Chemistry - Chemical Kinetics Question with Solution | TestHub

$$\frac{[B]}{[C]}=\frac{K_1}{K_2}=\frac{1}{2}\Rightarrow [C]>[B]$$

Hence, after a long time, the solution will be dextrorotatory.→

Now, $K=K_1+K_2=6.93\times 10^{-2}+13.86\times 10^{-2}=3\times 6.93\times 10^{-2}{\min }^{-1}$

Therefore, ${\mathrm{t}}_{1/2}=\frac{\ln 2}{\mathrm{K}}=\frac{0.693}{3\times 6.93\times 10^{-2}}=\frac{10}{3}\min$

$\begin{array}{llclc}& A& \to B& A& ⟶C\\ t=0& 2M& 0& 2M& 0\\ t=t& 2-(x+y)M& xM& 2-(x+y)M& yM\\ & & & & \\ & & & & \end{array}$

From the question, $\mathrm{x}+\mathrm{y}=1.5$ and $\frac{\mathrm{x}}{\mathrm{y}}=\frac{1}{2}$.

Therefore, $\mathrm{x}=0.5,\mathrm{y}=1.0$.

Hence, total rotation $=0.5\times 60^{\circ }+0.5\times \left(-72^{\circ }\right)$

$$+1.0\times 42^{\circ }=36^{\circ }$$