The liquid-phase reaction A + B → C follows an elementary rate law and is carried out isothermally in a flow system. The concentrations of A and B feed streams are 2 M before mixing. The volumetric flow rate of each stream is 5 dm3 /min and the entering temperature is 300 K. The streams are mixed immediately before entering. Two reactors are available: One is a gray 200.0 dm3 CSTR that can be heated to 77°C or cooled to 0°C, and the other is a white 800.0 dm3 PFR operated at 300 K that cannot be heated or cooled but can be painted red or black. (Note: k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol.) How long would it take to achieve 90% conversion in a 200 dm3 batch reactor with CA ° = CB ° = 1 after mixing at a temperature of 70°C?

1.887 minutes

Explanation:

We are given k = 0.07 dm3 /mol*min at 300 K and E = 20 kcal/mol = 20000 cal/mol

To solve this, first of all let's calculate the rate constant(k);

For this question, The formula is;

K(t) = k(300K) × exp[(E/R)((1/300) - (1/T2))]

R is gas constant = 1.987 cal/mol.K

For temperature of 70°C which is = 70 + 273K = 343K, we have;

K(343) = 0.07 × exp[(20000/1.987)((1/300) - (1/343))]

K(343) = 4.7 dm³/mol.min

The design equation is;

dX/dt = -(rA/C_Ao) = K•(C_Ao)²•(1 - X)²/(C_Ao) = (KC_Ao)(1 - X)²

Since there is no change in volume by cause of the state at which the reaction is carried out, that is liquid. Thus, integrating and solving for time for a 90% conversion we obtain;

(0.9,0)∫dX/(1 - X)².dX = (KC_Ao)((t, 0)∫dt

So, we'll get;

0.9/(1 - 0.9) = 4.77 × 1 × t

t = 9/4.77

t = 1.887 minutes

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