In a first order of reaction the concentration of reactant decreases from `800 "mol"//d m^(3)` to`50 "mol"//d m^(3)` in `2 xx 10^(2) sec`. The rate constant of reaction in `"sec"^(-1)` is

Options

A. `2 xx 10^(4)`
B. `3.45 xx 10^(-5)`
C. `1.386 xx 10^(-2)`
D. `2 xx 10^(-4)`

Answer

C

Explanation

`k = (2.303)/(t)"log"_(10)(a)/(a - x), t = 2 xx 102, a = 800, a - x = 50`
`k = (2.303)/(2 xx 10^(2)) "log"_(10) (800)/(50) = (2.303)/(2 xx 10^(2)) log_(10) 16`
`= (2.303)/(2 xx 10^(2))log_(10) 2^(4) = (2.303)/(2 xx 10^(4)) xx 4 xx 0.301`
`= 1.38 xx 10^(-2) s^(-1)`