Question

An irreversible reaction gave 90.% conversion in a batch reactor at 40.°C in 10.0 minutes and only required 3.0 minutes at 50.°C. a) You don't know what order this reaction is... At what temperature can 90.% conversion be reached in 1.0 minute in the same batch reactor and starting concentration? (DO THIS LAST) b) If this reaction is 1st order, find the times needed to reach 99% conversion at 40.°C and 50.°C. c) If this reaction is 2nd order, find the times needed to reach 99% conversion at 40.°C and 50.°C if CA. = 10 mol/L.

Answer 1

Let the reaction be

A = B

Conversion = 90%

Temperature T1 = 40°C + 273 = 313 K

Time t1 = 10.0 min

Temperature T2 = 50°C + 273 = 323 K

Time t2 = 3.0 min

For Batch reactor

dNA/dt = rA * V

d(NA/V)/dt = kCA

dCA/dt = kCA

Integrate the above equation

CA = CA0 exp(-kt)

CA0(1-X) = CA0 exp(-kt)

ln (1-X) = - kt

At X = 0.90

t1 = 10 min

ln (1-0.9) = - k * 10

k1 = 0.2303 min-1

At X = 0.90

t2 = 3 min

ln (1-0.9) = - k * 3

k2 = 0.7675 min-1

From the Arrhenius equation

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

ln (0.7675/0.2303) = (Ea/8.314) (1/313 - 1/323)

Ea = 101180.13 J/mol

Part a

Temperature T =?

Time t = 1.0 min

At X = 0.90

t = 1 min

ln (1-X) = - kt

ln (1-0.9) = - k * 1

k = 2.303 min-1

From the Arrhenius equation

ln (k/k1) = (Ea/R) (1/T1 - 1/T)

ln (2.303/0.2303) = (101180.13/8.314) (1/313 - 1/T)

2.303 = 12169.85 (1/313 - 1/T)

(1/313 - 1/T) = 0.00018924

T = 332.71 K

T = 59.71 °C

At 59.71 °C, conversion will be 90% in 1 min.

Part b

First order reaction

T1 = 40 + 273 = 313 K

Rate constant k1 = 0.2303 min-1

X3 = 0.99

Time t3 =?

ln (1-X) = - k1*t3

ln (1-0.99) = - 0.2303 * t3

t3 = 20 min

T2 = 50 + 273 = 323 K

Rate constant k2 = 0.7675 min-1

X3 = 0.99

Time t4 =?

ln (1-X) = - k2*t4

ln (1-0.99) = - 0.7675 * t4

t4 = 6 min

Part c

For 2nd order reaction

1/CA = (1/CA0) + kt

CA0 = 10 mol/L

X = 0.99

CA = CA0(1-X) = 10*(1-0.99) = 0.10 mol/L

T1 = 313 K

Rate constant k1 = 0.2303 min-1

1/CA = (1/CA0) + k1*t

1/0.10 = (1/10) + 0.2303*t

t = 43 min

T2 = 323 K

Rate constant k2 = 0.7675 min-1

1/CA = (1/CA0) + k1*t

1/0.10 = (1/10) + 0.7675*t

t = 13 min