Chemical Reaction Equilibria: a couple examples 11-14-16

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%

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)

Products – Reactants = Value All on this slide have the order of Methane + Ethylene - Propane ΔH°f298: -74250 + 52510 - (-104680) = 82670 J/mol (Table C.4) ΔG°f298: -50460 + 68460 - (-24290) = 42290 J/mol (Table C.4) Cp: (Table C.1) A: 1.702 + 1.424 - 1.213 = 1.913 B: 9.081 + 14.394 – 28.785 = -5.31 X 10-3 C: -2.164 - 4.392 + 8.824 = 2.268 X 10-6

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)) =(42290-82670)/8.314/298) + (82670/8.314/625) + (1/8.314/625*1.913*(625- 298))-(1/8.314*1.913*ln(625/298)) = -.439 K= exp(.439) K = 1.5236

Dr. Price’s way ΔG°f625 = ΔH°f625 – T* ΔS°f625 ΔS°f298 = (82670-44290)/298= 135.435 J/mol.K ΔS°f625 = 135.435 J/mol.K +8.314(A*ln(624/298)+B(625-298)… = 135.62 J/mol.K 82576 J/mol – (625 K)*(135.62 J/mol.K) = -2187.9 J/mol K = exp(ΔG°f625 /RT) = exp(-2187.9 J/mol/8.314 J/mol.K/625 K) = 1.5236

K = ε2/(1 – ε)(1 + ε) 1.5236 = ε2/(1 – ε)(1 + ε) ε = .777 at equilibrium Part B K = .852/(1 – .85)(1 + .85) K = 2.60 ΔG°=8.314 J/mol.K * T * ln(2.6) Iterate equation 13.18 at different T until you get ΔG°/RT value that equals K=2.6 T = 646.8 K

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

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))

Thermodynamic property changes: All on this slide in the order of: Water + Chlorine – HCl – Oxygen ΔH°f298: 2*(-241818) + 2*(0) - 4*(-92307) - (0) = -114408 J/mol (Table C.4) ΔG°f298: 2*(-228572) + 2*(0) - 4*(-95299) - (0) = -75948 J/mol (Table C.4) ΔCp (Table C.1) A: 2*3.470 + 2*4.442 - 4*3.156 - 1*3.639 = -.439 B: 2*1.450 + 2*.089 – 4*.623 – 1*.506 = 8*10-5 C: 0

Equation 13.18 ΔG°/RT = ((-75948 -(-114408)/8.314*298) + (-114408/8.314*773)+ (1/8.314*773*-.439*(773-298))-(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