`T_(50)` of first order reaction is 10 min. starting with 10 `"mol"^(-1)` rate after 20 min is :

Options

A. `0.0693molL^(-1)min^(-1)`
B. `0.0693xx2.5molL^(-1)min^(-1)`
C. `0.0693xx5molL^(-2)min^(-1)`
D. `0.0693xx10molL^(-1)min^(-1)`

Answer

B

Explanation

`T_(50)=10min`
`k=0.693//10=0.0693min^(-1)`
Concentration at starting =10M
Concentration after 20min (two half-lives) `=2.5m`
`((dx)/(dt))=k[A]=0.0693xx2.5molL^(-1)min^(-1)`.