In a first order reaction, the concentration of the reactant decreases form `800 "mol dm"^(-3)` to `50 "mol dm"^(-3)` in `2 xx 10^(4) s`. The rate constant of the reaction (in `s^(-1)`) is

Options

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

Answer

C

Explanation

`k=(2.303)/(2xx10^(4)s)"log"(800)/(50)`
`=(2.303log 16s^(-1))/(2xx10^(4))`
`=(2.303xx4log2)/(2xx10^(4))`
`=4.606 xx 0.3010 xx10^(-4)s^(-1)`
`=1.86 xx 10^(-4)s^(1)`.