The inversion of cane sugar proceeds with half life of 50 minutes at pH = 5 for any concentration of sugar. However if $pH = 6 .$ the half life changed to 500 minutes. The law expression of sugar inversion can be written as:

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$r = K [sugar]^{1} {[H^{+}]}^{0}$

$r = K [sugar]^{1} {[H^{+}]}^{1}$

$r = K [sugar]^{0} {[H^{+}]}^{0}$

$r = K [sugar]^{2} {[H^{+}]}^{0}$

answer is C.

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Detailed Solution

Here, the half time is,

$\begin{array}{l} t_{1 / 2} \propto (a)^{1 - n} \\ \frac{50}{500} = {(\frac{10^{- 5}}{10^{- 6}})}^{1 - n} \\ \Rightarrow 10^{- 1} = 10^{1 - n} \\ \Rightarrow n = 2 \end{array}$

$t_{1 / 2} =$ constant for any [sugar]

$\Rightarrow$ Order w.r.t. sugar= 1

$r = k [sugar]^{1} {[H^{+}]}^{1}$

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