The inverison of cane sugar proceeds with half life of `500 min` at `pH 5` for any concentration of sugar. However, if `pH = 6`, if the half life changes to `50 min`. The rate law expresison for the sugar inverison can be written as

Options

A. `r = k["sugar"]^(2)[H]^(6)`
B. `r = k["sugar"]^(1)[H]^(0)`
C. `r = k["suagr"]^(0)[H^(o+)]^(6)`
D. `r = k["sugar"]^(0)[H^(o+)]^(1)`

Answer

B

Explanation

Since `t_(1//2)` does not depends upon the sugar concentration means it is first order w.r.t. [suagr]
`:. t_(1//2) prop ["Sugar"]^(1)`
`t_(1//2) prop a^(n-1) = k`
`((t_(1//2))_(1))/((t_(1//2))_(2)) = ([H^(o+)]_(1)^(1-n))/([H^(o+)]_(2)^(1-n))`
`(500)/(50) = ((10^(-5))/(10^(-6)))^(1-n)`
`10 = (10)^(1-n) rArr n = 0`.