Two `1^(st)` order reactions have half`-` lives in the ration `3:2`. Then the ratio of time intervals `t_(1):t_(2)`, will be gt Where `t_(1)` is the time period for `25%` completion completion of the first reaction and `t_(2)` is time required for `75%` completion of the second reaction . `[log 2=0.3 , log 3=0.477]`

Options

A. `0.199 :1`
B. `0.420 :1`
C. `0.273:1`
D. `0.311:1`

Answer

4

Explanation

Time taken for `25%` completion.
`C_(1)C_(0)theta^(-k_(1)t)." "implies(3)/(4)=e^(-k_(1)t_(1))`
as `n(3//4)=-k_(1)b_(1)`so`t_(1)ln((4)/(3))`
time taken for `75%` completion `t_(2)=2.(ln2)/(k_(2))`
so required ration `(t_(1))/(t_(2))=(ln(4//3)k_(2))/(k_(1).2ln_(2))=(3)/(2)xx((ln4-ln3))/(ln4)=0.311:1`