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

A

$r = k [s u g a r]^{2} [H]^{6}$

B

$r = k [s u g a r]^{1} [H]^{0}$

C

$r = k [s u g a r]^{0} [H^{\oplus}]^{6}$

D

$r = k [s u g a r]^{0} [H^{\oplus}]^{1}$

Answer

Correct option is

B

$r = k [s u g a r]^{1} [H]^{0}$

Given,
Since $t_{1 / 2}$ does not depends upon the sugar concentration means it is first order w.r.t [sugar]
$∴ t_{1 / 2} \propto [s u g a r]^{1}$
$t_{1 / 2} \times a^{n - 1} = k$
$\frac{( t _{1} / 2 ) _{1}}{( t _{1 / 2)_{2}}} = \frac{[ H ^{\oplus} ] _{1}^{1 - n}}{[ H ^{\oplus} ] _{2}^{1 - n}}$

$\frac{5 0 0}{5 0} = (\frac{1 0 ^{- 5}}{1 0 ^{- 6}})^{1 - n}$

10 = $(10)^{1 - n} \Rightarrow n = 0$

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The inversion of cane sugar proceeds with a constant half-life of $500$ minute at $p H = 5$ for any concentration of sugar. However, if $p H = 6$ , the half-life changes to $50$ minute. Derive the rate law for inversion of cane sugar.

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

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