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CY 6151- UNIT II CHEMICAL THERMODYNAMICS APPLIED CHEMISTRY/SVCE Terminology of thermodynamics Second law Entropy - entropy change for an ideal gas, reversible and irreversible processes ,entropy of phase transitions. Clausius inequality. Free energy and work function: Helmholtz and Gibbs free energy functions Criteria of spontaneity Gibbs-Helmholtz equation Clausius-Clapeyron equation Maxwell relations Van’t Hoff isotherm and isochore . APPLIED CHEMISTRY/SVCE • The term thermodynamics means flow of heat. • In general it deals with the inter conversion of various kinds of energy in physical and chemical systems. • Thermodynamics – – Predicts the feasibility of a physical process or chemical reaction under given condition of temperature and pressure. – Predicts whether a chemical reaction would occur spontaneously or not under a given set of conditions – Helps to determine the extent to which a reaction would takes place. APPLIED CHEMISTRY/SVCE Terms used in thermodynamics APPLIED CHEMISTRY/SVCE Types of System • • • • • Homogeneous system Heterogeneous system Isolated system Open system Closed system APPLIED CHEMISTRY/SVCE Properties of a system • Extensive properties: These are thermodynamic properties which depend on the quantity of matter specified in the system e.g. mass, volume energy etc. • Intensive properties: These are thermodynamic properties which depend on characteristics of matter but independent of its amount e.g. pressure, temperature, viscosity,density m.p, b.p etc. APPLIED CHEMISTRY/SVCE Thermodynamic process: • • • • • • Isothermal process Adiabatic process Isobaric process Isochoric process Reversible process Irreversible process APPLIED CHEMISTRY/SVCE Reversible Process • Driving force and opposing force differ by small amount. • It is a slow process • The work obtained is more • It is am imaginary process • It consists of many steps • It occurs in both the directions • It can be reversed by changing thermodynamic variables Irreversible process • Driving force and opposing force differ by a large amount • It is a rapid process • The work obtained is less • It is a real process • It has only two steps i.e initial and final • It occurs in only one direction • It can’t be reversed. APPLIED CHEMISTRY/SVCE Internal Energy: U or E • The energy stored in a substance by virtue of its constituent atoms and molecules is called Internal energy. • It is the sum of vibrational energy, rotational energy, electronic energy etc. • Internal Energy change ((∆E)) • It is the difference in the internal energies of initial and final states of the system. ∆ E = E final – E initial But for the chemical reactions APPLIED CHEMISTRY/SVCE Enthalpy or Heat content of a system (H) “The heat content of the system” or “ sum of internal energy and pressure volume change work done” H = E + PV Unit KJ mol-1 Enthalpy change – It is the difference in the enthalpy of initial and final stages of the system. For the chemical reaction ΔH = H final –H initial ΔH = ΔE + P ΔV At constant Volume ΔH = ΔE APPLIED CHEMISTRY/SVCE First law of thermodynamics: The law of conservation of energy. Energy can be neither created nor destroyed, but it can be converted from one form to another. The mathematical form of First law of thermodynamics is ΔE = q – w where ΔE, q and w represent respectively the change in internal energy, quantity of heat supplied and work done. For a small change, dE = dq – dw ----- (1) Work done (dw) can be represented as PdV in terms of pressure volume changes for the gas. i.e dw = PdV ----- (2) APPLIED CHEMISTRY/SVCE APPLIED CHEMISTRY/SVCE Second law of thermodynamics : • Clausius statement : It states that heat cannot flow itself from a cold body to a hot body spontaneously without the intervention of an external energy. • Kelvin statement: It is impossible to take heat from a hot body and convert it completely into work by a cyclic process without transferring a part of heat to cold body. • II law in terms of entropy: A spontaneous process is always accompanied by an increase in entropy of the universe. • Other statements: • All spontaneous process are irreversible • It is not possible to construct a machine functioning in a cycle which can convert heat completely into equivalent amount of work without produces changes elsewhere. APPLIED CHEMISTRY/SVCE ENTROPY • Clausius introduced a new thermodynamic function called entropy. It is a measure of degree of disorder or randomness in a molecular system. It is also considered as a measure of unavailable form of energy. • The change in entropy of equal to the ratio of heat change to the temperature (T) of the reversible cyclic process. qrev S T APPLIED CHEMISTRY/SVCE Significance of entropy • Measure of disorder of the system: All spontaneous process are accompanied by increase in entropy as well as increases in the disorder .Increase in entropy implies increase in disorder. • Measure of probability: An irreversible process tend to proceed from less probable state to more probable state. Since entropy increases in a spontaneous process, entropy may be defined as a function of probability of thermodynamic state. • Entropy and unavailable energy: When heat is supplied to the system, some portion of heat is used to do some work. This portion of heat is available energy. The remaining portion is called unavailable energy. Hence entropy is defined as unavailable energy per unit temperature. APPLIED CHEMISTRY/SVCE Entropy change for a reversible ( non spontaneous)process the system absorbs q amount of heat from the surroundings at temperature T , the increase in entropy of the system is given by If 𝑞 ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑇 But the entropy of the surroundings decrease , because the surroundings loose the same −𝑞 of heat q i.e ∆𝑆 = 𝑇 Hence, the net change in the entropy is given by ∆𝑆𝑇𝑜𝑡𝑎𝑙 = ∆𝑆𝑆𝑦𝑠𝑡𝑒𝑚 + ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 𝑞 ∆𝑆𝑇𝑜𝑡𝑎𝑙 =𝑇 + −𝑞 𝑇 ∆𝑆𝑇𝑜𝑡𝑎𝑙 = 0 i.e in a reversible isothermal process, there is no net change entropy. APPLIED CHEMISTRY/SVCE Entropy change for a irreversible ( spontaneous) process: Consider a system maintained at higher temperature T1 and its surrounding maintained at a lower temperature T2.If q amount of heat passes irreversibly from the system to surroundings. then, Decrease in entropy of the system , ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = −𝑞 𝑇1 Increase in entropy of the surroundings. ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 = +𝑞 𝑇2 Net change in entropy is given by ∆𝑆𝑇𝑜𝑡𝑎𝑙 = ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 + ∆𝑆𝑠𝑢𝑟𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔𝑠 ∆𝑆𝑇𝑜𝑡𝑎𝑙 = −𝑞 𝑇1 + 𝑞 𝑇2 =q 𝑇1 − 𝑇2 𝑇1 𝑇2 Since T1 > 𝑇2 : 𝑇1 − 𝑇2 is positive ∆𝑆𝑇𝑜𝑡𝑎𝑙 > 0 APPLIED CHEMISTRY/SVCE Entropy change accompanying change of phase Solid to liquid: Let us consider the process of melting of 1 mole of substance being carried out reversibly .It would absorb molar heat of fusion at temperature equal to its melting point. Where ∆ Hf -molar heat of fusion Tf - fusion temperature. APPLIED CHEMISTRY/SVCE Liquid to vapour One mole of a substance changes from liquid to vapour state reversibly at its boiling point Tb Under constant pressure . ∆H V Molar heat of vapourisation. APPLIED CHEMISTRY/SVCE CLAUSIUS INEQUALITY OR THEOREM APPLIED CHEMISTRY/SVCE CLAUSIUS INEQUALITY OR THEOREM Claussius theorem is a mathematical explanation for II law of thermodynamics. It states that the cyclic integral of Wh ere is always less than or equal to zero q = Differential heat transfer at the system boundary during a cycle. T = Absolute temperature at the boundary. = Integration over the entire cycle. The Clausius inequality is valid for all cyclic, reversible or irreversible process. APPLIED CHEMISTRY/SVCE Helmholtz free energy or work function A The work function A is defined as 𝐀 = 𝐄 − 𝐓𝐒 E- Energy content of the system T- Absolute temperature S- Entropy APPLIED CHEMISTRY/SVCE ΔA= ΔE-TΔS By definition Δ S = qrev / T --------(1) According to I law ΔE= qrev – W-------(2) Using 1 and 2 ΔA= -Wrev - ΔA = W max APPLIED CHEMISTRY/SVCE . Gibbs free energy W useful APPLIED CHEMISTRY/SVCE Gibbs –Helmholtz equation (in terms of free energy and enthalpy) Free energy (G) is related with enthalpy (H) as G = H-TS Enthalpy (H) is related with internal energy (E) as H= E+ PV ∴ G = E+ PV – TS Upon differenciation dG dE PdV VdP TdS SdT The I law of thermodynamics equation for an infinitesimal change may be written as. dq dE dw If work done .dw is only due to expansion then dq dE PdV TdS dq dE PdV APPLIED CHEMISTRY/SVCE Combining equation 1 and 2 dG VdP SdT At constant Pressure dP = O and the above equation becomes dG SdT Or G S T P G1 - Initial - T1 at T+ dT G1+ dG1 G2 - Final - F1 at T+ dT G2+ dG2 dG1 = - S1dT : dG2 = - S2dT dG2 - dG1 = - S2dT –(- S1dT) d(∆G)= -∆S dT APPLIED CHEMISTRY/SVCE d(∆G)= -∆S dT (G ) T S ∆G = ∆H -T∆S (G ) H T P G T (G ) H T P G T Gibbs- Helmoltz Equation APPLIED CHEMISTRY/SVCE Gibbs –Helmholtz equation Applications Calculation of enthalpy change for the cell reaction If a Galvanic cell produces nF coulombs of electricity in a reversible manner, it must be equal to the decrease in the free energy J Where n- no of electrons involved in the process. F- Faraday 96500 coulombs. E = EMF in V Gibbs –Helmoltz equation = - = - = APPLIED CHEMISTRY/SVCE 𝑛𝐹𝐸=∆𝐻 − 𝑛𝐹𝑇 𝜕𝐸 𝜕𝑇 𝑃 Or 𝜕𝐸 ∆𝐻 = −𝑛𝐹𝐸 + 𝑛𝐹𝑇 𝜕𝑇 ∆𝐻 = −𝑛𝐹 𝜕𝐸 𝜕𝑇 𝐸−𝑇 𝑃 𝑃 Calculation of emf of the cell -𝑛𝐹𝐸=∆𝐻 − 𝑛𝐹𝑇 ∆𝐻 𝐸=𝑛𝐹 + 𝑇 𝜕𝐸 𝜕𝑇 𝑃 𝜕𝐸 𝜕𝑇 𝑃 Calculation of entropy change (∆𝑺) ∆𝐺 = ∆𝐻 − 𝑇∆𝑆 ∆𝐺= ∆𝐻 + 𝑇 𝜕(∆𝐺) 𝜕𝑇 𝑃 Comparing the above two equations −∆𝑆 = 𝜕 (∆𝐺) 𝜕𝑇 𝑃 -----------A APPLIED CHEMISTRY/SVCE Spontaneous and Non spontaneous process • The physical or chemical changes which proceed by themselves without the intervention of any external agents are known as spontaneous process. • All natural process are spontaneous. • All spontaneous process proceed in one direction and are thermodynamically irreversible. • Examples: Heat flow from a hotter to a colder body till they attain thermal equilibrium. Water flows by itself from a higher level to a lower level. The expansion of gas into an evacuated space. • The process which proceed in both directions are called spontaneous or reversible process. APPLIED CHEMISTRY/SVCE Spontaneous process (Click to see) Free energy and spontaneity ∆G = ∆H – T∆S The Sign of ∆G depends on magnitude of ∆H and ∆S For a spontaneus process dq orTdS dq T I law dq dE PdV TdS dE PdV or TdS ( dE PdV ) 0 dE PdV TdS O d ( E PV TS ) 0 dG 0 G Ve dS Thus spontaneous process involve a decrease in free energy. APPLIED CHEMISTRY/SVCE • The free energy change (∆G)is the criterion for predicting spontaneity or feasibility of a reaction. • The sign of ∆G depends on sign and numerical value of ∆H and T∆S APPLIED CHEMISTRY/SVCE Conditions for spontaneity ΔH ΔS ΔG = ΔH - TΔS NATURE OF PROCESS -Ve (EXOTHERMIC) + Ve - Ve spontaneous -Ve (EXOTHERMIC) - Ve - Ve (low T) spontaneous -Ve (EXOTHERMIC) - Ve + Ve (High T) Non spontaneous +Ve (ENDOTHERMIC) + Ve + Ve (low T) Non spontaneous +Ve (ENDOTHERMIC) + Ve - Ve (High T) spontaneous +Ve (ENDOTHERMIC) - Ve + Ve Non spontaneous APPLIED CHEMISTRY/SVCE MAXWELL RELATIONSHIP APPLIED CHEMISTRY/SVCE APPLIED CHEMISTRY/SVCE = APPLIED CHEMISTRY/SVCE APPLIED CHEMISTRY/SVCE 𝐝𝐄 = 𝐪 − 𝐏𝐝𝐕 𝐇 = 𝐄 + 𝐏𝐕 𝐆 = 𝐇 − 𝐓𝐒 𝛛𝐓 𝛛𝐕 𝛛𝐓 𝛛𝐏 𝛛𝐏 =− 𝛛𝐒 𝐒 = 𝐒 𝛛𝐒 𝛛𝐏 𝐓 𝐀 = 𝐄 − 𝐓𝐒 𝛛𝐒 𝛛𝐕 =− = 𝐓 𝛛𝐕 𝛛𝐒 𝐕 𝐏 𝛛𝐕 𝛛𝐓 𝐏 𝛛𝐏 𝛛𝐓 APPLIED CHEMISTRY/SVCE 𝐕 Clausius-Clapeyron Equation Rudolf Clausius 1822-1888 German Mathematician / Physicist Benoit Paul Emile Clapeyron 1799-1864 French Engineer / Physicist “Discovered” the Second Law Introduced the concept of entropy Expanded on Carnot’s work CLAUSIUS –CLAYPERON EQUATION Consider a system consisting of only 1 mole of substance in two phases A and B The free energies of the substance in two phases A and B be GA and GB Let the temperature and pressure of the system be T and P respectively. The system is in equilibrium ,so there is no change in free energy GA = GB If the temperature of the system raised to T + dT and Presure becomes P + dP GA + d GA = GB + d GB G = H – TS G= E + PV –TS Diffentiating the above equation. dG = dE + PdV + Vdp-TdS –SdT dG = VdP-SdT APPLIED CHEMISTRY/SVCE dGA = VAdP –SA dT--------(a) dGB = VBdP –SB dT---------(b) Where VAand VB are the molar volume of phases A and B respectively. SA and SB are molar entropies. Since GA = GB dGA = dGB Substituting in equation (a) and (b) VAdP –SA dT = VBdP –SB dT Clapeyron Equation APPLIED CHEMISTRY/SVCE Let us consider the following equilibrium Solid ↔ vapour Vv ›› Vs Liquid↔ vapour Vv ›› Vl PV = RT ; V= RT /P APPLIED CHEMISTRY/SVCE Integrating between the limits. P1 and P2 coressponding to T1 to This is Claussius –Clapeyron equation. APPLIED CHEMISTRY/SVCE Van’t Hoff isotherm It gives the quantitative relation ship between the free energy ∆ G and equilibrium constant K Consider a general reaction aA + bB ⇄ cC +dD We know G = H –TS = E+ PV –TS Or dG = dE +PdV +VdP-TdS-SdT But dq = dE + PdV and dS = dq/T ∴ dG = VdP-SdT At Constant temperature dG = VdP PV = RT or V = RT /P for one mole of a gas. APPLIED CHEMISTRY/SVCE On integration G = G0 + RT ln P --(2) G0 - standard free energy Let the free energies of A,B, C and D at their respective pressure PA ,PB ,PC and PD are GA ,GB,GC and GD respectively . Then the free energy change for the above reaction is given by ∆G = G Products – G reactants. ∆G = (cGC + dGD ) –(aGA + bGB) From (2) aGA = aG0 A+ RT ln PA bGB = bG0 B+ RT ln Pb cGA = CG0 C+ RT ln PC dGA = dG D+ RT ln PD APPLIED CHEMISTRY/SVCE ∆G = (cGC + dGD ) –(aGA + bGB ) = ( CG0 C+ RT ln PC + dG D+ RT ln PD ) –( aG0 A+ RT ln PA + bG0 B+ RT ln Pb ) Where ∆ G0 = standard free energy change of the reaction. At equilibrium ∆ G =0 Van’t Hoff isotherm APPLIED CHEMISTRY/SVCE Van’t Hoff isochore The effect of variation of equilibrium constant with temperature ∆G 0 = - RT ln Keq ln Keq = -∆G0 / RT = -(∆H0-T∆S0) /RT = - ∆H0/RT + ∆S0/R Vant’t Hoff isochore APPLIED CHEMISTRY/SVCE PROBLEMS APPLIED CHEMISTRY/SVCE Calculate the change in entropy acompanying the isothermal expansion of 4 moles of an ideal gas at 300K until its volume has increased three times ∆S = 2.3030 nR log (V2/V1) JK -1 = 2.303 Xx 8.314 log 3 = 36.54 JK-1 APPLIED CHEMISTRY/SVCE Calculate the entropy change when 100 g of ice converted into water at 0oC. Latent heat of fusion of ice is 80Cal/g. For 1 g of ice ∆S= 0.293 cal K-1 g-1 For 100 g of ice = 29.3 cal K-1 g-1 APPLIED CHEMISTRY/SVCE Gibbs free energy of a reaction at 300K and 310 K are -29 k Cal and 29.5 k.cal respectively. Determine its ∆H and ∆S at 300 K d(∆G) = - 29.5-(-29) = -0.5 = T2-T1 = 10 K ∆H = - 14kcal = 0.05 k.cal K-1 APPLIED CHEMISTRY/SVCE The equilibrium constant Kp for a reaction is 3.0 at 673 K and 4.0 at 773 K. Calculate the value of ∆H0 for the reaction. (R=8.314 JK-1) K1= 3.0 K2= 4.0 T1 = 673 T2 =773 ∆H= 12.490 kJ APPLIED CHEMISTRY/SVCE Zeroth Law If two systems say A and B are in thermal equilibrium with a third System C separately (that is A and C are in thermal equilibrium: B and C are in thermal equilibrium ) then they are in thermal equilibrium with each other (that is A and B are in thermal equilibrium) B A C APPLIED CHEMISTRY/SVCE