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 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)

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