Week 8 Ch 12 Thermodynamics Enthalpy: H=U + PV,  H=  U +  (PV) accounts for expansion work -P  P=const in thermodynamics processes, e.g., reactions, phase transitions, etc. Both U and H are State Functions and are path independent Heat of Reaction:  V=const and  P= const for both chemical and physical reactions. Heat Capacity: C=q/  T size dependent(extensive) Material property specific heat capacity c s =C/m size independent(Intensive), mostly used for non-pure substances, like allo phases). c P and c V are the molar heat capacities for P and V constant, respectively. It represents the ability of a substance to store energy in its rotations, vibrations, translation and electronic degrees of freedom, e.g., kT/2 for 1D trans and kT for per vibrations( =kT/2 and =kT/2) Standard State: The Thermodynamically stable state for pure liquids and solid, for gases ideal gas behavior, for solutions 1molar concentration of the dissolved species at P= 1atm and some specified T in each case Midterm Friday: Ch 9, 10, , 11.5, 18, One side of 1 page notes(must be hand written), closed book Review Session 2-3 pm, in FRANZ 1178

Equations of states for fixed amount of a pure substances, e.g., 1.0 mols of H 2 O P=F(V,T) State Functions are only defined in Equilibrium States, does not depend on path ! Equations of State P=nRT/V or P H2O =nRT/(V-nb H2O ) - a H2O (n/V) 2

Applies to Chemical as well as Physical Changes  V= V B – V A =V 2 – V 1  P= P B – P A =P 2 – P 1 Fast P, V and T are Thermodynamic State Variables and defines the Thermodynamic States (A and B) They do not depend on the path of the process Equilibrium State A Equilibrium State B

Equations of states for fixed amount of a pure substances, e.g., 1.0 mols of H 2 O 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

Fig. 12-3, p. 491 The difference in State Properties are independent of path e.g., like P(V,T) and Altitude! Non-state properties like Heat(q), work(w) the or the distance travelled depend on the path

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

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

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

First Law of Thermodynamics  U= q + w 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 Thermodynamic process at constant Volume V=const  V=0 so, work=0 w=0  U= q + w q V =  U Thermodynamic process at constant Volume V=const  V=0 so, work=0 w=0  U= q + w q V =  U

Reaction A  B  U=U A – U B can occur via 2 different paths, e.g., catalytic and non-catalytic,  U is the same via either path since it is a state function At V=const  U=U A – U B = q since  U= q V q 0 endothermic, q = 0 thermo-neutral Path(1) A  B  U=U A – U B A  B (1) A  B (2) a catalyst  U=U A – U B U is State function Path Independent Path(2) B A  U=q

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 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 H=U + PV !

What about when V=const what is q for a the reaction  U= q + w = q - P ext  V 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 does work (-P ext  V) against the P ext q>0

For Chemical Reactions A  B  H=H A – H B = H prod – H reac Path(a) A  B  H=H A – H B 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 A – H B H is a State function Path Independent

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

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

Thermodynamic Processes inno reactions/phase Transitions isotherm q in q out  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 ) P ext Ideal Gas  U= nc V  T  H=  U +  (PV)  H =nc V  T + nR  T  H=n(c V +R)  T For P=const  H=q=nc P  T c P =(c V + R) for all ideal gases c V = (3/2)R atomic gases c V >(3/2)R for Polyatomic gases

Heat required to Change n mole of ice to steam at 1 atm H 2 O P-T Phase Diagram T1T1 T2T2 q= q ics + n  H fus + q wat + n  H fus + q ste q ice =nc p (s)  T, q wat =nc p (l)  T and q st =nc p (g)  T