For a first order reaction, the ratio of the time taken for `7//8^(th)` of the reaction to complete to that of half of the reaction to complete is

Options

A. `3 : 1`
B. `1 : 3`
C. `2 : 3`
D. `3 : 2`

Answer

A

Explanation

`k = (2.303)/(t) "log" (a)/(a - x)`
for 7/8 of the reaction to complete, `t = t_(7//8)`
a - x = a - 7a/8 = a/8
`:. T_(7//8) = (2.303)/(k) "log" (a)/(a//8) (2.303)/(k) log 8`
For half of the reaction to complete , `t = t_(1//2)`
x = a - a/2 = a/2
`:. t_(1//2) = (2.303)/(k) "log" (a)/(a//2) = (2.303)/(k) log 2`
`:. (t_(7//8))/(t_(1//2)) = (log 8)/(log 2) = (log 2^(3))/(log 2) = (3 log 2)/( log 2 = 3`