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### Lecture17A

Course: CHEM 444, Spring 2012
School: University of Illinois,...
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Word Count: 804

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Energies: (Free) Helmholtz and Gibbs(Chapter 22.1,22.2) For an isolated system, dS&gt;0 spontaneous process occurs. But how is an open beaker isolated? It isnt, so we must consider S of the surroundings as well. Then, Suniverse= Ssurr+ Ssys&gt;0 will determine spontaneity. This is a lot of work to determine if a process is spontaneous. Is there an easier way? Is there an easier way? Consider a real (not...

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Energies: (Free) Helmholtz and Gibbs(Chapter 22.1,22.2) For an isolated system, dS>0 spontaneous process occurs. But how is an open beaker isolated? It isnt, so we must consider S of the surroundings as well. Then, Suniverse= Ssurr+ Ssys>0 will determine spontaneity. This is a lot of work to determine if a process is spontaneous. Is there an easier way? Is there an easier way? Consider a real (not ideal!) system under isothermal conditions and at dU = q + w constant V: constant V: Since: w = P dV = 0 ( constant V ) ext and Then: q TdS The Helmholtz energy (based solely on the system) can tell us whether a process will occur spontaneously! The Helmholtz energy will continue to decrease for any spontaneous process until it reaches equilibrium, where A will be a minimum. because ( because dS q / T ) dU = q TdS so dU TdS at constant V dU TdS 0 at constant V at constant Fig 22.1 Lecture 17 S + (more disorder) A - Reversible ( equilibrium ) : A = U T S = 0 1 Spontaneous? Always + + ? At high T, S controls - - ? At low T, U controls + - + Never Now for a reversible isothermal process: for reversible isothermal process: A = U T S = U qrev = 0 for reversible process Lecture A Spontaneous irreversible: 17 = U T S < 0 2 If A is positive, wrev represents the min amount of work required to cause the process to reversibly occur in the system. Actual work required for the process to occur is greater! ogous sys An analogous thermodynamic function exists for systems under constant pressure. Recall: dH = q p TdS ( = if reversible, < if spontaneous ) dH TdS 0 or d ( H TS ) 0 for isothermal, isobaric processes G H TS = Gibbs (free) energy (free) energy = U + PV - TS However, if V is not constant, then A = wrev What does this mean? dG<0 = Spontaneous process. Will continue to decrease until system reaches equilibrium when dG=0. If A is negative, A represents the max work that can be done by system on surroundings. G = H T S 0 ( at constant T and P ) A is negative for a spontaneous process, so work done is less, since since wrev is the maximum work that can be done. the maximum work that can be done Lecture 17 A = U - T S 0 Isothermal: d (U TS ) 0 at constant T U - (energy decreases) Let's define A U TS = Helmholtz (free) energy where dA = 0 at equilibrium dA 0 and dA < 0 for a spontaneous process 3 Lecture 17 4 H - (enthalpy decreases) + S + (more disorder) G - + ? Consider 2O H ( l ) H 2O ( g ) vap G = G ( H 2O ( g ) ) G ( H 2O ( l ) ) Spontaneous? Always 0 At high T, S controls - - ? At low T, H controls + - + Never 0 0 0 at 373.15K, vap H =40.65kJ/mol and vap S =108.9J/molK vap G 0 = 40.65kJ / mol 373.15 ( 0.1089 ) kJ / mol = 40.65 40.64 0 So vapor and liquid are in equilibrium at 373.15K. vapor and liquid are in equilibrium at 373 Example from text: NH 3 ( g ) + HCl ( g ) NH 4Cl ( s ) r H 0 = 176.2kJ / mol , r S 0 = 0.285 kJ / mol K r G 0 = 176.2 298.15 ( 0.285 ) = 91.2 kJ / mol = spontaneous! If r H 0 and r S 0 were temperature independent, at what temperature would the reaction be at equilibrium? H 176.2 r G 0 = 0 H = T S T = = = 618 K S 0.285 Within these restrictions, for T>618K, the reverse reaction would be spontaneous but it requires large change in temperature be spontaneous, but it requires a large change in temperature. Lecture 17 vap G 0 = vap H T vap S 0 0 Over a small temperature range vap H , vap S are ~ constant. So at 363.15K vap G = +1.12kJ / mol ( g l is spontaneous ) 0 0 at 385 at 385.15K vap G = 1.08kJ / mol 5 Lecture 17 There is also a concept of maximum work that we can associate with the Gibbs energy. Start with G = U TS + PV with dG = dU TdS SdT + PdV + VdP 1 Consider Na ( s ) + Cl2 ( g ) NaCl ( s ) 2 is spontaneous ( l g is spontaneous ) = (TdS + wrev ) TdS SdT + PdV + VdP = SdT + VdP + wrev + PdV wrev = PdV + wrev ,other ( i.e. electrical, gravitational, torsional ) electrical gravitational torsiona dG = SdT + VdP + ( PdV + wrev ,other ) + PdV dG = SdT + VdP + wrev ,other dG = wrev ,other at constant T and P f G 0 = 394kJ / mol Electrolysis (which represents non-PV work) of molten NaCl is used to produce Cl2(g) and Na metal. If we supply sufficient electrical work, we can drive the reverse reaction: 1 NaCl ( s ) Na ( s ) + Cl2 ( g ) r G 0 = 394kJ / mol = wrev ,elect . 394 2 work/energy for charged systems = Q V = charge voltage 1e- is transferred from Cl- to Na+ in this process: Q = qe N A = 1.602 1019 G = wrev ,other As we argued for the Helmholtz energy: G<0=Spontaneous process. So wrev,other is the maximum amount of non-pressure-volume work that can be done by system. Lecture 17 6 7 C e C 6.02 1023 = 9.64 104 e mol mol G 394 103 J / mol = = 4.08Volts Q 9.64 104 C / mol V=4.08 volts is the minimum voltage needed for electrolysis. Q V = G and V = Lecture 17 8
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