Question

Question 3. A batch chemical reactor achieves a reduction in

concentration of compound A from 100 mg/L to 5 mg/L in one hour. If
the reaction is known to follow zero-order kinetics, determine the
value of the rate constant with appropriate units. Repeat the
analysis if the reaction is known to follow first-order
kinetics.

Answer 1

Rate constant for zero-order kinetics: 1, 58 [mg/L.s]

Rate constant for first-order kinetics: 0,05 [1/s]

Step-by-step explanation:

The reaction order is the relationship between the concentration of species and the rate of the reaction. The rate law is as follows:

`r = k [A]^(x) [B]^(y)`

where:

• [A] is the concentration of species A,
• x is the order with respect to species A.
• [B] is the concentration of species B,
• y is the order with respect to species B

• k is the rate constant

The concentration time equation gives the concentration of reactants and products as a function of time. To obtain this equation we have to integrate de velocity law:

`v(t) = -(d[A])/(dt) = k [A]^(n)`

For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.

Rate Law: rate = k

Concentration-time Equation: [A]=[A]o - kt

where

• k: rate constant [M/s]

• [A]: concentration in the time t [M]
• [A]o: initial concentration [M]
• t: elapsed reaction time [s]

For first-order kinetics, we have:

Rate Law: rate= k[A]

Concentration -Time Equation: ln[A]=ln[A]o - kt

where:

• K: rate constant [1/s]
• ln[A]: natural logarithm of the concentration in the time t [M]
• ln[A]o: natural logarithm of the initial concentration [M]
• t: elapsed reaction time [s]

To solve the problem, wee have the following data:

[A]o = 100 mg/L

[A] = 5 mg/L

t = 1 hour = 60 s

As we don't know the molar mass of the compound A, we can't convert the used concentration unit (mg/L) to molar concentration (M). So we'll solve the problem using mg/L as the concentration unit.

Zero-order kinetics

we use: [A]=[A]o - Kt

we replace the data: 5 = 100 - K (60)

we clear K: K = [100 - 5 ] (mg/L) /60 (s) = 1, 583 [mg/L.s]

First-order kinetics

we use: ln[A]=ln[A]o - Kt

we replace the data: ln(5) = ln(100) - K (60)

we clear K: K = [ln(100) - ln(5)] /60 (s) = 0,05 [1/s]