Answer:
(a) To determine the space time needed to treat a feed of A0 = 100 mmol/liter in a plug flow reactor, we can use the following equation:
τ = V/Q
where V is the volume of the reactor and Q is the volumetric flow rate of feed.
To achieve 80% conversion, we need to find the corresponding concentration of A:
CA = CA0 (1-X)
CA = 100 mmol/liter (1-0.8) = 20 mmol/liter
From the data provided, we can see that the reaction rate follows a first-order kinetics. The rate constant can be calculated from any of the runs:
k = ln(1/0.2)/τ = ln(5)/60 sec⁻¹ = 0.0292 sec⁻¹
Now we can use the equation for a first-order reaction:
CA/CA0 = exp(-kτ)
We can rearrange this equation to solve for τ:
τ = ln(1/CAR)/k
where CAR is the desired concentration of R at 80% conversion:
CAR = CA0 - CA = 100 mmol/liter - 20 mmol/liter = 80 mmol/liter
τ = ln(1/0.8)/0.0292 sec⁻¹ = 33.5 seconds
Therefore, the space time needed in a plug flow reactor to treat a feed of A0 = 100 mmol/liter to 80% conversion is 33.5 seconds.
(b) To determine the space time needed to achieve the same conversion in a mixed flow reactor, we can use the equation for a continuous stirred-tank reactor:
X = (CA0 - CA)/CA0 = kτ/(1+kτ)
We can rearrange this equation to solve for τ:
τ = X/(k(1-X))
τ = 0.8/(0.0292 sec⁻¹(1-0.8)) = 115.8 seconds
Therefore, the space time needed in a mixed flow reactor to treat a feed of A0 = 100 mmol/liter to 80% conversion is 115.8 seconds.