The value of t 1/2 for the above process at 12 7 ∘ C is | TestHub

Passage / Comprehension

When sucrose is hydrolysed, it produces dextrorotatory "glucose" and laevorotatory "fructose," whereas the reactant itself is a dextrorotatory compound. The above conversion process follows first-order kinetics. Similarly, an optically active compound A is hydrolysed as follows:

$A + H_{2} O H^{+} 2 B + C$

The observed rotations of compounds A, B, and C are $6 0^{\circ}$ , $5 0^{\circ}$ , and $- 8 0^{\circ}$ per mole, respectively. The angles of rotation after 40 minutes and after the completion of the reaction were $2 6^{\circ}$ and $1 0^{\circ}$ , respectively. At $27$ ℃, the activation energy for conversion is $27 k J m o l^{- 1}$ .

The value of $t_{1/2}$ for the above process at $12 7^{\circ} C$ is

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Answer:

Solution:

$lo g \frac{k _{2}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{T _{1}} - \frac{1}{T _{2}})$

$lo g \frac{k _{2}}{5.57 \times 1 0 ^{- 3}} = \frac{27 \times 1 0 ^{3}}{2.303 \times 8.314} (\frac{1}{300} - \frac{1}{400})$

$lo g \frac{k _{2}}{5.57 \times 1 0 ^{- 3}} = 1.175$

$\frac{k _{2}}{5.57 \times 1 0 ^{- 3}} = 14.97$

$(k_{2})_{127} = 83.36 \times 1 0^{- 3}$

$(t_{1/2})_{127} = \frac{0.693}{k _{2}} = 8.4 min$

Chemistry - Chemical Kinetics Question with Solution | TestHub

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