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The inversion of cane sugar proceeds wit...

The inversion 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 expression for the sugar inversion can be written as

A

`r=K ["sugar"]^(2)[H^(+)]^(0)`

B

`r=K ["sugar"]^(1)[H^(+)]^(0)`

C

`r=K ["sugar"]^(1)[H^(+)]^(1)`

D

`r=K ["sugar"]^(0)[H^(+)]^(1)`

Text Solution

Verified by Experts

The correct Answer is:
b
At `pH=5`, the half-life is `500 min` for all concentrations of sugar, i.e, `t_(1//2) prop [sugar]^(0)`.
Thus, the reaction is I order w.r.t. sugar
Now rate`=K[sugar]^(1)[H^(+)]^(m)`
Also for `[H^(+)], t_(1//2) prop [H^(+)]^(1-m)`
`:. 500 prop [10^(-5)]^(1-m)`
`:. 50 prop[10^(-5)]^(1-m)`
`:. 10=(10)^(1-m)`
or `(1-m)=1 :. m=0`
Therefore, rate=`K[sugar]^(1) [H^(+)]^(0)`
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