I. An irreversible aqueous reaction, A B, gave 90% conversion in a batch reactor

Question

I. An irreversible aqueous reaction, A B, gave 90% conversion in a batch reactor at 40 °C in 10 min and required 3 min for this conversion at 50 °C. (18 points) (a) What are the rate constants at 40 °C and at 50 °C (k4o and k50), assuming rA--kCA? (b) What is the activation energy Ea for this reaction? (c) At what temperature can 90% conversion be obtained at 1 min? (d) Find the rate constant at the temperature obtained from (c) (e) Find the times for 99% conversion at 40 °C and at 50 °C. (f) Find the temperature to obtain 99% conversion in a time of 1 minute. Answer: (a) k40-0230 mini, kso = 0.768 min' (b) Ea = 1.015 x 10s J/mol (c) 33265 K (d) 2.30 min1 (e) 20 min at 40 °C and 6 min at 50 °C (f) 339.07K

Explanation / Answer

(a)

Given rate equation = r = kCA 1st order reaction

-dCA/dt = kCA

integrating it we get

ln(CAo/CA)= kt

CAo = initial concentration

CA = final concentration

at 40 oC, k = ln(CAo/CA) / t

CA = (1-XA) CAo = (1-0.9)CAo = 0. 1CAo

putting this value in equation

k1 = ln(CAo / 0.1CAo) / 10 min = 0.230 min -1

similarly at 50 oC, k2 = ln(CAo / 0.1CAo) / 3 min =0.768 min-1

b)

According to arrhenius equation

k = koe-EA/RT

taking ln both side

lnk = lnko + -EA/RT

for two values k1 and k2

ln(k2/k1) = EA/R[1/T1- 1/T2]

T1 = 40 oC = 40+273 = 313 K

T2 = 50 oC = 50 + 273 = 323 K

ln(0.768 min-1/0.230 min-1) = EA/8.314J/mol.K[1/313K - 1/323K]

EA = ln(0.768 min-1/0.230 min-1) / 0.00001189715 = 101344.47 J/mol ~ 1.015 x 105 J/mol

ko can be calculated by putting values

ko = k / e-EA/RT = 0.768 min-1 / e-1.015 x 10^5 J/mol / 8.314 J/mol.K x 323K = 1.88 x 1016 min-1

c)

for 90% conersion in 1 min

k = ln(CAo/0.1CAo) / 1min = 2.303 min-1

2.303 min-1 = 1.88 x 1016 min-1 e-EA/RT

-36.64 = -1.015 x 105 J/mol/(8.314 J/mol.K x T)

T = -1.015 x 105 J/mol/(8.314 J/mol.K x -36.64) = 333.19 K ~ = 60 oC

d)

k= koe-EA/RT

putting all the above values obtained we get k = 2.303 min-1 ( already calculated in c by using rate equation)

e)

for 99% conversion

CA= 0.01 CAo

T = 40 oC

t = ln(CAo/CA) / k = ln100 / 0.230 min-1 = 20 min

at 50 oC

t = ln100 / 0.768 min-1 = 5.99 min ~ = 6 min

f)

k = ln(CAo/0.01CAo) / 1min = 4.606 min-1

4.606 min-1 = 1.88 x 1016 min-1 e-EA/RT

-35.945 = -1.015 x 105 J/mol/(8.314 J/mol.K x T)

T = -1.015 x 105 J/mol/(8.314 J/mol.K x -35.945) = 339.63 K ~ = 339.07 K

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