1. 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 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  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. 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 1 st Law @ P=const. q P =  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 ! Energy transferred as heat @ constant pressure

3. A Flame: CH 4 + 2O 2  CO 2 + 2H 2 O(l) combustion gives off energy that is transferred as heat(q) to the gas in the piston which can do work against the P ext but since V is held, no pressure volume q>0 Energy transferred as heat @ constant pressure q P =  H

4. For Chemical Reactions A  B  H=H B – H A = H prod – H reac Path(a) A  B  H=H B – H A P=const  H=H A – H B = q V q 0 endothermic, q = 0 thermo-neutral Path(b) B A  H=q P=const A  B (1) A  B (2) a catalyst  U=H B – H A H is a State function Path Independent

5. Thermodynamic Processes no reactions/phase Transitions Ideal Gas expansion and compression  U= nc V  T &  H=nc P  T  H=  U +  (PV)  H =nc V  T + nR  T  H=n(c V +R)  T For P=const q P =  H=nc P  T c P =(c V + R) for all ideal gases c V = (3/2)R atomic gases c P = (5/2)R= 20.79 Jmol -1 K -1 c V >(3/2)R for and diatomic gases Polyatomic gases

6. Thermodynamic Processes no reactions/phase Transitions Ideal Gas expansion and compression  U= nc V  T &  H=nc P  T  H=  U +  (PV)  H =nc V  T + nR  T  H=n(c V +R)  T For P=const q P =  H=nc P  T c P =(c V + R) for all ideal gases c V = (3/2)R atomic gases c P = (5/2)R=(5/2)(8.31) c P =20.79 Jmol -1 K -1 c V >(3/2)R for and diatomic gases Polyatomic gases

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

8. Thermodynamic Processes no reactions/phase Transitions  U AC = q in + w AC q in = n c P (T B – 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 CA +  U CB = n c P (T B – T A ) - P ext  V + n c V (T C – T B ) 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

9. Thermodynamic Processes no reactions/phase Transitions  U AC = q in + w AC q in = n c P (T B – 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 CA +  U CB = n c P (T B – T A ) - P ext  V + n c V (T C – T B ) 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

10. H 2 O P-T Phase Diagram and phase transitions at P=const Melting Point: heat of fusion H 2 O(s)  H 2 O(l)  H fus = q= 6 kJmol -1 Boiling point; heat of vaporization H 2 O(s)  H 2 O(l)  H vap = 40 kJmol -1

11. For Phase Transitions at P=const: A(s)  A(l)  H fus = q Heat of Fusion A(l)  A(g)  H vap = q Heat of Vaporization A(s)  A(g)  H vub = q Heat of Sublimation NaCl(s)  Na + ( l )+ Cl - ( l ) Molten liquid T M = 801 °C Na + ( l )+ Cl - ( l )  Na(g) + Cl(g) T B = 1413 °C

12. Heat transfer required to Change n mole of ice to steam at 1 atm H 2 O P-T Phase Diagram T1T1 T2T2 q= q ice + n  H fus + q wat + n  H evap + q vap q ice =nc p (s)  T, q wat =nc p ( l )  T and q vap = nc p (g)  T For example If a piece of hot metal is placed in a container with a mole of water that was initially @ Temperature T 1 and The water and metal came to equilibrium At temperature T 2 Physical Reaction/phase transitions

13. 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  

14. 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)  = ?

15. 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

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

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  (rxn)   f (prod) -   f (react)  (rxn) = f   f (F) + e   f (E) – {a   f (A) + b   f (B)} Elements in their standard states energy   f (reactants)  (rxn   f (product)

18. State function U, H, P=F(V,T) State Functions are only defined in Equilibrium States, does not depend on path ! Equations of State Surface P=nRT/V or P H2O =nRT/(V-nb H2O ) - a H2O (n/V) 2

19. Hot q(T 1 )Cold (T 2 ) T 1  T 2 for T 2 < T 1 Heat flows from hot to cold? For the hot system q < 0 And for the cold system q > 0 The process is driven by the overall Increase in entropy! At V=const  U=q

20. Fig. 12-7, p. 495 Equivalence of work and heat (Joule’s Experiment)  h work= w=-mg  h -h 0 q in = 0 Since q=0 and  U=w=-mg  h=mgh But T changes by  T!So the energy transferred as work would Correspond to a heat transfer q=C  T w=mgh

21. w = - (force) x (distance moved) Gas P ext Gas h1h1 h2h2 w = -F(h 2 -h 1 )= P ext A (h 2 -h 1 )=P ext (V 2 - V 1 ) w = - P ext  V V =hA and P=F/A w 0 w > 0: work done on the system: increases U;  V <0 A A