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Step-by-Step Solution
Step 1: Identify the Type of Reaction
This is a first-order reaction:
$A \rightarrow \text{products}$
Step 2: Write the Integrated Rate Law
For a first-order reaction, the rate constant $k$ can be calculated using the relation:
$k = \frac{2.303}{t} \, \log \left(\frac{[A]_0}{[A]}\right)$
where $[A]_0$ is the initial concentration and $[A]$ is the concentration at time $t$.
Step 3: Substitute the Given Values
From the problem, the concentration changes from $0.1 \,\text{M}$ to $0.025 \,\text{M}$ in $40\,\text{minutes}$. Hence:
$[A]_0 = 0.1 \,\text{M}, \quad [A] = 0.025 \,\text{M}, \quad t = 40\,\text{minutes}.$
So,
$k = \frac{2.303}{40} \, \log \left( \frac{0.1}{0.025} \right).$
Step 4: Evaluate the Logarithmic Expression
$ \log \left(\frac{0.1}{0.025}\right) = \log (4) = 0.6020 \text{ (approximately) }.$
Therefore,
$k = \frac{2.303 \times 0.6020}{40}.$
Step 5: Calculate the Rate Constant $k$
$k = \frac{2.303 \times 0.6020}{40} \approx 3.47 \times 10^{-2} \,\text{min}^{-1}.$
Step 6: Determine the Rate at the Requested Concentration
The rate of a first-order reaction is given by:
$\text{Rate} = k [A].$
When $[A] = 0.01\,\text{M}$:
$\text{Rate} = 3.47 \times 10^{-2} \times 0.01 = 3.47 \times 10^{-4} \,\text{M} \,\text{min}^{-1}.$
Step 7: State the Final Answer
The rate of the reaction when the concentration is $0.01\,\text{M}$ is
$3.47 \times 10^{-4}\,\text{M} \,\text{min}^{-1}$.
