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(D) For a first-order reaction, the integrated rate equation is: $\ln \frac{[A]_0}{[A]_t} = kt$.Converting to base $10$ logarithm: $2.303 \log \frac{[

Vedclass · Accepted Answer

$0.868$

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(D) For a first-order reaction, the integrated rate equation is: $\ln \frac{[A]_0}{[A]_t} = kt$.Converting to base $10$ logarithm: $2.303 \log \frac{[A]_0}{[A]_t} = kt$.Given $[A]_0 = 1.0 	ext{ mol dm}^{-3}$, $k = 2 	imes 10^{-2} 	ext{ s}^{-1}$, and $t = 100 	ext{ s}$.Substituting the values: $2.303 \log \frac{1.0}{[A]_t} = (2 	imes 10^{-2} 	ext{ s}^{-1}) 	imes (100 	ext{ s})$.$2.303 \log \frac{1}{[A]_t} = 2$.$\log \frac{1}{[A]_t} = \frac{2}{2.303} \approx 0.868$.

For a first-order reaction $A \rightarrow \text{product}$, the rate constant is $k = 2 \times 10^{-2} \text{ s}^{-1}$. If the initial concentration of $A$ is $[A]_0 = 1.0 \text{ mol dm}^{-3}$, find the value of $\log \frac{1}{[A]_t}$ after $t = 100 \text{ s}$.

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