Consider the scheme of elementary reactions : Feed consists of A and inerts, C A

ID: 1065924 • Letter: C

Question

Consider the scheme of elementary reactions :

Feed consists of A and inerts, CA0 =1 mole/liter, and the operable temperature range is between 7 and 770 C.

a) What is the maximum amount of S obtainable per mole of A, and at what temperature and in what type of reactor is this obtained?

b) Find the minimum holding time to produce 99% of CS, max.

c) Repeat part (b) if k1 = 107 e -3500/T , all else remaining unchanged.

k k2 > R--------> S desired k1= 10 e-3500T, sec-1 k2-1012 e-10500T, sec-1 Kg Kg= ks= 108 e-7000T sec-1 T C-1 cec ese eee 000 Sk A

As notes reaction contant values are in 1/s, they are first reactions

rate = rate constant x [reactant]a

mole/ Ls = 1/s x [mole / l]a

a =1

Write down reaction rates for change in concentration for all species at time t

dCA/ dt = - k1[A] (1)

dCR/ dt = - k2 [R] - k3 [R] + k1 [A] (2)

dCS/ dt = - k2 [R] (3)

On dividing (3)/ (1)

Final S concentration - Initial S concentration/ (Final A concentration - Initial A concentration) = - k2 [R]/ k1 [A]

Assuming final conentration A =1

S - S0/A - A0 = - k2 [R]/ k1 [A]

Initial S0 concentration = 0 (no product is formed)

Intial A0 = 1

Final A = 0 (whole A is consumed)

Assume [A] = [R] for maximum S concentration at this both A & R are consumed only S & T are present

1012 e-10500/T/ 10 e-3500/T = Smax/ 1

Smax = 1011 * (e-10500/T + 3500/ T)

We will get maximum value of right hand side when T = 770 + 273 = 1043 K

Smax = 1.22 x 108 mole/ L

When we put T = 7 + 273 = 280 K

S = 1.38 mol/ L

Hence Smax =1.22 x 108 mole/ L

b) 99 % of Smax = 1.2 x 108 mole / L

del S/dt = k2 [S]

t = 23169 s