For a first-order reaction $A ightarrow ext{Products}$, initial concentration of A is $0.1 ext{ M}$, whi

Question

For a first-order reaction $A ightarrow 	ext{Products}$, initial concentration of A is $0.1 	ext{ M}$, which becomes $0.001 	ext{ M}$ after $5 	ext{ minutes}$. Rate constant for the reaction in $	ext{min}^{-1}$ is

Accepted Answer

For a first-order reaction, the integrated rate equation is given by:
$k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}$

Given:
Initial concentration, $[A]_0 = 0.1 	ext{ M}$
Final concentration, $[A] = 0.001 	ext{ M}$
Time, $t = 5 	ext{ mins}$

Substituting the values in the equation:
$k = \frac{2.303}{5} \log \left(\frac{0.1}{0.001}ight)$
$k = \frac{2.303}{5} \log (100)$
We know that $\log(100) = \log(10^2) = 2 \log(10) = 2 	imes 1 = 2$
$k = \frac{2.303}{5} 	imes 2$
$k = \frac{4.606}{5} = 0.9212 	ext{ min}^{-1}$

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