How long would it take for the concentration of the reactant to drop to one-forurth its initial value?
initial concentration of reactant=0.64mol/L
half-life=30s
0.64 to 1/4(0.64) = 0.16
0.64/0.16 = 4 so it must go through 2^n = 4 which makes n = 2 half lives.
If the half life is 30 sec, then it will take 60 seconds for it to go through 2 half lives.
To determine the time it takes for the concentration of the reactant to drop to one-fourth its initial value, we need to use the concept of half-life. By knowing the half-life and the initial concentration, we can calculate the time required.
The half-life is the time it takes for half of the reactant to decay. In this case, the half-life is given as 30 seconds.
To find the time required for the concentration to drop to one-fourth, we need to find how many half-lives it will take. Since one-fourth is half of one-half, it will take two half-lives for the concentration to drop to one-fourth.
Now, let's calculate how much time is needed for two half-lives to pass:
First Half-Life:
The initial concentration is 0.64 mol/L. After one half-life (30 seconds), the concentration would be halved to 0.32 mol/L.
Second Half-Life:
Since half of the reactant was consumed in the first half-life, the remaining half needs to be halved again. So, after the second half-life (another 30 seconds), the concentration would be halved once more to 0.16 mol/L.
So, it would take a total time of 30 seconds + 30 seconds = 60 seconds for the concentration of the reactant to drop to one-fourth (0.16 mol/L) of its initial value (0.64 mol/L).