Uses of Maxwell Relations

Use of Maxwell Relations: • Wh What at is the the In Inte tern rnal al Pr Pres essu sure re ?

  U     T =   V   T  dU  = − PdV  + T dS dS dU  dV 

= − P + T 

dS  dV 

  U     S     P     = − P + T    = − P + T      V  T    V  T    T  V         dU  = C V  dT  +    T  T  − P dV  = TdS  − PdV        

• What is

   H      ?   P  T 

dH  = VdP + T dS dH  dP

= V  + T 

dS  dP

   H     S     V      = V  + T    = V  − T      P  T    P  T    T  P    H   = V  1 − T  ( )      P  T  dH  = VdP + TdS  = C P dT  + V (1 − T )dP

• Determination of CP -CV

− C V  =   H    − C V    T   P   U   +   ( PV )  − C  C P − C V  =         T   P   T   P V     V     T       P    V     +  T    − P    + P   − C V  C P − C V  = C V    T  P     T  V     T  P   T  P   P    V   C P − C V  = T       = T V    T  V    T  P C P

2

C P

− C V  = VT 

Properties of the Gibbs Free Energy • Temperature Dependence of the Gibbs Free Energy

Gm

A

Constant Pressure Single Component System

B

= V m dP − S m dT    G m      = − S m   T   P dG m

solid

liquid Tm

gas Tb

T

Sm (S) < Sm (L) V m (S) then Tm increases with Pressure (general) – if Vm (L) < Vm (S) then Tm decreases with increasing pressure (water, etc..)

• Two phases α and β are in equilibrium at temperature T and pressure P, if Gmα (P,T) = Gm β (P,T) Let us change P by dP and find out what is the corresponding dT required to preserve the equilibrium between the two phases. Changes of T by dT and P by dP lead to a change of G by dG d Gmα = -Smα dT + Vmα dP for the α phase and d Gmβ = -Smβ dT + Vmβ dP for the β phase To preserve equilibrium conditions, Gmα + d Gmα = Gm β + d Gmβ which implies d Gmα = -Smα dT + Vmα dP = d Gmβ = -Smβ dT + Vmβ dP (Smβ -Smα ) dT = (Vmβ - Vmα ) dP dP dT 

∆ S  ∆ H  = = ∆V  T ∆V 

Clapeyron Equation β coexistence curve) (describes the α

Note at equilibrium, ∆G = ∆H - T ∆S = 0 for the transition

• Solid - Liquid Equilibrium

∆ H fus (T , P ) ∆ H  fus (T *, P *)   T      ln = ⇒ P = P* + dT  T ∆ V  fus ( T , P) ∆V  fus (T *, P *)  T  *   dP

if ∆Hfus and ∆Vfus can be assumed independent of T and P • Liquid - Vapor Equilibrium

∆ H vap( T , P ) ∆ H vap (T , P ) ∆ H vap (T , P ) = ≅ = TRT  dT  T ∆ V vap (T , P) TV vap (T , P) dP

P

⇒

dP P

=

∆ H vap (T , P) dT   R

T 2

P   −∆ H vap  1 1   ⇒ ln  =  −     P *   T  T  *   R

if ∆Hvap can be assumed independent of T and P.

• Solid - Vapor Equilibrium: dP dT 

=

∆ H sub (T , P) ∆ H sub (T , P) ∆ H sub ( T , P) ≅ = TRT  T ∆V sub ( T , P ) TV sub (T , P) P

⇒

dP P

=

∆ H sub (T , P)

dT 

 R

T  2

P   −∆ H sub  1 1      −   ⇒ ln   =  P *    T  T  *    R

if ∆Hsub can be assumed independent of T and P •

Close to the triple point ∆Hsub = ∆Hfus + ∆Hvap Because H is a State Function

Properties of Simple Mixtures •

  X     X  J  =  Partial molar value of extensive property X   n J      n I , P,T  is an Intensive property of a component in a mixture

• Chemical Potential: Partial molar Free Energy: µJ Activity: aJ

  G   =  J  =      n J   n I , P,T 

∅ + RT  ln(a )  J 

J 

Standard State (P = 1 bar) • Ideal Binary A-B Liquid Solution: Raoult’s Law PA = xA PA* where aA = xA = nA / (nA + nB) mole fraction of A ∆Gmix = (nA+nB)RT(xA ln(xA) + xB ln(xB)) ∆Hmix = 0 ∆Vmix = 0 ∆Smix = - (nA+nB) R (xA ln(xA) + xB ln(xB))

Phase Diagrams • Gibbs Rule of Phase: The number of independent intensive state variables necessary to fully define the state of a system is called the variance of the system(or the # of degrees of freedom) and is given by F = C - P + 2 where C = # of independent constituents (# of species if no reaction) P = # of phases 2 accounts for T and P • F = 0 the system is invariant (triple point) F = 1 the system is monovariant (L/S, L/V, S/V coexistence) F = 2 the system is bivariant (single phase & single component) F = 3 the system is trivariant (single phase & 2 components)

Chemical Equilibrium • Consider the reaction 2A + 3B C + 2D If we define the extent of reaction by ζ, when the reaction advances by dζ , the amount of reactants changes by: dnA = -2 dζ dnB = -3 dζ In general, for a reaction written as dnC = dζ ΣνJ J = 0, then dnJ = νJ dζ dnD = 2 dζ At constant temperature and pressure: dG = µA dnA + µB dnB + µC dnC + µD dnD dG = (- 2µA - 3µB + µC + 2µD ) dζ In general dG =

ΣµJ dnJ = (ΣµJ νJ) dζ

  G   ∆ R G =       P,T 

• Define the Reaction Free Energy as:

• Define the Chemical Potential of Each Species by:

   J  =   

  =   n J    n I , P,T  G

∅ + RT  ln a (  J  )

J 

where aJ is PJ /P° or f J /P° for gases and unity for pure liquid and pure solids.

∆RG = ∆RG° +RT lnQ

(Q is called the reaction quotient)

For the example above ∆RG° = - 2µA° - 3µB° + µC° + 2µD° 1

and

Q=

a a

2

C   D

3 2 a B a A

In general, ∆RG° = ΣνJ µJ°

and Q =

∏ a J  J   J 

• At equilibrium, ∆RG = 0

In general,

 a 1 a 2   C   D   K  =   a 3 a 2      B  A  eq

   J    K  =  ∏ a J      J   eq

RT ln K = - ∆RG° and K can be calculated at a given temperature from the free energies of formation for each of  the reactants and products

∆ R G ∅ = ∑  J 

∅( J ) G ∆ J  f 

Standard Gibbs Free Energy of Formation of 1 mole of  Substance J at temperature T and 1 bar found in thermodynamic tables.

Response of Chemical Equilibria to Conditions • Pressure Dependence: K depends on ∆G°(T) and T and NOT on pressure. • Temperature Dependence:

   ∆G        T     −∆ H     = 2 T     T       P         

 ∆ RG ∅        T         −∆ R H ∅   = 2 T  T       P

    

ln K  

T 

  =  P

∆ R H ∅  RT 

2

where ∆ RH° is given by:

∆ R H ∅ = ∑

∅ ( J ) H  ∆ J  f 

 J 

Note that the enthalpic term will also depend on T !!!

If the standard reaction enthalpy can be calculated at T* (generally at 298K) and if the molar heat capacity of reactants and products is known as a function of temperature, an expression for the temperature dependence of the reaction enthalpy can be obtained. Then the equilibrium constant can be calculated at any temperature (T1) in terms of the equilibrium constant at T* (the latter being calculated from the standard free energy of formation of products and reactants). K ( T 1)

∫ d ln K  =

K (T *)

T  1 1

∫ 

 R T *

∆ R H ∅ (T ) 2

dT 

T 

T  ∅ ∅ ∆ R H  (T ) = ∆  R H  (T  *) + ∫ ∆ R C P (T )dT  T *

∆ R C P (T ) = ∑  J 

J  C  p,m ( J , T )