13 CHAPER SPONANEOUS PROCESSES AND HERMODYNAMIC EQUILIBRIUM 13.1 he Nature of Spontaneous Processes 13.2 Entropy and Spontaneity: A Molecular Statistical Interpretation 13.3 Entropy and Heat: Macroscopic Basis of the Second Law of hermodynamics 13.4 Entropy Changes in Reversible Processes 13.5 Entropy Changes and Spontaneity 13.6 he hird Law of hermodynamics 13.7 he Gibbs Free Energy 571 he reaction of solid sodium with chlorine gas proceeds imperceptibly, if at all, until the addition of a drop of water sets it off.
13.1 HE NAURE OF SPONANEOUS PROCESSES 572 First Law of hermodynamics ~ Cannot predict the directionality of spontaneous processes. Second Law of hermodynamics Entropy, S S universe > 0 for a spontaneous process Gibbs free energy, G G system < 0 for a spontaneous process at constant P and 573 (1) (2) (3) Fig. 13.1 A bullet is hitting a steel plate: (1) (2) (3). he reverse process is exceedingly unlikely.
573 hot Expansion of a gas cold 13.2 ENROPY AND SPONANEIY: A MOLECULAR SAISICAL INERPREAION Free adiabatic expansion 575 Fig. 13.2 Free expansion of a gas into a vacuum. he half of the gas is found in each bulb, at equilibrium, after the stopcock is opened. Distribution of 2 molecules (N L =2, N R =0) or (N L =0, N R =2) 1 2! 1 1 Probability (P) 2C0 2 0! 2! 4 4 (N L =1, N R =1) 2 1 2! 1 2 Probability (P) 2C1 2 1! 1! 4 4 2
Distribution of 4 molecules 576 576 Distribution of N A = 6.0 10 23 molecules (1 mol) (N L = N A, N R =0) or (N L =0, N R = N A ) 23 23 NA 610 1.810 1 1 1 Probability N C A 0 0 2210 Statistical fluctuation: N 1 O N N 0 as N Random, statistical behavior of a large number of particles Directionality of spontaneous change
Entropy and Molecular Motions 578 Microstate ~ Microscopic, mechanical states available to N molecules in the system Number of microstates, (E,V,N) ~ Increasing the volume increasing available values of position increasing (E,V,N) Entropy, S ~ Measure of the number of available microstates Boltzmann s statistical definition of entropy S = k B ln (E,V,N) Free expansion of a gas Spontaneous process ~ increasing (E,V,N) ~ increasing S A299 EXAMPLE 8.7 Calculate the entropy of a tiny solid made up of four diatomic molecules of a compound such as carbon monoxide, CO, at = 0 when (a) the four molecules have formed a perfectly ordered crystal in which all molecules are aligned with their C atoms on the left and (b) the four molecules lie in random orientations, but parallel. (a) 4 CO molecules perfectly ordered: (b) 4 CO in random, but parallel: (c) 1 mol CO in random, but parallel:
579 EXAMPLE 13.3 Free expansion of 1 mol of a gas from V/2 to V. S =? Number of states available per molecule = cv Number of states available for N-molecules system = = (cv) N Entropy is an extensive quantity, S = S() = S[(cV) N ] N S ln S = N 0 k B ln (cv) -N 0 k B ln (cv/2) = N 0 k B ln 2 > 0 Entropy and Disorder Ordered state Disordered state : S sys > 0 gas expansion, melting, boiling, diffusion, Entropy is a measure of disorder (randomness). A302 8.31 List the following substances in order of increasing molar entropy at 298 K: H 2 O(l), H 2 O(g), H 2 O(s), C(s, diamond). Explain your reasoning.
A302 8.35 Without performing any calculations, predict whether there is an increase or a decrease in entropy for each of the following processes: (a) Cl 2 (g) + H 2 O(l) HCl(aq) + HClO(aq); (b) Cu 3 (PO 4 ) 2 (s) 3 Cu 2+ (aq) + 2 PO 3-4 (aq); (c) SO 2 (g) + Br 2 (g) + 2 H 2 O(l) H 2 SO 4 (aq) + 2 HBr(aq). 580 13.3 ENROPY AND HEA: MACROSCOPIC BASIS OF HE SECOND LAW OF HERMODYNAMICS Background of the Second Law of hermodynamics Efficiency of heat engines heat work hermodynamic efficiency of the Carnot cycle w q h l 1 h fundamental limit of an engine
581 Equivalent Formulations of the Second Law of hermodynamics Rudolf Clausius here is no device that can transfer heat from a colder to warmer reservoir without net expenditure of work. Lord Kelvin here is no device that can transfer heat withdrawn from a reservoir completely into work with no other effect. hermodynamic Definition of Entropy 581 Carnot s analysis Efficiency for a reversible heat engine cycle Clausius s analysis of Carnot s work q h h q l 0 stat l q is a e function dq rev : independent of path in any reversible process (state function) S Sf Si f i dq Clausius s thermodynamic definition of entropy rev
13.4 ENROPY CHANGES IN REVERSIBLE PROCESSES 582 S sys for Isothermal Processes Compression / Expansion of an ideal gas = V 2 S nrln V1 Phase ransitions q rev Sfus f H f fus 582 routon s rule S vap = 88 5 J K 1 mol 1 for most liquids Exception: Water, S vap = 109 J K 1 mol 1 ~ ordering due to hydrogen bonds
S sys for Processes with Changing emperature 584 B S dq A rev For a reversible adiabatic process (q = 0), S = 0. (isentropic) For a reversible isochoric process, dq rev = nc V d = 1 = = (Const V) For a reversible isobaric process, dq rev = nc P d = 1 = = (Const P) 579 EXAMPLE 13.5 (a) 5.00 mol argon expands reversibly at a constant = 298 K from a P = 10.0 to 1.00 atm. S =? = = +95.7 (b) 5.00 mol argon expands reversibly and adiabatically at an initial = 298 K from a P = 10.0 to 1.00 atm. hen the gas is heated at constant P back to 298 K. S =? For the first adiabatic process, S = 0; 119 K (Ex 12.11) For the second, = = = +95.7
emperature dependence of S vap o - For the entropy of vaporization of water at 25 o C, Heat the liquid to b ; allow it to vaporize; cool the vapor to 25 o C. For 1 mol of gas, = = 8.43 Calculate the standard entropy of vaporization of water at 85 o C, given that its standard entropy of vaporization at 100 o C is 109.0 J K -1 mol-1 and the molar heat capacities at constant pressure of liquid water and water vapor are 75.3 J K -1 mol-1 and 33.6 J K -1 mol-1, respectively, in this range. For 1 mol of gas, (85 o C) 85 o C 85 o C = = (85 o C) (85 o C) (85 o C)
13.5 ENROPY CHANGES AND SPONANEIY Entropy change for surroundings q H P surr sys (const ) S surr S tot for the thermodynamic universe Stot Ssys Ssurr 0 H sys surr 586
586 EXAMPLE 13.6 10.0 g of ice melts with a piece of Ni at 100C. S tot =? c s,p (Ni) = 0.46 J K 1 g 1, c s,p (H 2 O) = 2.09 J K 1 g 1, H fus (ice) = 334 J g 1 System (Ni), Surroundings (ice-water bath at 0C) (1) Calculate the mass of Ni heat lost by Ni = heat gained by ice-water bath = heat used in melting ice M Ni c s,p (Ni) = M Ice(melt) H fus (ice) M Ni (0.46)(373.15 273.15) = (10.0) (334) M Ni = 73 g Ice-water bath at 0C, 1 atm with 20 g ice. 586 (2) Calculate S sys (=S Ni ) S = nc P ln ( 2 / 1 ) = Mc s,p ln ( 2 / 1 ) S Ni = (73 g)( 0.46 J K 1 g 1 ) ln (273.15/373.15) = 10 J K 1 (3) Caculate S surr S surr = H sys / surr = [ M ice H fus (ice)] / bath = (10.0 g)(334 J g 1 )/273.15 K = 12 J K 1 (4) Caculate S tot S tot = S sys + S surr = 10 + 12 = +2 J K 1 S tot > 0 spontaneous process!
Irreversible Expansion of an Ideal Gas In an irreversible expansion, P P. wirrev Pext dv PdV wrev w w, q q irrev rev irrev rev q rev S q irrev ext ( U w q w q ) irrev irrev rev rev 587 Fig. 13.5 Work done by a system in reversible and irreversible expansions. Clausius inequality 588 For the same pair of initial and final states, q rev > q irrev q q rev irrev S q S : Clausius inequality For an isolated system, q = 0 S > 0 In a spontaneous process, the entropy of the universe increases. he Second Law of hermodynamics S tot = S sys + S surr > 0 (Irreversible process, Spontaneous) = 0 (Reversible process)
A312 EXAMPLE 8.12 Calculate S, S surr, and S tot for (a) the isothermal, reversible expansion and (b) the isothermal, free expansion of 1.00 mol of ideal gas molecules from 8.00 L to 20.00 L at 292 K. Explain any differences between the two path. (a) Isothermal reversible expansion at 292 K (b) Isothermal free expansion 292 K 13.6 HE HIRD LAW OF HERMODYNAMICS he hird Law of hermodynamics (a) S 0 as 0 K for any pure substance in its equilibrium. (b) he absolute zero temperature can not be obtained by finite processes. 590 Absolute molar entropy S o at 298.15 K and 1 atm =. + Fig. 13.6. A graph of c P / vs. for Pt.
A306 at 25 o C 13.7 HE GIBBS FREE ENERGY 592 S surr H sys S S S S H / ( H S )/ tot sys surr sys sys sys sys Hsys Ssys Stot Gibbs free energy: G H S Stot G sys S tot > 0 spontaneous (irrev) S tot = 0 reversible S tot < 0 impossible At constant and P, G sys < 0 spontaneous G sys = 0 reversible G sys > 0 nonspontaneous
595 Competition between H and S G = H S In general, H < 0 & S > 0 G < 0 In freezing, H 2 O(l) H 2 O(s) H < 0 in favor of freezing ~ dominates when < f S < 0 in disfavoring freezing Fig. 13.8 Plots of H and S vs. ~ dominates when > f temperature for the freezing of water. - G decreases as its is raised at constant P. G = H S ; H and S vary little with, S > 0 A317 - Decreasing rate of G m : vapor >> liquid > solid S m (vapor) >> S m (liquid) > S m (solid) - hermodynamic origin of phase transition no stable liquid phase CO 2
597 Effects of emperature on G o G = H S (1) H o < 0, S o > 0 G o < 0 at all (2) H o > 0, S o < 0 G o > 0 at all (3) G o = 0 at * = H o /S o a) H o < 0, S o < 0 G o < 0 at < * b) H o > 0, S o > 0 G o < 0 at > * Fig. 13.11 Spontaneous processes from competition between H o and S o. A324 EXAMPLE 8.16 Estimate at which it is thermodynamically possible for carbon to reduce iron(iii) oxide to iron under standard conditions by the endothermic reaction. (using H f0 and S m0 ), above 565 o C
Summary for Reversible Processes in Ideal Gas Isochoric Process: V = 0 w =-P ext V = 0 q = q v = nc v U = q v H = U +(PV) = 1 = = Isobaric Process: P = 0 w =-P ext V=-PV q = q p = nc p U = w + q H = q p = 1 = = Isothermal Process: = 0 = = U = (3/2)nR = 0, q = w H = U + (PV) = U + (nr) = 0 = = = Adiabatic Process: q = 0 P 1 V 1 = P 2 V 2 S = 0 U = w = nc v H = nc p
Optional: Carnot Cycles, Efficiency, and Entropy 597 Carnot Cycle Fig. 13.12 Stages of the Carnot cycle. A B: isothermal expansion C D: isothermal compression B C: adiabatic expansion D A: adiabatic compression all reversible processes 598 Path AB : Isothermal expansion (at h ) W AB = q AB = nr h ln (V B /V A ) Path BC : Adiabatic expansion q BC = 0, W BC = nc V ( h l ) Path CD : Isothermal compression (at l ) W CD = q CD = nr l ln (V C /V D ) Path DA : Adiabatic compression q DA = 0, W DA = nc V ( h l ) W net = W AB + W BC + W CD + W DA = nr h ln (V B /V A ) + nr l ln (V C /V D )
For an adiabatic process, 2 / 1 = (V 1 /V 2 ) 1 h / l = (V C /V B ) 1 for path BC h / l = (V D /V A ) 1 for path DA 599 (V C /V B ) 1 = (V D /V A ) 1 or V C /V B = V D /V A V B /V A = V C /V D W net = nr ( h l ) ln (V B /V A ) Efficiency of Carnot engine, = = =1 <1