At what temperature half-life of above process is equal to 31.1 minutes? | 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 $27 k J m o l^{- 1}$ .

At what temperature half-life of above process is equal to 31.1 minutes?

Options:

Answer:

Solution:

$(k)_{t} = \frac{0.693}{31.1}$

$lo g \frac{k _{t}}{k _{1}} = \frac{E _{a}}{2.303 R} (\frac{1}{300} - \frac{1}{T _{2}})$ ; $T = 345 K$ .

Alternate solution:

( $t_{1/2}$ ) at 27°C = $\frac{0.693}{5.57 \times 1 0 ^{- 3}} = 124.4$ .

( $t_{1/2}$ ) at 127°C = $8.4$ .

Hence, the temperature at which $t_{1/2} = 31.1 min$ should lie between 27°C and 127°C.

In option (c), only the temperature is between 27°C and 127°C.

Chemistry - Chemical Kinetics Question with Solution | TestHub

Options:

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