Rate constant for a reaction is `10^(-3) S^(-1)`. How much time is required to reduce the initial concentration of reactant to 25%

Options

A. 693 s
B. 1386 s
C. 6930 s
D. 2029 s

Answer

B

Explanation

`t=(2.303)/(K)"log"(a)/(a-x)`
where," "k=rate constant=`10^(-3)s^(-1)`
a=I"initial amount "=100
a-x= amount left after time, t =25
t=time required to reduce the initial concentration of reactant to 25% reaction
`therefore " " t=(2.303)/(10^(-3))"log"(100)/(25)=(2.303)/(10^(-3))`"log" 4
=` (2.303xx0.6020)/(10^(-3))=1386s`