentiating we getdG = dE + PdV +VdP -TdS -SdTTdS – dE – PdV = -dG + VdP - SdT

Substituting in equations (ix) we get-dG + VdP – SdT ≥0Or dG - VdP + SdT ≤ 0At constant pressure in an isothermal process( T is also constant) this equation becomesdG ≤ 0dG < 0 for an irreversible change (spontaneousdG = 0 for a reversible change (equilibrium)

Summary of Conditions for Spontaneity and equilibrium

Conditions Irreversible process (spontaneous)

Reversible process (equilibrium)

At constant E, V ds > 0 ds = 0 At constant S, V - dE > 0 -dE = 0 At constant S, P dH < 0 dH = 0At constant P,T dG < 0 dG = 0

1.Two moles of an ideal gas are allowed to expand reversibly and isothermally at 300K from a pressure of 1 atm to a pressure of 0.1 atm. What is the change in Gibbs free energy?ANS: -11.4882kJ2. For the reactionN2 (g) + 3H2(g) = 2NH3

The free energy changes at 25oC and 35oC are -33.089 and – 28.018kJ respectively. Calculate the heat of reaction at 35oC ANS: -184.176 kJ3. Calculate the standard entropy of formation,∆So

f of CO2 (g). Given the standard entropies of CO2 (g), C(s), O2(g), which are 213.6, 5.740 and 205.0JK-1 Ans: 2.86JK-1

Practise Questions

4. Show that Gibbs free energy is given in terms of the state parameterdG = VdP –SdT, Hence derive the expression()p = -S and()T = V