Given that for a reaction of nth order, the integrated rate equation is:
`K=1/(t(n-1))[1/C^(n-1)-1/C_(0)^(n-1)]`, where `C` and `C_(0)` are the concentration of reactant at time `t` and initially respectively. The `t_(3//4)` and `t_(1//2)` are related as `t_(3//4)` is time required for C to become `C_(1//4`) :

Options

A. `t_(3//4)=t_(1//2)[2^(n-1)+1]`
B. `t_(3//4)=t_(1//2)[2^(n-1)-1]`
C. `t_(3//4)=t_(1//2)[2^(n+1)+1]`
D. `t_(3//4)=t_(1//2)[2^(n+1)-1]`

Answer

a

Explanation

The ratio `t_(3//4)//t_(1//2)` using given equation,
`t_(3//4)/t_(1//2)=((4)^(n-1)-1)/(2^(n-1)-1)`
`=((2^(n-1)-1)(2^(n-1)+1))/((2^(n-1)-1))=(2^(n-1)+1)`