Question

Consider an elementary first order reaction found to have an activation energy (EA) of 250 kJ/mole and a pre-exponential (A) of 1.7 x 1014 s–1.

(a) Determine the rate constant at T = 750°C.

(b) At what temperature will the reaction be three times as fast as at 750°C?

(c) What fraction of the reaction will be completed in a batch reactor at 300°C over a period of 10 minutes?

Answer 1

activation energy (EA) = 250 kJ/mole x 1000J/kJ

= 250*1000 J/mol

pre-exponential factor (A) = 1.7 x 10^14 s-1

Part a

T = 750 + 273 = 1023 K

From Arrhenius equation

Rate constant k = A x exp(-Ea/RT)

= 1.7 x 10^14 x exp(-250*1000/8.314*1023K)

= 1.7 x 10^14 x 1.72 x 10^-13

= 29.16 s-1

Part b

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

ln 3 = (250*1000/8.314) (1/1023 - 1/T2)

1.0986 = 30069.76 (1/1023 - 1/T2)

(1/1023 - 1/T2) = 3.65 x 10^-5

1/T2 = (1/1023) - 3.65 x 10^-5

T2 = 1062.72 K

Part c

Time t = 10 min x 60s/min = 600 s

T = 300 + 273 = 573 K

Rate constant k = A x exp(-Ea/RT)

= 1.7 x 10^14 x exp(-250*1000/8.314*573K)

= 1.7 x 10^14 x 1.62 x 10^-23

= 2.75 x 10^-9 s-1

For the first order reaction

CA = CA0 exp(-kt)

CA = CA0 exp(-2.75*10^-9*600)

CA = CA0 x 0.999

CA/CA0 = 0.999