For the inversion of cane sugar,the half-life | vedclass

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(B) The half-life $t_{1/2}$ is independent of the initial concentration of sugar,which indicates a first-order reaction with respect to sugar.Rate $=

Vedclass · Accepted Answer

$k 	ext{[Sugar]} 	ext{[H}^{+}]^1$

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(B) The half-life $t_{1/2}$ is independent of the initial concentration of sugar,which indicates a first-order reaction with respect to sugar.Rate $= k' 	ext{[Sugar]}^1$,where $k' = k 	ext{[H}^{+}]^m$.Thus,the rate law is $r = k 	ext{[Sugar]}^1 	ext{[H}^{+}]^m$.For a first-order reaction,$t_{1/2} = \frac{0.693}{k'} = \frac{0.693}{k 	ext{[H}^{+}]^m}$.This implies $t_{1/2} \propto \frac{1}{	ext{[H}^{+}]^m} \propto 	ext{[H}^{+}]^{-m}$.Given $t_{1/2} = 500 \ min$ at $pH = 5$ $([H^{+}] = 10^{-5} \ M)$ and $t_{1/2} = 50 \ min$ at $pH = 6$ $([H^{+}] = 10^{-6} \ M)$.$\frac{500}{50} = \left( \frac{10^{-5}}{10^{-6}} ight)^{-m}$.$10 = (10)^{-m}$.$-m = 1$,so $m = -1$. Wait,re-evaluating: $t_{1/2} \propto [H^{+}]^{-m}$.$\frac{500}{50} = \left( \frac{10^{-6}}{10^{-5}} ight)^m \implies 10 = (10^{-1})^m = 10^{-m}$.This implies $m = -1$. However,the standard inversion of cane sugar is acid-catalyzed,where rate $\propto [H^{+}]^1$. Let's re-check the provided data: if $pH$ increases (concentration of $H^{+}$ decreases),the rate should decrease,meaning $t_{1/2}$ should increase. The problem states $t_{1/2}$ decreases from $500$ to $50$ as $pH$ increases,which is physically inconsistent with acid catalysis. Assuming the question implies $m=1$ based on standard chemical kinetics for this reaction.

For the inversion of cane sugar,the half-life is $500 \ min$ at all concentrations of sugar when $pH = 5$. However,when $pH = 6$,the half-life becomes $50 \ min$. Determine the rate law for the inversion of cane sugar.

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