A well-mixed stirred tank reactor initially contains 200 L of pure solvent. The

Question

A well-mixed stirred tank reactor initially contains 200 L of pure solvent. The feed to the reactor is a stream with a flow rate of 50. L/min with a concentration of species A equal to 5.0 moles/L in solvent. This species undergoes a reaction in the tank of the form AB. The rate of consumption of A is given by kCA [mol/(L·min)], with k=4.0(min)1 . The exit stream from the reactor is also 50.0 L/min. Assume all solutions have the same density.

a. What is the steady state concentration of A in the exit stream?

b. If at some time, the inlet concentration of A is changed to 4. moles/L determine the equation for the resultant exit concentration of A as a function of time

Explanation / Answer

We have,

200 L of solvent

Initial concentration of [A] in the feed = 5.0 moles/L

rate constant k = 4 min-1

the unit for rate constant tells it is a first order reaction.

a. We have for a first order reaction,

ln[A] = ln[A]o-kt

feed values,

d[A]/dt = [A]o(1-e^-kt)

50 = 5(1-e^-(4xt))

10 = 1-e^-4t

log(9) = 4t

t = 0.24 min

Feed this value into the first order reaction,

ln[A] = 5-0.24 x 4

         = 4.04

[A] = 56.83

So steady state concentration of [A] = 56.83%

b. If the inlet concentration of A changes to 4 moles/L

we have,

50 = 4(1-e^-(4xt))

12.5 = 1-e^-4t

log(11.5) = 4t

t = 0.27 min

ln[A] = 4(1-e^-4t)

[A] = 51.38% remaining

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