© Dario Bressanini S(V piccolo) < S(V grande) S(p piccola) < S(p grande) S(p,V)

© Dario Bressanini Entropy: Qualitative More disordered = higher S For given substance: S gas > > S liquid > S solid For given substance: S gas > > S liquid > S solid Same substance: higher T  higher S Temperature Entropy (S) Melting Point Boiling Point Solid Liquid Gas Entropy of fusion Entropy of vapourisation Fig 20.5

© Dario Bressanini Relative Entropy Entropy is a measure of disorder: Entropy is a measure of disorder:  In a phase change: »Solid Liquid Gas »Highly orderedLess orderedVery disordered Similarly, a gas is more probable than a solid: Similarly, a gas is more probable than a solid:

© Dario Bressanini Entropy, S S (gases) > S (liquids) > S (solids) S o (J/Kmol) H 2 O(liq)69.91 H 2 O(gas)188.8 S o (J/Kmol) H 2 O(liq)69.91 H 2 O(gas)188.8

© Dario Bressanini mole of an ideal gas initially at 8.00 atm and E C expands adiabatically against a constant external pressure of 1.50 atm until its pressure is 2.00 atm and its temperature is 21.3 E C. What is  S for the gas? Is the process reversible or irreversible? Can we evaluate  S for the gas by calculating:  S gas = I dq irr / T = I 0 / T = 0 Is  S for the gas greater than, equal to, or less than zero? To calculate  S for the gas we have to construct a hypothetical reversible path between the inital and final states: reversible isothermal expansion of an ideal gas dq = 0 (8.00 atm, E C) (2.00 atm, 21.3 E C) irreversible (2.00 atm, E C) reversible constant pressure cooling of an ideal gas  S irreversible =  S reversible expansion +  S reversible cooling = n R ln (P 1 / P 2 ) + n Cp ln (T 2 / T 1 ) = (1.000 mole) (8.314 J/mole K) [ln (8.00 atm / 2.00 atm) + (5/2) ln (251.9 K / K)] = J / K J / K = J / K

© Dario Bressanini  S E vap, 298 K H 2 O (l) > H 2 O (g) 298 K, 1.00 bar mole of liquid water is vaporized at 25.0 E C and 1.00 atm: Is this phase change reversible and, if not, why not? An alternate reversible path via which we can calculate the  S E 298 K is sketched below: SE1SE1 SE3SE3  S E vap, K = ? H 2 O (l) > H 2 O (g) K, 1.00 atm  S E vap, K = ? H 2 O (l) > H 2 O (g) K, 1.00 atm Step 1: In this step the liquid water is reversibly heated from K to K at a constant pressure of 1.00 atm:  S E 1 = K I K n C p, H 2 O (l) dT / T = (1.000 mole) (18.07 cal / mole K) ln (373.2 K / K) = cal / K

© Dario Bressanini  S E 3 = K I K n C p, H 2 O (g) dT = (1.000 mole) K I K [ x10 -3 T + 8x10 +3 T -2 ] dT / T = (1.000 mole) [ ln (298.2 K / K) x10 -3 (298.2 K K) - (8x10 +3 / 2) [1 / (298.2 K) / (373.2 K) +2 ] = cal / K The entropy change for the irreversible vaporization at K and 1.00 atm is therefore:  S E vap, =  S E 1 +  S E vap,  S E 3 = ( cal / K) + ( cal / K) + ( cal / K) = cal /K Step 2: In this step the liquid water is reversibly vaporized at K and 1.00 atm:  S E vap, K =  H E vap, K / K = + 9,720 cal  / K = cal / K Why did we choose to vaporize the water at K? Step 3: In this step the water vapor is reversibly cooled from K to the initial temperature of K at 1.00 atm:

© Dario Bressanini Otto Cycle Approximation of gasoline engine. a-b adiabatic compression *V  > rV b-c constant volume heat addition *Q H c-d adiabatic expansion d-a constant volume heat release a-e, e-a exhaust, intake QHQH QCQC W e b c d a V rV p

© Dario Bressanini Idealized Diesel Cycle Approximation of diesel engine. a-b adiabatic compression *rV  > V b-c constant pressure heat addition *Q H c-d adiabatic expansion d-a constant volume heat release a-e, e-a exhaust, intake QHQH QCQC W c e b d a V rV p

© Dario Bressanini Spontaneous Reactions reaction happening or arising without apparent external cause; self- generated reaction happening or arising without apparent external cause; self- generated

© Dario Bressanini The Solution Process For the dissolution of KCl (s) in water For the dissolution of KCl (s) in water KCl (s)  K + (aq) + Cl - (aq) Low entropyHigh entropy The formation of a solution is always accompanied by an increase in the entropy of the system! Entropy is the reason why salts like NaCl (s), KCl (s), NH 4 NO 3 (s) spontaneously dissolve in water. Entropy is the reason why salts like NaCl (s), KCl (s), NH 4 NO 3 (s) spontaneously dissolve in water.

© Dario Bressanini Entropy, S Entropy usually increases when a pure liquid or solid dissolves in a solvent.

© Dario Bressanini Entropy, S negative sign indicates system is more ordered negative sign indicates system is more ordered reverse the reaction and sign changes reverse the reaction and sign changes  S o 298 = kJ/K where the + sign indicates system is more disordered  S o 298 = kJ/K where the + sign indicates system is more disordered

© Dario Bressanini Entropy, S Example 15-15: Calculate  S o 298 for the reaction below. Use appendix K. Example 15-15: Calculate  S o 298 for the reaction below. Use appendix K.

© Dario Bressanini Entropy, S Changes in S are usually quite small compared to  E &  H. Changes in S are usually quite small compared to  E &  H.

© Dario Bressanini Fig. 20.9

© Dario Bressanini The Standard Entropy of Reaction, S o rxn For many chemical reaction:S o = S o products - S o reactants > 0 Standard entropy of reaction, S o rxn : ∑mS o products -∑nS o reactants S o rxn = But for reactions in which the moles of product substances decrease, particularly gases which have very high entropy, we predict that the entropy of the products is less than that of the reactants and the entropy decreases during the reaction: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) S o = S o products - S o reactants < 0 S o rxn = (2 mol NH 3 x S o of NH 3 ) - [(1 mol x J/mol K) + (3 mol x J/mol K)] S o rxn = -197 J/K.. As we predicted, S o < 0

© Dario Bressanini Calculating the Standard Entropy of Reaction, S o rxn –I Problem: Calculate the S o rxn for the oxidation of one mole of S 8 to form either SO 2 (g), or SO 3 (g) at 25 o C: S 8 (s) + 8 O 2 (g) 8 SO 2 (g) or S 8 (s) + 12 O 2 (g) 8 SO 3 (g) Plan: To determine S o rxn, we apply Equation We predict the sign of S o rxn from the change in the number of moles of gas: 8 = 8 or 12 = 8, so the entropy will decrease ( S o rxn < 0). Solution: Calculating S o rxn. From Appendix B values, Rx #1 S o rxn = ( 8 mol SO 2 x S o of SO 2 ) - [(1 mol S 8 x S o of S 8 ) + ( 8 mol O 2 x S o of O 2 )] = ( 8 mol x J/mol K) - [(1 mol x J/mol K) + (8 mol x J/mol K)] = (1,985.6 J/K) - [( J/K) + (1,640.0 J/K)] = J/K - 2, J/K = J/K...

© Dario Bressanini Calculating the Standard Entropy of Reaction, S o rxn - II Rx #2 S o rxn = ( 8 mol SO 3 x S o of SO 3 ) - [(1 mol S 8 x S o of S 8 ) + ( 12 mol O 2 x S o of O 2 )] = ( 8 mol x J/mol K) - [(1 mol x J/mol K) + (12 mol x J/mol K)] = (2, J/K) - [( J/K) + (2,460.0 J/K)] = 2, J/K - 2, J/K = J/K Summary: For Reaction #1 (SO 2 is product) S o rxn = J/K For Reaction #2 (SO 3 is the product) S o rxn = J/K As we predicted they are both negative, but Rx#1 is close to zero, and we would also predict it would be close to zero, since the number of moles of gaseous molecules did not change from reactant to product....