Question

When sucrose is hydrolysed, it produces dextrorotatory "glucose" and laevorotatory "fructose" whereas the reactant itself is a dextrorotatory compound. The above a conversion process follows first order kinetics. Similarly, an optically active compound $A$ is hydrolysed as follows.
$A + H _{2} O \xrightarrow{ H ^{+}} 2 B + C$
The observed rotation of compound $A, B$ and $C$ are $60^{\circ}$, $50^{\circ}$ and $-80^{\circ}$ per mole respectively. The angles of rotation after $40$ minutes and after the completion of reaction were $26^{\circ}$ and $10^{\circ}$ respectively. At $27^{\circ} C$ activation energy for conversion is $27\, kJ\, mol ^{-1}$. (Use: $\log 1.25=0.0969, \log 14.97=$ 1.175)
The value of $t _{1 / 2}$ for the above process at $127^{\circ} C$ is

• A. 1.2 hours
• B. 8.4 minutes
• C. 1.5 hours
• D. 5.68 minutes

Answer 1

Answer: (B)

$\log \frac{ k _{2}}{ k _{1}}=\frac{ E _{ a }}{2.303 R }\left[\frac{1}{ T _{1}}-\frac{1}{ T _{2}}\right]$
$\log \frac{ k _{2}}{5.57 \times 10^{-3}}=\frac{27 \times 10^{3}}{2.303 \times 8.314}\left[\frac{1}{300}-\frac{1}{400}\right]$
$\log \frac{ k _{2}}{5.57 \times 10^{-3}}=1.175$
$\frac{ k _{2}}{5.57 \times 10^{-3}}=14.97$
$\left( k _{2}\right)_{127}=83.36 \times 10^{-3}$
$\left( t _{\frac{1}{2}}\right)_{127}=\frac{0.693}{ k _{2}}=8.4\, \min$