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

Question

An aqueous, irreversible reaction gave 90.% conversion in a batch reactory at 40. degree C in 10. minutes and required only 3.0 minutes at 50. degree C. What is the activation energy for this reaction? At what temperature can 90.% conversion be obtained in one minute? Find the rate constant assuming first-order kinetics. Assuming first order kinetics, find the times for 99% conversion at 40. and 50. degree C. Assuming second order kinetics and C_A0 = 1.00 mole/liter, find the times for 99% conversion at 40. and 50. degree C.

Explanation / Answer

(1)

First we determine the rate constant as follows:

rate constant(k) at 400C = ln{[A0]/[A]}/t

= ln{100/10}/10.0 min

= 0.230 min-1

rate constant at 50 0C = ln(100/10)/3.0 min

= 0.768 min-1

T1 = 400C = 313 K

T2 = 500C = 323 K

Now, using Arrhenius Equation we get,

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

ln(0.768/0.230) = (Ea/8.314)*[(1/313) - (1/323)]

ln(3.34) = (Ea/8.314)*[(3.19*10^-3) - (3.10*10^-3)]

1.21 = (Ea/8.314)*[9.0*10^-5]

Ea= 1.21*8.314) / [9.0*10^-5]

or, E a= 111777 J = 111.8 kJ

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.