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Princessirisperez0
14.09.2019 •
Chemistry
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.
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Ответ:
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:
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 constantThe 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:
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]
Ответ: