The decomposition of a substance follows first order kinetics. If its concentration is reduced to 1/8 th of its initial value in 12 minutes, the rate constant of the decomposition system is

Options

A. `((2.303)/(12) log (1)/(8)) "min"^(-1)`
B. `((2.303)/(12) log 8) "min"^(-1)`
C. `((0.693)/(12)) "min"^(-1)`
D. `((1)/(12) log 8) "min"^(-1)`

Answer

B

Explanation

`k = (2.303)/(12) "log" (a)/(a - x)` (for first order)
`k = (2.303)/(12) "log" (1)/(1//8) = ((2.303)/(12) "log" 8) "min"^(-1)`