Second Law of Thermodynamics:Second Law of Thermodynamics:∆∆∆∆SUNIV= ∆∆∆∆SSYS+ ∆∆∆∆SSURR≥≥≥≥ 0The net entropy will increase or stay the same. It will never decrease.The net entropy will increase or stay the same. It will never decrease.∆∆∆∆SSYS< 0q∆∆∆∆SSURR> 0∆∆∆∆SSURR≥≥≥≥qT∆∆∆∆SUNIV= 0 only for a reversible process∆∆∆∆SUNIV> 0 for all other processesAtkins: The entropy of the universe tends to increase.Atkins: The entropy of the universe tends to increase.ReversibleexpansionReversibleexpansionMM∆∆∆∆Sunivis pathway dependent:w = - P2 ∆∆∆∆V = -1.42 kJP1= 15 atmV1= 1 LT = 298 KP2= 1 atmV2= 15 LT = 298 Kw = - nRT ln()V2V1PressureVolumeP1= 15 atm, V1= 1 LP2= 1 atm, V2= 15 L= -4.12 kJIrreversibleexpansion Irreversibleexpansion q = -w = 1.42 kJqrev= -wrev= 4.12 kJ∆∆∆∆Ssys = 13.8 J K-1∆∆∆∆Ssys = 13.8 J K-1∆∆∆∆Ssurr= -13.8 J K-1∆∆∆∆Ssurr= -13.8 J K-1∆∆∆∆Suniv= 0 J K-1∆∆∆∆Suniv= 0 J K-1∆∆∆∆Ssys = 13.8 J K-1∆∆∆∆Ssys = 13.8 J K-1∆∆∆∆Ssurr= -4.8 J K-1∆∆∆∆Ssurr= -4.8 J K-1∆∆∆∆Suniv= 9 J K-1∆∆∆∆Suniv= 9 J K-1PressureVolumeP1= 15 atm, V1= 1 LP2= 1 atm, V2= 15 LReversiblecompressionReversiblecompressionMMIsothermal compression is pathway dependent:w = - P1 ∆∆∆∆V = 21.3 kJP1= 15 atmV1= 1 LT = 298 KP2= 1 atmV2= 15 LT = 298 Kw = - nRT ln()V1V2= 4.12 kJIrreversiblecompression Irreversiblecompression q = -w = -21.3 kJqrev= -wrev= -4.12 kJ∆∆∆∆Ssys = -13.8 J K-1∆∆∆∆Ssys = -13.8 J K-1∆∆∆∆Ssurr= 13.8 J K-1∆∆∆∆Ssurr= 13.8 J K-1∆∆∆∆Suniv= 0 J K-1∆∆∆∆Suniv= 0 J K-1∆∆∆∆Ssys = -13.8 J K-1∆∆∆∆Ssys = -13.8 J K-1∆∆∆∆Ssurr= 71.4 J K-1∆∆∆∆Ssurr= 71.4 J K-1∆∆∆∆Suniv= 57.6 J K-1∆∆∆∆Suniv= 57.6 J K-1T1T2T1> T2Corollary:Corollary:For an isolated system ∆∆∆∆SSYS≥≥≥≥ 0SpontaneousWill not happenWLeft= (0.25)20= 9.09××××10-13WRight= (1)20= 1number of moleculesin a truly macroscopic system N ≈≈≈≈1023q?A gas will not spontaneously compress.Heat will not flow spontaneously from a cooler to a warmer object.q small ⇒⇒⇒⇒T1, T2 ≈≈≈≈constant∆∆∆∆S1=qT1∆∆∆∆S2=-qT2∆∆∆∆SSYS= ∆∆∆∆S1+ ∆∆∆∆S2= + = q < 0 qT1-qT2T2-T1T1 T2∆∆∆∆SSYS< 0 violating the second law∴∴∴∴it will not happen∆∆∆∆SSYS< 0 violating the second law∴∴∴∴it will not happen∆∆∆∆S=27.7 k∆∆∆∆S=-27.7 kSecond Law of Thermodynamics:Second Law of Thermodynamics:“The second law of thermodynamics has as much truth as saying that, if you poured a glass of water into the ocean, it would not be possible to get the same glass of water back again”James Clerk Maxwell (1831-1879)In a constant pressure process In a constant pressure process (most of real life)(most of real life)::∆∆∆∆H = qp∆∆∆∆HT∴∴∴∴∆∆∆∆S =Consider the entropy change associated with the melting of an ice cube at T = 273.15.∆∆∆∆S =n ∆∆∆∆fusHT∅∆∆∆∆fusHT∅=∆∆∆∆S n=6.01 kJ mol-1273.15 K∆∆∆∆S = 22.0 J K-1mol-1What if the temperature changes during an expansion, reaction orphase transition?dHTdS =dH = dqp= CpdTCpdTT∴∴∴∴∆∆∆∆S =∫T1T2Sample Problem:Sample Problem:Sample Problem:1. Calculate the final temperature assuming adiabatic mixing (i.e. no heat transfer to or from surroundings).2. Calculate the entropy change for A, B and the whole system.Sample Problem:Sample Problem:Sample Problem:The heat capacity of chlorine gas is given by:Cp= (31.0 + 0.008 T ) J K-1mol-1Calculate ∆∆∆∆S when 2.00 moles of Cl2(g) are heated from 300K to 400K at constant pressure.A: 160 g H2O T = 86.0 oCB: 35.0 g H2O T = 25.0 oCCp= 75.3 J K-1mol-1∆∆∆∆SUNIV= ∆∆∆∆rS - ∆∆∆∆rH ≥≥≥≥ 0T ∆∆∆∆SUNIV= ∆∆∆∆rS - ∆∆∆∆rH ≥≥≥≥ 0T Spontaneity:Spontaneity:When will a chemical reaction occur spontaneously?∆∆∆∆SSYS= ∆∆∆∆rS ∆∆∆∆SSURR= qpT heat absorbed from or released to the surroundings= -∆∆∆∆rHT Endothermic, exothermic and energy neutral processes all may occur spontaneously. ∴∴∴∴∆∆∆∆HSYSand ∆∆∆∆USYSdo not control spontaneity.∆∆∆∆SUNIV= ∆∆∆∆SSYS+ ∆∆∆∆SSURR≥≥≥≥ 0Second Lawa reaction is spontaneous if ∆∆∆∆SUNIV> 0∆∆∆∆rS > ∆∆∆∆rHT A reaction is spontaneous if and only if:∆∆∆∆rH> 0∆∆∆∆rH< 0∆∆∆∆rH> 0∆∆∆∆rH< 0∆∆∆∆rS< 0∆∆∆∆rS> 0∆∆∆∆rS> 0∆∆∆∆rS< 0EnthalpyEntropy Exothermic?Spontaneous?EndothermicEndothermicExothermicExothermic“heat required”“heat required”“heat released”“heat released”NOYES∆∆∆∆SUNIV< 0∆∆∆∆SUNIV> 0∆∆∆∆rS > ∆∆∆∆rHT IFEntropy DrivenEnthalpy DrivenIF−−−−∆∆∆∆rH > -∆∆∆∆rST But ...Energy must be conservedEnergy must be conservedFirst LawEntropy Rules!Entropy Rules!Second Law∆∆∆∆RS=∑∑∑∑S(Prod.)-∑∑∑∑S(React.)∅∅∅∅∅∅∅∅∅∅∅∅Standard Entropy of Reaction:Standard Entropy of Reaction:Sample Problem:Sample Problem:Sample Problem:Calculate the change in entropy and change in enthalpy for the combustion of ten grams of table sugar (sucrose) at 298 K. How does this compare with the information on the label?Sample Problem:Sample Problem:Sample Problem:Consider the following reaction: 3/2 O2(g) + 2 Fe (s) →→→→Fe2O3(s)Calculate ∆∆∆∆SSYS, ∆∆∆∆SSURR, and