1. Chemical Reaction Equilibria
Part III

2. 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

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

4. 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

5. 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

6. 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

7. 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 = J/mol
DHo = J/mol

8. Get A, B, C, D for each component and calculate DA, DB, DC, and DD.
Calculate DG at 925K using equation ; DG = 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 = =0.617

9. 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?

10. 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 = J/mol
DHo = J/mol
Get A, B, C, D for each component and calculate DA, DB, DC, and DD.
Calculate DG at 300K using equation ; DG = J/mol
Calculate K = exp (- DG/RT) =679.57

11. K =(y3)/(y2)3/2(y1)1/2
you can show (see problem 13.9) that
eeq =1-( KP/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-( KP/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

12. (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-( KP/Po)-1/2 = 2/3
K =  at what T, K has this value?
K = exp (- DG (T)/RT) =
 solve for T=577.6 K

13. 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-( K/1.184xP/Po)-1/2 = 2/3
iii) solve for K =0.0729; iv) since K (Tguess) = must be=exp (- DG (T)/RT), evaluate exp (- DG (Tguess)/RTguess). Is it equal to ? If not, change Tguess and go to (i)
Solution after convergence T = K