What volume of PFR (report in m3) is required for steady state removal of 95% of a substance that decays according to first order kinetics with a rate constant of 0.07 d-1

The volume of the PFR is 4.28m3

Explanation: A Plug Flow Reactor (PFR) is a type of reactor that has a steady state flow. Therefore the flow rate is a constant, which is always measured in liters.

In a PFR, the rate of conversion is always dependent on the rate of flow.

For any PFR to have a conversion rate of 90% and above, that means the flow rate should be 100l/min or 100l/day depending on the rate constant.

Therefore since our conversation is 95% and our rate constant is 0.07per day, that means our flow rate is 100liters/day

The equation for a first order PFR is

dFa/dV = Ra(1)

To calculate volume, the equation becomes

(R/K) ln(Cao/Ca) = V(2)

R = flow rate = 100l/day = 0.1m3/day

K= rate constant = 0.07/day

Cao= initial conversion rate

Ca = conversion rate = 95%

V= volume

STEP 1: FIND Cao,

Cao + Ca = 100%

Cao = 100% - 95%= 5% = 0.05

Therefore

0.05 = Ca/ Cao

Ca = 0.05Cao(3)

STEP3 : Find the volume.

Using equation 2 and substituting equation 3 in equation 2.

(0.1m3/day ÷ 0.07/day)ln(Cao/0.05Cao) =

1.43m3 × ln20 = 4.98m3

Therefore the volume of the PFR is

4.98m3

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.