The value of rate constant of the above reaction at 27ºC 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 rate constant of the above reaction at 27ºC is

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Answer:

Solution:

$A ⟶ 2 B + C$

$t = 0$ , $a$ $0$ $0$

$t = 40$ , $a - x$ $2 x$ $x$

$t = \infty$ , $0$ $2 a$ $a$

$(a - x) \times 60 + 2 x \times 50 + x \times (- 80) = 26$

$60 a - 40 x = 26 . \dots (1)$
$2 a \times 50 + a \times (- 80) = 10.. \dots (2)$

By (1) and (2)

Therefore, $a = \frac{1}{2}$ and $x = \frac{1}{10}$

$K_{27℃} = \frac{2.303}{40} lo g \frac{\frac{1}{2}}{\frac{1}{2} - \frac{1}{10}}$

$K_{27℃} \approx 5.57 \times 1 0^{- 3} mi n^{- 1}$

Chemistry - Chemical Kinetics Question with Solution | TestHub

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