$\mathrm{t}_{1 / 4}$ can be taken as the time taken for the concentration of a reactant to drop to…

$\mathrm{t}_{1 / 4}$ can be taken as the time taken for the concentration of a reactant to drop to $\frac{3}{4}$ of its initial value. If the rate constant for a first order reaction is $\mathrm{K}$, the $\mathrm{t}_{1 / 4}$ can be written as
  1. $0.10 / \mathrm{K}$
  2. $0.29 / \mathrm{K}$
  3. $0.69 / \mathrm{K}$
  4. $0.75 / \mathrm{K}$

Solution

$ t_{1 / 4}=\frac{2.303}{K} \log \frac{1}{1-\frac{1}{4}}=\frac{0.29}{K} $

Asked in: JEE Main 2005

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