The half-life period of a 3rd order reaction is found to be 1 hour when we start with a concentration of `0.3" mol L"^(-1)` of the reactant. If we start with `0.1" mol L"^(-1)`, the half-life period in hours will be

Answer

9

Explanation

`t_(1//2)=k1/([A]_0^(n-1))`
For 3rd order reaction, `t_(1//2)=k1/([A]_0^2)`
`1=k1/((0*3)^2)" or "k=0*09`.
When `[A]_0=0*1molL^-1`,
`t_(1//2)=(0*09)1/((0*1)^2)=9"hours"`.