THERMODYNAMICS – ENTROPY AND FREE ENERGY 3A-1 (of 14) Thermodynamics studies the energy of a system, how much work a system could produce, and how to predict the spontaneous changes a system will undergo

3A-2 (of 14) FIRST LAW OF THERMODYNAMICS – The change in energy of a system is equal to heat that enters the system plus the work done on the system ΔE = q + w 1847 HERMANN VON HELMHOLTZ Presented a mathematical argument for the Law of Conservation of Energy

ENTHALPY (H) – The total energy of a system plus the pressure-volume product of a system H = E + pV q p = ΔH At constant pressure the enthalpy change of a system is equal to the heat that enters the system 3A-3 (of 14)

1865 RUDOLF CLAUSIUS Proposed that there is always some thermal energy not available to be converted into work He called this ENTROPY (  ΔT ) 3A-4 (of 14)

ENTROPY (S) – A measure of the number of arrangements available to a system in a given state 1890 LUDWIG BOLTZMANN Proposed a statistical meaning for entropy 3A-5 (of 14)

S = k log W Boltzmann calculated the entropy of a system by S = entropy k= Boltzmann Constant (R/N A, or 1.38 x J/moleculeK) W= number of ways particles can be arranged in a given state while keeping the total energy constant 3A-6 (of 14)

Calculate the entropies of the following samples of CO 2 (g) and CO 2 (s) S solid < S liquid < S gas The state of matter indicates the relative entropies of different substances W = 20 x 19 = 380 W = 2 x 1 = 2 (1.38 x J/K) log 380 (1.38 x J/K) log 2 3A-7 (of 14) = 3.56 x J/K= 4.15 x J/K

OTHER FACTORS IN PREDICTING S FOR CHEMICAL SUBSTANCES After states, S is higher for substances with 1) larger masses 2)delocalized electron systems 3)weaker bonds 4)more internal complexity C 6 H 6 > C 6 H 12 S is higher for delocalized pi systems Fe > C diamond S is higher for delocalized metallic bonding Cl 2 > F 2 S is higher for larger masses C 3 H 8 > C 2 H 6 S is higher for larger mass K > CrS is higher for weaker metallic bonding CO > N 2 S is higher for more internal complexity 3A-8 (of 14) Cl 2 > Br 2 S gas is higher than S liquid

THIRD LAW OF THERMODYNAMICS – The entropy of a perfect crystal at 0 K is 0 At 0 K W = 16 C C CC OO OO CCCCOOOO CC C C OO OO CC C C OO OO CC C C OO OO CCCCOOOO CO C C OC OO C C C C OO OO Because of the 3 rd Law, all substances have entropy values > 0 3A-9 (of 14) W = 1 (1.38 x J/K) log 1 = 0 J/K (1.38 x J/K) log 16 = 1.66 x J/K

STANDARD STATE – Gaseous reactants and products are at 1 atm, dissolved reactants and products are at 1 M, elemental reactants and products are in their natural state at room temperature and pressure Thermodynamic data is tabulated under specified conditions called STANDARD CONDITIONS Third Law Entropies of substances in their standard states at 25ºC (Sº) are found in the appendix of your textbook and on the class website Standard thermodynamic data is identified with a “º”, such as ΔHº 3A-10 (of 14) CALCULATING ΔS FOR CHEMICAL CHANGES

All Sº values are positive ΔSº reaction = Sº products - Sº reactants Be (s) + O 2 (g) + H 2 (g) → Be(OH) 2 (s) Sº values are not ΔSº f values, they are the absolute entropy (the innate disorder) of each substance 3A-11 (of 14)

Calculate the standard entropy change for 2NiS (s) + 3O 2 (g) → 2SO 2 (g) + 2NiO (s) Sº for NiS (s) = 53 J/molK Sº for O 2 (g) = 205 J/molK 2 mol SO 2 (248 J/molK SO 2 )+ 2 mol NiO (38 J/molK NiO) - 2 mol NiS (53 J/molK NiS)- 3 mol O 2 (205 J/molK O 2 ) = -149 J/K This says the system is becoming more ordered Sº for SO 2 (g) =248 J/molK Sº for NiO (s) =38 J/molK 3A-12 (of 14)

PREDICTING ΔS FOR CHEMICAL CHANGES CaCO 3 (s) → CaO (s) + CO 2 (g) 0 gas molecules → 1 gas molecule 2SO 2 (g) + O 2 (g) → 2SO 3 (g) 3 gas molecules → 2 gas molecules ΔS is + for reactions that have 1) more gaseous products than reactants 2)solid reactants dissolving 3)dissolved gases coming out of solution ΔS = + ΔS = – Ag + (aq) + Cl - (aq) → AgCl (s) dissolved ions precipitateΔS = – H + (aq) + HCO 3 - (aq) → H 2 O (l) + CO 2 (g) gas evolved from solutionΔS = + 3A-13 (of 14)

Predict the sign of the entropy change for each: Dissolving of solid sugar Deposition of iodine vapor ΔS is positive ΔS is negative 3A-14 (of 14)

SPONTANEITY SECOND LAW OF THERMODYNAMICS – In any spontaneous process there is always an increase in total entropy ΔS sys + ΔS surr = ΔS univ ΔS univ > 0 SPONTANEOUS PROCESS – One that proceeds without outside intervention 3B-1 (of 20)

Energy + H 2 O (l) ⇆ H 2 O (g) ΔS univ = ? ΔS sys > 0 To calculate ΔS univ, one must know ΔS sys and ΔS surr 3B-2 (of 20) ΔS sys = 1 mol (189 J/molK) – 1 mol(70. J/molK) = 119 J/K= kJ/K (1)ΔS sys

Energy + H 2 O (l) ⇆ H 2 O (g) The sign of ΔS surr depends on q sys for endothermic rxns: q sys is positive, and this makes ΔS surr negative for exothermic rxns: q sys is negative, and this makes ΔS surr positive The magnitude of ΔS surr is inversely proportional to T ΔS surr = -q sys ______ T For a process at constant pressure ΔS surr =-ΔH sys ________ T 3B-3 (of 20) (2)ΔS surr

Energy + H 2 O (l) ⇆ H 2 O (g) T = 298 K ΔS surr =-44 kJ ________ 298 K = kJ/K 3B-4 (of 20) ΔS univ = ΔS sys + ΔS surr ΔS univ = kJ/K kJ/K = kJ/K ∴ this process is nonspontaneous at 298 K ΔH sys = 1 mol (-242 kJ/mol) – 1 mol(-286 kJ/mol) = 44 kJ

Energy + H 2 O (l) ⇆ H 2 O (g) T = 398 K ΔS surr =-44 kJ ________ 398 K = kJ/K 3B-5 (of 20) ΔS sys =0.119 kJ/K ΔS univ = kJ/K kJ/K = 0.01 kJ/K ∴ this process is spontaneous at 398 K ΔH sys and ΔS sys are temperature independent

1873 JOSIAH WILLARD GIBBS Proposed a way to determine the maximum amount of energy that could be converted into work Gibbs Free Energy (G) – The maximum energy of a system that can be converted into work 3B-6 (of 20)

G = H - TS ΔG sys = ΔH sys - TΔS sys -ΔG sys = -ΔH sys + ΔS sys ________ ________ T T ΔS surr -ΔG sys = ΔS surr + ΔS sys ________ T ΔS univ 3B-7 (of 20)

-ΔG sys = ΔS univ ________ T ΔS univ > 0 for a spontaneous process  ΔG sys < 0 for a spontaneous process 3B-8 (of 20) The change in Gibbs Free Energy (ΔG) indicates if a process is spontaneous or nonspontaneous

Spontaneous processes (1)proceed to lower energy states (exothermic changes)  Decreases in enthalpy favor spontaneity (2)spontaneous processes proceed to the states with the highest positional probability  Increases in entropy favor spontaneity Spontaneous processes are favored by a decrease in enthalpy and an increase in entropy ΔG = ΔH - TΔS 3B-9 (of 20)

CONTRIBUTIONS OF ΔH AND ΔS TO SPONTANEITY ΔG = ΔH - TΔS ΔHΔH ΔSΔS ΔGΔG - + ? ? exothermic endothermic exothermic more disorder more order more disorder more order spontaneous nonspontaneous spontaneous at high T spontaneous at low T = (-) - (+)= (-)= (+) - (-)= (+)= (+) - (+)= (?)= (-) - (-)= (?) 3B-10 (of 20) While ΔH and ΔS are temperature independent, ΔG is temperature dependent

For the reaction: H 2 O (s) ⇆ H 2 O (l) ΔHº fus = 6,038 J/mol ΔSº fus = 22.1 J/molK Find ΔGº at -20.0ºC and 20.0ºC and tell if melting or freezing is spontaneous. 253 ΔGº = ΔHº - TΔSº = 6,038 J/mol - (253.2 K)(22.1 J/molK)= 442 J/mol reaction is nonspontaneous  reverse reaction is spontaneous  H 2 O (l) freezes at -20.0ºC 3B-11 (of 20)

For the reaction: H 2 O (s) ⇆ H 2 O (l) ΔHº fus = 6,038 J/mol ΔSº fus = 22.1 J/molK Find ΔGº at -20.0ºC and 20.0ºC and tell if melting or freezing is spontaneous. 293 ΔGº = ΔHº - TΔSº = 6,038 J/mol - (293.2 K)(22.1 J/molK)= -442 J/mol reaction is spontaneous  H 2 O (s) melts at 20.0ºC 3B-12 (of 20)

For the reaction: H 2 O (s) ⇆ H 2 O (l) ΔHº fus = 6,038 J/mol ΔSº fus = 22.1 J/molK Find ΔGº at 0.0ºC. 273 ΔGº = ΔHº - TΔSº = 6,038 J/mol - (273.2 K)(22.1 J/molK)= 0 J/mol The temperature of a phase change (like melting) is when the two phases are in equilibrium 3B-13 (of 20) For any process at equilibrium, ΔG = 0(For any process at equilibrium and under standard conditions, ΔGº = 0)

For the reaction: H 2 O (l) ⇆ H 2 O (g) ΔHº vap = 44,400 J/mol ΔSº vap = 119 J/molK Find the normal boiling point of water ΔGº = ΔHº - TΔSº = 44,400 J/mol _________________ 119 J/molK = 373 K 0 = ΔHº - TΔSº - ΔHº = - TΔSº ΔHº = T _____ ΔSº = 100. ºC 3B-14 (of 20) (values at 1.00 atm pressure) At a phase change temperature there is an equilibrium between the 2 phases

THERMODYNAMIC FUNCTIONS (1) Enthalpy (H) – The total energy plus pV product of a system Change in Enthalpy (ΔH) – Indicates if a process is exothermic or endothermic at constant pressure 3B-15 (of 20) (3) Gibbs Free Energy (G) – The maximum energy of a system that can be converted into work Change in Gibbs Free Energy (ΔG) – Indicates if a process is spontaneous or nonspontaneous (2) Entropy (S) – The energy of a system not available for work, related to the number of arrangements available to a system in a given state Change in Entropy (ΔS sys ) – Indicates if a process is becoming more ordered or more disordered Change in Entropy (ΔS univ ) – Indicates if a process is spontaneous or nonspontaneous

PREDICTING ΔG FOR STANDARD STATE CHEMICAL REACTIONS 1) ΔG = ΔH - TΔS 2)Standard Free Energies of Formation at 25ºC (ΔGº f ) found in the appendix of the textbook ΔGº f of elements in their standard states = 0 or, in standard states, ΔGº = ΔHº - TΔSº ΔGº reaction = ΔGº f (products) - ΔGº f (reactants) 3B-16 (of 20)

Calculate the standard free energy change at 25°C for 2CH 3 OH (g) + 3O 2 (g) → 2CO 2 (g) + 4H 2 O (g) 298 ΔGº f for CH 3 OH (g) = -163 kJ/mol 298 ΔGº f for O 2 (g)= 0 kJ/mol 2 mol CO 2 (-394 kJ/mol CO 2 )+ 4 mol H 2 O (-229 kJ/mol H 2 O) - 2 mol CH 3 OH (-163 kJ/mol CH 3 OH) - 3 mol O 2 (0 kJ/mol O 2 ) = kJ This says, when starting with a container having all of the reactants and products in their standard states, the forward reaction is spontaneous 298 ΔGº f for CO 2 (g) =-394 kJ/mol 298 ΔGº f for H 2 O (g) =-229 kJ/mol 3B-17 (of 20)

Calculate ΔHº, ΔSº, and ΔGº at 25ºC for 2SO 2 (g) + O 2 (g) → 2SO 3 (g) SO 2 (g) O 2 (g) SO 3 (g) 2 mol SO 3 (-396 kJ/mol SO 3 ) - 2 mol SO 2 (-297 kJ/mol SO 2 ) - 1 mol O 2 (0 kJ/mol O 2 ) = -198 kJ ΔHº f (kJ/mol)Sº (J/molK) ΔHº ∴ reaction is exothermic 3B-18 (of 20)

Calculate ΔHº, ΔSº, and ΔGº at 25ºC for 2SO 2 (g) + O 2 (g) → 2SO 3 (g) SO 2 (g) O 2 (g) SO 3 (g) 2 mol SO 3 (257 J/molK SO 3 ) - 2 mol SO 2 (248 J/molK SO 2 ) - 1 mol O 2 (205 J/molK O 2 ) = -187 J/K ΔHº f (kJ/mol)Sº (J/molK) ΔSº ∴ reaction becomes more ordered 3B-19 (of 20)

Calculate ΔHº, ΔSº, and ΔGº at 25ºC for 2SO 2 (g) + O 2 (g) → 2SO 3 (g) SO 2 (g) O 2 (g) SO 3 (g) ΔGº = ΔHº - TΔSº = -198 kJ - (298.2 K)( kJ/K) = -142 kJ ΔHº f (kJ/mol)Sº (J/molK) ΔGº ∴ when all reactants and products are present and start in their standard states, the forward reaction is spontaneous 3B-20 (of 20)

ΔGº predicts spontaneity when reactants and products are in standard states ΔG predicts spontaneity when reactants and products are not in standard states ΔG = ΔGº + RT ln Q 3C-1 (of 14) PREDICTING ΔG FOR NON-STANDARD STATE CHEMICAL REACTIONS

When relating thermodynamic functions (ΔG) with Q, Q must be expressed in ACTIVITIES instead of molarities or pressures ACTIVITY (a) – The effective pressure or concentration of a substance due to interactions between molecules, atoms, or ions The ratio of the pressure of a gas to its standard state pressure, or, the ratio of the concentration of a dissolved substance to its standard state concentration 3C-2 (of 14)

Calculate the free energy change for 2CO (g) + O 2 (g) → 2CO 2 (g) in a container that is atm CO, atm O 2, and 3.0 atm CO 2 at 25ºC. The ΔGº f for CO (g) is -137 kJ/mol and the ΔGº f for CO 2 (g) is -394 kJ/mol. ΔG = ΔGº + RT ln Q Q = a CO2 2 __________ a CO 2 a O2 = (3.0) 2 _________________ (0.010) 2 (0.020) = 4.5 x 10 6 ΔG =-514,000 J+ (8.314 J/K)(298.2 K) ln 4.5 x 10 6 = -476,000 J 3C-3 (of 14) ΔGº = 2 mol CO 2 (-394 kJ/mol CO 2 ) - 2 mol CO (-137 kJ/mol CO) - 1 mol O 2 (0 kJ/mol O 2 ) = -514 kJ

For a reaction at equilibrium ΔG = 0 ΔG = ΔGº + RT ln Q 0 = ΔGº + RT ln Q 0 = ΔGº + RT ln K eq -RT ln K eq = ΔGº ΔGº = -RT ln K eq 3C-4 (of 14) ΔGº is a valuable quantity because it is related to a reaction’s equilibrium constant

K eq 1 >1 <1 ΔGº RELATIONSHIP BETWEEN K eq AND ΔGº ΔGº = -RT ln (1) ΔGº = -RT ln (>1) ΔGº = -RT ln (<1) 3C-5 (of 14) ∴ if ΔGº is 0 it means K eq = 1 ∴ if ΔGº is - it means K eq > 1 ∴ if ΔGº is + it means K eq < 1 The magnitude of the K eq is indicated by the sign of ΔGº

HF (aq) + H 2 O (l) ⇆ H 3 O + (aq) + F - (aq) The K eq (K a ) for the above reaction is 7.2 x at 25ºC. Calculate ΔGº. ΔGº = -RT ln K eq ΔGº = 18,000 J ΔGº = -(8.314 J/K)(298.2 K) ln (7.2 x ) 3C-6 (of 14)

4NO (g) ⇆ 2N 2 O (g) + O 2 (g) An equilibrium mixture at 25ºC has p NO = atm, p N2O = 3.50 atm, and p O2 = 1.74 atm. Find K p and ΔGº for the reaction ΔGº = -RT ln K eq ΔGº = -28,200 J ΔGº = -(8.314 J/K)(298.2 K) ln (8.731 x 10 4 ) K p = p N2O 2 p O2 ___________ p NO 4 = x 10 4 = (3.50 atm) 2 (1.74 atm) __________________________ (0.125 atm) 4 3C-7 (of 14)

N 2 O 4 (g) ⇆ 2NO 2 (g) ΔGº = -RT ln K eq ΔGº = ln K eq _____ -RT K eq = e -(6,000 J)/(8.314 J/K)(298.2 K) Find ΔGº and K eq at 25ºC for the reaction If the ΔGº f for N 2 O 4 (g) is 98 kJ/mol and the ΔGº f for NO 2 (g) is 52 kJ/mol. e -ΔGº/RT = K eq ΔGº = 2 mol NO 2 (52 kJ/mol NO 2 ) - 1 mol N 2 O 4 (98 kJ/mol N 2 O 4 ) = 6 kJ = C-8 (of 14)

HNO 2 (aq) + H 2 O (l) ⇆ H 3 O + (aq) + NO 2 - (aq) = e -(19,000 J)/(8.314 J/K)(298.2 K) Find ΔGº and K a at 25ºC for the reaction If the ΔGº f for HNO 2 (aq) is -54 kJ/mol, the ΔGº f for H 2 O (l) is -237 kJ/mol, the ΔGº f for H 3 O + (aq) is -237 kJ/mol, and the ΔGº f for NO 2 - (aq) is -35 kJ/mol. = 4.7 x mol NO 2 - (-35 kJ/mol NO 2 - )1 mol H 3 O + (-237 kJ/mol H 3 O + ) - 1 mol HNO 2 (-54 kJ/mol HNO 2 ) = 19 kJ ΔGº = e -ΔGº/RT = K eq 3C-9 (of 14) - 1 mol H 2 O (-237 kJ/mol H 2 O) How would you find ΔGº and K a at 35ºC for the reaction? 308 ΔGº = 298 ΔHº + T 298 ΔSº e -ΔGº/R(308) = K eq

Solid NH 4 Cl is placed in a 5.00 L flask at 25ºC and at equilibrium the total pressure is atm. Find K p and ΔGº for the reaction NH 4 Cl (s) ⇆ NH 3 (g) + HCl (g) xx Initial atm’s Change in atm’s Equilibrium atm’s 00 + x p total =p NH3 + p HCl atm = x + x K p = p NH3 p HCl atm = x = x 2 = (0.125) 2 = C-10 (of 14)

Solid NH 4 Cl is placed in a 5.00 L flask at 25ºC and at equilibrium the total pressure is atm. Find K p and ΔGº for the reaction NH 4 Cl (s) ⇆ NH 3 (g) + HCl (g) = 10,300 J ΔGº = -RT ln K eq = -(8.314 J/K)(298.2 K) ln C-11 (of 14)

2HCl (g) ⇆ H 2 (g) + Cl 2 (g) g HCl (g) are placed in a 3.00 L container and partially decomposes. The equilibrium pressure of HCl (g) is atm at 25ºC. Find K p and ΔGº for pV = nRT p = nRT _____ V = ( mol)( Latm/molK)(298.2 K) _________________________________________________________ (3.00 L) = atm g HCl = mol HCl x 1 mol HCl ________________ g HCl 3C-12 (of 14)

2HCl (g) ⇆ H 2 (g) + Cl 2 (g) g HCl (g) are placed in a 3.00 L container and partially decomposes. The equilibrium pressure of HCl (g) is atm at 25ºC. Find K p and ΔGº for x x 2HCl (g) ⇆ H 2 (g) + Cl 2 (g) Initial atm’s Change in atm’s Equilibrium atm’s x+ x x K p = p H2 p Cl2 _________ p HCl 2 = (0.014) 2 __________ (0.310) 2 = 2.0 x x= – 2x = C-13 (of 14)

2HCl (g) ⇆ H 2 (g) + Cl 2 (g) g HCl (g) are placed in a 3.00 L container and partially decomposes. The equilibrium pressure of HCl (g) is atm at 25ºC. Find K p and ΔGº for = 15,000 J ΔGº = -RT ln K eq = -(8.314 J/K)(298.2 K) ln 2.04 x C-14 (of 14)