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  yiˆi 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)   yii    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