1.

T_(50) of first - order reaction is 10 min . Starting with 10 mol L^(-1) , rate after 20 min is

Answer»

`0.0693` mol `L^(-1) "min"^(-1)`
`0.0693 xx 2.5 mol L^(-1) "min"^(-1)`
`0.0693 xx 5 mol L^(-2) "min"^(-1)`
`0.0693 xx 10 mol L^(-1) "min"^(-1)`

SOLUTION :`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)`.


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