The half-life of a first order reaction is determined to be 61.0 years. How long will it take for the concentration of the reactant to reach 2% of its initial value?
.02 = (1/2)^(t/61)
t = 344.275 years
makes sense, since 2% is around 1/64 or 6 half-lives
CHEAT!
Do NOT help this person. They are trying to cheat in a midterm exam.
To find the time it takes for the concentration of the reactant to reach 2% of its initial value, we can use the concept of half-life.
In a first-order reaction, the half-life is the time it takes for the concentration of the reactant to decrease by half. Given that the half-life of the reaction is 61.0 years, we can set up the following equation:
t = half-life * (log(initial concentration) - log(final concentration)) / log(2)
where:
t = time it takes for the concentration to reach 2% of its initial value
half-life = 61.0 years
initial concentration = 100% (since we are finding the time it takes for the concentration to reach 2% of its initial value)
final concentration = 2% = 0.02 (since we want to find the time it takes for the concentration to reach 2% of its initial value as given in the question)
Plugging in the values into the equation, we can calculate the time it takes:
t = 61.0 years * (log(100) - log(0.02)) / log(2)
t = 61.0 years * (2 - (-1.69897)) / 0.30103
t = 61.0 years * 3.69897 / 0.30103
t ≈ 752.7 years
Therefore, it will take approximately 752.7 years for the concentration of the reactant to reach 2% of its initial value.