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

Question

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 (A) 0.10/K (B) 0.29/K (C) 0.69/K (D) 0.75/K

Tardigrade Admin · Accepted Answer

k =(2.303/t) log (a/a -x) or k =(2.303/t1/4) log (4a/3a) =(2.303/t1/4) log (4/3) or k = (2.303 ×0.125/t1/4) =(0.29/t1/4)

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

$0.10/ K$

$0.29/ K$

$0.69/ K$

$0.75/ K$

$k = \frac{2.303}{t}$ log $\frac{a}{a - x}$
or $k = \frac{2.303}{t _{1/4}}$ log $\frac{4 a}{3 a} = \frac{2.303}{t _{1/4}}$ log $\frac{4}{3}$
or $k = \frac{2.303 \times 0.125}{t _{1/4}} = \frac{0.29}{t _{1/4}}$

Q. 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

0.10/K

0.29/K

0.69/K

0.75/K