Question

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?

Answer 1

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:
t_1/2 = 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

Answer 2

Thanks Boss

Answer 3

oops. typo.
That should be 2.25E-4 M