Consider the irreversible, liquid phase isomerization reaction carried out in a
ID: 536152 • Letter: C
Question
Consider the irreversible, liquid phase isomerization reaction carried out in a solvent containing dissolved catalyst at 25°c in a batch reactor A -->^k_a B
The apparent first-order reaction rate constant, k_a, decreases with time because of catalyst deterioration. A chemist colleague of yours has studied the catalyst deactivation process and has proposed that it can be modeled by k_a = k/ 1+ k_d t
in which k is the fresh catalyst rate constant and k_d is the deactivation rate constant.
(a) Write down the mole balance for this reactor
(b) solve the mole balance for c_a(t). Sketch your solution.
(c) If it takes two hours to reach 50% conversion and the fresh catalyst has a rate constant of 0.6 hr^-1, what is k_d?
(d) How long does it take to reach 75% conversion?
Explanation / Answer
Answer:
a) Mole balance
-rA = k_a CA
-rA = (k/1+ k_d*t) CA
ln Co/CA = K_a* t
b) mole balance CA = Co exp(-K_a* t) = Co exp[(-k/1+k_d*t)* t]
c) Assuming 1 mole of A present
for 50% conv Co = 1 mol
CA = 0.5 mol
ln Co/CA = [(k/1+k_d*t)* t]
ln (1/0.5) = [ (0.6/1+ k_d * 2)* 2] = 1.2/ 1+ 2 K_d
0.693 = 1.2/ 1+ 2 K_d
k_d = 0.366 h-1
d) for 75% conversion
Co = 1 mol
CA = 0.25 mol
ln Co/CA = [(k/1+k_d*t)* t]
ln (1/0.25) = [ (0.6/1+ 0.366 * t)* t]
t = 4 hr
= rxn takes 4 hrs to reach 75%