the rate constant for the first order reaction decomposition of N2O5 at temperature t°C is 2.2×10^-1. if the initial concentration of N2O5 is 3.6 ×10^-3 mol/dm^3. what will be the concentration after four halve-lives?
You don't have any units for rate so I'm using seconds.
You can do this three ways.
First way:
(1/2)^4 = 0.0625 = fraction after 4 half lives
0.0625 x 3.6E-3 = 2.25E-4 amount remaining after 4 half lives
2nd way:
t1/2 = 0.693/k = 0.693/0.22 = 3.15 s
4 half lives = 3.15*4 = 12.60 s
ln(No/N) = kt
ln(0.0036/N) = 0.22*12.60
ln(0.0036/N) = 2.77
N = 2.25E-4
3rd way:
begin amount no half lives remaining
------------------------------------------------------------
0.0036............................0......................0.0036 M
........................................1.....................0.0018
.........................................2.....................0.0009
..........................................3.....................0.00045
...........................................4....................0.000225 = 2.25E4 M
oops. typo.
That should be 2.25E-4 M
Thanks Boss
To find the concentration of N2O5 after four half-lives, we need to know the rate constant for the first-order reaction and the initial concentration of N2O5.
The half-life of a first-order reaction can be determined using the equation:
t1/2 = (0.693 / k)
Where:
t1/2 is the half-life
k is the rate constant
Given that the rate constant (k) for the first-order reaction of N2O5 at temperature t°C is 2.2×10^-1, we can calculate the half-life (t1/2):
t1/2 = (0.693 / 2.2×10^-1)
t1/2 ≈ 3.15
Since we know the half-life, we can now calculate the concentration of N2O5 after four half-lives. Four half-lives is equal to:
(4 * t1/2) = (4 * 3.15) = 12.6
To calculate the concentration after four half-lives, we can use the formula for the concentration of a first-order reaction:
[A] = [A0] * (2^(-n))
Where:
[A] is the final concentration
[A0] is the initial concentration
n is the number of half-lives
Given that the initial concentration of N2O5 is 3.6×10^-3 mol/dm^3 and the number of half-lives is 12.6, we can calculate the concentration after four half-lives as follows:
[A] = (3.6×10^-3) * (2^(-12.6))
[A] ≈ 3.6×10^-3 * 6.5×10^-4
[A] ≈ 2.34×10^-6 mol/dm^3
Therefore, the concentration of N2O5 after four half-lives is approximately 2.34×10^-6 mol/dm^3.