The integrated rate equation for a first-orde | vedclass

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(A) For a first-order reaction,$A ightarrow 	ext{product}$,the rate is given by: $	ext{Rate} = -\frac{d[A]}{dt} = k[A]$.Rearranging and integratin

Vedclass · Accepted Answer

$k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t}$

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(A) For a first-order reaction,$A ightarrow 	ext{product}$,the rate is given by: $	ext{Rate} = -\frac{d[A]}{dt} = k[A]$.Rearranging and integrating the equation: $\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -\int_0^t k dt$.This yields: $\ln \frac{[A]_t}{[A]_0} = -kt$.Thus,the integrated rate equation is $k = \frac{1}{t} \ln \frac{[A]_0}{[A]_t}$.Alternatively,using base $10$ logarithms,$k = \frac{2.303}{t} \log_{10} \frac{[A]_0}{[A]_t}$.Both options $A$ and $B$ represent the correct integrated rate equation for a first-order reaction.

The integrated rate equation for a first-order reaction,$A \rightarrow \text{product}$,is

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