The rate constant of a reaction is `0.0693 min^(-1)`. Starting with `10 mol`, the rate of the reaction after `10 min` is

Options

A. `0.693 M "min"^(-1)`
B. `0.0693 xx2 M "min"^(-1)`
C. `0.0693xx5 M "min"^(-1)`
D. `0.693xx(5)^(2) M "min"^(-1)`

Answer

C

Explanation

`[A]_(0)=10 M`
t = 10 min
`therefore t=(2.303)/(k)"log"([A]_(0))/([A])`
`10=(2.303)/(0.693)"log"(10)/([A])`
`(0.693xx10)/(2.303)="log"(10)/([A])`
`(0.693xx10)/(2.303)="log"(10)/([A])`
`0.3010="log"(10)/([A])`
`therefore (10)/([A])=2`
`[A]=10//2 = 5M`
Rate `=k[A]=0.693xx5"M min"^(-1)`