The rate constant for a certain reaction is kkk = 3.50×10−3 s−1s−1 . If the initial reactant concentration was 0.450 MM, what will the concentration be after 19.0 minutes?

A zero-order reaction has a constant rate of 3.50×10−4 M/sM/s. If after 65.0 seconds the concentration has dropped to 3.50×10−2 MM, what was the initial concentration?

The rate law is defined as the amount of reactant present in the initial concentration and used up during the course of the reaction. The rate depends on the following:-

According to the question, the expression of the rate is . As the data are given in the question, the rate constant is , the initial concentration and time is 19 minutes that is the 1140s.

We have to find the final concentration,  Integrated rate law for a first-order reaction is given as;

For the second problem, the data given is as follows:-

We have to find the Initial Concentration, after solving the equation.

Hence, the final concentration is 0.5775m.

For more information, refer to the link:-

link

Explanation:

Question 1

Rate constant = 3.50×10−3 s−1

Initial Concentration [A]o =  0.450 M

Final concentration [A] = ?

time = 19 minutes = 1140 s (upon conversion to seconds)

Integrated rate law for a first order reaction is given as;

ln[A] = ln[A]o - kt

ln[A] = ln (0.450) - (3.50×10−3)1140

ln[A] = -0.799 - 3.99

ln[A] = -4.789

[A] = 0.0083M

Question 2

Rate constant, k = 3.50×10−4 M/s

time = 65 s

Final Concentration [A] = 3.50×10−2 M

Initial Concentration [A]o = ?

Integrated rate law for a first order reaction is given as;

[A] = [A]o - kt

[A]o = [A] + kt

[A]o = 3.50×10−2 + (3.50×10−4) * 65

[A]o = 3.50×10−2 + 0.02275

[A]o = 0.05775 M

it is a solid

explanation: