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

Options

A. `K=1/tlog_@.(r_1-r_oo)/(r_0-r_oo)`
B. `K=1/tIn.(r_0-r_oo)/(r_t-r_oo)`
C. `K=1/tIn.(r_oo-r_0)/(r_t-r_oo)`
D. `K=1/tIn.(r_oo-r_t)/(r_oo-r_0)`

Answer

To solve the question regarding the first-order rate equation for the inversion of sugar, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Reaction**:
The inversion of sugar (like cane sugar) is a first-order reaction where sucrose is hydrolyzed into glucose and fructose.

2. **Define the Rotations**:
- \( r_0 \): 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. **Identify the Change in Rotation**:
The change in rotation can be expressed as:
- At \( t = 0 \): \( r_0 \)
- At \( t = t \): \( r_t \)
- At \( t = \infty \): \( r_{\infty} \)

4. **Calculate the Amount of Reactant and Product**:
- Let \( A \) be the initial amount of cane sugar.
- At \( t = 0 \), the concentration of glucose and fructose is 0.
- At time \( t \), if \( x \) moles of cane sugar have reacted, the remaining cane sugar is \( A - x \), and the amount of glucose and fructose formed is \( x \).
- At \( t = \infty \), all cane sugar has reacted, so \( x = A \).

5. **Relate the Rotations to Concentrations**:
The change in rotation can be related to the amount of sugar reacted:
- \( r_0 - r_{\infty} \) corresponds to the initial rotation minus the final rotation.
- \( r_0 - r_t \) corresponds to the initial rotation minus the rotation at time \( t \).

6. **Apply the First-Order Rate Law**:
The first-order rate equation is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{A}{A - x} \right) \]
Substituting the values:
\[ k = \frac{2.303}{t} \log \left( \frac{r_0 - r_{\infty}}{(r_0 - r_{\infty}) - (r_0 - r_t)} \right) \]
Simplifying gives:
\[ k = \frac{2.303}{t} \log \left( \frac{r_0 - r_{\infty}}{r_t - r_{\infty}} \right) \]

7. **Convert to Natural Logarithm**:
Since the options are in natural logarithm (ln), we convert the logarithm:
\[ k = \frac{1}{t} \ln \left( \frac{r_0 - r_{\infty}}{r_t - r_{\infty}} \right) \]

8. **Identify the Correct Option**:
Comparing with the provided options, we find that:
\[ k = \frac{1}{t} \ln \left( \frac{r_0 - r_{\infty}}{r_t - r_{\infty}} \right) \]
matches with **Option B**.

### Final Answer:
The correct expression for the first-order reaction is:
\[ k = \frac{1}{t} \ln \left( \frac{r_0 - r_{\infty}}{r_t - r_{\infty}} \right) \]
**Option B is the correct answer.**