Question

For a reaction, A→B + C, it was found that at the end of 20 minutes from the start, the total optical rotation of the system was 60° and when the reaction is complete it was 120°. Assume that only B and C are optically active and dextro rotatory, the rate constant (in $hr^{-1}$) of this first order reaction would be: (Use, Inx = 2.3 logx, log2 = 0.3)

Answer 1

2.07

Answer 2

The reaction is a first-order reaction: A → B + C.
We are given that only products B and C are optically active. This implies that the initial optical rotation ($r_0$) at $t=0$ is $0^\circ$, as only reactant A is present.
The optical rotation at time $t$ ($r_t$) is due to the formation of products B and C.
The optical rotation when the reaction is complete ($r_\infty$) corresponds to the maximum amount of products B and C formed.

For a first-order reaction where the measured property (optical rotation in this case) is proportional to the concentration of products formed, and the initial value of the property is zero, the rate constant ($k$) can be calculated using the formula:

$k = \frac{1}{t} \ln \left( \frac{r_\infty}{r_\infty - r_t} \right)$

Given values:

• Time, $t = 20$ minutes
• Optical rotation at $t=20$ min, $r_t = 60^\circ$
• Optical rotation when the reaction is complete, $r_\infty = 120^\circ$

Substitute these values into the formula: $k = \frac{1}{20 \text{ min}} \ln \left( \frac{120^\circ}{120^\circ - 60^\circ} \right)$ $k = \frac{1}{20} \ln \left( \frac{120}{60} \right)$ $k = \frac{1}{20} \ln (2)$

We are given $\ln x = 2.3 \log x$ and $\log 2 = 0.3$.
Therefore, $\ln 2 = 2.3 \times 0.3 = 0.69$.

Now, substitute the value of $\ln 2$ into the equation for $k$: $k = \frac{0.69}{20} \text{ min}^{-1}$ $k = 0.0345 \text{ min}^{-1}$

The question asks for the rate constant in $hr^{-1}$. To convert from $min^{-1}$ to $hr^{-1}$, multiply by 60 (since 1 hour = 60 minutes): $k (\text{in } hr^{-1}) = 0.0345 \times 60 \text{ hr}^{-1}$ $k = 2.07 \text{ hr}^{-1}$