(07.06 MC) During an experiment, ice and water are placed in a perfectly insulated thermos flask at 0° C. Describe this system when it phase reaches equilibrium. (10 points)

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 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:

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]