At $\mathrm{T}(\mathrm{K})$, if the rate constant of a first order reaction is $4.606 imes 10^{-3} \mathrm{~s}^{-1}$, the time to reduce the initial concentration of the reactant to $1 / 10$ in seconds is :

Question

At $\mathrm{T}(\mathrm{K})$, if the rate constant of a first order reaction is $4.606 	imes 10^{-3} \mathrm{~s}^{-1}$, the time to reduce the initial concentration of the reactant to $1 / 10$ in seconds is :

Aicharya · Accepted Answer

To find the time taken for the concentration of the reactant to reduce to $1 / 10$ th, we use the formula:
$t=\frac{\ln (10)}{k}$
Given $k=4.606 \times 10^{-3} \mathrm{~s}^{-1}$, we have:
$t=\frac{\ln (10)}{4.608 \times 10^{-5}}$
Solving for $t$, we get approximately 500 seconds.

At $\mathrm{T}(\mathrm{K})$, if the rate constant of a first order reaction is $4.606 \times 10^{-3}…

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