An irreversible aqueous reaction gave 80% conversion in a batch reactor at 35 o

Question

An irreversible aqueous reaction gave 80% conversion in a batch reactor at 35oC in 12 min and required 4 min for this conversion at 50oC. (a) What is the activation energy for this reaction? (b) At what temperature can 80% conversion be obtained in 1 minute? (c) Find the rate coefficient assuming first order kinetics. (d) Assuming first order kinetics, find the times for 99% conversion at 35 and at 50oC. (e) Assuming first order kinetics, find the temperature to obtain 99% conversion in 1 minute. (f) Assuming second order kinetics, find the times for 99% conversion at 35 and 50oC. (g) Assuming second order kinetics, find the temperature to obtain 99% conversion in 1 minute. (h) What is the space-time required in a plug flow reactor for 80% conversion at 50oC?

Explanation / Answer

(a) rate constant(k) at 350C = ln{[A0]/[A]}/t = ln{100/20}/12 = 0.134 min-1

rate constant at 50 0C = ln(100/20)/4 = 0.402 min-1

Now, using Arrhenius Equation we get,

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

or, ln(0.402/0.134) = (E/8.314)*[(1/308) - (1/323)]

or, E = 60578 J = 60.578 kJ

b) rate constant,k = ln(100/20)/1 = 1.609 min-1

Now,let the temperature be T

Thus, ln(k2/k1) = (E/R)*[(1/T1)-(1/T)]

or, ln(1.609/0.402) = (60578/8.314)*[(1/323) - (1/T)]

or, 344.16 K = T = 71.16 0C

d) time at 350C for 99% conversion = ln(100/1)/0.134 = 34.37 min

time at 50 0C for 99% conversion = ln(100/1)/0.402 = 11.455 min

time for 99% conversion at 71.16 0C = ln(100/1)/1.609 = 2.862 min

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