An aqueous, irreversible reaction gave 90.% conversion in a batch reactory at 40

ID: 476710 • Letter: A

Question

An aqueous, irreversible reaction gave 90.% conversion in a batch reactory at 40.C in 10. minutes and required only 3.0 minutes at 50.C.

1. What is the activation energy for this reaction?

2. At what temperature can 90.% conversion be obtained in one minute?

3. Find the rate constant assuming first-order kinetics.

4. Assuming first order kinetics, find the times for 99% conversion at 40. and 50.C.

5. Assuming second order kinetics and CA0 = 1.00 mole/liter, find the times for 99% conversion at 40. and 50.C.

1.Higher the temperature lower the time for conversion and hence rate constant= 1/time

Hence from Arhenius equation ln (K2/K1)= (Ea/R)*(1/T1-1/T2)

K2= 1/3=0.333 and K1=1/10= 0.1 and T1= 40+273=313K, T2= 50+273=323K

R= 8.314 Joules/mole.K, ln(0.333/0.1)= (Ea/8.314)*(1/313-1/323)

Ea= 101114 joules/mole Activation energy

2. let the temperature be T2. K2= 1/1 =1 /min, K1=0.1/min , T1= 313K

Hence ln(1/0.1) =(101114/8.314)*(1/313-1/T2), T2= 333K =333-273= 60 deg.c

3.For 1st order reaction –ln(1-XA)= Kt

At 40 deg,c, -ln(1-0.9)= K*10, K= 0.2302/min

At 50 deg,c , K= -ln(1-0.9)/3 = 2.302/3 = 0.77/min

At 60 deg,c K= -ln(1-0.9)/0.1 = 23.02/min

4. At 40 deg.c, for 99% conversion -ln(1-0.99)= 0.2302*t, t = 4.605/0.2302 = 20 min

At 50 deg.c , time = 4.605/0.77 = 6 min

5. for second order reaction, KCAOt= XA/(1-XA), XA/(1-XA)= 0.99/(1-0.99)= 99

Hence time, t= 99/(KCAO)

At 40 deg.c, t= 99/(0.2302*1)=430 min and at 50 deg., t= 99/(0.77*1)= 128 min

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