The half-life of a first order reaction is determined to be 65.0 years. How long will it take for the concentration of the reactant to reach 2% of its initial value?
366.85
65.0 times 5.64386 = answer
Nice Short cut...!!
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This is a question from the second midterm of 3.091x by edx. Stop cheating, it's silly and pointless.
To calculate the time it takes for the concentration of the reactant to reach a certain value, we can use the formula for the half-life of a first-order reaction:
t = (0.693 / k)
Where:
t = time it takes for the concentration to reach a certain value
k = rate constant of the reaction
In this case, we are given the half-life, which is the time it takes for the concentration to decrease by half. So, we can rewrite the formula as:
t1/2 = (0.693 / k)
We can rearrange the formula to solve for the rate constant (k):
k = 0.693 / t1/2
Now, we can substitute the value of the half-life (t1/2 = 65.0 years) into the formula to find the rate constant:
k = 0.693 / 65.0
k ≈ 0.01066154 per year
Now that we have the rate constant, we can use it to calculate the time it takes for the concentration to reach 2% of its initial value. To do that, we can modify the formula for the half-life as follows:
t = (ln(C0 / Cf)) / k
Where:
t = time it takes for the concentration to reach a certain value
C0 = initial concentration of the reactant
Cf = final concentration of the reactant (in this case, 2% of C0)
Since we want to find the time it takes for the concentration to reach 2%, the final concentration (Cf) will be 0.02 * C0. Substituting these values into the formula:
t = (ln(C0 / (0.02 * C0))) / k
t = (ln(1 / 0.02)) / k
Simplifying further:
t = ln(50) / k
Now we can substitute the rate constant (k) we calculated earlier to find the time it takes for the concentration to reach 2% of its initial value:
t = ln(50) / 0.01066154
t ≈ 152.6 years
Therefore, it will take approximately 152.6 years for the concentration of the reactant to reach 2% of its initial value.