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8 th Week Chap(12-13) Thermodynamics and Spontaneous Processes

8 th 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

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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. 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 !

  3. 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

  4. 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

  5. Thermodynamic Processes no reactions/phase Transitions Ideal Gas expansion and compression U= ncVT & H=ncPT H=U + (PV) H =ncVT + nRT H=n(cV +R) T For P=constqP=H=ncPT 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

  6. Thermodynamic Processes no reactions/phase Transitions Ideal Gas expansion and compression U= ncVT & H=ncPT H=U + (PV) H =ncVT + nRT H=n(cV +R) 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

  7. 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)

  8. 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)

  9. 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)

  10. 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

  11. 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

  12. 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

  13. 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

  14. Example of Hess’s Law C(s,G) + O2(g)  CO2(g) DH1 = -393.5 kJ CO2((g ) CO(g) + ½O2(g) DH2 = + 283 kJ C(s,G) + O2(g)  CO(g) + ½ O2(g) DH = ?

  15. Example of Hess’s Law C(s,G) + O2(g)  CO2(g) DH1 = -393.5 kJ CO(g) + ½O2(g)  CO2((g) DH2 = -283 kJ C(s,G) + O2(g)  CO(g) + ½ O2(g) DH =DH1+DH2= -110.5 kJ

  16. 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° =DH°f (CO) + 1/2 DH°f(O2) – {DH°f (C(s,G) + DH°f(O2)} = -110.5 kJmol-1 Standard Enthalpy Change DH°

  17. In General for a reaction, with all reactants and products are at a partial pressure of one atm and/or concentration of 1 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

  18. 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 !

  19. 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!

  20. 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 Correspond to a heat transfer q=CDT w=mgh 0 Dh work= w=-mgDh qin= 0 -h Fig. 12-7, p. 495

  21. 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

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