The inversion of cane sugar proceeds with the half-life of 600 minutes at $p H = 5$ for any concentration of sugar. However at $p H = 6$ , the half life changes to 60 minute. The rate law expression for sugar inversion can be written as: (if it is first order with respect to sugar)

Question

The inversion of cane sugar proceeds with the half-life of 600 minutes at

p H = 5

for any concentration of sugar. However at

p H = 6

, the half life changes to 60 minute. The rate law expression for sugar inversion can be written as: (if it is first order with respect to sugar)

Toppr Admin · Accepted Answer

Since t1-2 does not depend upon the sugar concentration- it is first order w-r-t -sugar-x2234-t1-2-x221D-sugar-1t1-2-xD7-an-x2212-1-k-t1-2-1-t1-2-2-H-x2295-1-x2212-n1-H-x2295-1-x2212-n2

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:r=k[sugar]1[H]0r=k[sugar]2[H]6None of theser=k[sugar]0[H⊕]6

Since t1/2 does not depend upon the sugar concentration, it is first order w.r.t [sugar]∴t1/2∝[sugar]1t1/2×an−1=k(t1/2)1(t1/2)2=[H⊕]1−n1[H⊕]1−n2

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