1. Chemical Reaction Equilibria: a couple examples

2. Problem 13.16 (p. 536) C3H8 (g)  C2H4 (g) + CH4 (g)
Find the fractional conversion of propane at 625K
Find the temperature with a fractional converstion of 85%

3. Determining the reaction coordinate
Assume starting basis of 1 mol C3H8
Change in component per extent of reaction
C3H8 : 1 staring mole, -1 for reaction: 1-ε
C2H4 : 0 starting moles, +1 for reaction: ε
CH4 : 0 starting moles, +1 for reaction: ε
Total: 1- ε + ε + ε = 1+ ε
C3H8 : (1 - ε)/(1 + ε)
C2H4 : ε/(1 + ε)
CH4 : ε/(1 + ε)
Overall for this reaction: K = ε2/(1 – ε)(1 + ε)

4. Products – Reactants = Value
All on this slide have the order of Methane + Ethylene - Propane
ΔH°f298: ( ) = J/mol (Table C.4)
ΔG°f298: (-24290) = J/mol (Table C.4)
Cp: (Table C.1)
A: = 1.913
B: – = X 10-3
C: = X 10-6

5. ΔCp = A(T-To) + B/2(T2-To2) + C/3(T3-To3)
ΔG°/RT = -ln K
My way (Truncating ΔCp at A for simplicity’s sake)
ΔG°/RT = ((ΔGo°-ΔHo°)/RTo) + (ΔHf298°/RT)+ (1/RT*ΔCp°*(T-To))- (1/R*ΔCp°*ln(T/To))
=( )/8.314/298) + (82670/8.314/625) + (1/8.314/625*1.913*( ))-(1/8.314*1.913*ln(625/298))
= K= exp(.439) K =

6. Dr. Price’s way
ΔG°f625 = ΔH°f625 – T* ΔS°f625
ΔS°f298 = ( )/298= J/mol.K
ΔS°f625 = J/mol.K (A*ln(624/298)+B( )… = J/mol.K
82576 J/mol – (625 K)*( J/mol.K) = J/mol
K = exp(ΔG°f625 /RT) = exp( J/mol/8.314 J/mol.K/625 K) =

7. K = ε2/(1 – ε)(1 + ε)
= ε2/(1 – ε)(1 + ε)
ε = .777 at equilibrium
Part B
K = .852/(1 – .85)( )
K = 2.60
ΔG°=8.314 J/mol.K * T * ln(2.6)
Iterate equation at different T until you get ΔG°/RT value that equals K=2.6
T = K

8. Problem 13.11
4HCl (g) + O2 (g)  2H20 (g) + 2Cl2 (g)
This reaction occurs at 500°C (773K) and 2 bar. There are 5 mols of HCl for each mol of O2. What is the composition of this system at equilibrium?
Assume ideal gas

9. Determining reaction coordinates
HCl: 5 starting moles, -4 for reaction: 5-4ε
Oxygen: 1 starting mole, -1 for reaction: 1- ε
Water: 0 starting moles, +2 for reaction: 2ε
Cl2: 0 starting moles, +2 for reaction: 2ε
Overall change (denominator of reaction coordinate): 5-4ε + 1- ε + 2ε + 2ε = 6- ε
K = (2ε/5-4ε)4*((6- ε)/(1- ε))

10. Thermodynamic property changes:
All on this slide in the order of: Water + Chlorine – HCl – Oxygen
ΔH°f298: 2*( ) + 2*(0) - 4*(-92307) - (0) = J/mol (Table C.4)
ΔG°f298: 2*( ) + 2*(0) - 4*(-95299) - (0) = J/mol (Table C.4)
ΔCp (Table C.1)
A: 2* * * *3.639 = -.439
B: 2* *.089 – 4*.623 – 1*.506 = 8*10-5
C: 0

11. Equation 13.18
ΔG°/RT = (( ( )/8.314*298) + ( /8.314*773)+ (1/8.314*773*-.439*( ))-(1/8.314*(-.439)*ln(773/298)) = -1.97
K = exp(-ΔG°/RT)= exp(1.97) = 7.18
Set the previously determined reaction coordinate equal to 7.18 and you get a ε of .793
Determining equilibrium composition
HCl: (5-4*.793)/(6-.793) = .3508
Oxygen: (1-.793)/(6-.793) = .0397
Water: 2*.793/(6-.793) = .3048
Chlorine: 2*.793/(6-.793) = .3048