1. Tardigrade
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  3. Chemistry
  4. t1/4 can be taken as the time taken for concentration of reactant to drop to 3/4 of its initial value. If the rate constant for a first order reaction is k, then t1/4 can be written as

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Q. $t_{1/4}$ can be taken as the time taken for concentration of reactant to drop to $3/4$ of its initial value. If the rate constant for a first order reaction is $k$, then $t_{1/4}$ can be written as

Chemical Kinetics

Solution:

$k =\frac{2.303}{t}$ log $\frac{a}{a -x}$
or $k =\frac{2.303}{t_{1/4}}$ log $\frac{4a}{3a} =\frac{2.303}{t_{1/4}}$ log $\frac{4}{3}$
or $k = \frac{2.303 \times0.125}{t_{1/4}} =\frac{0.29}{t_{1/4}}$