The inversion of a sugar follows first order rate equation which can be followed by noting the change in the rotation of the plane of polarization of light in the polarimeter. If `r_(oo), r_(f)` and `r_(0)` are the rotations at `t = oo, t = t`, and `t = 0`, then the first order reaction can be written as
(a)`k = (1)/(t)log.(r_(1)-r_(oo))/(r_(o) -r_(oo))`
(b) `k = (1)/(t)ln.(r_(0)-r_(oo))/(r_(t)-r_(o))`
(c) `k = (1)/(t)ln.(r_(oo)-r_(o))/(r_(oo)-r_(t))`
(d)`k = (1)/(t)ln.(r_(oo)-r_(t))/(r_(oo)-r_(0))`

Options

A. `k = (1)/(t)log.(r_(1)-r_(oo))/(r_(o) -r_(oo))`
B. `k = (1)/(t)ln.(r_(0)-r_(oo))/(r_(t)-r_(o))`
C. `k = (1)/(t)ln.(r_(oo)-r_(o))/(r_(oo)-r_(t))`
D. `k = (1)/(t)ln.(r_(oo)-r_(t))/(r_(oo)-r_(0))`

Answer

To solve the problem regarding the first-order reaction of sugar inversion and its relationship with the rotation of polarized light, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
The inversion of sugar (sucrose) into glucose and fructose can be represented as:
\[ \text{Cane Sugar} \xrightarrow{\text{H}_2\text{O}} \text{Glucose} + \text{Fructose} \]
This reaction follows first-order kinetics.

2. **Define the Variables**:
- \( r_0 \): Rotation at time \( t = 0 \) (initial rotation).
- \( r_t \): Rotation at time \( t \).
- \( r_\infty \): Rotation at time \( t = \infty \) (final rotation when the reaction is complete).

3. **Determine the Change in Rotation**:
The change in rotation can be defined as:
- The amount of sugar that has reacted at time \( t \) can be represented as \( x \).
- Therefore, the rotation at any time \( t \) can be expressed in terms of the initial and final rotations:
\[ x = r_0 - r_t \]
At \( t = \infty \):
\[ x = r_0 - r_\infty \]

4. **Apply the First-Order Rate Law**:
For a first-order reaction, the rate constant \( k \) can be expressed as:
\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right) \]
where \( [A]_0 \) is the initial concentration and \( [A] \) is the concentration at time \( t \).

5. **Relate Concentrations to Rotations**:
Using the relationship between the concentrations and the rotations:
\[ [A]_0 = r_0 - r_\infty \quad \text{and} \quad [A] = r_0 - r_t \]
We can substitute these into the rate equation:
\[ k = \frac{2.303}{t} \log \left( \frac{r_0 - r_\infty}{r_0 - r_t} \right) \]

6. **Convert Logarithm to Natural Log**:
Since the options are given in natural logarithm (ln), we can use the conversion:
\[ \log x = \frac{\ln x}{2.303} \]
Thus, we can rewrite the equation as:
\[ k = \frac{1}{t} \ln \left( \frac{r_0 - r_\infty}{r_0 - r_t} \right) \]

7. **Identify the Correct Option**:
After analyzing the options provided:
- The correct expression matches option (b):
\[ k = \frac{1}{t} \ln \left( \frac{r_0 - r_\infty}{r_t - r_\infty} \right) \]

### Conclusion:
The correct answer is:
**(b)** \( k = \frac{1}{t} \ln \left( \frac{r_0 - r_\infty}{r_t - r_\infty} \right) \)