QUESTION1 2 points Save Answer The liquid-phase reaction A+B->R+S takes place in

Question

QUESTION1 2 points Save Answer The liquid-phase reaction A+B->R+S takes place in a 100 L continuously stirred tank reactor operating isothermally and at steady state. The rate law is given by -rA CA*CB and the rate constant, k, is 7 L mol min. Two feed streams, one containing 2 mol/L of A and the other containing 1.4 mol/L of B are introduced in equal volumes into the reactor. (Note that this means that you need to calculate the stating concentrations of A and B in the reactor.) Assume that the density of the materials in the system remains constant. What is the space time (in utes) necessary to convert 99% of the limiting reactant? OA 23 OB. 1.6 46 OD. 2.2 OE. 14

Explanation / Answer

The given reaction

A + B = R + S

Rate law r = - k*CA*CB

Limiting reactant = B (minimum number of moles)

Design equation for CSTR in terms of B

V/FB0 = XB / (-r)

100 L / FB0 = 0.99/(k*CA*CB)

100 L / FB0 = 0.99/[k*(CA0 - CB0*XB)*CB0(1-XB)]

100 L / FB0 = 0.99/[7 mol-1 min-1 *(2 - 1.4*0.99)mol/L*1.4mol/L(1-0.99)]

100 L / FB0 = 0.99/0.060172

FB0 = 6.077 mol/min

CB0 x v = 6.077 mol/min

Volumetric flow rate v = 6.077/1.4 = 4.341 L/min

Space time = volume / Volumetric flow rate

= 100/4.341

= 23 min

Option A is the correct answer

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