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 rate constant of the above reaction at $27^{\circ} C$ is

• A. $1.94 \times 10^{-3} \min ^{-1}$
• B. $1.37 \times 10^{-6} \min ^{-1}$
• C. $6.13 \times 10^{-3} \min ^{-1}$
• D. $5.57 \times 10^{-3} \min ^{-1}$

Answer 1

Answer: (D)

$( a - x ) \times 60+2 x \times 50+ x (-80)=26$
$60 a -40 x =26\,\,\,$...(i)
$2 a \times 50+ a (-80)=10\,\,\, $...(i)
By (i) and (ii)
$a =\frac{1}{2} ; x =\frac{1}{10}$
$k _{27^{\circ} C }=\frac{2.303}{40} \log \frac{1 / 2}{\frac{1}{2}-\frac{1}{10}}$
$k _{27^{\circ} C }=5.57 \times 10^{-3} \min ^{-1}$