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Step-by-Step Solution
Step 1: Identify the half-life and rate constant
The given half-life of the reaction is 1 minute. For a first-order reaction, the relationship between the half-life t_{1/2} and the rate constant k is:
k = \frac{\ln 2}{t_{1/2}}
Substituting t_{1/2} = 1 min, we get:
k = \frac{\ln 2}{1} = \ln 2 \approx 0.69 \text{ min}^{-1}
Step 2: Determine the fraction of reactant remaining
The reaction is 99.9% complete, which means only 0.1% of the reactant is left unreacted. In decimal form, 0.1% is 0.001. Thus the fraction remaining is 0.001 .
Step 3: Use the first-order integrated rate law
For a first-order reaction, the amount of reactant remaining after time t is given by:
\ln \left( \frac{[A]_0}{[A]} \right) = kt
Here, [A] is the concentration remaining at time t , and [A]_0 is the initial concentration. In our case:
\frac{[A]}{[A]_0} = 0.001 \quad \Rightarrow \quad \ln \left( \frac{1}{0.001} \right) = kt
Thus,
kt = \ln(1000)
Step 4: Solve for the time t
We know \ln 10 = 2.3 , so \ln(1000) = \ln(10^3) = 3 \ln(10) = 3 \times 2.3 = 6.9. Hence:
k t = 6.9
Substitute k = 0.69 \text{ min}^{-1} :
0.69 \times t = 6.9 \quad \Rightarrow \quad t = \frac{6.9}{0.69} = 10 \text{ minutes.}
Final Answer
The time required for 99.9% completion of the reaction is 10 minutes.
