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The reaction 3A  B + C occurs in water, with a known pre-exponential rate factor...

The reaction 3A  B + C occurs in water, with a known pre-exponential rate factor of 0.12 L/mol sec and an activation energy of 4.3 kcal/mol. The inlet feed to a reactor is 2.5M A, flowing at 15 L/min. The outlet stream contains 0.65M B. If the reactor is run at 55°C, what is the residence time for a: Pftr and Cstr

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Expert Solution

Reaction :

Feed to the reactor is 2.5 M of A at flow rate of 15 L/min

Outlet stream has 0.65 M of B.

Pre-exponential factor,k₀ = 0.12 L/(mol sec)

Activation energy, E_a = 4.3 kcal /mol

converting kcal to joule; 1 kcal = 4184 joule

Activation energy, E_a = 4.3 kcal /mol * 4184 joule/kcal = 17991.2 joule/mol

R = 8.314 J/mol K

T = 55 C = 328 K

from the units of pre-exponential factor, it is evideng that it is a second order reaction. Thus,

-r_A = KC_A² = k₀e^-E_a/RT * C_A²

a) PFR

-r_A = KC_A² = k₀e^-E_a/RT * C_A² =0.12*e^{-17991.2/(8.314* 328)} * C_A² = 1.63*10^-4*C_A²

C_A0 = 2.5 M

Given that 0.65 M of B has formed. From reaction for each mole B formed, 3 moles of A is consumed. Thus for 0.65 moles B, 0.65*3 = 1.95 moles of A will consumed.

C_A = C_A0 - 1.95 = 2.5 - 1.95 = 0.55 M

On integrating,

1.63*10^-4* = [1/C_A]_2.5^0.55

this gives,

= 8700 min

b) For MFR

From previous calculation, K = 1.63*10^-4

C_A0 = 2.5 M

C_A = 0.55 M

1.63*10^-4 = (2.5 - 0.55)/(0.55²)

on solving,

= 39547.736 min

Amanda K answered 1 year ago

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