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

c) for first order reaction,
`K=(2.303)/t log N_(0)/N = (2.303)/(2 xx 10^(4)s) log (800 "mol"//dm^(3))/(50 "mol"//dm^(3))`
`=(2.303 xx 1.204)/(2 xx 10^(4)s) = 1.386 xx 10^(-4)s^(-1)`.