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}](https://tex.z-dn.net/?f=r%20%3D%20k%20%5BA%5D%5E%7Bx%7D%20%5BB%5D%5E%7By%7D)
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
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}](https://tex.z-dn.net/?f=v%28t%29%20%3D%20-%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%20%3D%20k%20%5BA%5D%5E%7Bn%7D)
For the kinetics of zero-order, the rate is apparently independent of the reactant concentration.
<em>Rate Law: rate = k</em>
<em>Concentration-time Equation: [A]=[A]o - kt</em>
where
- [A]: concentration in the time <em>t</em> [M]
- [A]o: initial concentration [M]
- t: elapsed reaction time [s]
For first-order kinetics, we have:
<em>Rate Law: rate= k[A]</em>
<em>Concentration -Time Equation: ln[A]=ln[A]o - kt</em>
where:
- K: rate constant [1/s]
- ln[A]: natural logarithm of the concentration in the time <em>t </em>[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]
