δs > 0 predicts spontaneous processes. U A + U B = constant U A = U B ds = ds A + ds B Case 1: TB > TA (-) (-) (+)(+) ds = C v(t )
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1 ---onight: Lecture 5 July 23 δs > 0 predicts spontaneous processes. ---Assignment 2 (do not include 1-3 done in-class): Due Friday July 30: Class time ---Assignment 3 posted ater class tonight. Whatever we don t cover in class tonight is due Aug 3 class time. --- Assignment 4 is nearly complete. Completion date: Aug 10 mailbox by 12:00 noon. U A + U B = constant U A = U B ds = ds A + ds B = du A A = du B A + du B B + du B B ds = du B 1 B 1 A S = S A + S B Analyze ds For: --Assignment 5 Due August 12 mailbox by Noon. ---resentation Assignment posted. Case 1: B > A ds > 0 Case 2: A > B (-) (-) (+)(+) ds increases or ds > 0 spontaneous low o heat δsuniverse > 0 predicts spontaneous processes. ersible ds > q irreversible ds q Irreversible ath ds > q irr A B Reversible ath Understood that equal is or reversible only! roblems Involving Entropy I Always start with the 1st Law or the enthalpy equations and involve restricting the path to eliminate variables. du = ds ext dv = C v d + dv A. Entropy Changes With Constant V du = ds ext dv = C v d + dv ds = C v( ) d ds = C v d S = 2 Is Cv constant and assumes no phase changes 1 C v d roblems Involving Entropy I 1 mole o methane is heated at constant volume rom 27.0 C to 63.0 C. What is the entropy change in methane (the system) at constant volume and (b) at constant pressure: provided that the heat capacity, Cv is given as and Cp = Cv + nr roblems Involving Entropy II Always start with the 1st Law or the enthalpy equations and involve restricting the path to eliminate variables. H dh = ds + Vd = C p d + d B. Entropy Changes With Constant H dh = ds + Vd = C p d + d ds = C p d ds = C p( ) 2 C p ( ) d S = d 1 Is Cv constant and assumes no phase changes
2 C. Isothermal Reversible Expansion o An Ideal Gas Case I. Entropy Changes ====> d = 0, du = 0 du =0=dS ext dv = C v d + dv du =0=dS ext dv = C v d + ds = dv = nr V dv S = nr ln 2 1 S = 2 1 dv int dv = S = nr ln 2 1 V2 V 1 nr dv V C. Entropy Changes With hase Changes More complicated to calculate as there are discontinuities in heat capacities and other heating involved. qi = m ice C ice + H usion + m water C water qi = m H2OC H2O + H vap + m steam C steam S = 2 d 1 C p C p (ice) d + H usion C p (H 2 O) d + H vap C p (H 2 O) d C p (steam) d D. Entropy Changes In Chemical Reactions Entropy o an element or compound is a unction o temperature and tabulated. S rxn = n i S i (products) m j S j(reactants) S is an inconvenient lab parameter or spontaneity. ds q irrev du = q dv 2nd Law o hermo Clausius Inequality (Clausius Inequ 1st Law o hermo substituting or qirrev Fe 2 O 3 (s) + 3 H 2 (g) 2 Fe(s) + 3 H 2 O(l) S rxn =? S rxn = 2S Fe +3S H 2O S Fe2O3 +3S H2 ds du + dv ds dh dg = dh ds 0 0 dh ds Recast criteria or spontaneous reactions he Gibbʼs Free Energy is a measure o spontaneity using and as variables. S universe = S Sys + S surr 0 = S Sys + q surr 0 = S Sys q sys 0 = S Sys H sys 0 S universe = H Sys S Sys 0 G universe = H Sys S Sys 0 he Gibb s Free Energy,!G universe is the unction that indicates whether a reaction will occur spontaneously. he sign o!g determines the reaction direction, the magnitude determines the maximum amount o work the system can do (i negative). emperature in Kelvin!Guniverse =!Hsys -!Ssys Gibb s Free Energy Change Enthalpy term Useless or dispersed energy!guniverse =!Hsys - }!Ssys Don t be conused by!gsys =!Hsys -!Ssys symbols. hese all mean the same thing and!grxn =!Hrxn -!Srxn pertain to the system!
3 !G represents the maximum amount o non-v work that can be done in a reaction. here are 3 Cases or the Sign o!g sys : 1.!G sys < 0 (!S univ > 0) or or a spontaneous process 2.!G sys > 0 (!S univ < 0) or a non-spontaneous process 3.!G sys = 0 (!S univ = 0) or or a process at equilibrium -!G < 0 is the maximum useul work that can be produced by a chemical reaction.!g = work max he w-term in the 1st Law has its greatest possible value, and the q-term is at its smallest under reversible conditions. - For!G > 0 (non-spontaneous process)!g is the minimum work that must be done on the system to make the process take place. Using Hess s Law and thermodynamic data, we can compute!g o rxn under standard state conditions in two dierent ways as well. Use able o!h o and!s o!gsys 1 2 Use able o!g o hermodynamic ables Substance!H!G S!H and!g have units o kj/ mol (energy/mol) S has units o J/mol K!G o rxn =!H o rxn -!S o rxn!h o rxn = " mi!h o i (products) - " ni!h o i(reactants)!s o rxn = " mis o i (products) - " nis o i(reactants)!g rxn = [ c!g (C) d!g (D) + ] - [ a!g (A) + b!g (B) ]!Grxn, the standard ree energy o a reaction can be computed directly rom its ormation data, or rom!h o and S o ormation data,!g aa + bb cc + dd!g rxn =?!G rxn = " ni!gi (products) - " mi!gj (reactants) = [c!g (C) + d!g (D)] [a!g (A) + b!g (B)]!G o rxn =!H o rxn -!S o rxn OR!H o rxn = " mi!h o i (products) - " ni!h o i(reactants) = [c!h (C) + d!h (D)] [a!h (A) + b!h (B)] he Fundamental Equations o hermodynamics du = ds dv dh = ds + Vd dg = Sd + Vd da = Sd dv!s o rxn = " mis o i (products) - " nis o i(reactants) = [cs (C) + ds (D)] [as (A) + bs (B)]
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