In 30 minutes, a first-order reaction is 50% complete. The time required to complete 87.5 percent of the reaction is ____________________*30 minutes60 minutes90 minutes120 minutes
Question
In 30 minutes, a first-order reaction is 50% complete. The time required to complete 87.5 percent of the reaction is ____________________*30 minutes60 minutes90 minutes120 minutes
Solution
The answer is 90 minutes.
Here's the step-by-step explanation:
A first-order reaction follows the equation:
ln([A]0/[A]) = kt
where [A]0 is the initial concentration, [A] is the concentration at time t, k is the rate constant, and t is the time.
When the reaction is 50% complete, half of the reactant has been used up, so [A]0/[A] = 2. Plugging this into the equation gives:
ln(2) = k*30 minutes
Solving for k gives k = ln(2)/30 minutes.
Now, when the reaction is 87.5% complete, 87.5% of the reactant has been used up, so [A]0/[A] = 1/(1-0.875) = 8. Plugging this into the equation gives:
ln(8) = k*t
Substituting the value of k from the first equation gives:
ln(8) = (ln(2)/30 minutes)*t
Solving for t gives t = 30 minutes * ln(8)/ln(2) = 90 minutes.
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