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.
$A+H_{2}O\stackrel{\mathrm{H}{}^{+}}{\to }2B+C$
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 $27kJmo{l}^{-1}$. (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)

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{1}{T_1}-\frac{1}{T_2}\right]$
$\log \frac{k_2}{5.57\times 10^{-3}}=\frac{27\times 10^3}{2.303\times 8.314}\left[\frac{1}{300}-\frac{1}{400}\right]$
$\log \frac{k_2}{5.57\times 10^{-3}}=1.175$
$\frac{k_2}{5.57\times 10^{-3}}=14.97$
${\left(k_2\right)}_{127}=83.36\times 10^{-3}$
${\left(t_{\frac{1}{2}}\right)}_{127}=\frac{0.693}{k_2}=8.4\min$