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11.04.2011

# 11.04.2011 - Today in Chemistry 260 Lecture 25...

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Unformatted text preview: Today in Chemistry 260 Lecture 25 Friday November 4, 2011 •  •  •  •  thermodynamic de;inition of entropy third law of thermodynamics absolute entropies second law of thermodynamics Next Week in Chemistry 260 •  reading: (Mon.) 13.5 (Wed.) 13.7 (Fri.) AD 5.1 ­5.5, 5.7 •  more spontaneity, the Gibbs energy, phase equilibria •  MS 8 due Friday, Nov. 11 Thermodynamic De<inition of Entropy Entropy is a state function; heat is not. Irreversible expansion B ΔS = SB - SA = rev Adding heat increases the entropy of the system. q>0 M P2 = 1 atm V2 = 15 L T = 298 K P2 = 1 atm, V2 = 15 L Volume Reversible expansion ȹ V ȹ w = −nRT lnȹ 2 ȹ = −4.12kJ ȹ V ȹ ȹ 1 Ⱥ The entropy of the system increases in an isothermal expansion. q>0 P1 = 15 atm, V1 = 1 L P1 = 15 atm V1 = 1 L T = 298 K q ∫A dT q = –w = 1.42 kJ w = –P2 ΔV = –1.42 kJ M Pressure Heat that would be transferred in a reversible process dS = dqrev T ΔSsys = qrev = –wrev = 4.12 kJ q rev 4.12 kJ = = 13.8 J K −1 T 298 K Entropy change in a system is calculated from reversible heat Thermodynamics meets statistics… Entropy Change Accompanying Heating Thermodynamic de;inition: ΔS = SB − SA = ∫ B A dq rev 1 B q = ∫ dq rev = rev T TA T (V ) isothermal ⇒ ΔU= q + w = 0 wrev = - nRT ln V2 1 (V ) ΔS = nR ln V2 1 qrev = nRT ln dS = (V ) V 2 1 entropy change in the isothermal expansion of an ideal gas. dqrev T CdT T Tf CdT !S = " Ti T W1∝ (V1 )N ( V ) N V2 W2∝ (V2 )N C= q !T macroscopic change !S = C ln Tf Ti ΔS = k ln (V2)N  ­ k ln (V1)N ΔS = k ln V2 1 N molecules of gas dq dT infinitesimal change O2 Statistical de;inition (counting states): S = k ln W V1 C= dS = ( N ) ( V ) = N k ln V2 1 V2 ( ) = N NA k ln V 1 n A R Ti ΔS Tf ȹ V ȹ ΔS = nR lnȹ 2 ȹ ȹ ȹ ȹ V1 Ⱥ 1 Most of life is at constant pressure… Absolute Entropy T2 ∴ Δ S = Δ H T Δ H = qp ΔS = ∫ T1 C (T ) dT T Because there is an absolute zero for temperature, the entropy for any state can be calculated ... if we know S(0). Consider the entropy change associated with the melting of an ice cube at T = 273.15 K. !S = !S = !H fus T !H fus 6.01 kJ mol-1 = T 273.15 K S(T) = ∫ C(T' ) dT ' + S(0) T' !S = 22.0 J K -1 mol-1 We also de;ined entropy statistically: What if the temperature changes during an expansion, reaction or phase transition? dH dS = T T 0 dH = dqP = CP dT ! "S = # T2 T1 S = k ln W or S(T) = k nW(T) l W(T) depends on T through Boltzmann population factors: CP dT T Third Law of Thermodynamics W(T) ∝ number of accessible states at T W ∝ probability of a given state being populated Third Law Entropy Pn ∝ e − ΔE / k BT Consider the absolute molar entropy of HCl at 1 atm, 298.15K S = 0 for all perfectly crystalline materials at T = 0 K If there is only one possible arrangement of atoms: (i.e. a perfect crystal): 98.36 K Solid I 1.3 + 29.5 S = k ln W = k ln 1 = 0 12.1 158.91 K Solid II 21.1 12.6 Liquid 9.9 188.07 K 85.9 Gas 13.5 S298.15 = 185.9 JK ­1mol ­1 Every substance has a <inite (nonzero) heat capacity. S(T) = ∫ T 0 Solid I C(T' ) dT ' + S(0) T' S = ∫ (Cp/T) dT from 16 K to 98.36 K using experimental Cp ∴ except for a perfect crystal at absolute zero, every substance has a <inite positive entropy. Standard absolute or third law molar entropies are tabulated, generally at 298 K. Residual Entropy The entropy of a substance may not go to zero  ­ even at 0 K. Consider CO O C μ = 0.12 D If there is no preferred orientation W = 2N For one mole: W = 2NA S = k ln W = k ln 2NA = NAk ln 2 = R ln 2 = 5.8 J K ­1 mol ­1 Consider CH3D S = ∫ (Cp/T) dT from 0 K to 16 K using Debye formula: Cp = aT3 Actual ca. 4.2 J K ­1 mol ­1 No dipole moment, same number of electrons No preferred orientation! W = 4N or for one mole W = 4NA S = R ln 4 = 11.5 J K ­1 mol ­1 S = ΔH/T for each phase transition Solid II Liquid Gas S = ∫ (Cp/T) dT from 98.36 K to 158.91 K using experimental Cp S = ∫ (Cp/T) dT from 158.91 K to 188.07 K using experimental Cp S = ∫ (Cp/T) dT from 188.07 K to 298.15 K using experimental Cp ConcepTest A small amount of pure liquid benzene is mixed with a large amount of pure liquid water. When the entropy due solely to mixing is subtracted, the residual entropy change is negative. This is inconsistent with which of the following conclusions? A.  The ordering of a benzene molecule that is surrounded by water molecules is greater than in pure benzene. B.  The ordering of water molecules in the vicinity of a benzene molecule is less than in pure water C.  Molecules of benzene and water form a complex Actual 11.7 J K ­1 mol ­1 2 The Second Law of Thermodynamics Feel The Second Law of Thermodynamics ΔSUNIV = ΔSSYS + ΔSSURR > 0 The net entropy will increase or stay the same. It will never decrease. ΔSSYS < 0 q ΔSSURR > 0 ΔSsurr ≥ q q , ΔSsurr = rev T T ΔSUNIV = 0 only for a reversible process ΔSUNIV > 0 for all other processes The entropy of the universe tends to increase. Feel The Second Law of Thermodynamics Feel The Second Law of Thermodynamics Stretching the rubber band Unstretching the rubber band Step 1: Hold rubber band in two hands Step 1: Hold rubber band stretched Step 2: Stretch it quickly and touch it to your upper lip Step 2: Quickly let it unstretch and touch it to your upper lip Step 3: Note how it feels Step 3: Note how it feels Feel The Second Law of Thermodynamics ΔSuniv is pathway dependent! ΔSsys = 22.5 J K ­1 irreversible expansion w =  ­P2ΔV =  ­1.42 kJ M q =  ­w = 1.42 kJ ΔS surr = qactual = − 1.42 kJ T P1 = 15 atm V1 = 1 L T = 298 K cold M reversible expansion ȹ V ȹ w = −nRT lnȹ 2 ȹ = −6.71 kJ ȹ V ȹ ȹ 1 Ⱥ qrev =  ­wrev = 6.71 kJ Pressure More confined ΔSsurr =  ­4.8 J K ­1 ΔSuniv = 17.7 J K ­1 P1 = 15 atm, V1 = 1 L P2 = 1 atm V2 = 15 L T = 298 K hot Less confined 298 K P2 = 1 atm, V2 = 15 L Volume ΔSsys = 22.5 J K ­1 ΔSsurr =  ­22.5 J K ­1 ΔSuniv = 0 J K ­1 3 ConcepTest Corollary: for an isolated system, ΔSSYS ≥ 0 number of molecules A gas will not spontaneously compress. ΔS=27.7 k spontaneous WLeft = (0.25)20 = 9.09×10 ­13 WRight = (1)20 = 1 will not happen ΔS= ­27.7 k Humpty Dumpty sat on a wall. Humpty Dumpty had a great fall. All the king's horses and all the king's men Couldn't put Humpty together again. Putting Humpty back together again would require which of the following? in a truly macroscopic system N α 1023 Heat will not <low spontaneously from a cooler to a warmer object T2 T1 q? T1 > T2 q small T1, T2 ~ constant !S1 = qrev T1 !SSYS = !S1 + !S2 = !S2 = " qrev T2 # T " T1 & qrev " qrev + = q% 2 <0 T1 T2 \$ T1T2 ( ' ΔSSYS < 0 violating the second law ∴ it will not happen A.  requires ΔSsurr > 0 B.  requires ΔSsurr ≤ ΔSsys C.  requires ΔSuniv < 0 D.  requires ΔSuniv > 0 E.  more than one of the above 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) 4 ...
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