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Step-by-Step Solution
Step 1: Write down the known information
• The reaction is first order.
• The half-life period, $t_{1/2}$, is given as 6.93 minutes.
• We want the time required for 99% completion of the reaction.
Step 2: Express the rate constant $k$ in terms of half-life for a first-order reaction
For a first-order reaction, the rate constant $k$ is related to the half-life period $t_{1/2}$ by:
$ k = \dfrac{0.693}{t_{1/2}} $
Substituting the given half-life, $t_{1/2} = 6.93\text{ min}$:
$ k = \dfrac{0.693}{6.93} = 0.1 \,\text{min}^{-1}
$
Step 3: Use the integrated rate law for first-order reactions to find the time for 99% completion
The integrated rate law for a first-order reaction in terms of concentrations is:
$ k = \dfrac{2.303}{t}\,\log \dfrac{[A]_0}{[A]} $
For 99% completion, the fraction of reactant remaining is $[A] = 1\%$ of the original concentration $[A]_0$. Thus:
$ \dfrac{[A]_0}{[A]} = \dfrac{[A]_0}{0.01[A]_0} = 100
$
Therefore,
$ k = \dfrac{2.303}{t}\,\log(100) = \dfrac{2.303}{t} \times 2
$
since $\log(100) = 2.$
Step 4: Substitute the value of $k$ and solve for $t$
From Step 2, $k = 0.1\text{ min}^{-1}$. So,
$ 0.1 = \dfrac{2.303 \times 2}{t} = \dfrac{4.606}{t}
$
Rearranging to find $t$:
$ t = \dfrac{4.606}{0.1} = 46.06\,\text{minutes.}
$
Step 5: State the final answer
The time required for the completion of 99% of this first-order reaction is 46.06 minutes.
