- #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?