For the reaction `R-X+OH^(-)toR-OH+X^(-1)`, the rate of reaction is given as, rate `=4.7xx10^(-5)[R-X][OH^(-)]+0.24xx10^(-5)[RX]`. What percentage of R-X react by `S_(N)2` mechanism when `[OH^(-)]=0.001` molar?

Options

A. 1.9
B. 4.7
C. 2.8
D. 4.9

Answer

To solve the problem, we need to analyze the given rate equation and determine the percentage of the reaction that proceeds via the \( S_N2 \) mechanism.

### Step-by-Step Solution:

1. **Identify the Rate Law Components**:
The rate of reaction is given as:
\[ \text{rate} = 4.7 \times 10^{-5} [R-X][OH^-] + 0.24 \times 10^{-5} [R-X] \]
Here, the first term corresponds to the \( S_N2 \) mechanism (second order) and the second term corresponds to the \( S_N1 \) mechanism (first order).

2. **Separate the Rate Contributions**:
- The rate contribution from the \( S_N2 \) mechanism is:
\[ \text{Rate}_{S_N2} = 4.7 \times 10^{-5} [R-X][OH^-] \]
- The rate contribution from the \( S_N1 \) mechanism is:
\[ \text{Rate}_{S_N1} = 0.24 \times 10^{-5} [R-X] \]

3. **Express the Total Rate**:
The total rate of the reaction can be expressed as:
\[ \text{Total Rate} = \text{Rate}_{S_N2} + \text{Rate}_{S_N1} \]

4. **Substitute the Given Hydroxide Ion Concentration**:
We are given that \([OH^-] = 0.001 \, \text{M}\). Substitute this value into the \( S_N2 \) rate expression:
\[ \text{Rate}_{S_N2} = 4.7 \times 10^{-5} [R-X] (0.001) \]
\[ \text{Rate}_{S_N2} = 4.7 \times 10^{-8} [R-X] \]

5. **Calculate the Total Rate**:
The total rate can now be expressed as:
\[ \text{Total Rate} = 4.7 \times 10^{-8} [R-X] + 0.24 \times 10^{-5} [R-X] \]
Convert \( 0.24 \times 10^{-5} \) to the same power of ten:
\[ 0.24 \times 10^{-5} = 2.4 \times 10^{-6} \]
Therefore, the total rate becomes:
\[ \text{Total Rate} = (4.7 \times 10^{-8} + 2.4 \times 10^{-6}) [R-X] \]
\[ \text{Total Rate} = (2.4 \times 10^{-6} + 0.000047) [R-X] \approx 2.4 \times 10^{-6} [R-X] \]

6. **Calculate the Percentage of \( S_N2 \) Mechanism**:
The percentage of the reaction that proceeds via the \( S_N2 \) mechanism is given by:
\[ \text{Percentage}_{S_N2} = \left( \frac{\text{Rate}_{S_N2}}{\text{Total Rate}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage}_{S_N2} = \left( \frac{4.7 \times 10^{-8} [R-X]}{(2.4 \times 10^{-6} [R-X])} \right) \times 100 \]
The \([R-X]\) cancels out:
\[ \text{Percentage}_{S_N2} = \left( \frac{4.7 \times 10^{-8}}{2.4 \times 10^{-6}} \right) \times 100 \]
\[ = \left( \frac{4.7}{2.4} \times 10^{-2} \right) \times 100 \]
\[ = \left( 1.9583 \times 10^{-2} \right) \times 100 \approx 1.92\% \]

### Final Answer:
The percentage of \( R-X \) that reacts by the \( S_N2 \) mechanism is approximately **1.92%**.