Chemical Reaction Equilibria Part III

Equilibrium constant K 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

K for gas phase reactions this is the equation that we used in the previous example

analysis of K for ideal gases 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 n, which is the change in the total number of moles of the reaction if n is negative => the total number of moles decreases: if P increases, the LHS must increase to keep K constant, => the equilibrium shifts to the right, eeq increases if n 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: the mole fractions of the other species in the product gas the mole fraction of steam required in the feed

solution C4H10(g)CH2:CHCH:CH2 (g) +2H2(g) n =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, what is the equilibrium mole fraction of ammonia at 1 bar and 300 K?

solution n = -1 no = 2 In order to calculate K, we need DG at 300 K ½ N2(g) + 3/2 H2(g)  NH3 (g) n = -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

for an ideal solution model, (I) (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, (I) i) Define a guess T. Start with Tguess =578 K (obtained in part a) and 100 bar. Use virial equation of state, for fi’s that are functions of temperature ii) Obtain new expression for K using the calculated fi’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