Chemical Reaction Equilibria Part III Equilibrium constant K i fˆi K o i fi For gas-phase reactions, fio = Po =1 bar At a given temperature, if P changes, the compositions at equilibrium will change in such a way that K remains constant fˆi yiˆi P i y ˆ i i i P o K P K for gas phase reactions y ˆ i i i i P o K P if ideal solution is assumed ˆi i if ideal gases yi i i P o K P i 1 this is the equation that we used in the previous example analysis of K for ideal gases y i i i P o K P At constant P, if the reaction is endothermic, DH0 >0, K increases when T increases, therefore the LHS term will increase, the reaction shifts to the right, eeq increases if the reaction is exothermic, DH0 <0, K decreases when T increases, therefore the LHS term will decrease, the reaction shifts to the left, eeq decreases effect of pressure (at constant T) • it depends on , which is the change in the total number of moles of the reaction • if is negative => the total number of moles decreases: y i i i P o K P – if P increases, the LHS must increase to keep K constant, => the equilibrium shifts to the right, eeq increases • if is positive => the total number of moles increases: – if P increases, the LHS must decrease to keep K constant, => the equilibrium shifts to the left, eeq decreases example • the production of 1,3-butadiene can be carried out by dehydrogenation of n-butane: C4H10(g)CH2:CHCH:CH2 (g) +2H2(g) Side reactions are suppressed by introduction of steam. If equilibrium is attained at 925 K and 1 bar and if the reactor product contains 12 mol% of 1,3 butadiene, find: (a) the mole fractions of the other species in the product gas (b) the mole fraction of steam required in the feed solution C4H10(g)CH2:CHCH:CH2 (g) +2H2(g) =2 no = 1+x y1 = (1-e)/(1+x+2e) y2 = e/(1+x+2e)=0.12 y3 = 2 y2=0.24 In order to calculate K, we need DG at 925 K First calculate DGo and DHo at 298 K, from tables appendix C DGo =235030 J/mol DHo = 166365 J/mol • Get A, B, C, D for each component and calculate DA, DB, DC, and DD. • Calculate DG at 925K using equation 13. 18; DG = 9.242 x103 J/mol • Calculate K = exp (- DG/RT) =0.30 • K =(y3)2(y2)/y1=(0.24)2(0.12)(1+x+2e)/(1-e)=0.3 • here there are two unknowns, x, and e • However since we know y2 we have another equation y2 = e/(1+x+2e)=0.12 • Therefore, solve for e=0.84 and x=4.31mol steam • Get y1 = (1-e)/(1+x+2e) =0.023 • ysteam= 4.31/5.31 =0.812 (in the feed) • yH2O at equilibrium = 1-0.24-0.12-0.023 =0.617 Ammonia synthesis reaction ½ N2(g) + 3/2 H2(g) NH3 (g) the equilibrium conversion of ammonia is large at 300K but decreases rapidly with increasing T. However, reaction rates become appreciable only at higher temperatures. For a feed mixture of hydrogen and nitrogen in the stoichiometric proportions, (a) what is the equilibrium mole fraction of ammonia at 1 bar and 300 K? solution ½ N2(g) + 3/2 H2(g) NH3 (g) = -1 no = 2 In order to calculate K, we need DG at 300 K First calculate DGo and DHo at 298 K, from tables appendix C DGo =-16450 J/mol DHo =- 46110J/mol • Get A, B, C, D for each component and calculate DA, DB, DC, and DD. • Calculate DG at 300K using equation 13. 18; DG = -16270 J/mol • Calculate K = exp (- DG/RT) =679.57 • K =(y3)/(y2)3/2(y1)1/2 • you can show (see problem 13.9) that eeq =1-(1+1.299KP/Po)-1/2=0.9664 • yNH3 =e/(2-e)=0.935 (b) At what T does the equilibrium mole fraction of ammonia equal 0.5 for a pressure of 1 bar? if yNH3 =0.5, eeq = 2/3 = 1-(1+1.299KP/Po)-1/2 K = 6.16 at what T, K has this value? K = exp (- DG (T)/RT) =6.16 solve for T; iterative; T=399.5 K (c) At what temperature does the equilibrium mole fraction of ammonia equal 0.5 at a pressure of 100 bar, assuming the equilibrium mixture is an ideal gas? For P =100 bar, eeq =1-(1+1.299KP/Po)-1/2 = 2/3 K = 0.06159 at what T, K has this value? K = exp (- DG (T)/RT) =0.06159 solve for T=577.6 K (d) at what temperature does the equilibrium mole fraction of ammonia equal 0.5 for a pressure of 100 bar, assuming the equilibrium mixture is an ideal solution of gases? for an ideal solution model, P i (I) yii o K P i i) Define a guess T. Start with Tguess =578 K (obtained in part a) and 100 bar. Use virial equation of state, for i’s that are functions of temperature ii) Obtain new expression for K using the calculated i’s in equation (I): 1-(1+1.299K/1.184xP/Po)-1/2 = 2/3 iii) solve for K =0.0729; iv) since K (Tguess) = 0.0729 must be=exp (- DG (T)/RT), evaluate exp (- DG (Tguess)/RTguess). Is it equal to 0.0729? If not, change Tguess and go to (i) Solution after convergence T = 568.6 K