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8 th Week Chap(12-13) Thermodynamics and Spontaneous Processes
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  1. 8th Week Chap(12-13) Thermodynamics and Spontaneous Processes • Standard State: The Thermodynamically stable state for pure liquids(Hg) and solid(graphite), for gases ideal gas behavior, for solutions 1.0 molar concentration of the dissolved species at P= 1.0 atm and some specified T in each case • Reversible and Irreversible processes: Processes that occur through a series of equilibrium states are reversible. Adiabatic(q=0) paths are reversible • Thermodynamic Universe= System + Surrounding Isolated : No Energy and Matter can go in or out Adiabatic: No heat goes in or out • Entropy (S): measure of disorder Absolute Entropy S=kBln Number of Available Microstate  • Second Law of Thermodynamics: Heat cannot be transferred from cold to hot without work S ≥ q/T In-quality of Clausius S = q/T for reversible (Isothermal) processes S > q/T for irreversible processes Midterm Friday: Ch 9, 10, 11.1-11.3, 11.5, 18, 12.1-12.6 One side of 1 page notes(must be hand written), closed book Review Session Today 6-7 pm, in FRANZ 1260

  2. Potential Energy part of the Internal Energy U = <PE> + <KE> Physical bonds intermolecular H-bonding Dipole-Dipole Dipole-induced Dispersion/vdW Chemical bonds Do not use Intra-molecular Covalent ionic

  3. <u>=(8kBT/πm)1/2 average speed <u2>=(3kBT/m) average square speed urms= (3kBT/m)1/2 root mean square speed ump = (2kBT/m)1/2 most probable speed <KE>=1/2 m <u2>=(3/2)kT KE part of the Internal Energy U Fig. 9-14, p. 384

  4. HCl potential curve H2 potential curve Energy H--H--Cl 0 H2 + Cl H + HCl H  HCl H2 Cl

  5. Reaction Path Diagram HD + Cl H + DCl Energy “Transition State” reactants before this point. products after it [H--D--Cl]# Activated Complex Ear Eaf H-D + Cl H + D-Cl Rxn Path

  6. The Boltzmann Factor- exp(-Ea/RT) Fraction of Molecules That will react at Temperature T

  7. First Order or Uni-molecular Rate=k[A] Rate=-d[A]/dt= k[A] ln[A] - ln[A]0 = -kt [A]=[A]0exp{-kt} [A]=[A]0/2 -In(2)=-kt1/2 t1/2= ln2/k Half-Life

  8. 2nd Order reactions Rate = k[A]2 Bimolecular are 2nd order reactions A + A  Products Rate=-(1/2)d[A]/dt = k[A]2 1/[A] = 1/[A]0 + 2kt [A]=[A]0/2 (1/[A]0)= 2kt1/2 t1/2= 1/2k[A]0 Half-life

  9. Kinetics of Approaching Equilibrium Equilibrium: Evaporation rate equal the condensation Rate  Equilibrium C2H4(l)  C2H4(g) Rate Law Rate=k1=const, , Order zero • C2H4(l)  C2H4(g) Evap • C2H4(g)  C2H4(l) Cond Rate Law Rate=k2[C2H4(g)] First Order Fig. 10-16, p. 459

  10. Elementary Reactions : Single Step Reactions as opposed to complex multi-step reactions Termolecular reaction: A + B + C  Products Rate = k[A][B][C] Recombination Reaction First Step in the formation of liquids as well 3rd body takes away E ≥ KE Energy free bound KE 3H2O  (H2O)2 + H2O D0 bond energy @ R>>Re where V(R)~0 E=KE=2.4 kJ mol-1 D0= 5.0 kJ mol-1 Collision takes away 3 kJ mol-1 E after collision E=-0.6 kJ mol-1

  11. A + B Collision Rate ZAB=[A][B]{(√2)<u>(πd2)}(N0)2 A + B Reaction Rate Reaction Rate= -d[A]/dt=-d[B]/dt ~ ZAB p exp{-Ea/kT} -d[A]/dt=-d[B]/dt =[A][B]{(√2)<u>(πd2)}(N0)p exp{-Ea/kT} -d[A]/dt=-d[B]/dt = k[A][B]

  12. Given the overall reaction: 2NO(g)  N2(g) + O2(g) • Propose a plausible 2 step mechanism • 2NO(g)  N2O(g) + O(g) (2) N2O(g) + O(g)  N2(g) + O2(g) Assume Step (2) is rate limiting • Forward Rate=k1[NO]2 and Reverse Rate =k-1[N2O][O] • Forward Rate =k2[N2O][O] Fast Equilibrium Forward Rate = Reverse Rate: k1[NO]2 =k-1[N2O][O] Overall Rate=k2[N2O][O] = k2(k1/k-1) [NO]2 = = k2K1[NO]2 Then this is also a plausible Mechanism where K=k1/k-1

  13. Enzyme Substrate Catalytic Reaction Steady State Approximation Enzyme(E), Substrate(S), the Complex(ES) and the Reaction Product(B) E + S ES (1) ES  P (2) The Overall Rate = d[P]/dt = k2 [ES] Since [ES]~const d[ES]/dt = k1[E][S] - k-1[ES] - k2[ES]~0 Now since [E]0= [E] + [ES] and [E]= [E]0 - [ES] [ES]=k1[E]0[S]/{k1[S] + (k-1 + k2)} d[P]/dt=k2k1[E]0[S]/{k1[S] + (k-1 + k2)} let KM=(k-1 + k2)/k1 d[P]/dt=k2[E]0[S]/{[S] + KM} Michaelis-Menten Eq.

  14. Max Rate =k2 [E]0 High [S] Max Rate d[P]/dt=k2[E]0[S]/{[S] + KM} Slope= k2[E]0/KM Low [S]

  15. Catalysis Rxn path Diagram Internal Energy Change @ V=const Difference in the Stored between Reactants/Products

  16. U= q + w 1st Law Thermodynamic process at constant Volume V=const V=0 so, work=0 w=0 U= q + w qV = U q>0 A Flame: CH4 + 2O2 CO2 + 2H2O(l) combustion gives off energy that is transferred as heat(q) to the gas in the piston which can do work against the Pext but since V is held, no pressure volume

  17. Heat flows from hot to cold? At V=const U=q Hot q(T1) Cold (T2) T1  T2 for T2 < T1 For the hot system q < 0 And for the cold system q > 0 The process is driven by the overall Increase in entropy!

  18. w = - (force) x (distance moved) Pext Pext Gas A A Gas h1 h2 w = -F(h2-h1)= PextA (h2-h1)=Pext(V2 - V1) w = - PextV V =hA and P=F/A w < 0: system (gas in cylinder) does work: reduces U; V >0 w > 0: work done on the system: increases U; V <0

  19. Equivalence of work and heat (Joule’s Experiment) Since q=0 and DU=w=-mgDh=mghBut T changes byDT!So the energy transferred as work would Corresponds to a heat transfer q=CDT w=mgh 0 Dh work= w=-mgDh qin= 0 -h Fig. 12-7, p. 495

  20. Energy transferred as heat @ constant pressure What is the heat of reaction when V is not constant: When the system can do work against and external pressure ! Use the Enthalpy H=U + PV Since H = U + (PV) if P=const and not V H = U + PV but w = -PV Therefore H = U - w but U = q + w by the 1st Law @ P=const. qP= H Note that the Enthalpy is a state function and is therefore Independent of path; It only depends on other state functions i.e., H=U + PV !

  21. Energy transferred as heat @ constant pressure qP = DH q>0 A Flame: CH4 + 2O2 CO2 + 2H2O(l) combustion gives off energy that is transferred as heat(q) to the gas in the piston which can do work against the Pext but since V is held, no pressure volume

  22. For Chemical Reactions AB H=HB – HA= Hprod – Hreac Path(a) AB H=HB – HA Path(b) AB (1) AB (2) a catalyst U=HB – HA H is a State function Path Independent A H=q P=const B P=const H=HA – HB = q Vq < 0 exothermic, q > 0 endothermic, q = 0 thermo-neutral

  23. Thermodynamic Processes no reactions/phase Transitions Ideal Gas expansion and compression U= ncVT & H=ncPT For P=constqP=H For V=constqV=U cP=(cV + R) for all ideal gases cV= (3/2)R atomic gases cP= (5/2)R= 20.79 Jmol-1K-1 cV>(3/2)R for and diatomic gases Polyatomic gases

  24. Thermodynamic Processes no reactions/phase Transitions Ideal Gas expansion and compression U= ncVT & H=ncPT For P=constqP=H=ncPT cP=(cV + R) for all ideal gases cV= (3/2)R atomic gases cP= (5/2)R=(5/2)(8.31) cP =20.79 Jmol-1 K-1 cV>(3/2)R for and diatomic gases Polyatomic gases

  25. Thermodynamic Processes no reactions/phase Transitions qin>0 Pext isotherm qout<0 UAC = qin + wAC qin= n cP(TB – TA) > 0 and wAC = - PextV UCB = qout + wCB qout= n cV(TC – TB) < 0 and wCB = - PV=0 UAB = UCA + UCB = n cP(TB – TA) - PextV + n cV(TC – TB)

  26. Thermodynamic Processes no reactions/phase Transitions wAC= - PextVAC work=-(area) Under PV curve qin>0 Pext isotherm qout<0 UAC = qin + wAC qin= n cP(TB – TA) > 0 and wAC = - PextV UCB = qout + wCB qout= n cV(TC – TB) < 0 and wCB = - PV=0 UAB = UCA + UCB = n cP(TB – TA) - PextV + n cV(TC – TB)

  27. Thermodynamic Processes no reactions/phase Transitions qin>0 Pext isotherm wDB= - PextVAC work=-(area) Under PV curve qout<0 UAC = qin + wAC qin= n cP(TB – TA) > 0 and wAC = - PextV UCB = qout + wCB qout= n cV(TC – TB) < 0 and wCB = - PV=0 UAB = UCA + UCB = n cP(TB – TA) - PextV + n cV(TC – TB)

  28. H2O P-T Phase Diagram and phase transitions at P=const Melting Point: heat of fusion H2O(s)H2O(l) Hfus= q= 6 kJmol-1 Boiling point; heat of vaporization H2O(s)H2O(l) Hvap= 40 kJmol-1

  29. For Phase Transitions at P=const: A(s)A(l) Hfus= q Heat of Fusion A(l)A(g) Hvap= q Heat of Vaporization A(s)A(g) Hvub= q Heat of Sublimation NaCl(s)Na+(l )+ Cl-(l ) Molten liquid TM = 801 °C Na+(l )+ Cl-(l )  Na(g) + Cl(g) TB= 1413 °C

  30. Heat transfer required to Change n mole of ice to steam at 1 atm q= qice + nHfus + qwat + nHevap+ qvap qice=ncp(s)T, qwat=ncp(l)T and qvap= ncp(g)T For example If a piece of hot metal is placed in a container with a mole of water that was initially @ Temperature T1 and The water and metal came to equilibrium At temperature T2 H2O P-T Phase Diagram Physical Reaction/phase transitions T1 T2

  31. State function U, H, Equations of State Surface P=nRT/V or PH2O=nRT/(V-nbH2O) - aH2O(n/V)2 P=F(V,T) State Functions are only defined in Equilibrium States, does not depend on path !

  32. Hess’s Law applies to all State Function DH A D • AD DH • AB DH1 • BC DH2 • CD DH3 DH1 DH3 C B DH2 DH= DH1 + DH2 + DH3 Fig. 12-14, p. 506

  33. The Standard State: Elements are assigned a standard heat of Formation DH°f= 0 Solids/liquids in their stable form at p=1 atm In species solution @ concentration of 1Molar For compounds the DH°f Is defined by its formation From its elements in their Standard states: DH°f = 0 C(s,G) + O2(g)  CO2(g) DH°f (CO2(g) = -393.5 kJ mol -1 CO(g) + ½O2(g)  CO2((g) DH= -283 kJ (enthalpy change) C(s,G) + O2(g)  CO(g) + ½ O2(g) DH° standard enthalpy change DH° =DH°f (CO) + 1/2 DH°f(O2) – {DH°f (C(s,G) + DH°f(O2)} = -110.5 kJmol-1 Standard Enthalpy Change DH°

  34. In General for a reaction, with all reactants and products at a partial pressure of 1.0 atm and/or concentration of 1.0 Molar aA + bB  fF + eE The Standard Enthalpy Change at some specified Temperature DH°(rxn) = DH°f(prod) - DH°f(react) DH°(rxn) = f DH°f(F) + eDH°f(E) – {aDH°f(A) + bDH°f(B)} energy Elements in their standard states DH°f(reactants) DH°f(product) DH°(rxn

  35. Equilibrium Vapor Pressure Pure H2O P-T Phase Diagram Vapor Pressure Lowering After dissolution Table 10-3, p. 460

  36. Dissolution of a Nonvolatile((zero partial pressure) solute, e.g., sugars or salts. Heterogeneous Phase equilibrium of a 2 component solution H-bonds Fructose C6H12O6 Hydrated Fructose C6H12O6

  37. Boiling Point Elevation Tb = boiling point T’b = elevated boiling pt. Vapor pressure Lowering With added solvent DTb=T’b – Tb=(1/S)X2 0.102 °C Fig. 11-11, p. 493

  38. As an example: NaCl(s) dissolves completely in water. NaCl(s) + H2O(l)  Na+(aq) +Cl-(aq) 1.0 mole NaCl(s) produces 2 moles of ions in soln Given 0.0584 grams of NaCl(s) is dissolved in 10 grams of H20(l) What is the boiling point(T’b) at p=1 atm? Tb=100°C for pure water DTb = Kbm ( m=molality of soln); Kb(H2O)= 0.512 K kg mol-1 m=(m2/M2)/(m1[1000gkg-1])=n2/(m1/[1000gkg-1]) n2=(2)x(0.0584 g/ 58.4 gmol-1) = 2.0 x 10-3 mols m1 =10 g m =2x10-3 mols/0.01 kg = 0.2 mols kg-1 DTb = (0.512)x(0.2) K = 0.1024 K Tb = 100 °C + 0.102 °C= 100.102 °C

  39. Freezing Point Depression only consider cases where the pure solvent crystalizes from solution, e.g., ice Crystalizes from salt water and NaCl(s) does not DTf = - Kfm ( m=molality of soln) DTf = T’f - Tf Melting point(Tf) lowered to keep the vapor pressure over the pure solid and liquid solution the same at Equilibrium! DP1= - X2P°1 S= -DP1/DTf (pure solvent) Solved for DTf DP1 DTf = 0.037 °C = 0.037 K Fig. 11-12, p. 496

  40. Example : again for 0.058 gmol-1 of NaCl in 10 g H2O(l) over H2O(s) The molality is m = 0.02 gkg-1 (grams of solute per kg of solvent) DTf = - Kfm ( m=molality of soln) DTf = - (1.86)x(0.02) K= - 0.0372 °C T’f = 0 +(-0.037)°C=-0.037°C Table 11-2, p. 494

  41. Osmotic Pressure π=[solute]RT Solute molar concentration Recall that pressure in the tube P=rghso π=rghVan’t Hoff proposed π=[solute]RT Which is similar to PV=nRT for an ideal gas. Note that is the solute Concentration but the Solvent mass density r(kg meter-3)! At Equilibrium the rate of The Solvent molecules Crossing the membrane from solution Is equal to the rate from the solvent Solute molecules Lowers the rate of Solvent molecules Crossing the Membrane From the solution Fig. 11-14, p. 498