Question

Question 3. A batch chemical reactor achieves a reduction in

Answer 1

Answer:

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

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

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:

$v(t) = -\frac{d[A]}{dt} = k [A]^{n}$

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]