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.

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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

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally