Metallurgical ThermodynamicsMT 2102 Credit:04Instructors: Dr. C. K. Behera and Mr. J. K. SinghMarks DistributionTotal Marks 100Sessional Test - I 15Sessional Test - II 15Assignments / Attdn. 10End Semester 60Grading SystemS 90 -100A 80-89B 70-79C 60-69DEF50-5940-49< 40Course ContentBasic Principles Extensive and intensive properties, thermodynamic systems and processes.First Law of Thermodynamics, enthalpy, Hess Law, heat capacity, Kirchhoffs law.Second Law of Thermodynamics, entropy, entropy change in gases, significance of sign change of entropy.Troutons and Richards rules. Driving force of a chemical reaction, combined statement of first and second laws of thermodynamics, Helmholtz and Gibbs free energies.Ellingham diagram, Equilibrium constants, vant Hoffs isotherm, Le Chatelier principle.Clausius-Clapeyron equation.Maxwells equations, Third Law of Thermodynamics.Solution Thermodynamics Solution, mixture and compound.Raoults law: activity, ideal solution, standard state.Partial molar quantities, Gibbs-Duhem equation, chemical potential, fugacity, activity and equilibrium constant.Free energy of mixing, excess and integral quantities.Regular solutions, -function.Dilute solutions: Henrys and Sieverts laws.Alternative standard states. Gibbs-Duhem integrationStatistical concept of entropy.Elements of Gibbs Phase Rule and its applications.Experimental Techniques Determination of thermodynamic quantities by different techniques, viz. calorimetry, chemical equilibria, vapour pressure and electrochemical: aqueous, fused and solid electrolytes; formation, concentration and displacement cells.Suggested Reading 1. D.R. Gaskell:Introduction to Metallurgical Thermodynamics, McGraw-Hill.2. L.S. Darken and R.W. Gurry: Physical Chemistry of Metals, McGraw-Hill.3. G.S. Upadhyaya and R.K. Dube: Problems in Metallurgical Thermodynamics and Kinetics, Pergamon.4. J. Mekowiak:Physical Chemistry for Metallurgists, George Allen & Unwin.5. J.J. Moore:Chemical Metallurgy, Butterworths.6. R.H. Parker:An Introduction to Chemical Metallurgy, Pergamon.Scope, Basic Concepts andDefinitions Thermodynamics is that branch of science which deals with the study of the transfer and conversion of energy from one form into other and its conversion to work.It deals with only conventional forms of energies like electrical, mechanical, chemical. etc.The non-conventional energy like nuclear energy related to atomic and sub-atomic particle forms has to be dealt separately because in that case all matter would have to be considered as per the famous Einsteins equation : E = mC2 .Here the subject matter of discussion is chemical and/or metallurgical thermodynamics alone.The systems under discussions consisting of large no. of particles i. e macro systems. ClassificationThermodynamics may be broadly classified intothree :Classical: it treats a substance as continuum, ignoring behavior of atoms and molecules. It consists of first, second and third laws of thermodynamicsScope, Basic Concepts andDefinitions Statistical thermodynamics: The application of probability theory, quantum theory and statistical mechanics allowed it to arrive at macroscopic thermodynamic relations from atomistic point of view.Irreversible Thermodynamics: Irreversible thermodynamics deals with the application of thermodynamics to irreversible processes such as chemical reactions.The thermodynamics generally means classicalthermodynamicsScope, Basic Concepts andDefinitions Chemical thermodynamics is based on the three laws of thermodynamics systematically applied to various physico-chemical processes in physical chemistry.Broadly speaking, the application primarily chemical thermodynamics to metals and materials lead to the development and growth of Metallurgical thermodynamics or its later generalization as Thermodynamics of materials.Processing of ceramics and metals is carried out primarily at high temperatures which led to the development of metallurgical thermodynamics.Scope, Basic Concepts andDefinitions It is the ability to do work. This is too mechanical an answer.The broader definition is : it is the capacity to bringabout changes in the existing materials as per the requirements.Forms of energiesMechanical: Kinetic, potential and configurational.Thermal:Heat exchanged.Electrical: Electrical energy = current x time x potential difference.Chemical:Chemical energy = no. of chemical bonds x bond strengthEnergy System and SurroundingAny portion of the universe selected for consideration is known as the system or thermodynamic system. A thermodynamic system must, of necessity be stable with respect to its chemistry during its study. If the system is undergoing continuously some chemical change cannot be considered as systemFor example a live animate body like tree and human being are not system. All inanimate aggregates are called systems as they have their fixed chemistryThe rest of the universe excluding system is called surrounding.Scope, Basic Concepts andDefinitions Classification of thermodynamic systemsThermodynamic SystemsIn terms of interaction with surroundingBased on Material DistributionBased on compositionIsolatedNeither matter nor energy is exchanged with surroundingOpenBoth matter and energy exchange occursClosed Can exchange energy not the matterHomogeneousHeterogeneousSingle componentMulti componentHomogeneousHeterogeneousHomogeneousHeterogeneousHomogeneous and Heterogeneous systemHomogeneous system consists of single phase only.Heterogeneous consists of more than one phase.State of a systemAs the position of a point in the space is described by its coordinates w.r.tsome prefixed axes, similarly the state of a system is described by some experimentally determinable parameter which can lead to the complete reproduction of the system.These parameters are temperature, pressure and volume.The minimum number of variable required to describe the state of the system are called independent state variables.Incase of a multicomponent system the independent state variables are i) composition ii) two of the three variables T, P, V.All other variables whose values get fixed with the specification of independent state variables are referred to as dependent state variables.State variables are also known asstate properties or state functions.State of a system (cont.)Extensive and Intensive State propertiesIf a state variable, whether dependent or independent, is a function of the mass of the system it is known as extensive state property. For example: Mass, Volume, weight, length, energy etc.If a state variable is independent of the mass or size of the system it is called intensive state property. For example: temperature, pressure, conductivity, density, colour, odor, malleabilty, hardness, specific heat, molar volume etc. Product of an intensive and an extensive state variable is also an extensive state variable.Ratio of two extensive properties yields an intensive properties. For example: Density = m / VThe general convention in chemical thermodynamics is to go for molar properties, which are intensive and become independent of quantity of matter and hence of more general applicability.Extensive and Intensive State propertiesEquation of StateThe state of the system can be described in the form of some mathematical equation involving some state variables. The analytical form as applicable to the system under consideration is known as equation of statee.g. for an ideal gas PV = nRT.The above relationship is an expression which correlates the P, T and V. In fact this is found to be true in case of solids as well as liquids though exact form of this relationship may not be known.The same can be described in generalized form:F(P,V,T) = 0Thermodynamic ProcessesA material system may under go a change, under externally or internally imposed constraints, in terms of their state variables from the existing one to some differentvalues. Such a change in the state of the system is known as thermodynamic processes.For example: Expansion of a gas from V1 to V2 may be called as a thermodynamic process. Many a times such processes are carried out under additionally imposed conditions and are named accordingly. Such processes are:Isothermal processes: These are the processes which proceed without any change in temperature of the system e. g. melting of ice or metal. dT = 0Isobaric process: These are the processes which proceed without any change in pressure of the system e.g. processes carried out in open atmosphere.dP = 0Isochoric Processes: These are the processes which proceed without any change in volume of the system e.g. the processes carried out in vessel of known volume . dV = 0Thermodynamic ProcessesAdiabatic processes: These are the processes which proceed without any exchange of heat by the system with its surrounding e. g. the system is completely insulated from the surrounding . dq = 0.for ideal gas PV = const. = Cp/CvPolytropic Processes: Those processes which obey equation PVn = const where n is any positive number between 1 and .Thermodynamic ProcessesPath and State FunctionsThe property whose change depends on only the initial and final states of the system not on the path adopted to bring about the change is called state function. Mathematically therefore if the property is a state function (X) then in a cyclic process, when system under goes a change and returns to original state then dX = 0 If Y is not a state function Y 0So Y is called a path functionProperties of State FunctionIf a system has two independent variables say x and y and any other function or property can be expressed in its total differential formdz = Mdx + Ndywhere M and N may be function of x and y then the function z is a state variable if and only if

For an ideal gas T = PV/R wherein P and V are independent variables and T as dependent variable. It can be expressed in total differential asYXxNyM

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.|Relationship among state variablesThermodynamic EquilibriumMechanical Equilibrium: if there is no pressure gradient in the system.Thermal Equilibrium: if there is no temperature gradient in the system.Chemical Equilibrium: if the rate of forward reaction is equal to rate of backward reaction.Complete thermodynamic equilibrium is thus that situation where the system is in equilibrium with respect to all such potentials like mechanical, thermal and chemical.Reversible and Irreversible ProcessesA process that can be reversed in its direction by an infinetesimal change in one or more of the state variables is said to be a reversible process.A classical example of this is the gas cylinder and piston. If the pressure of the gas is say P atm and (p+dp) is exerted from outside on the piston, the gas inside the piston shall be compressed. However, if the external pressure is (p-dp) then the gas shall expand.On the contrary a matchstick when it burns, the process can not be reversed by changing one or other parameters. Once burnt can not be reproduced by reversing the process. This is typically a Irreversible process.Other examples are mixing of two gases, mixing of two liquids to form a solution or flow of electric current through resistor.All natural processes are irreversible.Reversible and Irreversible ProcessesExperimental Evidence Leading to First LawFor number ofthermodynamic cycle eachconsisting of number of processesIf , howeverwork and heat measured in same unit then for a thermodynamic cycle W = q184 . 4 iiqWIt is impossible to produce energy of any kind or form without the disappearance of an equivalentamount of energy System goes from state I to state II by various path and returns to the initial state bypath r only. Then we can writeWa +Wr = qa + qror Wa qa = (qr - Wr)Simillarly along other paths Wb qb = (qr - Wr) Wc qc = (qr - Wr) Wd qd = (qr - Wr)Orqa Wa = qb Wb= qc Wc= qd Wd The difference between q and W shall remain constant as long as initial and final states are not changed.Experimental Evidence Leading to First LawFrom the previous discussion, the following conclusions can be drawn:i) The amount of heat exchanged and work done for taking the system from state I to II is different for different paths thus these functions are path functions. These are denoted by symbol q and w for infinetesimal change.ii) The difference between the heat input q and work done W can be equated to change inanother variable, say U i.e q - W= U Since q-W is independent of path, U is a state function. It has further been proved that for a thermodynamic cycle U = 0It is denoted by dU for infinetesimal change.Experimental Evidence Leading to First LawLet us consider that the system goes from state I to II by absorption of heat from the surrounding and doing work on the surrounding. As we know that q and W cannot be equal we can consider two distinct cases:i) q < W: system partly imparts energy for the work done.ii) q > W: The heat is partly retained by the system itself and partly returned to the surrounding in the form of work done.In both the cases system acts as a reservoir of energy. This stored energy in the system which can change during a thermodynamic process is called internal energy and denoted by symbol U.Experimental Evidence Leading to First LawInternal EnergyIt consists ofmacroscopic kinetic energy due to motion of the system as a whole.potential energy of the system due to its position in the force field.kinetic energy of atoms and molecules in the form of translation, rotation and vibration.energy of interaction amongst atoms and moleculescolumbic energy of interaction amongst electrons and nucleii in atomsenergy contents of the electrons and nucleii of atomsIn conventional chemical thermodynamics which we shall be concerned with only kinetic energies of atoms and molecules and interaction amongst atoms and molecules i. e. items (3) and (4) are considered to be important since changes occurringin them principally contribute to U.Absolute value of the energyis not known. All we can determine is change in internal energy.Internal energy will depend on temperature for a material of fixed mass, composition and structure.U is function of Temperature only.Internal EnergyStatement of First lawFor an infinetesimal process, the statement is: dU = q WSum of all forms of energy exchanged by a system with its surrounding is equal to the change in internal energy of the system which is a function of state.Energy can neither be created nor can be destroyedSignificance of the First LawIt is based on the law of conservation of energy.It brought in the concept of the internal energy.It separates heat and work interactions between the system and surroundings as two different terms.It treats internal energy as a state property is an exact differential.Internal energy in terms of Partial DerivativesFor given homogeneous system consisting of given amount of substance of fixed composition;U = F(P,V,T)If any two variables are independent third will be automatically fixed. This also therfore can be stated as:U = F(P,V); U = F(V,T) ; U = F(P,T)From the theorem of partial derivativesInternal energy in terms of Partial DerivativesEnthalpyIf pressure is maintained constant during change of system from state I to state II the work doneFrom the first law: UII UI = q P(VII VI )On rearranging these terms (UII + PVII ) - (UI + PVI ) = q Or HII - Hi = H = q Where H = U + PV is called entalpy. The heat content at constant pressure is called enthalpy.( )I IIIIIIIIV V P dV P PdV w Enthalpy is a state property but heat is not. As we know dH = dU + PdV + VdPAs U is a function of state so one can writeHence = M dP+N dV(Say)V P U H + dVVUdPPUdUP V

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.|T TPUEnthalpyInternal Energy Vs EnthalpyInternal energy is all that energy stored in the system. But, What does enthalpy mean? This can best illustrated by the example of calcination of calcium carbonate.CaCO3 = CaO + CO2The enthalpy change of the process will be equal to invested bond energy to break the bond between CaO and CO2 or internal energy provided both CaO and CO2 are solids. However if CO2 is allowed to form gas then breaking of one mole of CaCo3 nearly 22.4 ltrs. of CO2 will be formed. The expansionin volume will take place.In this process of expansion the system will do work equivalent to PdV as mechanical work on the surrounding.Hence in addition to the requirement of energy for breaking the bond of Cao-CO2 additional energy equivalent to PdV will have to be supplied to the system making a total of U + PdV and this is the enthalpy change of the system on calcination.For chemical processes where there is no significant change in volume as in2CaO3. SiO2 = 2CaO + SiO2 The PdV is practically absent and U and H are almost the same.In chemical and metallurgical world, even if term PdV is absent, it worthwhile to refer to H which is more appropriate.Internal Energy Vs EnthalpyHeat CapacityIn chemical and metallurgical processes, materials get heated or cooled and therefore it is necessary to know the amount of heat required to heat or amount of heat liberated on cooling a material over a certain temperature range.Different materials require different amounts of heat to get heated through the same temperature rise. This is because the materials have different heat capacities.This is so because of the variation in the crystal structure of the materials and their related parameters. The heat capacity of a substance is the amount of heat required to raise its temperature by one degree.In thermodynamics the molar heat capacity (C) i.e. heat capacity of 1 g-mole of a substance is most widely employed. ThusThe molar heat capacity at constant volume is given by Since at constant volume q = dU.TqCV VVTUTqC

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.| PT RVInterrelationship of CP and CVApplication of First law to Thermodynamic ProcessesWith the help of the first law, one is able to calculate the changes taking place in internal energy and enthalpy of a system during a thermodynamic process.Processes which are frequently studied and to which this law will be applied include: i) Isothermal process ii) Adiabatic process iii) polytropic processes.In all these cases working substance of the system shall be considered to be an ideal gasi) Isothermal Process: The process is carried out when dT = 0 thus dU = 0 . Henceq = W = PdVIt is also true that in this caseIfVf > Vi i.e. for expansion of gas q is positive i.e. the system will absorb heat from its surroundings and produce an equivalent amount of work.

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.| fifififiVVifVVVVVVVVT R V d T RVdVT R dVVT RPdV w qln lnApplication of First law to Thermodynamic ProcessesApplication of First law to Thermodynamic Processes IfVf < Vi i.e. for compression of gas q is negative i.e. the system will impart heat to the surroundings at the same time absorbing mechanical energy from them in the form of mechanical work done on it.The change in enthalpy of the system will be:Tf = Ti in the isothermal process.Thus in an isothermal process with an ideal gasinternal energy and enthalpy remains unchanged and work done is equal to the heat exchanged.0 ) () ( + i fi i f f i fT T RV P V P U H H HApplication of First law to Thermodynamic Processesii) Adiabatic Process: There is no exchange of heat between the system and surrounding i. e. q = 0. Hence first law takes the form dU = - W = - P dVOr dU = CV dT = - P dVThis indicate that system will perform work at the cost of its internal energy and therefore the lowering of temperature of the system will result.Adiabatic work done (w) Change in Internal energy = CV (Tf Ti) ( )i f VTTVT T C dT Cfi Application of First law to Thermodynamic ProcessesRelationship between P and V and T and V in adiabatic process.

Or Integration of this equation under the limiting condition V = Viat T = TiAnd V = Vf at T = Tf VdVT R dT CV VdVTdTRCV Adiabatic ProcessAnd substituting the expression for R inabove ExpressionOrwhere OrOr Also since by definitionH = U + P VOrdH = dU + P dV + V dPAs dU = - P dVSo dH = V dPififV PVVVTTC C Cln ln 1 1 i i f fV T V TVPCC Const V T 1 Const V P Adiabatic ProcessEnthalpy change in a adiabatic process is As for such process. This on integration lead to the expression fifiVVPPdV P dP V H Const V P ]]]]

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.| 11 nfi i i PVVR V P CHPolytropicProcessSummary of ThermodynamicProcessesProcess characteristics P-V-T relationshipWork done Heat exchangeIsothermal dT = 0PV = const RT ln (Vf / Vi) RT ln (Vf / Vi)Isochoric dV = 0 P/T = const 0 Cv (Tf Ti)Isobaric dP = 0 V/T = const Pi(Vf Vi) Cp (Tf Ti) + Pi(Vf Vi)Adiabatic q = 0 PV = const (PfVf- Pi Vi) / (1- ) 0polytropic - PVn = Const (PfVf- Pi Vi) / (1- n) Cv (Tf Ti).( - n)/(1-n)Thermo chemistry It may be defined as the branch of science which deals with the study of heat exchanges associated either with chemical reaction or physical changes in the state of matter such as melting, sublimation or evaporation etc.If heat is produced by a chemical reaction it is denoted by ve sign (Exothermic reaction)If heat is absorbed during a chemical reaction it is denoted by +ve sign (endothermic reaction).Hesss LawIt states that the total heat exchanged in a given chemical reaction, which may take place under constant pressure, volume or temperature is the same irrespective of the fact whether it is made to take place over a path involving formation of number of intermediate products or over the one involving the formation of final product from the reactant directly in one stage.

H =H1+H2+Hx+HyHesss LawVariation of heat Capacity with TemperatureExperimental data consists of CP as function of temperature.For the liquid and solid, The P V term is very small. Hence H is taken as equal to U and CP as equal to CV.In other words no distinction is made between CP and CV so far as applications are concerned.It has been found that experimental CP Vs. T data for elements and compounds fit best with an equation of type:CP = a + b T + c T-2 Where a, b, c are empirically fitted constant and differ from substance to substance.The last term is the smallest and therefore often ignored.In some cases, such as liquid metals, both bT and cT-2 are usually ignored.The above expression is also valid for diatomic and polyatomic gases as well.Variation of heat Capacity with TemperatureVariation of Enthalpy with TemperatureChange in enthalpy during the course of process (H) is equal to the heat supplied to the system (q) at constant pressure.The constant pressure restriction is mostly not important for example H is not function of P for ideal gases.Energies of solid and liquids are hardly affected by some changes in pressure due to their very small molar volumes. In other words, the VdP term is negligible.In most metallurgical and materials processing, gases are ideal and pressure is maximum a few atmosphere. Therefore H = q approximation is quite all right.Classification of Enthalpy changeAbsolute value of enthalpy change of a substance is not known. Only we can measureare changes of enthalpy.Enthalpy change occur due to various causes.Sensible heat: enthalpy change due to change of temperature of a substance is known as sensible heat. It is divided into:i) change in enthalpy without any change in aggregation of the substance: As a universal convention, thermo chemical data books take sensible heat at 298 K (250C) as zero for any substance. Hence sensible heat at temperature T, per mole of a substance is given as: 298 K is known as reference temperaturedT C H HTP T 298298Classification of Enthalpy changePutting the expression for heat capacity, we getWhere A, B, C and D are lumped parameters and functions of empirical constants a, b, c( )( ) ( ) ( )D T C T B T Ac T T adT T c T b a H HTbTT+ + + + + 1 22981 12 222982298298 298Classification of Enthalpy changeii) Enthalpy change due to changes in state of aggregation of substance :These are isothermal processes. By convention enthalpy change s for all isothermal processes are designated by H. For example, Hm = enthalpy change of one mole of solid due to melting (i. e. latent heat of fusion per mole )Hv = enthalpy change of one mole of liquid due to vaporization (i. e. latent heat of vaporization per mole )Classification of Enthalpy changeConsider a pure substance A, which is undergoing following changes during heating from 298to T KA (Solid) A (Liquid)A (Gas) A (Gas) at 298 K at Tmat Tb at TTm and Tb are the melting and boiling points of A.ThenCp(s), Cp(l) and Cp(g),are for solid liquid and gaseous A. it is only applicable for pure substance.dT g C HdT l C H dT s C H HmbbmmTTP VTTP mTP T + ++ + ) () ( ) (298298Classification of Enthalpy changeHeat of reaction (H): This is the change of enthalpy that occurs when a reaction takes place. By convention reaction is considered to be isothermal. Consider the following reaction occurs at temperature T:A (pure) + BC (pure) = AB (pure) + C (pure)Then H (at T) = HAB (at T) + HC (at T) - HA (at T)- HBC (at T)Where HAB , HBC, HAand Hc are molar enthalpies of pure Ab, BC, A and C, respectively.Classification of Enthalpy changeHeat of mixing (Hmix): This is the change of enthalpy that occurs when a substance is dissolved in solvent.This process is generalized as A (pure)=A (in solution)This process is also accompanied by a change ofenthalpy(Hmix), where (Hmix) = HA (in solution) HA(pure)Again by convention, the process is assumed to be isothermal for thermodynamic calculations Classification of Enthalpy changeSome comments:For calculation of enthalpy change, reactions, dissolutions and phase transformations has been assumed to occur isothermally. Since enthalpy is a state property, it depends only on initial and final states and not the path.For a reversible isothermal process, the temperature remains constant all through. If the process is not reversible, then temperature at the beginning and end of a process would be same. In between, temperature can vary significantly. Sign convention for H:The process accompanied by liberation of heat are called exothermic. This happens if the enthalpy in the final state (state 2) is lower than the initial state (State 1) i. e. H2 < H1 so for the processstate 1 state 2we have H = H2 H1< 0Therefore H is negative. The opposite is an endothermic process which is characterized byabsorption of heat and positive value of H.Standard state of enthalpy The stable state of a substance changes with the temperature. The stable state of H2O is ice which is below 0 oC, liquid water 0-100 oC and a stable gas at 1 atm pressure above 100 oC. Considering all these points a standard state has been defined as a pure element or compound at its stable state at the temperature under consideration and at 1 atm pressure. Thus the standard state of H2O at 50 oC is pure water at 1 atm pressure. By convention enthalpy changes at standard state are denoted by subscript 0 e. g. 0 0T TH and H As already mentioned, 298 K is the universal reference temperature for compilation of sensible heats. By this convention, sensible heat at standard state of a substance is arbitrarilytaken as zero at298 K.This is solely for calculation of sensible heats not H0 for a process occurring at 298 KStandard state of enthalpyKirchoffs LawUtility:It allows us to calculate the enthalpy changes at various temperatures, provided the enthalpy is known at some other temperature, and the heat capacity of the reactants and products are known in the range of temperatures under consideration. Derivation:Consider a chemical reaction at temperature T1 whose enthalpy change is H1. Calculate the enthalpy change H2 at temperature T2.The product of this reaction at T2 can be obtained in many different ways. Let us, however, consider the reaction to be carried out first by reacting the reactants (x+y) at T1 and then raising the temperature of the products from T1 to T2 along ABC.Kirchoffs LawThe heat absorbed by this process will beThe second way of obtaining the same result is to raise the temperature of the reactantsfrom T1 T2 and then react them together at T2 i. e. along ADC.The heat absorbed by this process is + 211TTP dT C Hz( )+ + + 21y xTTP P 2dT C C HKirchoffs LawAccording to Hesss law the heats absorbed during the two ways of producing z must be the same since the initial conditions of the reactants and final conditions of the products are in each case the same.= Or OrOr ( )+ + + 21y xTTP P 2dT C C H+ 211TTP dT C Hz( ) [ ]+ 211 2TTP P PdT C C C H Hy x z 211 2TTP dT C H H + 211 2TTP dT C H HKirchoffs LawSince all the terms are known in the right hand of the equation, H2 can be calculated. We know OrThis is the statement of the Kirchoffs law. It means the rate of change of enthalpy of a process or a reaction with temperature is given by the difference of the heat capacity at constant pressure of products and reactants taking part in the reaction 211 2TTP dT C H HPPCTH

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.| Kirchoffs LawProblem-1Problem-2Enthalpy change: H (a c) = U (ac) + (Pc Vc Pa Va) = - 9.13 + n R (Tc Ta) = - 9.13 + 4.09 x 8.3144 x (119 - 298) = - 9.13 - 6.0870 kJ = - 15.2170 kJProblem-3 Problem - 5Calculate the heat of the following reaction at 1000KFe2O3(s) + 3C (s) = 2Fe(s) + 3CO(g) from the following data:Cp( , Fe) =4.18 + 5.92 x 10-3 Tcaldeg-1mol-1Cp(CO) = 6.79 + 0.98 x 10-3T 0.11 x 105 T-2 caldeg-1mol-1 Cp(Fe2O3) = 4.10 + 1.02 x 10-3 T 2.10 x 105T-2caldeg-1mol-1The heats of formation of Fe2O3 and CO at 298 K are -197000 and 26400 cal/mol. respectively.Solution Cp = Cp (product) Cp (reactant) = 2.Cp( -Fe)+ 3 Cp(CO) Cp(Fe2O3) 3Cp(C) 2 x Cp( -Fe) = 8.36 + 11.84 x 10-3T3 x Cp(CO) =20.37 + 2.94 x 10-3 T -0.33 x 105 T-2__________________________________________Cp (product) = 28.73 + 14.78 x 10-3T 0.33 x 105 T-2Cp = 23.49 + 18.60 x 10-3T 3.55 x 105 T-23Cp(C) = 12.30 + 3.06 x 10-3T 6.30 x 105 T-2__________________________________________Cp (reactant) = 35.79 + 21.66 x 10-3T -4.85 x 105 T-2 Cp = -7.06 - 6.88 x 10-3T + 9.52 x 105 T-2Solution= 117800 7.06(1000-298) -6.88. (10002 2982) x 10-3 9.52 x 105 (1000-1 298-1)=117800 5850= 111950 cals.dT C H Hp.1000298298 1000 + ( )10002982 5 3dT ) T 10 x 52 . 9 + T 10 x 88 . 6 06 . 7 + 117800 =Problem-6The standard heat of formation, of ammonia gas is -11.03 kcal/mol. at 250C utilizing the data given below derive a general expression for heat of formation applicable in the temperature range 273 1500K.1 1 2 6 33deg 10 728 . 0 10 787 . 7 189 . 6 , + mol cal T x T x NHCp1 1 2 6 32 ,deg 10 0808 . 0 10 414 . 1 450 . 6 + mol cal T x T x N Cp1 1 2 6 32 ,deg 10 4808 . 0 10 2 . 0 947 . 6 + mol cal T x T x N CpSolutionThe reaction is = -7.457 + 7.38 x 10-3T 1.409 x 10-6T2 ( ) ( ) ( ) mol cal H g NH g H g N / 1103023210298 3 2 2 +223,2321H C C C Cp N pNH pp 298273p 273 298dT . C + H = H dT C - H = H 298273p 298 273( )2982732 6 3dT T 10 x 409 . 1 T 10 x 38 . 7 + 457 . 7 - -11030Solution=-11030 + 7.457 (298 273) - x 7.38 x 10-3 (2982-2732) +x 10-6 (2983-2733)= -11030 + 186.4 52.7 +2.9 = -10893 cal.= -10893 7.457 T + 3.69 x 10-3T2 0.47 x 10-6T3 (-7.457 x 273 + 3.69 x 10-3 x 2732 0.47 x 10-6 2733)= -10893 +2036 275 + 10 7.457 T + 3.69 x 10-3T2 0.47 x 10-6T3= - 9122 7.457T + 3.69 x 10-3T2 0.47 x 10-6 T3=-9122 7.457T + 3.69 x 10-3T2 0.47 x 10-6 T3 cal/molFor H1000K solution T = 1000 in the above equation H1000K= -9122 -7457 + 3690 470 = -13359( ) + + TTdT T x T x H H2732 6 327310 409 . 1 10 38 . 7 457 . 7Solution= - 11030 7.457 (1000-298) + 3.69 x 10-3 (10002-2982) 0.47 x 10-6 (10003-2983)= -11030 5234.8 + 3362.3 457.6 = -13360.1 cal/mol1000298p 298 1000dT C + H = H ( )dT T 10 x 409 . 1 T 10 x 38 . 7 + 7457 . 7 + 11030 - =10002983 6 3Problem- 4ProblemCT-ISolutionSolution for Q-2: Fe + O2 = FeOCal 63200 - = H = H FeO 298 112310332 p tr10332981 p 298 1123dT C + H - dT C + H = H 2 5 32 O p mag Fe p Feo p 1 pT 10 47 0 - T 10 42 4 - 9 3 = C21- C - C = C . . ., , , ,2 5 32 O p nonmag Fe p Feo p 2 pT 10 47 0 - T 10 5 1 + 92 0 - = C21- C - C = C . . ., , , ,kcal 208 63 - = H 1123.CT-ISolutionSolution for Q-3:(a)As U, T, ans S are dependent state functionsWe can express G in the following mannerdG = dU + PdV + VdP TdS SdT( ) ( )dV + dP = dUPV U VP U ( ) ( )dV + dP = dSPV S VP S ( ) ( )dV + dP = dTPV T VP T CT-ISolution

( ) ( ) ( ) ( ) ( ) ( )NdV + MdP =dP S - T - V + + dV S - T - P + = dGVP T VP S VP u pV T pV S pV u ] [ ] [( ) ( ) ( ) ( ) ( )) (PVT pV T vP S PVS pV S vP T PVU vP M 2 2 2S - - T - - 1 + =( ) ( ) ( ) ( ) ( )) (PVT pV T vP S PVS pV S vP T PVU PV N 2 2 2S - - T - - 1 + =So G is a function of stateCT-ISolution(b)dH = TdS + VdP( ) [ ] ( )NdV + MdP =dV T + dP V + T = dHPVS vPS ( ) ( ) ( )1 + T + =PVS VP S PV T PV M 2) (( ) ( ) ( )) (PVS PV S VP T VP N 2T + =So H is not a function of stateSecond law of ThermodynamicsSystem which are away from equilibrium, upon initiation, shall move towards equilibrium and such processes are referred to as natural or spontaneous or irreversible processes.Examples: heat flow from hotter to colder body or depressurisation of inflated tube in a low pressure surroundings or free fall under gravity and so on.IntroductionThe spontaneous change from an existing state to equilibrium state is possible because in the existing state the system happens to be at higher potential which is the driving force for the change to occur.Higher temperature is therefore the higher thermal potential which makes heat flow from higher to a lower temperature. Similarly higher pressure is a higher mechanical potential and so on. Second law of ThermodynamicsIf the system is in equilibrium and if it is to be moved in the reverse direction in the above examples it would be termed as unnatural or non-spontaneous processes.Therefore water can not be raised to an overhead tank unless energy in the form of motor pump set is provided to the system from the surroundings. Similarly heat can not flow from a colder to a hotter body unless aided by the surroundings in the form of compressor energy as in the refrigerator.But in chemical processes it is not readily possible to assess as to what is a natural or unnatural process. This can only be evaluated from the equilibrium state of the system.The knowledge of equilibrium CO / CO2 ratio in contact with Fe or FeO can only guide us as to how to prevent oxidation of iron or effect reduction of iron oxide by providing a suitable CO / CO2 gas mixture as surroundings.Heat EnginesHeat engines operate in a cycle, converting heat to work then returning to original state at end of cycle.A gun (for example) converts heat to work but isnt a heat engine because it doesnt operate in a cycle.In each cycle the engine takes in heat Q1 from a hot reservoir, converts some of it into work W, then dumps the remaining heat (Q2) into a cold reservoirHOTCOLDEngineQ1Q2WThis whole process for conversion of heat to work necessarily produces a permanent change in the cold reservoir by way of release of heat into it. This is called compensation. It means the entire heat that is absorbed initially is not converted to useful work but a part of it is rejected to a cold reservoir as of necessity. In other words in a cyclic process it is just not possible to convert all heat into mechanical work. In a non-cyclic process it may be possible.HOTCOLDEngineQ1Q2WEfficiency of a heat engineDefinition:cycle perinputheat cycle perdone work efficiency 1 QWBecause engine returns to original state at the end of each cycle, U(cycle) = 0, so W = Q1 - Q2Thus:12112 1QQQQ Q Efficiency of a heat engineAccording to the first law of thermodynamics (energy conservation) you can (in principle) make a 100% efficient heat engine.BUT.The second law of thermodynamics says you cant:Kelvin Statement of Second Law:No process is possible whose SOLE RESULT is the complete conversion of heat into workWilliamThomson,Lord Kelvin (1824-1907) HOT COLDHOTTER COLDERWARM WARMQHOT COLDQBoth processes opposite are perfectly OK according to First Law (energy conservation)But we know only one of them would really happen Second LawHeat flowClausius Statement of Second Law:No process is possible whose SOLE RESULT is the net transfer of heat from an object at temperature T1 to another object at temperature T2, ifT2 > T1Rudolf Clausius(1822-1888) How to design a perfect heat engine1)Dont waste any workSo make sure engine operates reversibly (always equilibrium conditions, and no friction).2)Dont waste any heatSo make sure no heat is used up changing the temperature of the engine or working substance, ie ensure heat input/output takes place isothermally Sadi Carnot(1796-1832)HotSourceT1Q1PistonGasabWorking substance (gas) expands isothermally at temperature T1, absorbing heat Q1 from hot source. The Carnot Cycle (I): isothermal expansionT1PVT1T1abQ1Q1The Carnot Cycle (II): adiabatic expansionGas isolated from hot source, expands adiabatically and temperature falls from T1 to T2.PistonGasbcGas isolated from hot source, expands adiabatically, and temperature falls from T1 to T2PVT1T1abT2T2cThe Carnot Cycle (III): isothermal compressionPistonGasdcQ2ColdSinkT2Gas is compressed isothermally at temperature T2 expelling heat Q2 to cold sink.T2VPT1abT2cdPT1T1abT2T2cdQ2Q2The Carnot Cycle (IV):adiabatic compressionGas is compressed adiabatically, temperature rises from T2 to T1 and the piston is returned to its original position. The work done per cycle is the shaded area.PistonGasadGas is compressed adiabatically, temperature rises from T2 to T1 and the piston is returned to its original position. Work done is the shaded area. VPT1abT2cdWEfficiency of ideal gas Carnot engine12112 1QQQQ Q We can calculate the efficiency using our knowledge of the properties of ideal gasesVPT1abT2cdWQ2Q1VPT1abT2cdWVPT1T1abT2T2cdWWQ2Q2Q1Q1ababVVnRT W Q ln1 1 Isothermal expansion (ideal gas)PVT1T1abQ1Q1VaVbIsothermal compression (ideal gas)dccdVVnRT W Q ln2 2 VPT1abT2cdQ2VPT1abT2cdQ2VPT1abT2cdPT1T1abT2T2cdQ2Q2abdcVVnRTVVnRTQQlnln1 11212 (2)(1)12111211 d ac bV T V TV T V TAdiabatic processes) 2 ( ) 1 ( dcabVVVV2211121TQTQTT Efficiency of ideal gas Carnot engineVPT1abT2cdWQ2Q1VPT1abT2cdWVPT1T1abT2T2cdWWQ2Q2Q1Q1So.. hcTT 1 0 hhcchhccTQTQTQTQConservation of Q/TFor all Carnot Cycles, the following results hold:What about more general cases?????0 hhcchhccTQTQTQTQWas derived from expressions for efficiency, where only the magnitude of the heat input/output matters. If we now adopt the convention that heat input is positive, and heat output is negative we have:0 +hhccTQTQThe expressionIn other words0cycleTdQFor any reversible cycle:The Carnot CycleThe Importance of the Carnot Engine1. All Carnot cycles that operate between the same two temperatures have the same efficiency.2. The Carnot engine is the most efficient engine possible that operates between any two given temperatures.EntropyTo emphasise the fact that the relationship we have just derived is true for reversible processes only, we write:0 cyclerevTdQWe now introduce a new quantity, called ENTROPY (S) TdQdSrevEntropy is conserved for a reversible cycle Is entropy a function of state?PVABpath1path2For whole cycle:0 cyclerevTdQS2 path 1 path2 path 1 path0

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.| BArevBArevABrevBArevTdQTdQTdQTdQA B BA BAS S S S 2) (path 1) (pathEntropy change is path independent entropy is a thermodynamic function of stateReversible cycleIrreversible processesCarnot EnginehcrevhccTTQQ

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.| 1 1 Irreversible EnginehcirrevhcTTQQ

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.| T2Entropy change of vessel I = - q1 / T1Entropy change of vessel I =q / T2Total change of entropy will be the sum of entropy changes of the two vessels. Thus When q is +ve and T1 T2 is +ve then the entropy change for a real irreversible process in a closedsystem must also be positive.( )2 12 1ocess Pr1 22 1 ocess PrT TT T qSTqTqS S S

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.| + T1 > T2 ; Sirr > 0 (+ve) : The entropy change of a real process is greater than zero(+ve). T1 = T2 ; Srev = 0 (zero): Dynamic equilibrium exists between two vessels and there is no heat transfer.T1 < T2 ; Sirr < 0 (-ve). : Entropy change is negative and the process proceeds in the reverse direction.Sign of entropy change shows the direction of flow of heat energy.Calculation of entropy changeEntropy is a state property. Hence the basis and procedure for calculation of entropy changes are similar to those for the enthalpy changes.Hesss law and Kirchoffs law are applicable here too.A pure substance at its stablest state also constitutes standard state for entropy at that temperature which is designated as.0TSFollowing significant difference are to be noted between entropy and enthalpy.Entropy changes are to be calculated only through the reversible path. This restriction is not there for any other state property, including enthalpy.Absolute value of entropy can be determined. Thermodynamic data sources provide the values of entropy of a substancefor pure substance. This is in contrast with the energy where only changes are available in the data sources. Entropy changes associated with phase transformationFor a pure substance, reversible phase changes (melting, boiling etc.) at a constant pressure occurs at a constant temperature. Therefore,for melting for boiling

in general for phase trans.m0m 0mTHS V0V 0VTHS tr0tr 0trTHS A (Solid) A (Liquid)A (Gas) A (Gas) at 298 K at Tmat Tb at TTm and Tb are the melting and boiling points of A. Then( ) ( )( )dTTg CTHdTTl CTHdTTs CS STT0PV0VTT0Pm0mT2980P02980Tbbmm ++++ Entropy changes in chemical processesBy nature, reactions and mixing are irreversible. Like enthalpy changes, entropy changes are considered only for the isothermal process. However, they cannot be calculated from enthalpy as done for the phase transformation in view of irreversibility.Consider the reactionA + BC = AB + CThe entropy change of the reaction at temperature T) T at ( S ) T at ( S ) T at ( S ) T at ( S S0BC0A0C0AB0T + The entropy change of the reaction with temperature can be calculated as ( ) ( )( )( )dTTCSdTTCdTTCSS S S S S S211212111 2 1 2 1 2TTP0TTTt tan reacPTTproductP0Tt tan ac Re0T0Tproduct0T0T0T0T + + + Various interpretations of entropy for an infinitesimal, isothermal reversible process.Entropy is times arrow i. e. a fundamental indicator of time.Entropy has the relationship with heat not available for work.Entropy is a measure of disorder of a system.TqdSrevCombined statement of first and second lawIf a system is capable of doing only mechanical work the first law equation can be put as:dU = q P dV. It is not exact differential. Holds true for the reversible processes.The second law for the reversible process gives dS = q rev / T or q = T dSThe two laws therefore can lead to dU = TdS P dVor TdS = dU + PdVThis is the combined statement of first and second law.It includes only those terms which are state functions only and hence is exact differential equation.For irreversible process second law gives dS > 0Or therefore, TdS > dU + PdVFor a unnatural or non spontaneous processdS < 0Or therefore TdS < dU + P dVThermodynamic PotentialsA system by itself, in isolated state or in contact with surrounding shall stay in equilibrium unless acted upon by some constraints.If a system tends to move as a result of being not in equilibrium or as a result of external constraint, there must be a driving force making the system move from within itself or under the applied constraints. This driving force is referred to as thermodynamic potential driving the system to change to a new state.Is this driving force or potential the same under all conditions of the system or it varies from condition to condition of the system ?Heat flow s from higher to lower temperature. The flow is possible due to higher thermal potential.Similarly the higher pressure is the driving mechanical potential.Also higher electrical voltage is driving the electrons under the influence of electrical potential as voltage.However the question remains as to what is that potential, forcing a thermodynamic process to take place?It is easy to imagine that a system with higher associated energy will be relatively unstable as compared to the one having lower energy.In other words, when a natural changes take occurs the system moves from higher to lower energy levels or moves from higher to lower thermodynamic potentials.It also means that the energy is a potential driving the system for change to occur.It also means that for equilibrium to exist the potential of all the systems or sub-systems within it must be at the same potential. So far the internal energy, enthalpy and entropy have been evolved as energy parameters but it is not yet known as to whether these act as potential in driving a particular process?It must be noted that for natural process the system moves from lower to a higher entropy level hence entropy can not be considered as a driving force.Can internal energy and enthalpy qualify as potentials capable of driving the process towards equilibrium if already at higher levels ?If internal energy and enthalpy are potential terms then changes in them will have to be zero at equilibrium and these will have to decrease for natural changes and increase for unnatural changes.Let us therefore see if criteria can be evolved mathematically to evaluate what constitute as thermodynamic potential and if so under what conditions?Potentials under constant volume and constant entropy conditionsMathematically these conditions are expressed asdV = 0 and dS = 0i) TdS > dU + P dVfor natural processSo under above condition we can write dU < 0ii) TdS = dU + P dVfor the equilibrium processSo under above conditions we can write dU = 0iii) TdS < dU + P dV for unnatural processSo under above conditions we can writedU > 0As internal energy decreases for natural process, remains same at equilibrium and increases for unnatural process thus fully qualifies to be called as thermodynamic potential under the conditions of constant volume and entropy. As a corollary, the internal energy must be the function of entropy and volume can be expressed: U = F (S, V) Total differential of internal energy is given by

dVVUdSSUdUS V

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.|We know: dU = TdS P dVOn comparison this leads to andBy specifying U as function of S and V it is possible to evaluate T and P thereby describing the system fully. By describing U as a function of any other two variables, it is not possible to describe the system fully.TSUV

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.|Potentials under constant pressure and constant entropy conditionsMathematically these conditions are expressed asdP = 0 and dS = 0i) TdS > dU + P dVfor natural processSo under above condition we can write dU + d (PV ) < 0Or d (U + P V) < 0Or dH < 0ii) TdS = dU + P dVfor the equilibrium processSo under above conditions we can write dU + d (PV ) = 0Or dH = 0iii) TdS < dU + P dV for unnatural processSo under above conditions we can writedU + d (PV ) > 0Or dH > 0So on the whole dH > = < 0 for unnatural, equilibrium and natural processes as the case may be. Therefore, the term enthalpy qualifies as being called a potential term under constant pressure and entropy conditions.Enthalpy can therefore be described mathematically as:H = F (P, S)Total differential of enthalpy can be written as Also by definition:dH = d (U + P V)= dU + P dV + V dP= T dS + V dP On comparison we getdPPHdSSHdHS P

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.|Mathematically these conditions are expressed asdV = 0 and dT = 0i) TdS > dU + P dVfor natural processSo under above condition we can write dU - TdS < 0 Or d ( U T S) < 0Let us define new mathematical termA = U T Sthen dA 0Or dA > 0The newly defined function qualifies for being referred as a potential term under constant volume and temperature conditionsA being the function of all state variables A also must be state function. Expressing as before A= F (V, T)Total differential of ANow A = U T S dA = dU TdS S dTAnd since, TdS = dU + P dVThereforedA = - S dT P dVOn comparison we getdVVAdTTAdAT V

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.|This function A was first defined by Helmholtz and therefore is known as Helmholtz free energy.We knowdA = dU T dS And for reversible processqrev = T dSAn thereforeqrev = dU dAFrom the First law dU = qrev WAnd hence qrev = dU + WComparing the two equations for qrev we get- dA = WOr dA = - WThat the change in Helmholtz free energy in a process is equal to the amount of work done by the system on the surrounding or is equal to the amount of work the system is capable of doing.These are by far the most commonly adoptedconditions in chemical and metallurgical engineering practices. It meansdP = 0 and dT = 0i) TdS > dU + P dVfor natural processSo under above condition we can write dU + d (P V) - TdS < 0 Or d ( U + P V T S) < 0Let us define new mathematical termG = U + P V T Sthen dG 0Or dG > 0The function defined above thus qualifies for being called as a potential term under constant pressure and temperature conditionsG is a function of all state variable. soit is a state function.The function G is referred to as Gibbs free energy.NowG = U + P V T SAnd H = U + P VThen G = H T SOrG = H T Sfor a finite changeOrdG = dH T dS SdTin differential form At constant pressure dP = 0 and dH = q = TdSAnd hence dG = - SdTOr ST dG dP

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.|Again dG = dH T dS SdT On puttingdH = dU + P dV + V dP dG = dU + P dV + VdP T dS SdT Putting in this dU + P dV = T dS dG = T dS + V dP T dS S dT = V dP S dTAt constant temperature when dT = 0 ; dG = V dPOrSince G is a state function any change in G can be represented as G = U T S + (P V)= - Wrev+ (P V)The change of Gibbs free energy during a process is equal and opposite in sign to the net reversible work obtainable from the process occurring under isothermal condition when correctedfor change in volume under isobaric condition.VP dG dT

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.|Since G is a state function any change in G can be represented as G = U T S + (P V)= - Wrev+ (P V)The change of Gibbs free energy during a process is equal and opposite in sign to the net reversible work obtainable from the process occurring under isothermal condition when correctedfor change in volume under isobaric condition.Important thermodynamic relationsFor closed and isolated systems have fixed mass and composition and thereversible work is done against pressure. dU = T dS - PdV dH = T dS + V dP dA = - S dT P dV dG = - S dT + V dPCriteria for thermodynamic equilibriaDifferential form Finite difference form(dU)S, V = 0(dH)S, P = 0(dA)T, V = 0(dG)T, P = 0( U)S, V = 0( H)S, P = 0( A)T, V = 0( G)T, P = 0Since it is easy to maintain temperature and pressure constant the Gibbs free energy criteria is employed in chemical and metallurgical processes. However, in other areas, other criteria are also employed.Maxwells equationsYXxNyM

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P TV TP SV STVPSTPVSSVPTSPVT

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.|Maxwells relations are used frequently in thermodynamics for the calculation of changes in thermodynamic variables for different processes.The Driving Force of a Chemical Reaction Why does a chemical reaction take place?It wasbelieved that the energy change accompanying a reaction can be measured directly by the enthalpy change at constant pressure, or change in intrinsic energy ( U) at constant volume. The reasoning behind this would be apparent if a system losses energy as a result of a chemical reaction, that reaction will take place spontaneously and the greater the quantity of heat lost, the greater the driving force behind the reaction.C (s)+ O2 (g)= CO2(g) with evolution of heat energy, i.e. cal/mol which happens to be a driving force in this case.2 Fe (s) + O2 (g) = 2FeO(s) and3FeO(s) + 2 Al (l) = 2Fe(l) + Al2O3(s)Both have negative heats of reaction at 16000C and both take place spontaneously On the other hand, if we consider the reaction; ZnO(s) + C(s) = Zn(g) + CO (g) andThis reaction will not take place at 250C but if the system is heated to 11000C, carbon will reduce zinc oxide to produce zinc metal. 940500298 H8320001373+ H 568000298+ HThe Driving Force of a Chemical Reaction (cont.)we cannot use heats of formation as a criterion of their tendency to take place.We must therefore search for a more consistent rule for the driving force of a reaction.Consider the reaction: ZnO(s) + C(s) = Zn(g) + CO (g)At 11000C, i.e above the boiling point of Zn (9070C), the reaction will take place between phases indicated above.At 250C, zinc is a solid (melting point 419.60C) so that the reaction isZnO(s) + C(s) = Zn (s) + CO (g)The Driving Force of a Chemical Reaction (cont.)The most obvious difference in these reactions apart from their temperature, is the difference in the physical state, means a difference in the state of order of a system and consequently in the entropy of the system. S0 at 250C should be much less than that 11000C, because at 11000C two molecules of gas are being produced from two solid molecules whereas at 250C two solid moles only produce one gaseous mole. The Driving Force of a Chemical Reaction (cont.)A solid to gas transformation means a comparatively large entropy increase and the following figures for the reduction of ZnO by C: Why not use entropy change of a reaction as a measure of the driving force behind a chemical reaction?The idea is immediately contradicted if we consider the formation of the oxide of any metal:We know that this reaction is spontaneous an oxide film forms readily on iron at room temperature. Thus a positive entropy change is not the criterion of a chemical reaction.deg mol / cal 68 = S and deg mol / cal 46 = S 1373 298mol cal 17 - = S s FeO = g O21+ s Fe0298 2deg / , ) ( ) ( ) (The Driving Force of a Chemical Reaction (cont.)The Second Law of thermodynamics states that a spontaneous process is always accompanied by an increase in entropy of the system and its surroundings.The surroundings receive e a quantity of heat - H at constant temperature and pressure entropy increase of the surroundings = If S is the entropy change of the system, total entropy change of the system and surroundings = S For the 2nd Law to be obeyed ( S -) must be +veTH TH TH ve + =TH - S TThe Driving Force of a Chemical Reaction (cont.)we can say that ( H T S) must always be ve for a reaction to proceed spontaneously in order that the total entropy change of the system and surroundings can be +ve, when the reaction proceeds.We should examine the factor ( H T S) for the above three reactions in question.For the reduction of ZnO by C at 250C The reaction cannot proceed because this factor is +ve, and the total entropy change of the system and its surroundings is ve This is negative so that the reaction will take place spontaneously at 11000C.This explains why zinc oxide smelting must be carried out at temperature of the order of 11000C in order that reduction of the oxide by carbon can proceed.( )mol cal 43100 + = 46 298 - 56800 = S T - H 298 298deg /( )cal 10200 - = 68 1373 - 83300 = S T - H C 1100 at1373 13730The Driving Force of a Chemical Reaction (cont.)For the oxidation of iron at 250C Again it is negative, and the reaction takes place spontaneously at 250C.Consistent RuleThe driving force of a reaction can be calculated as ( H - T S); the more negative this factor, the greater the driving force and if the factor is +ve, the reaction will not proceed spontaneously. G = H - T Scalled Gibbs Free energy( ) . 57140 17 298 63200298 298cal S T H The Driving Force of a Chemical Reaction (cont.)It is the maximum work available from a system at constant pressure other than that due to a volume change.Most metallurgical processes work at constant pressure.We had already seen the fundamental importance of the factor ( H-T S), so that G is a measure of the driving force behind a chemical reaction.For a spontaneous change in the system, G must be negative, the more negative, the greater the driving force.The Driving Force of a Chemical Reaction (cont.)Thermodynamic Mnemonic SquareIn order to memorize important thermodynamic relationship involving the different thermodynamic potentials Max Born suggested a diagram known as Thermodynamic Mnemonic Square.Four sides of the square are leveledin alphabetical order starting from the top and moving in clock-wise directionwith four thermodynamic potentialsnamely A, G, H, U. The other four primary functions namely V, T, P, S are placed at the four corners in such a way that each thermodynamic potential is surrounded by the condition under which it actsUVTS P HGAWe can write the total differentials of each ofpotentials with help of this diagram as follows:Differential of any potential is equal to the sum of the differentials of its adjoining variables with their coefficients equal to diagonally opposite variables.This coefficients are taken to be positive if arrow points away from the variable and negative if it points towards the variable. dG = - S dT + V dPMaxwells relations can also be read from the diagram as follows:It is required to consider the corners onlyThe square is rotated in anti clock-wise direction in such a way that the function whose partial differential is to be arrived it appears on the top left hand corner of the square.The partial derivative of this variable w.r. t variable on the bottom left hand corner keeping the lower right hand corner variable as constant is equal to the partial derivative of the top right hand corner variable w.r.t bottom right hand corner variable with bottom left hand corner variable as constant.Negative sign on the right hand side of the equation is put if the arrows are placed unsymmetrically with respect to vertical axis drawn from centre of the square on rotation.V SSPVT

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.|Standard state of free energyStandard state is pure element or compoundat 1 atm pressure and at its stablest state at the temperature under consideration.Free energy change of the reaction are generallycalculatedwhen the reactants and products arepresent in their standard states is called the standard free energy change.The standard free energy is designated by G0T.Like enthalpy, we cannot measure the absolute value of free energy but change in free energy is quite possible to measure.Therefore, there must be some reference point with respect to which the actual values of various substance can be calculated.The free energies of stable form of the elements at 298 K and 1 atm pressure are arbitrarily assigned as zero value.The free energy of formation of compound are calculated on the basis of the above assumption and the value is described as standard free energy of formation.This quantity is generally reported at 298 K and for compound say MO, it would be written as G0298, MO.The Hesss law is applicable for calculation of free energy change as it is a state property.The standard free energy of formation of compound and the standard free energy of compound are same. M + O2 = MO G0298 = G0298,MO- G0298,M - G0298, O2 G0298,MO = G0298,MO00known be must products and t tan reac of energies free dard tan s thereaction a of change energy free dard tan s the calculate to inorderG b G a G d G cG G Gby given is K 298 at G change energy free dard tan s theD d C c B b A areaction the For0B , 2980A , 2980D , 2980C , 2980t tan ac Re , 2980oduct Pr , 29802980 + + + Some thermodynamic relationshipsBy definition G = H TS = U + PV TSdifferentiating dG =dU + PdV + VdP TdS S.dT (1)Assuming a reversible process involving work due only toexpansion at constant pressureAccording to First Law: dU = dq PdV andfrom Second Law:dq = TdSdU = TdS PdV(2) From (1) and (2) we getdG = (TdS PdV) + PdV + VdP TdS SdTor dG = VdP SdT(3)Some thermodynamic relationshipsAt constant pressure, dP = 0, so that= - Sand at constant temperature, dT = 0 and dG = VdPfrom PV = RT for ideal gas, dG = Integrating between the limits PA and PB at constant Temperature if GA is the free energy of the system in its initial state and GB the free energy in its final state when the system undergoes a change at constant pressurePTG

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.|dPPRT) 4 ( ln BPAP ABA BPPRTPdPRT G G Gbut G = GB GA and S = SB SA and so that d ( G) = - S.dTon substitutionSome thermodynamic relationshipsdT S dG and dT S dGB B A A .( ) ( )dT S - S = G - G dA B A BSTGP ]]]

( )T PTGH S T H G,]]]

+ known on Gibbs Helmholtz equationCalculation of G0 at high temperatureIt is possible to calculate G0of a reaction at high temperature from the H0 and S0values at 298 K (available in the literature ) in this wayIt must be remembered that if any transformation takes place between 298 and T K in reactants and products must be introduced in the above equation.]]]

+ ]]]

+ T298P 0298T298P02980T0T0T0TdTTCS T dT C H G orS T H GFugacityThe variation of Molar Gibbs free energy of a closed system of fixed composition, with pressure at constant temperature is given bydG = V dPFor one mole of ideal gas dG = (RT / P) dP = RT d ln PFor isothermal change of pressure from P1 to P2 at T G (P2, T) G (P1, T) = RT ln (P2 / P1) Fugacity contdAS Gibbs free energy do not have absolute values, it is convenient to choose an arbitrary reference state from which changes in Gibbs free energy can be measured. This reference state is called standard state and chosen as being the state ofone mole of pure gas atone atm pressure and the temperature of interest.The Gibbs free energy of the ideal gas in the standard state G (P = 1, T) is designated as G0 (T).Fugacity contd..Thus the Gibbs free energy of 1 mole of gas at any other pressure P is given as G (P, T) = G0 (T) + RT ln PSimply G = G0 + RT ln PSo the molar Gibbs free energy of the ideal gas is a linear function of the logarithm of the pressure of the gasFugacitycontdIf the gas is not ideal then the relation of molar free energy and logarithm of pressure is not linear.A function is invented which when used in place of pressure gives a linear relationship.This function is called Fugacity ( f ) is partially defined as dG = RT dln f In addition, as the real gas and the ideal gas behave the same at very low pressure, it is obvious that(f / P) 1 as P 0.after integration from standard state to any arbitrary state we can write G - G0=RT ln (f / f0)Where G0 is the Gibbs free energy of the real gas in the standard state. fis the fugacity at aspecified state and f0 is standard state fugacity.Fugacity is the measure of escaping tendency of a gas. Fugacity is the effective pressure corrected for non-idealityFugacitycontdActivityActivity (a) is defined asSo we can write

0iii0ffa orffa i0ii0a ln RT G G Or a ln RT G G Partial molar free energy of component i in the state of interest.free energy of component I in the standard state.The standard state of a substance are generally chosen as the pure solid or liquid form of the substance at 1 atm pressure and temperature under considerationor as the gases at 1 atm pressure at the temperature under consideration.The activity of a substance in its standard state is seen to be unity.Equilibrium ConstantLet us consider a general chemical reaction at constant temperature and pressureCapital letters for chemical element and small letters for the number of gram mole. The general free energy change of the reaction may be written as + + + + R r Q q M m L l + + + M L R QG m G l G r G q GThe standard free energy change can be written asSubtracting these two equations,OrOr + + + + 0M0L0R0Q0G m G l G r G q G( ) ( )( ) ( ) + + M0M L0LR0R Q0Q0G G m G G lG G r G G q G G + + + M LR Q0a ln RT m a ln RT la ln RT r a ln RT q G GJ ln RTa aa aln RT G GmMlLrRqQ 0 The parameter J is called activity quotientOr Which is known as vant Hoff isotherm.Let us consider the equilibrium in which all the reactant and products are in equilibrium. In this case the activity product defined as K, the thermodynamicequilibrium constant.At equilibrium G = 0,Or eqmMlLrRqQa aa aK

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.| [ ] K ln RT J ln RT G. eq at0

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.| KJln RT GJ ln RT G G0+ Since the standard state of a substance is at 1atm pressure the G0 of a reaction is function of temperature only. Therefore, the equilibrium constant K also function of temperature only.(J / K) < 1 reaction is in forward direction.(J / K) > 1 reaction is in backward direction.For this process, mass, composition and pressure are assumed to be constant only variation of temperature is considered.Derivation of Gibbs Helmholtz EquationP2 22PPTGT1THTGobtain we T by bothsides DevidingTGT H S T H G r 0STG

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.|+

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.|+

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.|Gibbs Helmholtz EquationThis is one of the form of Gibbs Helmholtz Equation.The alternative form can be derived as follows.( )2P2 2PTHTT GorTHTGTGT1or ]]]

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.|( )( ) ( )( )]]]]

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.| +

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.| ]]]]]]

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.|+

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.| +]]]

]]]]]]

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.|2P2P2PP PPTGTGT1T GTGT1TT1T GorGT 1TTGT1GT 1GT1T1T GThis is the alternative form of Gibbs Helmholtz Equation .We can write Gibbs Helmholtz Equation for a process as( )HTHTT1T G22P

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.| ]]]]]]

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.|( )( )HT1T GTHTT GP2P ]]]]]]

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.| ]]]

Utility Evaluation of enthalpy of reaction fromfree energy change. Evaluation of free energy change from calorimetric data. ( ). e temperatur one at Hand G or es temperatur two at G either of knowledge the fromed min er det are These . t tan cons egration int are I and H WhereT IT12cT2bT ln T a HT dTc2bTaTHTT d T d T c T b aT1THTT d T d CT1THT G orT dTHTG00 00203 2022 20P2 200T20T0T + ]]]

++ ]]]

+ + + ]]]

+ Troutons and Rechards RulesRechards rule: it states that the ratio of latent heat of melting to the temperature of the normal melting point of the F.C.C metal is approximately 9.61 J / K and that of BCC metal isapproximately 8.25 J / K. i.e Troutons Rule: it states that the ratio of latent heat of boiling to the temperature of the normal boiling point of the F.C.C metal is approximately 87 J / K. metal C . C . B for K / J 25 . 8THmetal C . C . F for K / J 61 . 9THmmmmK / J 87THVVProblemThe initial state of one mole of an ideal gas is P = 10 atm and T = 300 K. Calculate the entropy change in the gas fora) Reversible isothermal decrease of pressure to 1 atm.b) Reversible Adiabatic decrease of pressure to 1 atm.c) Reversible constant volume decrease of pressure to 1 atm.Solution. ltrs 62 . 241462 . 2 10PV PV V P V Pprocess isothermal per as K 300 T T and atm 1 P ) a. ltrs 462 . 210300 08207 . 0PT RV T R V PK 300 T and atm 10 P21 12 2 2 1 11 2 2111 1 1 11 1 K / J 14 . 19462 . 262 . 24ln 314 . 8VVln RdVVRdVTPTdV P U dTqS12VVVV2121rev2121

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.|

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.| + process adiabatic for 0 q as 0TqdS ) brevrev K / J 71 . 2830030ln 314 . 823TTln CdTTCTU dTdV P U dTqSK 3008207 . 0462 . 2 1RV PT , R23C ., atm 1 P ) c12VTTV212121rev2 22 V 221

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.| + ProblemCalculate the entropy change of the universe in isothermal freezing of 1 gm-mole of super cooledliquid gold at 1250 K from the following data for gold. Tm = 1336 K, H0m = 12.36 kJ/mol,CP(s) = 23.68 + 5.19 x 10-3 T J / mol / K, CP(l) = 29.29 J / mol / Kg Surroundin System UniverseS S SSolution + . property state a is entropyce sin choose we path reversible which to as difference no makes It. path reversible a along only calculated be to is S , But. e temperatur freezing m equilibriu the at occuring not is it ce sin, le irreversib but isothermal is process freezing case this InS ) iSystemSystem) 2 State ( ) 1 State (K 1250 at K 1336 at K 1336 at K 1250 at) s ( Au ) s ( Au ) l ( Au ) l ( Auis path reversible simplest TheCooling Freezing Heating ( )11250133631336125012501336Pm0m13361250P1 2 SystemK J 327 . 9T dTT 10 19 . 5 68 . 23133612360T dT69 . 29T dT) s ( CTHT dT) l ( CS S S ++ + Hence . K 1250 at reversibly freezing during released heat absorbs ng Surrroundi. capacity heat inite inf has it and K 1250 at is g surroundin the if only possible is This. K 1250 at be to assumed been have system the of state final and initial the BothS ) iig SurroundinTHTqSSystemg Surroundin ( )J 12460T d T 10 19 . 5 68 . 23 12360 T d 69 . 29T d ) s ( C H T d ) l ( C H H H1250133631336125012501336P0m13361250P 1 2 System + + + 1Systemg SurroundinK J 967 . 9125012460THS 1UniverseK J 64 . 0 967 . 9 327 . 9 S + Suniverse is positive since the process is irreversible.Problemone gram of liquid ThO2at 2900 0C is mixed with 5 gm of ThO2at 3400 0C adiabatically. a) what is the final temperatureb) What is the entropy change of the system and surrounding Combined c) is the process spontaneous? Assume CP to beIndependent of temperature.( ) ( )( ) ( ) fHfCTTP HTTP Cff P f Pcold P Hot PTT dC 5 ) S ( body hot of change entropyTT dC ) S ( body cold of change entropyK 3589 TT 3673 C 1 3173 T C 5T d C m T d C mbody colder by absorbed heat body hotter by released Heat ) aSolutionPCfPHfPTTPTTPC H SystemC 00752 . 0TTln CTTln C 5TT dCTT dC 5S S ) S ( system the of change entropy Total ) bfCfH

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.|+

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.|+ + eous tan spon is process the So0 S S S. lly adiabatica out carried is process aszero is g surroundin the of change Entropy ) cg Surroundin System Universe> + ProblemCalculate the standard entropy change of the following reaction at 1000 K. Pb (l) + 0.5 O2(g) = PbO (s)and also calculate the entropy change of the Universe.Given: Tm, Pb = 600K, H0m, Pb = 4812 J mol-1, H0PbO,298 = -219 kJ S298, PbO = 67.78 J K-1, S298, Pb = 64.85 J K-1, S298, O2 = 205.09 J K-1CP, PbO(s) = 44.35 + 16.74x10-3 T J K-1mol-1CP,Pb(s) = 23.55 + 9.75X10-3T J K-1mol-1CP,Pb(l)= 32.43 3.09X10-3 T J K-1mol-1CP,O2= 29.96 + 4.184X10-3T- 1.67x 105 T-2 J K-1mol-110O , 2980Pb , 2980PbO , 29802982P0298201000K J 615 . 99 09 . 2052185 . 64 78 . 67S21S S S) 2 .......( .......... ) s ( PbO ) g ( O21) s ( Pbreaction the for values C and S from) 1 ......( .......... ) s ( PbO ) g ( O21) l ( Pbreaction the for Sevaluation of consists physically oblem Pr TheSolution2 + +1 1 2 5 3P1 1 2 5 3Pmol K J T 10 836 . 0 T 10 89 . 4 81 . 5 ) 2 ( Cmol K J T 10 836 . 0 T 10 72 . 17 05 . 3 ) 1 ( C + + + + ( )( )12 2532 253 02981000600Pm0m600298P029801000K J 079 . 9907 . 0 956 . 1 96 . 2 02 . 8 35 . 0 35 . 5 13 . 2 615 . 99600110001210 836 . 0600 1000 10 89 . 46001000ln 81 . 5600481229816001210 836 . 0298 600 10 72 . 17298600ln 05 . 3 ST dT) 1 ( CTHT dT) 2 ( CS Swrite can we , K 600 at melts Pb that Noting + + + + +

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.| +

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.|+

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.| +

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.| ++ ( ) ( )( )( )kJ 244 . 21873 . 55 80 . 1564 2324 4812 20 . 141 79 . 2402 10 . 21 9 10 21960011000110 836 . 0 600 1000210 89 . 4600 1000 81 . 5 48122981600110 836 . 0298 600210 72 . 17298 600 05 . 3 HH H T d ) 2 ( C H T d ) 1 ( C H His K 1000 at reaction of heat or reaction the by released Heat35 2 2352 2302980PbO , 29802981000600P0m600298P029801000 + + + + +

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.| + +

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.| + + + 1g Surroundin System Universe101000g surroundinK J 119 218 079 . 99S S SK J 2181000244 . 218THS. capacity heat e arg l having K 1000 at is ounding surr the that g min Assu + + ProblemFrom the data given for different phases involved, calculate the change in entropy of one mole of manganese when it is heated from 298 K to 1873 K under one atmospheric pressure.(CP) -Mn = 5.7 + 3.4 x 10-3 T 0.4 x 105 T 2cal / deg/ mol(CP) -Mn = 8.3 + 0.73 x 10-3 Tcal / deg/ mol(CP )-Mn = 10.70 cal / deg/ mol(CP )-Mn = 11.3cal / deg/ mol(CP )Mn, l = 11.0 cal / deg/ molTransformation Temperature / K Latent heat inCal / mol. l9901360141015175355254303500The total change in entropy will be equal to the sum of the changes in entropy due to the rise of temperature and those due to the phase transformation occurring on heating. Thus( ) ( )( ) ( )( )dTTC15173500dTTC1410430dTTC1360525dTTC990535dTTCS18731517Mn l P15171410Mn P14101360Mn P1360990Mn P990298Mn P + ++ + ++ + + Mn l Mn Mn Mn Mn1517 1410 1360 990 dTT113071 . 2 dTT3 . 113049 . 0dTT7 . 103860 . 0 dTTT 10 7 . 0 3 . 85404 . 0 dTTT 10 4 . 0 T 10 4 . 3 7 . 5S187315171517141014101360136099039902982 5 3 + + + ++ + +++ + . mol / K / cal 3655 . 19 3188 . 2 3071 . 2 8265 . 03049 . 0 3863 . 0 3860 . 0 8945 . 2 5404 . 0 4010 . 9 S + + ++ + + + + ProblemUsing the following values, calculate the standard free energy change per mole of the metal at 1000 K for the reduction of molybdic oxide and chromic oxide by hydrogen:. mol / kcal 5 . 45 G. mol / kcal 0 . 120 G. mol / kcal 5 . 205 G0) g ( O H , 10000) s ( MoO , 10000) s ( O Cr , 1000233 2 mol / kcal 5 . 45 G ; O H O21Hmol / kcal 0 . 120 G ; MoO O23Mo. mol / kcal 5 . 205 G ; O Cr O23Cr 2Solution01000 2 2 201000 3 201000 3 2 2 + + +( ) ( ) [ ]( ) ( ). ve is change energy free the as e temperaturthis at reduced be can molybdenum , But . e temperaturthis at hydrogen by reduced be not can it , ve is oxidechromium of reduction the for change energy free the ce sin. mol / kcal 5 . 16 120 5 . 45 3 GO H 3 Mo H 3 MoO. mol / kcal 5 . 3425 . 205 5 . 45 3GO H 3 Cr 2 H 3 O Cras written be can reaction reduction The010002 2 3010002 2 3 2+ + + + +ProblemThe standard heat of formation of solid HgO at 298 K is 21.56 kcal / mol. The standard entropies of solid HgO, liquid Hg and O2 at 298 K are 17.5, 18.5 and 49.0 cal /deg/mol, respectively. Assuming that H0 and S0 are independent of temperature, calculate the temperature at which solid HgO will dissociate into liquid Hg and O2./ mol deg/ / cal 49 S/ mol deg/ / cal 5 . 18 S/ mol deg/ / cal 5 . 17 Smol / kcal 56 . 21 HSolution0) g ( O , 2980) l ( Hg , 2980) s ( HgO , 2980) s ( HgO , 2982 C 572 K 8455 . 2521560T T 5 . 25 21560 0mol / cal T 5 . 25 21560 S T H G0 G when dissociate to start will HgO solid TheS T H G write can WeT of t independen are S and H ce sinK J 5 . 25 49215 . 18 5 . 17S21S S S. mol / kcal 56 . 21 H H) s ( HgO ) g ( O21) l ( Hgas written be can reaction The0029802980T0029802980T0 01o) g ( O , 298o) l ( Hg , 298o) s ( HgO , 29802980) s ( HgO , 298029822 + + +ProblemFor the reaction WO3 + 3H2(g) =W(s) + 3H2O(g)a)Calculate G0 and K at 400, 700 and 1000Kb)What is the maximum moisture content of H2 needed for thereaction given:WO3(s) = W(s)+ 3H2+Solution: WO3(s) = W(s)+ H2+WO3(s) + 3H2(g)W(s) + 3H2O(g), cals T T T G g O 7 . 91 log 2 . 10 201500 ), (2302 + . 1 . 13 58900 , ) ( ) (2102 2cals T G g O H g O + cals T T T G 4 . 52 log 2 . 10 248000 + cals T T T G g O 7 . 91 log 2 . 10 201500 ), (2302 + . . , ) ( ) ( cals T 3 39 + + 176700 - = G g O H 3 = g O2302 2Problem-2a) i) ii) iii)K400= 1.26x10-8K700= 2.5x10-2K1000= 0.22b)K at G0 0400 . 14460 400 4 . 52 400 log 400 . 10 248000400cals x x G + K at G0 0700 . cals 5122 = G 0700K at G0 01000 . cals 3000 = G 01000RT Ge K orRTGK e i K RT G000ln . . ln 322

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.|HO HppKsolution + = 1 (0.00247+1) = 1 % moisture = 0.24 %At 700K, K700 = 2.317x10-3( )3 -3 18 - 3 1400HO H10 x 472 2 = 10 x 253 1 = K =pp22. .O Hp2 2Hp2Hp. ..atm 9975 0 =002472 11= p2Hatm 002466 0 = p andO H2.( )296 0 = 10 x 5 2 =pp3 12 -HO H22. .solution (0.296 +1)= 1at 1000K K1000 = 0.22 pH2O= 0.3776 % H2O at 1000K = 37.76 %2Hp2283 0 = p and 7716 0 =132 11= pO H H2 2. ..% . % 83 22 = K 700 at O H 2( )6067 0 = 22 0 =pp3 1HO H22. .6223 0 =6067 11p= H2..Problem-3What is the maximum partial pressure of moisture which can be tolerated in H2 H2O mixture at 1atmospheric total pressure without oxidation of nickel at 7500C?SolutionEquationunder question Ni + H2O = NiO(s) +H2 = 450 + 10.45 T cals.. . ) ( ) ( ) ( cals T 55 23 + 58450 - = G s NiO = g O21+ s Ni01 2. . ) ( ) ( ) ( cals T 1 13 + 58900 - = G g O H = g O21+ g H02 2 2 202010rG - G = G solution G 01023 = 450 + 10.45 x 1023= 11140= - RT lnKln k = - =-5.48006K = + 0.0041669 =from 1 & 2 : 1023 987 . 111140O HHpp22(1)0041669 . 02 2O H Hp p (2)12 2 +O H Hp p1 = p + p 0041669 0O H O H2 2.99585 . 00041669 . 11

2 O Hp or00415 . 02 HpHence maximum tolerable partial pressure of moisture is 0.99585 atm problemThe standard free energychange for reaction at 1125K NiO(s) + CO(g) = Ni(s) + CO2(g) is -5147 cals. Would an atmosphere of 15% CO2, 5% CO and 80% N2 oxidise nickel at 1125K?Solution: G0 = -RTlnK = -4.575 x 1125 log K = -5147 K = 10K for the above reaction is In the given atmosphere i.e J/K is solution cal 273 27300K atm cc 1180 22400 1180 1=. .K 3 101 =cal 273 27300K cal 024212 0 22400 1180 1180 1= ..( )738 . 65

,48 . 638 . 65 (ii)sls lffvvv v TLdTdP( )s lffv vLT dPdT .. 024212 . 0 1. 17400 . 738 . 6548 . 638 . 655 . 692 49cal atm cccalK atm cc

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.| Problem-2 ( )024212 0 34 9 - 089 1017405 692 49= . . ..024212 0 75 0 17405 692 49= . ..= 0.354 K Ans Problem-2The latent heat of vaporization of zinc is 27.3 k cal/mole at the boiling point of 9070C.Find the vapour pressure over pure zinc at 8500C.Solution:for l g transformation we have( ) PRTv v vv TLv v TLdTdPl ggeel gee >> ; .K T p K T atm pRTLPdTdP1123 , ? , 1180 ., 12211 2

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.| 2 11 2122ln1T TT TRLpporT RLPdPSolution ln p2 0 == -0.5909869 p2= 0.55378 atm

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.|1123 11801180 1123987 . 127300Lechateliers PrincipleStatementIt states that if an equilibrium in a system is upset, the system will tend to react in a direction that will reestablish the equilibrium.Factorsi) Concentration of reactant and productii) Pressureiii) TemperatureEffect of concentrationConsider the reactionA + B = C + DAt equilibrium for whichIf the concentration of any one of the reactant or product is altered the concentration of others must alter to keep K constant.If for example [D] is increased by adding more D to the system then [C] will decrease (by reactionwith D to form A and B) preserving the value of K.In general, If the product is added, the system will shift towards left to reestablish the equilibrium and converse is true for the reactant.[ ] [ ][ ] [ ] B AD CK If the reactant is removed, the equilibrium will shift towards reactant and same is true for the product.This shift of equilibrium may be used to advantage in some commercial processes since continuous removal of product will drive the reaction to completion. This is more easily achieved if one of the product is a gas e.g. CaCO3 = CaO + CO2In the lime kiln removal of CO2 in an air current drives the reaction towards right.Other example is the production of magnesium:2 MgO(s)+Si(s) = 2Mg (g) +SiO2(s) At the temperature employed the magnesium vapour can be evacuated.Effect of PressureChanges in pressure have no effect on the position of equilibrium when only solids or liquids are involved or in gaseous reactions involving no volume change e.g. FeO(s) + CO(g) = Fe(s) + CO2((g)Since equal volumes of gas appear in both sides.However, the equilibrium position for a reaction in which changes in gaseous volume occur may be displaced by pressureWe can consider following examplesN2(g)+ 3 H2(g) = 2 NH3(g)1 vol + 3 vol = 2vol Here in this case there is an overall volume decrease( 4 2).If the system is subjected to a pressure increase the system will move in such a direction as to lessen this increase in pressure .By moving to the right the volume diminution results in reduction in pressure i. e increasing pressure drives the reaction to the right.Ammonia is commercially produced at 350 atm. Effect of TemperatureK is unaffected by concentration and pressure change but dependent on temperature.Exothermic reaction:N2+3 H2 =2 NH3 H = - 92.37 kJIf the temperature is increased the system reacts in such a way as to oppose this constraint by removing heat. Which can be done by shifting to the left.Exothermic reactions are favoured by low temperatureEndothermic reaction:ZnO+C= Zn+CO H = +349 kJ Increase in temperature will again shift the equilibrium in the direction which absorbs heat i. e.to the right in this case.Endothermic reactions are therefore, favoured by high temperatures.Clausius-Clapeyron EquationIt is extremely important for calculating the effect of change of pressure on the equilibrium transformation temperature.DerivationLet us consider a single solid substance in equilibrium with its liquid at its temperature of meltingand under one atm pressure.There is a natural tendency for the molecule to pass from solid into liquid or vice versa. The number of atoms passing at any time from one state to other will depend on the temperature and pressure.Let us assume that the molar free Energy of the solid at constant T, PIs GA and of the liquid is GB; thenIf GA > GB the solid metal can decrease its free energy by dissolving, i. e. ifGA > GBSolid Liquid; solid melts since G = -veGA = GBSolid = Liquid; Equlibrium since G = 0GA < GBSolid Liquid; liquid solidifies since G = +veMe (s)GAMe (l)GBThe condition for a dynamic equilibrium between the solid and the liquid metal is GA = GB anddGA = dGBThe change in the free energy of either phase may caused by the change in temperature and pressure of the phases.dG = f ( T, P)dP .PGdT .TGdGTAPAA

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.|+

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.|dP .PGdT .TGdGTBPBB

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.|+

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.|At equilibrium dGA = dGBOr OrOr Or

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.|+

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.|dP .PGdT .TGTAPA dP .PGdT .TGTBPB

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.|+

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.|dP v dT S dP v dT SB B A A+ + ( ) ( ) dT S S dP v vA B A B dT . S dP . v trtrTHS asVST dP d ( )A B trtrv v THT dP dOrIf v is +ve as is normally in the case of melting and Htr will be +ve quantity i. e as the pressure increases the transition temperature should also increase. In other words the equilibrium melting point shall increase with pressure and vice versa.However, for ice water system v is ve and hence (dP / dT)is ve and which is fully exploited in the game of skating on ice wherein the skate pressure increase shall decrease the melting point of ice and hence help skating by providing fluid for lubrication.v THT dP dtrtrClausius Clapeyron Equation Applications of Clausius Clapeyron Equation

( ). ly respective liquid andvapor of volumes molar the are v and v andion vapourizat of heat latent the is H wherev v THT dP dEquilibia Vapour Liquidliq vapVliq vapVApplications of Clausius ClapeyronEquation

Contd.( )2V2VvapvapVliq vapT RHT dP ln dT RH PT dP dget we on substituti afterPT Rvgas idealan as behaves vapor the that g min Assuv THT dP dhence v v >>

t tan cons egration int is C whereCT RHP ln. eq above of egration intthen t tan cons is H g min AssuVV+ concerned erval int temp the over valuemean the be will method this by calculated metalliquid of ion vapourizat of heat The . C is ercept intandRHisT1. vs P ln plot the of slope TheV

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.|. e temperatur of t independenis H that g min Assu . ly respective T and Ttemps the to ing correspond P and P its lim thewithin done be can equation C C of n IntegratioV 2 12 1( ). known are ion vapourizat of heat ande temperatur another at pressure vapour the if e temperatur anyat pressure vapour the calculate to used be can equation ThisT1T1RHPPln orTT dRHP ln d1 2V12TT2VPP2121

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.| Solid Vapour (sublimation) EquilibriaOn the basis of assumptions similar to those madein liquid vapour equilibria, one can obtain the similar expression for solid vapour equilibria.Where Hs is the heat of sublimation

( )2ST RHT dP ln d Solid liquid (Fusion) equilibriaApplying C C equation toso solid liquid equilibria Hf is the molar heat of fusion, vliq and vsolid are the molar volumes of liquid and solid respectively.This equation may be applied to calculate the change in melting point of a metal with change of pressure.( )( )fsolid liq fsolid liq ffHv v TP dT dv v THT dP d Solid Solid equilibriaThe rate of change of transition temperature at which two crystalline forms of a solid are in equilibrium with pressure is given by following expressions:Where Htr is the molar heat of transition, v and v are the molar volume of the indicated forms of solid measured at Ttr.

trT( )( )trtrtrtrHv v TP dT dv v THT dP d Ellingham DiagramEllingham diagrams are basically graphical representation of G0 vs. T relations for the chemical reactions of chemical and metallurgical engineering interest.H. J. T. Ellingham in 1944, was first to plot the standard free energy of formation of oxides against temperature and these later became known as Ellingham diagram.Later on the same plotting was applied for sulphides, chlorides, fluorides etc.Oxide diagrams are mostly used in metallurgy.Features of Ellingham diagram1) Formation reaction fo