1. 9 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 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=k B ln  Number of Available Microstate  ~ # of quantum states~size 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 Third Law of Thermodynamics: S  as T  for a pure substance Reversible Processes:  S =∫q rev /T but not Isothermal From the 2nd Law S T = 0 ∫ T q rev /T Absolute Entropy

2. Thermodynamic Processes no reactions/phase Transitions isotherm P ext T>T T T<T Isotherms T>T>T P = nRT/V for T =const Isothermal Paths are always reversible, e.g. A  B and B  A Whereas ABC and ADB are not reversible since they occur in a finite number of steps. Though both paths must result in the same changes in U,  U=U B – U A, since U is a state function.

3. Thermodynamic Processes no reactions/phase Transitions  U AC = q in + w AC q in = n c P (T C – T A ) > 0 and w AC = - P ext  V  U CB = q out + w CB q out = n c V (T C – T B ) < 0 and w CB = - P  V=0  U AB =  U AC +  U CB = n c P (T C – T A ) - P ext  V + n c V (T C – T B ) isotherm q in >0 q out <0 P ext T>T T T<T

4. Thermodynamic Processes no reactions/phase Transitions isotherm q in >0 q out <0 P ext w AC = - P ext  V AC work=-(area) Under PV curve w AC = - P ext  V AC work=-(area) Under PV curve Area under A-C-B (-)work done by the system or (+)work done on the system

5. Thermodynamic Processes no reactions/phase Transitions isotherm q in >0 q out <0 P ext w DB = - P ext  V AC work=-(area) Under PV curve w DB = - P ext  V AC work=-(area) Under PV curve Area under A-D-C (-)work done by the system or (+)work done on the system In the irreversible path A-D-B-C-A more work is done on the system than by the system! Whereas the reversible path A  B the work is the same

6. Fig. 12-14, p. 506 Hess’s Law applies to all State Function A B D C              A  D  (1)A  B   (2)B  C   (3)C  D  

7. Example of Hess’s Law C(s,G) + O 2 (g)  CO 2 (g)   = -393.5 kJ CO 2( (g )  CO(g) + ½O 2 (g)   = + 283 kJ C(s,G) + O 2 (g)  CO(g) + ½ O 2 (g)  = ?

8. Example of Hess’s Law C(s,G) + O 2 (g)  CO 2 (g)   = -393.5 kJ CO(g) + ½O 2 (g)  CO 2( (g)   = - 283 kJ =========================== C(s,G) + O 2 (g)  CO(g) + ½ O 2 (g)  =     = -110.5 kJ

9. The Standard State: Elements in their most stables form are assigned a standard heat of Formation  H° f = 0 The standard state of solids/liquids is the pure phase @1 atm Gases it’s the ideal gas @ 1 atm For species in solution the Standard State is the concentration of 1.0 Molar @ P=1.0 atm  H° f = 0 for H + ions solution For compounds the  H° f is defined by formation of one mole of the compound from its elements in their Standard States @ 1 atm, 1 molar and some T   f = 0 C(s,G) + O 2 (g)  CO 2 (g)  H° f = -393 kJmol -1 @25 °C

10. In General for a reaction, with all reactants and products at a partial pressure of one 1 atm and/or concentration of 1 Molar aA + bB  fF + eE The Standard Enthalpy Change at some specified Temperature  (rxn)   f (prod) -   f (react)  (rxn) = f   f (F) + e   f (E) – {a   f (A) + b   f (B)} energy Elements in their standard states   f (elements)=0   f (reactants)  (rxn   f (product)

11. For Example consider the reaction: 1/2 O 2 (g)  O(g)   f [O(g)] (atomization energy/bond enthalpy) the formation of Ozone 3/2O 2 (g)  O 3 (g)   f [O 3 (g)] energy   f [O 2 (g)]=0   f (reactants)  (rxn)   f [O 3 ] O(g)   f [O(g)]>0 O(g) + O 2 (g)  O 3 (g)   =   f [O 3 ] -   f [O] -   f [O 2 ] O3O3 Standard Heat of rxn for

12. For example consider the combustion rxn(not balanced): CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O( l )    = standard  ? energy   f [O 2 (g)]=0   f [CH 4 ] +   f [O 2 ]  (rxn)   =   f [CO 2 ] +   f [H 2 O] -   f [CH 4 ] -   f [O 2 ] CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O( l )   f [CO 2 ] +   f [H 2 O]

13. The combustion rxn(balanced): CH 4 (g) + 2O 2 (g)  CO 2 (g) +2H 2 O( l )    = standard  ? energy   f [O 2 (g)]=0   f [CH 4 ] +   f [O 2 ]  (rxn)   =   f [CO 2 ] +2   f [H 2 O] -   f [CH 4 ] - 2   f [O 2 ] CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O( l )   f [CO 2 ] +   f [H 2 O]

14. For example consider the combustion rxn: CH 4 (g) + O 2 (g)  CO 2 (g) + H 2 O( l )    = standard  ?   =   f [CO 2 ] +   f [H 2 O] -   f [CH 4 ] -   f [O 2 ] energy   f [O 2 (g)]=0   f [CH 4 ] +   f [O 2 ]  (rxn) CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O( l )   f [CO 2 ] +   f [H 2 O]

15. Consider the Reversible Isothermal Expansion of an Ideal Gas: So T=const for the system.  U = q + w = 0 q = -w = (P ext  V) Heat must be transferred from the Surroundings to the System And the System does work on the surroundings If the process is reversible, P ext ~ P=nRT/V Which means done slowly and changes are infinitesimal  V  dV: The gas expands from V 1 to V 2 then  q ~ PdV= (nRT/V)dV = (nRT) dlnV since dlnV=dV/V Integrating from V 1 to V 2 q = nRT[lnV 2 - lnV 1 ]=nRTln(V 2 /V 1 ) (q) Heat transferred in the Reversible Isothermal Expansion of an Ideal Gas q = nRTln(V 2 /V 1 )

16. Reversible Irreversible q = nRTln(V 2 /V 1 )

17. Carnot will use both Isothermal and Adiabatic q = 0 processes to define the entropy change  S For an Ideal gas  U = q + w =w  U=nc V  T = -P ext  V Reversible Process P ext ~ P=nRT/V  T  dT and  V  dV nc V dT = - P dV

18. Carnot will use both Isothermal and Adiabatic q = 0 processes to define the entropy change  S Reversible Process P ext ~ P=nRT/V  T  dT and  V  dV nc v dT = - P dV nc v (dT/T)=-nRdV/V c v dln(T) =-R dln(V) c v ln(T 2 /T 1 )=Rln(V 1 /V 2 ) T2T2 T1T1 V1V1 V2V2

19. c v ln(T 2 /T 1 )=-Rln(V 1 /V 2 ) A simpler form cab be obtained c v ln(T 2 /T 1 )= ln(T 2 /T 1 ) C v =Rln(V 1 /V 2 )= ln(V 1 /V 2 ) R (T 2 /T 1 ) C v = (V 1 /V 2 ) R Recall that c p = c v + R (T 2 /T 1 ) C v = (V 1 /V 2 ) R = (V 1 /V 2 ) C p -C v let  =c p /c v (T 2 /T 1 ) = (V 1 /V 2 )  -1 this relationship is very important for adiabatic processes in engines

20. Carnot considered a cyclical process involving the adiabatic/Isothermal Compression/Expansion of an Ideal Gas: The so called Carnot cycle Which Yields (q h /T h ) –(q l /T l ) = 0 and therefore  S = q/T and that S is a State Function Since q = nRTln(V 2 /V 1 ) then  S = q/T= nRln(V 2 /V 1 ) ThTh TlTl q h ;T h ad q l: : T l ad