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Step 1: Recognize the Type of Reaction
The problem states that the reaction is a first-order reaction. For a first-order reaction, the rate is directly proportional to the concentration of the reactant.
Step 2: Write the Relevant Rate Expression
For a first-order reaction (A ā Products), the rate constant $k$ can be determined from the rate data at two different times using the formula:
$k \;=\; \dfrac{2.303}{t_{2} \;-\; t_{1}} \;\log \!\biggl(\dfrac{\text{rate}_{1}}{\text{rate}_{2}}\biggr)$
Step 3: Substitute the Given Data and Calculate the Rate Constant
At t = 10 s, the rate is 0.04 mol Lā1 sā1.
At t = 20 s, the rate is 0.03 mol Lā1 sā1.
Hence, $t_{2} - t_{1} = 20 - 10 = 10 \text{ s}$.
$\text{rate}_{1} = 0.04 \,\text{mol L}^{-1}\text{s}^{-1}$ and $\text{rate}_{2} = 0.03 \,\text{mol L}^{-1}\text{s}^{-1}$.
Plug the values into the formula:
$k \;=\; \dfrac{2.303}{10} \;\log \!\biggl(\dfrac{0.04}{0.03}\biggr)$
Calculate inside the logarithm:
$\dfrac{0.04}{0.03} = 1.3333...$
Then,
$k = \dfrac{2.303}{10} \;\log(1.3333...)$
Evaluate the logarithm (base 10):
$\log(1.3333...) \approx 0.1249$
So,
$k \approx \dfrac{2.303}{10} \times 0.1249 \approx 0.0287 \,\text{s}^{-1}$
Step 4: Determine the Half-Life
For a first-order reaction, the half-life ($t_{1/2}$) is given by:
$t_{1/2} = \dfrac{0.693}{k}$
Substitute $k = 0.0287\,\text{s}^{-1}$:
$t_{1/2} = \dfrac{0.693}{0.0287} \approx 24.1 \,\text{s}$
Thus, the half-life period of the reaction is approximately 24.1 s.
