Thermodynamics Lecture Notes PDF

What is the Thermodynamic Thermodynamics is the branch of science which studies the transformation of energy from one form to another. Chemical thermodynamics looks at the energy transformations, which occur as a result of chemical reactions. Chemical thermodynamics ,answer the important q...

What is the Thermodynamic Thermodynamics is the branch of science which studies the transformation of energy from one form to another. Chemical thermodynamics looks at the energy transformations, which occur as a result of chemical reactions. Chemical thermodynamics ,answer the important question “what determines the position of chemical equilibrium?”. Thermodynamics is the study of heat, work, energy, and the changes they produce in the states of systems  Thermochemistry Terminology Walls A system may be separated from its surroundings by various kinds of walls. Rigid wall A wall whose shape and position are fixed Permeable wall A wall that allows the passage of matter and energy  Thermochemistry Terminology Impermeable wall: It allows no matter passing through it Adiabatic wall Is one that does not conduct heat at all  Changes in Internal Energy The internal energy (E) The internal energy E of a system will change if, 1-Heat passes into or out of the system 2- Work is done on or by the system 3-Mass enters or leaves the system  Types of the system Matter Heat Heat Heat Heat Open Closed b Isolated a c system system system b c a Open system: The internal energy may change due to transfer of heat, mass and work between system and surroundings. Closed system: No transfer of mass is possible: internal energy may only change due to heat and work. Isolated system: No change in the internal energy is possible: heat, work and mass transfer are all impossible..  System properties Temperature (T) , Pressure (P),Volume (V) ,Mass (m) Extensive properties Intensive properties  System properties Extensive properties The properties which depend on the total amount of the substance of the system Ex. Mass, number of mole, volume, energy, electric charge ,entropy and heat capacity, Free energy, Enthalpy.  System properties System properties Intensive properties Do not depend on the amount of matter in the system. Ex. density, pressure ,temperature, partial molar quantities, chemical potential and refractive index, specific heat, boiling point, freezing points, Free energy per mole, surface tension, viscosity Some extensive and intensive properties Extensive property Intensive property Volume Molar volume No of moles Density, Refractive index Mass Surface tension, Viscosity Free energy Free energy per mole Enthalpy Specific heat, Pressure Heat capacity Temperature, Boiling point Freezing point  THERMODYNAMIC EQUILIBRIUM When properties of the system like T,P and V are constant with time the system is said thermodynamic equilibrium. Equilibrium divided into three kinds Thermal (no change in properties of system and surrounding when separated by thermally conducting wall) Mechanical (no acceleration and no turbulence) Chemical (concentration of substance constant with time)  THERMODYNAMIC PROCESSES It is a change in the state of the system over time, starting with a definite initial state and ending with a definite final state, 1-Isothermal process The temperature of the system remains constant during various operations. T = 0  THERMODYNAMIC PROCESSES 2- Adiabatic process No exchange of heat between system and surroundings (dq=0) 3-Isobaric process An isobaric process no change in pressure of the system. (dp =0)  THERMODYNAMIC PROCESSES 4- Isochoric Process No change in volume of the system (dV=0). 5- Cyclic process If state is change to another state and returns to its initial state. dq=0 dv=0 dp=0  THERMODYNAMIC PROCESSES (6) Reversible Process The process is reversed by a very small modification of variable.  (7) Irreversible process If the change is produced rapidly and the system does not have a chance to attain equilibrium then the process is called irreversible. All natural processes are irreversible and hence spontaneous. A state function is a property of a system depends only on the state of the system and not on how the system arrived at that state. By convention , State Function are always in CAPITAL LETTERS and path function are not.  Some Mathematical Techniques Used In Thermodynamics The state of a thermodynamic system generally is a function of more than one independent variable, example – the volume of a pure substance. This volume is a function of temperature and pressure of the substance, V  f ( P, T ) Equation for the total differential.  V   V  dV    dP    dT  P T  T  P  , Exact Differentials and STATE FUNCTIONS Let, Z = f (x , y) We can write for the total differential,  Z    Z  dZ    dx    dy  x  y  y  x  Mdx Ndy Where, x and y are independent variables of the system and d is exact differential, and  is partial differential).  z   z  M   , N     x  y  y  x If Z is a state function – its change must be independent of the path, and therefore  M   N    z    z             y  x  x  y y  x  y x  y  x The last equation is used in thermodynamics, and is called the (Euler’s theorem). Example If dZ = (68x 3 y+22y5 )dx+ (17x 4 +110xy4 )dy Prove that dz is an exact differential Solution  Z   Z     M  68x 3 y+22y 5    N  17x 4 +110xy 4  x  y  y x dz is an exact differential  Z x y 2  Z   Z     M = 2xy    N = x 2  x  y  y x   z   M    z   N       2x      2x y  x  y  y x x  y x  x  y  M   N       2Z 2Z Z is state function and  y  x  x  y  dZ is exact differential yx xy  10/4/2024 What is the Thermodynamic Thermodynamic Terminology Changes in Internal Energy Types of the system System properties THERMODYNAMIC EQUILIBRIUM THERMODYNAMIC PROCESSES Z = f (x , y)  Z   Z  dZ    dx    dy  x  y  y x  Z   Z  dZ    dx    dy  x  y   y x Z = f (x,y) When z is constant dz =0 1  Thermochemistry Terminology System The macroscopic part of the universe under study in thermodynamics Surrounding The remainder of the universe, which can in anyway affect or can be affected by the system, The Universe = The System + The Surroundings  10/4/2024 + ( ) ( ) ( ) = ( ) ( ) ( ) =− Cyclic rule Example For one mole of a gas ,pressure ,volume and temperature are related using the following equation A -Prove that dp is an exact differential a RT B - p is state function P+ = V2 V  V   P   T  A- Arrange the equation then, c        1  P T  T V  V P RT a P=  V V2 2  10/4/2024 P = RT (V -1 )  a (V 2 ) differentiate the last equation relative to V at constant T  P  -2 3   = -RT (V )  2 a (V )  V  T  2P R  2 *  T V V Differentiate the first equation with respect to T at constant V and then the differentiate the equation with respect V at constant T  2P R ** ( ) = = R (V-1 )   V  T V 2 From equations * and ** we prove that dp is an exact differential b- from the above prove , P is a state function  V   P   T   P   P  c        1 dP    dV    dT  P T  T V  V P P  f (V , T )  V T  T V  RT 2a  R dP    2  3  dV    dT  V V  V  3  10/4/2024 R R  RT 2a  dP   dT-  2  3  dV At constant volume dv =0 then dP    dT V  V V  V  a At constant pressure dP=0 then, R  RT 2a  0    dT-  2  3  dV V  V V  R  RT 2a   V  dT=  V 2  V 3  dV      T   RT 2a  V  b     2  3    V P  V V  R   RT 2a   RT 2a  dP    2  3  dV dP    2  3  dV At constant temp. dT=0  V V  V V   V  1    RT 2a c  P T  V2 V3 multiplying equations a,b,c,then  V     P T    R   RT 2a  V   1    *  2  3   *  RT 2a   V  V V  R    3   V2 V  4  10/4/2024 Chapter II First Law of Thermodynamics Chapter II First Law of Thermodynamics 1st law: energy can be neither created nor destroyed. A system at a given temperature and pressure which contains a definite amount of mass has a definite internal energy; this energy is a result of the motion of the molecules, the position of the molecules, intermolecular attraction Change in internal energy, E = E2 – E1 5  10/4/2024 E = E2 – E1 E = change in internal energy, E2 = internal energy of system in the final state, E1 = internal energy of system in the initial state. E depends only on the initial and final states of the system and is independent on the path Work done by the system (W) Increase in the energy content ( E) Heat absorbed by the system (q) q = E + w If heat (q) is added to a system, it will either raise the internal energy, or be converted into work (w) done by the system against the surroundings, or both. 6  10/4/2024 F P w = force x distance A Force = Pressure x Area = Pressure x m2 Work = Pressure x m3 = Pressure x VExpanded gas w = P (V2-V1) = PV Since, q = E +w E = q - PdV Therefore, q = E + PV From the first law it follows that: If the system absorbs heat from the surroundings, q is positive. If work is done by the system against the surroundings,(expansion) PdV is positive. Work done on the system (compressing a gas) is negative. The quantities q and w depend on the path taken between states 1 and 2. 7  10/4/2024 Work of Compression and Expansion of a gas at constant temperature Compression of a gas from P1, V1, T to P2, V2, T in a single step. Compression of a gas from P1, V1, T to P2, V2, T in two, three, and an infinity number of step. It is clear that by using more and more steps we arrive eventually at the diagram in Fig. c, which shows that the minimum amount of work is required in the limit of an infinite number of steps. 8  10/4/2024 Expansion of a gas from P1, V1, T to P2, V2, T in a single step. Expansion of a gas from P1, V1, T to V2, T in two, three, and an infinite number of steps. More work can be obtained on the surroundings by using two, three, or an infinite number of steps, as shown in Fig.c V2 V2 W   P ex dv   P gas dv V1 V1 9  10/4/2024 Heat Changes at Constant Volume and at Constant Pressure From first law E = q – P  V For a process under constant volume V = 0, then qv= E Constant pressure processes are more common in chemistry because most operations are carried out at atmospheric pressure E2 – E1= qp – P (V2 – V1) or qp = (E2 + PV2) – (E1 + PV1) H = E + PV Thus qp may be written as qp = H2 – H1 = H The enthalpy (H) is a state function and is independent of the path from the definition of H since E, P, V are all state functions. 10  10/4/2024 Heat capacity: The heat capacity of a system, C, is the heat required to raise the temperature of q the system 1C. C  T2  T1 qv ΔE At constant volume: CV   T2  T1 ΔT Cv is the heat capacity under constant volume. At constant pressure qp H Cp  T 2  T1  T Cp is the heat capacity under constant pressure. the rate of change of E with temperature at constant volume   E  Cv      T v the rate of change of H with temperature at constant pressure. H  C p      T p 11  10/15/2024 Z = f (x , y)  Z  dZ    Z  dx     dy 2Z 2Z   x y  y x  yx xy ( ) ( ) ( ) =− cyclic rule First Law of Thermodynamics E = q - w E = q - PdV Compression in a single step. Compression in more steps. Expansion in a single step. Expansion in more steps. V 2 V2 W   P ex dv   P gas dv qv= E qP= H V1 V1   E  H  Cv   C p       T v  T p 1  10/15/2024 Special case of work Free Expansion Work done by free expansion = 0 In a free expansion, gas is allowed to expand into a vacuum. This happens quickly, so there is no heat transferred. No work is done because the gas does not displace anything. There is no change in internal energy, so the temperature is constant Measurement of (E/V)T (The Joule experiment) Bulb A was filled with air and the bulb B was evacuated. The two bulbs were connected by a stop cock. The surrounding water bath was stirred and its temperature was recorded. The stop cock was then opened A B and the temperature was again recorded. No temperature change was observed 2  10/15/2024 In this process no work is done by the gas, as it expands into vacuum w = 0. and no heat exchanged with the surrounding water Joule detected no temperature change i.e. dq = 0. the first law of thermodynamics along with the results of Joule's experiment gives dE = dq – dw = 0 E = f (T, V) E E E dE= ( )V dT +( )T dV 0= CV dT +( )T dV T V V dT = 0 E  E  dE= CV dT +( )T dV E dV  0   0 V 0= ( )T dV  V T V Thus the energy is independent of volume. This means that the energy of the gas is a function of temperature only. This behavior is Joule's law which may be expressed as E = f (T). However, this is valid only for ideal gases in which internal pressure or force of attraction between the molecules of the gas is zero. 3  10/15/2024 Relation between CP and CV H = E + PV dH = dE + PdV +VdP 1 E  f (T , V )  E   E  dE    dT   dV 2  T V  V T Substitute Therefore dH equals, from 2 to 1  E   E  dH    dT    dV  PdV VdP  T  V  V T  E   E  dH    dT    dV  PdV  VdP  T  V  V  T Divide the last equation by dT at constant pressure we get  H   E   E   V   V   P  =        + + P  +V    T P  T V  V T  T P  T P  T P  E   H   P    =Cv ,   =Cp ,   =0   T V  T  P  T  P  E   V   V  CP =CV +     + P   +0  V T  T P  T P  E    V   E    V  C p = C v +   + P  C p  C v    + P      V T   T  P    V T   T  P 4  10/15/2024 Since (dV/dT)P > 0, the heat capacity at constant pressure CP is always larger than the heat capacity at constant volume.(N.B liquid water behaves abnormally between 0 oC and 4oC; the rate of volume change with temperature (dV/dT)P is negative and therefore in this range CP < CV ) For an ideal gas, PV = RT for one mole  E   V     R/ P    0 Joule E xp eriment  T P  V T  E    V  C  C v    +P  R C p  C v  0 + P    p    V T   T  P P CP – CV = R Isothermal reversible expansion of an ideal gas Consider that n moles of an ideal gas are enclosed in a cylinder fitted with an ideal piston. Let V1 = volume of the gas at constant T. P is the external pressure acting over the gas which is approximately equal to the pressure of the gas. The work done by the gas for the small volume change dV would be dw = pdv V2 w   pdV For an ideal gas PV = nRT nRT V1 P V 5  10/15/2024 V2 nRT V2 dV w  dV  nRT  V1 V V1 V V2  nRT[ln V]V 1 V2 V2 w  nRT ln w  2.303 nRT log V1 V1 P1V1 = P2V2 For ideal gas P P1 w  nRT ln 1 w  2.303 nRT log P2 P2 Internal energy change (E) For iso-thermal, process, the initial and final temperatures are equal or dT=0. So the internal energy of the ideal gas remains constant during expansion because E depends on temperature only. Thus, E = E2 – E1 = 0 (Joule's law) Mathematical form of the first law is q = E + w V2 V2 w  q  2.303 nRT log w  q  nRT ln V1 V1 6  10/15/2024 Enthalpy change (H) H  H2 - H1  H  E  PV  (E2 PV 2 2) (E1 P 1V1)  (E2  E1)  (PV 2 2  P1 V1 )  E  (PV 2 2  PV 1 1) Since E = 0 and P1V1 = P2V2 for an ideal gas. H  0  0  0 Example Calculate the minimum work that must be expanded in the compression of 1 kg of ethylene from 10-1 m3 to 10-2 m3 at a constant temperature of 3000 K, assuming the gas to be ideal. Solution The work done in the reversible isothermal compression of an ideal gas is given by V2 w  nRT ln V1 1000 n  35.65moles 28.05 10 2 w  35.65 * 8.314 * 300 * ln 1  2.047x105 J  204.7 kJ 10 7  10/15/2024 Example Calculate the maximum work obtainable by the isothermal expansion ,of 2 moles of nitrogen, assumed ideal, initially at 25oC, from 10 liters to 20 liters. When a gas is expanded reversibly, the work performed is: V2 w  nRT ln V1 V  w  nRT ln  2  max  V1   20  w  2*2*(273  25) ln    826.23 Cal max  10  Actual Expansion (Irreversible Case) Energy change E For Ideal gas and T is constant, ΔE  E 2  E 1  0 Heat change q q = E + w The pressure is sudden change from P1 to P2 nRT nRT q = w = P2 V = P2(V2-V1)  P2 (  ) P2 P1 P2 w = q = nRT(1- ) P1 8  12/17/2024 Third Law of Thermodynamics What is a Perfect Crystal? Perfect crystal at 0 K Crystal deforms at T > 0 K 273 Third Law of Thermodynamics The entropy of a pure crystalline substance at absolute zero kelvin is 0. The third law used in calculating absolute entropies. S = kB ln Ω Where S is entropy, k is the Boltzmann constant = 1.380649×10−23 J K-1 and Ω is the number of different ways in which the energy of the system can be achieved by rearranging the atoms or molecules among their available states. ‫عدد الطرق المختلفة التي يمكن من خﻼلها تحقيق طاقة النظام عن طريق إعادة ترتيب الذرات أو الجزيئات بين حاﻻتها‬.‫المتاحة‬ 1 274  12/17/2024 The absolute value of entropy is known at T= 0K ( S is zero ). ⇒ the absolute value of entropy can be determined at any temperature Statistical explanation to the 3rd Law At temperature T = 0 K all particles of a perfect crystal are at the lowest energy level: this is possible in a single way: Ω = 1 ⇒ ln 1 = 0, S = k ln Ω ⇒ S = 0 J/K 275 Mathematical Explanation of the Third Law As statistical mechanics, the entropy of a system can be expressed via the following equation: S – S0 = 𝑘B ln𝛀 Where,  S is the entropy of the system.  S0 is the initial entropy. Now, for a perfect crystal that has exactly one unique ground state, 𝛀 = 1. Therefore, the equation can be rewritten as follows S – S0 = 𝑘B ln(1) = 0 [because ln(1) = 0] When the initial entropy of the system is selected as zero, the following value of ‘S’ can be obtained: S – S0 = 0 ⇒ S = 0 Thus, the entropy of a perfect crystal at absolute zero is zero. 2 276  12/17/2024 Applications of the Third Law of Thermodynamics An important application of the third law of thermodynamics is that it helps in the calculation of the absolute entropy of a substance at any temperature ‘T’. These determinations are based on the heat capacity measurements of the substance. For any solid, let S0 be the entropy at 0 K and S be the entropy at T K, then 𝑻𝑪 𝒑 ∆𝑺 = 𝑺 − 𝑺𝟎 = 𝒅𝑻 𝟎 𝑻 According to the third law of thermodynamics, S0= 0 at 0 K, 𝐶 𝑆= 𝑑𝑇 𝑇 277 The simplified expression for the absolute entropy of a solid at temperature T is as follows: 𝑻𝑪 𝑻 𝒑 𝑺= 𝒅𝑻 = 𝑪𝒑 𝒍𝒏 𝑻 𝟎 𝑻 𝟎 S  CP ln T Here Cp is the heat capacity of the substance at constant pressure and this value is assumed to be constant in the range of 0 to T K. 3 278  12/17/2024 Thermodynamic Terminology Types of the system System properties THERMODYNAMIC EQUILIBRIUM THERMODYNAMIC PROCESSES  2Z 2Z Z = f (x , y)  yx xy  Z   Z  dZ    dx    dy   x y  y x 𝝏𝒛 𝝏𝒚 𝝏𝒙 ( )𝒙 ( )𝒛 ( )𝒚 = −𝟏 cyclic rule 𝝏𝒚 𝝏𝒙 𝝏𝒛 279 qv= E qP= H First Law of Thermodynamics E = q - w E = q - PdV Compression in a single step. Compression in more steps. W = PV W = P(Vf – Vi) W = L. atm 1 L atm = 101.3 J Expansion in a single step. Expansion in more steps. 4 280  12/17/2024 H   E    V    E  C    C    p   T p C p  C v    + P     V T   T  P v  T v Work done by free expansion = 0 CP – CV = R for ideal gad  E  Joule experiment   0  V T Work isothermal reversible process V2 P1 E= H = 0 q  w  n R T ln q  w  n R T ln V1 P2 Work in irreversible isothermal P2 w = q = nRT(1- ) E= H = 0 P1 281 in reversible adiabatic Decrease of internal energy results in the lowering of temperature in case of adiabatic q=0 E = - w expansion increase of temperature for adiabatic E = nCV (T2 – T1) = -w compression w= nCV (T1 – T2) H = nCp (T2 – T1) R /C T  V  V 1 2 1    T2  V1  T 1  V   2    T1  V2  T2 V2R / Cv  T1V1 1     T  P  2   2  R /C P T2 P 1     2  1  CP / CV  P2  V1       P1  V 2  PV 1 1 PV 2 2 T  P  T1  P1  1  1  C v  ( P 2 / P1 ) R Adiabatic Irreversible Expansion T 2  T1 Cp 5 282  12/17/2024 Comparison between isothermal and adiabatic process dP P Lower than slope in adiabatic   γ dV V dP P slope in isothermal  - dV V E = q- w The process is (vapour) If the system absorbs heat q is positive. (liquid) If heat lost q is negative w= PV = P(Vv – VL) if value of Vv and VL is If work is done by the system, is positive. given but if not given Work done on the system is negative. = PVv if Vv > VL then neglect VL w = RT 283 Thermochemistry aA+bB cC+ dD HReaction =  n (Hf )P -  n(Hf )R H = E + ngas RT  ΔH  ( )p  Δ CpG.R kirchoff s equation H - E = ngas RT T qp = qv + ngas RT Hess’s Law H  C p T HT  HT  Cp (T2  T1) 2 1 1 1 ΔH T2  ΔH T1  Δα(T 2  T1 )  Δβ(T 22  T12 )  Δγ(T 23  T13 ) ... 2 3 H = H1 + H2 + H3 + H4 BOND ENERGY Bond formation - Ve , ΔH =  bond formation +  Bond breaking Bond Braking + ve 6 284  12/17/2024 Second Law Of Thermodynamics dq rev Spontaneous Changes  T 0 Carnot Heat Engine 2 dq S  rev S2  S1 dS  dq rev Total work w cyc q  q1 T T    2 1 Heat absorbed q 2 q2 mathematical statements of the second law of T2  T1 T q   1 1 1 1 thermodynamics. T2 T2 q2 0 < < 1 285 V  P  reversible adiabatic ΔS  nRln  2  reversible ΔS  nR ln  1  processes are called  V1  Isothermal  T2  isentropic processes. Srev (System) = Sirr (System) (state function) But q rev not equal q irr Entropy Change on Heating qirr (sys) = nRT ( 1 – P2/P1) = - qsurr or Cooling of a Substance S surr = - q/T system in irriversible process Reversible Adiabatic Changes S  0 T  ΔS  nC V ln  2  ΔH trans  T1  ΔS trans  T  Phase changes. Ttrans ΔS  nC P ln  2   T1  Srev (Sys) = -Srev (surr) ΔSuniverse  ΔSsystem  ΔSsurroundings ΔSuniverse  ΔSsystem  ΔSsurroundings  0 Entropy Change for Isolated System (dS)rev = 0 and (dS)irr > 0 Or (S)rev = 0 and (S)irr > 0 7 286  12/17/2024 Combined Form of the First and Second Laws of thermo-dynamics dE = TdS – PdV  T P Maxwell equation(1)      VS SV dH = TdS + VdP T V Maxwel equation (2)     PS  SP A = E - TS 1-Helmholtz free energy A -(A)T = wrev P   S      Maxwell`s equation (3) dA= -PdV – SdT  T  v  V  T V  P   A  E A  nRT ln  2    nRT ln 1     2   V1  P2 T  T v T 287 2- Gibbs free energy G G = H - TS - (G)T,P = wnet  S   V       Maxwell's (4) dG = -SdT+VdP  P  T  T  P P2 V   (G / T )   H  G  nRT ln  nRT ln 1    P1 V2 T P T2 Application of the Gibbs-Helmholtz Equation G  H  T S Zn | Zn SO4 (soln)// CuSO4 (soln)| Cu  E o   E o  -G = nF Eo  nFE  H  nFT o  S  nF    T P  T P 8 288  12/17/2024 (A)system (G)system ΔSuniverse  ΔSuniverse T T Since for a spontaneous process the entropy change of the universe must be positive, (Asystem , Gsystem) must be negative. (i) Criteria for equilibria at constant T and V. ( ΔA) T, V  0 (ΔG) T,P  0 for a reversible process ( A)T , V 0 In an irreversible process (spontaneous changes) ( ΔA) T, V  0 (ii) Criteria for equilibria at constant T and P. for a reversible process (G)T , P  0 For irreversible (spontaneous) changes (ΔG)T , P  0 289 Property Sign Nature of Process (G)T,P or (A)T,V Negative Spontaneous Zero Equilibrium Positive Non-Spontaneous dP Sβ  S ΔS Clapeyron dP H dP   equation  HT V dT Vβ  Vα ΔV dT T  V dT H T ( P2  P1 )  ln 2 V T1 This form of equation is usually employed for processes involving condensed phases. 9 290  12/17/2024 dP ΔH vap dP ΔH fus dP  H sub    dT T(V V  VL ) dT T(V L  VS ) dT T  V dP Δ Htrans  The Clausius- Clapeyron Equation dT T (V  V ) dP ΔHvap P ΔHvap ΔT RT 2    dT TVV RT2 ΔP ΔH vap P  H vap ln P    C a plot of ln P versus 1/T should be linear with a slope of Hvap / R RT P2 Δ H vap  T2  T1  H vap 1 1 ln ( )   ln P2  ln P1   [  ] P1 R  T1T2  R T2 T1 291 G = -R T ln K Chemical Equilibrium  K 2 H  T2  T1 d ln K H van't Hoff equation. ln  ( )  K1 R T1 T2 dT RT 2 ΔH ο So that plot of ln K against 1/T should be a straight line of ln K    constant slope -H/ R. RT 10 292  12/10/2024 Application of the Gibbs-Helmholtz Equation Zn | Zn SO4 (soln)// CuSO4 (soln)| Cu -G = nF Eo  E o   E o   nFE  H  nFT o  S  nF   T P   T  P 249 (A)system ΔSuniverse  T (G)system ΔSuniverse T Since for a spontaneous process the entropy change of the universe must be positive, (Asystem , Gsystem) must be negative. ( ΔA) T, V  0 (ΔG) T,P  0 1 250  12/10/2024 (i) Criteria for equilibria at constant T and V. for a reversible process ( A)T , V 0 In an irreversible process (spontaneous changes) ( ΔA) T, V  0 (ii) Criteria for equilibria at constant T and P. for a reversible process (G)T , P  0 For irreversible (spontaneous) changes (ΔG)T , P  0 251 Property Sign Nature of Process (G)T,P or (A)T,V Negative Spontaneous Zero Equilibrium Positive Spontaneous in reverse direction dP Sβ  S ΔS Clapeyron dP H dP   equation  HT V dT Vβ  Vα ΔV dT T  V dT H T ( P2  P1 )  ln 2 V T1 This form of equation is usually employed for processes involving condensed phases. 2 252  12/10/2024 dP ΔH vap dP ΔH fus dP  H sub    dT T(V V  VL ) dT T(V L  VS ) dT T  V dP Δ Htrans  The Clausius- Clapeyron Equation dT T (V  V ) dP ΔHvap P ΔHvap ΔT RT 2    dT TVV RT2 ΔP ΔH vap P  H vap ln P    C RT H vap 1 1 P2 Δ H vap  T2  T1  ln P2  ln P1   [  ] ln ( )   R T2 T1 P1 R  T1T2  253 Example The heat of vaporization of water is 40820 Jmole-1, the molar volume of liquid water and of steam are 16.78 ml and 30.199 liters all at 100 oC. What would be the change in the boiling point of water at 100 oC if the atmospheric is changed by 1 mmHg. ( 1liter atm = 101.3 J ) ΔH dP vap 40820    3.62 JK 1L1 dT T  VV  VL  373*  30.199  0.01678  dP 3.62   0.0357 atm deg 1  0.0357*760  27.19 mmHg deg 1 dT 101.3 dT 1   0.0368 deg/mmHg dP 27.19 3 254  12/10/2024 Example An organic compound X melts at 80oC. If the vapor pressure of the liquid X is 10mmHg at 85.8oC and 40 mm at 119.3oC , Calculate Heat of vaporization of the liquid, the boiling point and the entropy of vaporization at the boiling point P2 ΔHvap  T2  T1  ln ( )   P1 R  T1T2  40 ΔHvap  392.3  358.8  ΔHvap  48475J / mole  48.47kJ / mole ln ( )  358.8 *392.3  10 8.314 At the boiling point Tb the vapor pressure (P2) is one atm or 760mm Hg. P2 Hvap 1 1 760 48475 1 1 Tb = 489 K ln  [  ] ln  [  ] P1 R T2 T1 10 8.314 Tb 358.81 ΔSvap = ΔHvap /Tb =48475/489 =99.13 J/deg/mole 255 Chemical Equilibrium In the study of chemical processes generally two questions are importance. First is that how fast the reaction proceeds to form the product and the second question is that how far the reaction will proceed in a given direction? The answer of the first question constitutes the domain of the chemical kinetics and second answer is provided by the chemical equilibria. 4 256  12/10/2024 In this Chapter we shall consider the thermodynamic aspects of the chemical equilibria. Thermodynamic Derivation of the Law of Mass Action: The law of mass action as such cannot be derived ‫ﻻداﻋﻰ‬thermodynamically, what we can do is that we can deduce an expression for the equilibrium constant by thermodynamic method. This method for deriving the expression of equilibrium constant has the great merit of not requiring any postulated reaction mechanism. 257 Let us consider the general reaction Reactants = products a A + bB = c D+ dD in which the reactants and the products are ideal gases. The free energy of any mixture can be calculated from the following equation. 5 258  12/10/2024 dP  dG nRT P G = n RT ln p + constant Constant = G (the standard free energy)  G = G  + nRT ln p the free energy of any mixture of a mole of A and b moles of B is given as follows. 259 GA  GA  a RT ln PA GB  GB  b RT ln PB Similarly the free energy of the products is given as, GC  GC  c RT ln PC GD  GD  d RT ln PD 6 260  12/10/2024 GReaction = GP - GR     GοC  cRTlnPC  GοD  dRTln PD   G  aRTln P   G ο A A ο B  bRTln P B at equilibrium there is no free energy change G= 0 and equation reduces to,     GοC  cRTln PC  GοD  dRTln PD   G  aRTln P   G ο A A ο B   dRTln PB  0 261 G  G   G  G  RT ο c ο D ο A ο B  ln PCc  RT lnP Dd  RT ln P  RT ln P   0 a A b B PCc PDd ΔGο  RT ln 0 PAa PBb Pcc PDd or ΔGο   RT ln Pcc PDd PAa PBb  a b  K eq = equilibrium constant PA PB G = -R T ln K (75) 7 262  12/10/2024 Variation of Equilibrium Constant with Temperature : G = -R T ln K ΔGο ln K   d ln K 1 d (ΔGο/ T) RT . dT R dT (G/T) H   2  T P T d ln K H   dT RT 2 This equation is van't Hoff equation. 263 Integration of Van't Hoff Equation: If the change of heat content H were independent of temperature, general integration of the van't Hoff equation would give. d ln K H  H   dT RT 2  d ln K   RT 2 dT ΔHο ln K    constant RT 8 264  12/10/2024 So that plot of ln K against 1/T should be a straight line of slope -H/ R. Integration between the temperature limits of T1 and T2, the corresponding values of the equilibrium constant being K1 and K2, gives K 2 H  T2  T1 ln  ( ) K1 R T1 T2 265 Free Energy and Temperature By knowing the sign (+ or -) of S and H, we can get the sign of G and determine if a reaction is spontaneous. 9 266  12/10/2024 Example One mole of a perfect gas at 27oC expands isothermally and reversibly from10 atm to 1 atm against a pressure that is gradually reduced. Calculate q, w, and each of the thermodynamic quantities ΔE, ΔH, ΔG, ΔA, and ΔS. P  W  RTln  1  w  10   1.987*300.15*ln    1373cal mole 1 P  max  2  1 ΔE = 0 ΔE = q-w ΔH = ΔE + Δ(PV) =0 +0 =0 q = w =1373 cal mole-1 q 1373 ΔS rev  4.57 cal K1mole1 ΔA = -Wmax = -1373 cal mole-1 T 300 ΔHΔG 0(1373) P  1 alsoΔS  4.57 cal K1mole1 G  nRTln  2   1.987 * 300.15ln   1373 cal mole  1 T 300 P   10   1 267 Example For the synthesis of ammonia, Kp = 1.64*10-4 at 673 K when the reaction is represented by N2 (g) +3H2 (g) = 2NH3 (g) Calculate ΔGo for the reaction at the given temperature. ΔG o   RTlnK P ΔG o  8.314*673*ln(1.64*104 ) ΔG o  48775.6J  48.842 kJ 10 268  12/10/2024 Calculate ΔGo and the equilibrium constant for the following reaction at 25oC C  2H (g)  CH (g) (graphite) 2 4 using the given data: Standard enthalpy of the reaction ΔHo298 = -74830 J Sographite = 5.68 JK-1mole-1, SoHydrogen = 130.59 JK-1mole-1 , Sometane = 186.19 JK-1mole-1 ΔSo  So  2*So  So methane hydrogen graphite ΔSo  186.19  2*130.59  5.68  80.67 JK 1 ΔG o  ΔH o  TΔSo ΔG o  74830  298*(  80.67)  50790 J 50790 ΔG o   RTlnK  8.314*298*lnK lnK   20.4999 P P P 8.314*298 K  7.97*108 P 269 Example Calculate the value of Kp at 25oC and 800oC for the water gas reaction using the following data: CO (g)  H 2 O (g)  H 2 (g)  CO 2 (g) Substance H2 (g) CO (g) H2O (g) CO2 (g) ΔGof (KJ) 0.0 -137.27 -228.59 -394.38 ΔHof (KJ) 0.0 -110.52 -241.83 -392.51  Products   ΔGo reactants ΔG o   ΔG o ΔG o  (ΔG o (H )  ΔG o (CO )   (ΔG o (CO)  ΔG o (H O)   f 2 f 2   f f 2  ΔG o  0  (  394.38)  (  137.27)  (  228.59)  28.52 KJ     11 270  12/10/2024 ΔG o   RTlnK P 28.52 lnK   11.511 8.314*103*298 P K  9.98*104 At 25oC P ΔHo   ΔHo  P ro d u c t s    Δ H o re a c ta n t s Δ H o   ( Δ H o ( H )  Δ H o ( C O )    (Δ H o (C O )  Δ H o ( H O )   f 2 f 2   f f 2  Δ H o   0  (  3 9 2.5 1 )    (  1 1 0.5 2 )  (  2 4 1.8 3 )    4 0.1 6 k J K 2 H  T2  T1 ln  ( ) K1 R T1 T2  K  -4 0.1 6 1073  298  ln  2    9.9 8 * 1 0 4  8. 3 1 4 * 1 0  3  1 0 7 3 * 2 9 8    K  0.8 2 a t 8 0 0 o C 2 271 12  12/4/2024 Combined Form of the First and Second Laws of thermo-dynamics dE = TdS – PdV  T P Maxwell equation(1)      VS SV dH = TdS + VdP T V     Maxwel equation (2) PS  SP 1-Helmholtz free energy A A = E - TS -(A)T = wrev  P   S  dA= -PdV – SdT     Maxwell`s equation (3)  T  v  V  T 214 V  P   A  E A  nRT ln  2    nRT ln 1     2  V1  P2 T  T  v T 2- Gibbs free energy G G = H - TS - (G)T,P = wnet dG = -SdT+VdP  S   V        P  T  T  P Maxwell's (4) P2 V   (G / T )   H  G  nRT ln  nRT ln 1  P1 V2  T  P T2 1 215  12/4/2024 Application of the Gibbs-Helmholtz Equation Relation between electrical and chemical energy. Consider Daniell`s cell Zn | Zn SO4 (soln) // CuSO4 (soln)| Cu 216 If the e.m.f. of the reversible cell is Eo volts and the quantity of electricity passing is n Faradays (or nF Coulomb) then the electrical work done by the system is equal to n x F x Eo joules. Hence -G = nF Eo This relationship may be substituted in the Gibbs-Helmholtz equation to give  E o   E o  S  nF    nFE o  H  nFT   T P  T P 2 217  12/4/2024 Here (Eo/T)P is the rate of change of e.m.f. with temperature. Example For Daniell cell calculate G if Eo = 1.10 volt at 25o C and n=2; F = 96500 Coulomb Solution. For this case have G = -nF E = -2x (96500 Coulomb) x 1.10 Volt = -212300 Volt Coulomb ( Joule) 218 Example: For the Weston standard cell calculate G, H and S using the following data: Eo = 1.01463 Volts at 25o C; n=2; F=96500 (dEo /dT)P = -5.0 x 10-5 Volts/deg Solution. G = -nFEo  E o  = -195815 Joules  nFE o   H  nFT     T P -195815   H  2 * 96500 * 298 * -5.0 x 10 -5  E  o S  nF   -195815   H  2875.7  T P ΔH =-198690.7 Joule  S  2 * 9 6 5 0 0 * -5.0 x 1 0 -5  S = - 9.6 5 J 3 219  12/4/2024 Free Energy of the System and Entropy of Universe: (1) (A) System and (S) Universe Consider an isolated system consisting of system and the surroundings (thermostat) as shown in Fig. (a). spontaneous changes (a) at System is a vessel with a fixed volume and constant containing a substance which can undergo a temperature and chemical reaction volume (A)T,V < 0 The surrounding which is a thermostat is considered to have a large heat capacity so that any heat given to it or taken from it by the system will not change the bath temperature 220 Suppose that the contents of the vessel (system) undergo the volume of the vessel is constant no work is done by the system and w=0. first law for this process gives Eq where q is the heat absorbed by the system from the surroundings. Ssurrounding = -(qsystem/T)= - (Esystem/T) where T is the temperature of the thermostat 4 221  12/4/2024 Suniverse = Ssystem + Sunrroundings ΔEsystem  ΔSsystem T  1 T   ΔEsystem  T ΔSsystem  A= E - TS A= E - TS at constant T (A) system Suniverse   T 222 This proves that the change in the Helmholtz free energy of the system accounts automatically for the entropy changes of the system and surroundings. Since for spontaneous the entropy change of the universe must be positive, then (A) for the system must be negative. 5 223  12/4/2024 (2) (G) System and (S) Universe (G)system ΔSuniverse T 224 For spontaneous change at constant T and P (G)system 0 6 225  12/4/2024 Free Energy of the System and Entropy of Universe: (1) (A) System and (S) Universe  (  A ) system  S un iverse  T Since for spontaneous the entropy change of the universe must be positive, then (A) for the system must be negative. (A)system  0 (2) (G) System and (S) Universe - (  G ) system Δ S u n iverse  T Since for a spontaneous process the entropy change of the universe must be positive, (Gsystem) must be negative. (  G ) system  0 226 (i) –(A/T) is the entropy change of the universe when the change takes place at constant volume, (ii) –(G/T) is the entropy change of the universe when the system undergoes a change at constant pressure. (i) Criteria for equilibria at constant T and V. A = E – TS At constant temperature and volume for a reversible process (dA)T,V = 0 dA= - SdT- PdV For finite changes we will have ( A)T , V 0 7 227  12/4/2024 In an irreversible process (spontaneous changes) (dA) T,V  0 For finite changes ( ΔA) T, V  0 (ii) Criteria for equilibria at constant T and P. For this case we consider the Gibbs free energy function G = H – TS For a reversible change dG = -SdT + VdP 228 At constant temperature and pressure dT = 0 = dP and (dG)T,P = 0 For finite changes (G)T,P  0 For irreversible (spontaneous) changes (dG)T,P < 0 (ΔG)T , P  0 For finite changes 8 229  12/4/2024 The thermodynamic criteria of equilibrium are summarized below: Property Sign Nature of Process (G)T,P or (A)T,V Negative Spontaneous Zero Equilibrium Positive Non-spontaneous 230 Physical Equilibria Involving Phase Transitions: The Clapeyron-clausius equation: Clapeyron equation A thermodynamics relation between the changes of pressure with change of temperature of a system at equilibrium is called Clapeyron equation. This relationship can be derived as follows: 9 231  12/4/2024 Let us consider the equilibrium transition of one mole of a substance from one phase () to another phase () occurring at constant pressure and temperature. Phase () = phase (), at constant T and P. At equilibrium the free energy of the substance in both the phases must be identical, since (G)T,p = 0. That is G G 232 Let the temperature and pressure are changed to the new values (T + dT) and (P + dP) in such a manner that the phases  and  are still at equilibrium. Now, (G + dG) and (G + dG), then at equilibrium (G  dG) (G  dG) α β dG  SdT VdP  (dG)  (dG) α β  SdT  V dP   Sβ dT  Vβ dP 10 233  12/4/2024 (Sβ  Sα ) dT  (Vβ Vα ) dP dP Sβ  S ΔS Clapeyron   equation dT Vβ  Vα ΔV S = H/T, where H is the change in enthalpy for the reversible transformation at temperature T. Substituting this values of  S into Clapeyron equation we get, dP H Clapeyron dP  HT V dT T  V equation dT 234 dP H  dT T  V Above equation is integrated on the assumption that  H and  V are constant, P2 T2 H dT  dP   V  T P1 T1 H T2 (P2  P1 )  ln V T1 This form of equation is usually employed for processes involving condensed phases.(Solid change to liquid without present vapour phase) 11 235  12/4/2024 dP H Applications of Clapeyron Equation  Clapeyron dT T  V equation Liquid-vapour equilibria (Vaporization process) Liquid = Vapour Phase () = Phase ()  H =  Hvap = Enthalpy of vaporization  V = V - V = (Vv - VL) dP ΔHvap  dT T(VV  VL ) 236 Since  Hvap is positive and V > V dP dT will always be positive i.e., with increase of P, there must be increase in T. 12 237  12/4/2024 (ii) Solid-liquid equilibria (Fusion process). Solid = Liquid Phase () = Phase () H = Hfus = Enthalpy of fusion  V = Volume of liquid (VL) – Volume of solid (Vs) Clapeyron equation can be written as dP ΔH fus  dT T(V L  VS ) 238 For the conversion of solid into liquid heat is required and  Hfus is always positive. But the volume of liquid may be greater or less than that of solid. So the sign of dP/d T will depend on the nature of the system. For example, the volume of liquid water is less than the volume of ice, so V is negative and hence dP/dT, is negative. 13 239  12/4/2024 That is, with increase of pressure, the melting point would decrease for ice-water equilibrium. The same is true for bismuth metal. 240 (iii) Solid-vapour equilibrium (sublimation process) Solid = Vapour Phase () = Phase () H = Hsub = Enthalpy of sublimation.  Hsub  Hfus  Hvap V = volume of vapour phase – volume of solid phase. dP  H sub Hsub is always positive  (64) dT T  V V, the solid-vapour equilibria have always a positive value 14 241  12/4/2024 (iv) Equilibrium between two crystalline forms S o li d ( )  S o l i d (  ) dP Δ Htrans  (65) dT T (V  V ) where  Htrans = Enthalpy of transition, V and V are the molar volumes of the indicated form. The sign of dP/dT will depend upon the signs of  Htrans and the densities of the two forms. 242 The Clausius- Clapeyron Equation Clausius-Clapeyron equation is applied in case liquid vapour equilibria dP ΔH vap  The Clapeyron equation, dT T(V V  V L ) Vv >> VL and Vv = RT/P. dP Δ H vap  dT T V V dP ΔHvap P ΔHvap (66)   dT TVV RT2 15 243  12/4/2024 This differential equation may be employed if the changes in temperature and pressure are small. ΔT RT 2  (67) ΔP ΔH vap P 244 Integration of Clausius-Clapeyron Equation dP ΔHvap dT  (68) P R T2 Integration on the assumption on that  Hvap is independent of temperature, we get, ΔHvap ln P  T 2 dT Const. H vap R ln P    C (69) RT 16 245  12/4/2024 -. C being the integration constant. a plot of ln P versus 1/T should be linear with a slope of Hvap / R Thus, the heat of vaporization can be calculated,  H vap using (in Joules per mole) = Slope  8.314. 246 It is more convenient to integrate equation between suitable initial and final limits to get, P2 ΔHvap T2 2 Hvap 1 1  ln P  R  T dT lnP2 lnP1  [  ] R T2 T1 P1 T1 or P2 ΔHvap  T2  T1  ln ( )   P1 R  T1T2  This equation can be used to calculate the heat of vaporization if the vapour pressure data at two different temperatures are known. 17 247  11/28/2024 dq rev  T 0 2 dq dq rev rev S  S2  S1 dS  1 T T mathematical statements of the second law of thermodynamics. T2 V V  Isothermal S=nCVln  nRln 2 ΔS  nRln  2  T1 V1  V1  S=nCPln T2 P  nRln 1 P  T1 P2 ΔS  nR ln  1   T2  reversible adiabatic Reversible Adiabatic Changes  S  0 processes are called ΔH trans isentropic processes. Phase changes. ΔS trans  Ttrans Entropy Change for Isolated System (dS)rev = 0 and (dS)irr > 0 Or (S)rev = 0 and (S)irr > 0 1  11/28/2024 Entropy Change on Heating or Cooling of a Substance T  T  ΔS  nC V ln  2  ΔS  nC P ln  2   T1   T1  ΔSuniverse  ΔSsystem  ΔSsurroundings Example 1 mole of an ideal gas is compressed isothermally at 298oK. If the gas is compressed from 1 atm to 10 atm reversibly and irreversibly against an external pressure of 10 atm, calculate the entropy change for both cases. ΔS For the reversible isothermal process P 1 S system  nR ln  1   1*8.314 ln    19.14 EU  P2   10  ΔS (surroundings) = +19.14 EU ΔS (universe) = ΔS (system) + ΔS (surroundings) = -19.14 + 19.14 = 0 2  11/28/2024 ΔS For the irreversible isothermal process ΔS (system) in the irreversible = reversible ΔS (system) Because the Entropy is a state function ΔSsystem,irreversible= -19.14 EU  P   10  q  nRT 1  2   1* 8.314 * 2981    1* 8.314 * 298 * 9 irreversib le  P   1  1 q  22298.15 J irreversib le This is the heat lost by the system and gained by the surrounding qsurroundings,irreversible = + 22298.15 J ΔSsurroundings,irreversible = qsurr/T= +22298.15/298= +74.9 EU ΔS (universe) = ΔS (system) + ΔS (surroundings) =-19.14 +74.9 = +55.83 EU Example What is the change in entropy when argon at 25oC and 1 atm pressure is expanded isothermally from 500 cm3 to 1000 cm3 and simultaneously heated to a temperature of 100oC, CV=3/2 * R PV 1* 0.5 n   0.020 moles RT 0.082 * 298 V  T  S  nR ln  2   nCV ln  2   V1   T1   1000  3  373  S  0.020 *8.314* ln    0.020* *8.314 ln    500  2  298 