The irreversible, elementary, liquid-phase reaction D + E --> F is run in a batc

Question

The irreversible, elementary, liquid-phase reaction D + E --> F is run in a batch reactor. At the temperature in the reactor, the forward rate constant is 4.82 L/(mol*hr). The batch reactor initially contains 40 L of liquid containing 15 moles of D and 6 moles of E.

a. Determine the limiting reactant and show the complete stoichiometric table.

b. If the reaction proceeds for 30 minutes, calculate the conversion of the limiting reactant and the concentration of each species.

Now assume the reacton is elementary and reversible, with Kc = 2.5 L/(mol*hr):

c. Calculate the equilibrium conversion.

d. Repeat part (b) for the reversible reaction.

Explanation / Answer

As per the reaction, D+E---->F, one mole of D requires 1 mole of E to form one mole of F. Stoichimetric ratio of D to E =1:1

given ratio of reactants = 15:6 =2.5 :1

So excess is D and limiting is E.

CDO= Concentration of D at t=0 is = 15/40 =0.375 moles/L and CEO= 6/40 =0.15 moles/L, concentration of reactant E at t=0. M= CDO/CEO = 15/6= 2.5

the rate of reaction -r = K[D] [E]

-dCE/dt = K(CEO-CEOXE)*(CDO-CEOXE)

-dCE/dt = KCEO2*(1-XE)*(M-XE)

when integrated the expression is

ln[ (M-XE)/M*(1-XE)] = CEO*(M-1)*Kt

XE= conversion of E, the limiting reactant

at 30 minutes =30/60 hrs =0.5 hrs

ln [(2.5-XE)/ (2.5*(1-XE) = 0.15*(2.5-1)*4.82*0.5=0.54

when this equatino is solved for XE, the conversion is 55%.

c)

At equilibrium let the conversin be XE

hence K1CEO*(1-XE)*(CDO-CEOXE) = K2 (CFO+CEOXE)

since initially CFO=0

K1 and K2 are rate constant for forward and backward reactions rate constants

K1/K2= KC= Equilibrium constant =2.5

Kc= 2.5 = XE/ [(1-XE)*(M-XE)*CEO]

M= 2.5 and CEO = 0.15

2.5 = XE/[(1-XE)*(2.5-XE)*0.15

XE/ [(1-XE)*(2.5-XE) = 2.5*0.15 =0.375

when solved using excel , XE = 0.436, XE= Equilibrium conversion

d)

dCE/dt = K1CEO2*(M-X)*(1-X) – K2CEOX

where X= conversion at any time, t.

but K1/K2= CEOXE/ [(CEO2*(M-XE)*(1-XE)]

= XE/ (CEO*(M-XE)*(1-XE) = 2.5

K1= 2.5K2. K2= K1/2.5 =4.82/2.5=1.93

Given

dCE/dt = 4.82*CEO2*(M-X)*(1-X) – 1.93CEOX

   = CEO*[CEO*4.82*(M-x)*(1-X) – 1.93X]

dX/dt= 0.15*4.82*(M-X)*(1-X)- 1.93X =0.723*(2.5-X)*(1-X )-1.93X

=0.723*( 2.5-3.5X+X2) -1.93X

dX/dt = 1.81-2.53X+0.723X2-1.93X =0.723X2 – 4.46X+1.81

dX/ (0.723X2- 4.46X+1.81)= dt

dX/(5.95-X)*(0.21-X)= dt

(1/5.74)*[ 1/(0.21-x) – 1/(5.95-x)] dX= dt

0.174 *[ 1/(0.21-x) – 1/(5.95-x)] dX= dt

0.174 [-ln(0.21-x) + ln (5.95-x)] =t + K, where   K is integration constant

0.174* ln [(5.95-x)/(0.21-x)] = t+K

At t=0 X=0

Hence 0.174* ln (5.95/0.21)= K

0.174*3.34= K

K= 0.58

Hence 0.174* ln [(5.95-x)/(0.21-x)] = t+0.58

At t=30 min

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