Respuesta :
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
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:-
- Amount
- Pressure
- Volume
According to the question, the expression of the rate is[tex]Rate = k[A][B][/tex] . As the data are given in the question, the rate constant is [tex]3.50×10^{-3}1/s[/tex] , the initial concentration[tex][A]_o = 0.450 M[/tex] 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;
[tex]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[/tex]
For the second problem, the data given is as follows:-
- Rate constant ,[tex]k = 3.50*10^-^4 M/s[/tex]
- Time = 65 s
- Final Concentration [A] = [tex]3.50*10^-^2 M[/tex]
We have to find the Initial Concentration, after solving the equation.
[tex][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[/tex]
Hence, the final concentration is 0.5775m.
For more information, refer to the link:-
https://brainly.com/question/14945022
