- #1

utkarshakash

Gold Member

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## Homework Statement

The inversion of cane sugar proceeds with a constant half life of 500 minute at pH=5 for any concentration of sugar. When pH is changed to 4, the half life becomes 50 minutes. Deduce the rate law expression for inversion of sugar.

## Homework Equations

[itex]t_{1/2}*a^{n-1}[/itex] is constant for reaction of any order

## The Attempt at a Solution

Since pH=5

Therefore, [itex][H^{+}]=10^{-5}[/itex] (Initial Concentration)

Similarly Final [itex][H^{+}]=10^{-4}[/itex]

Using first equation I have

500*[itex](10^{-5})^{n-1}[/itex]=50*[itex](10^{-4})^{n-1}[/itex]

10=[itex]10^{n-1}[/itex]

n=2

So it must be second order wrt [itex][H^{+}][/itex]

Since the rate does not depend of sugar so it must be zero order wrt sugar

Therefore the rate law expression should be

Rate=k[Ester][itex][H^{+}]^{2}[/itex]

But the answer in my book says

Rate=k[Ester][itex][H^{+}][/itex]

Whats wrong?