Question

`T_(50)` of first - order reaction is 10 min . Starting with 10 mol `L^(-1)` , rate after 20 min is
A. `0.0693` mol `L^(-1) "min"^(-1)`
B. `0.0693 xx 2.5 mol L^(-1) "min"^(-1)`
C. `0.0693 xx 5 mol L^(-2) "min"^(-1)`
D. `0.0693 xx 10 mol L^(-1) "min"^(-1)`

Answer 1

Correct Answer - b
`T_(50) = 10` min
k = 0.693/10 = 0.0693 `"min"^(-1)`.
Concentration at starting = 10 M
Concentration after 20 min (two half-lives) = 2.5 M
`((dx)/(dt)) = k[A] = 0.0693 xx 2.5` mol `L^(-1) "min"^(-1)`.