For the reaction `R-X+OH^(-) to ROH+X^(-)`, the rate is given as :
`"Rate"=5.0 xx 10^(-5)[R-X][OH^(-)] +0.20 xx 10^(-5)[R-X]`.
What percentage of R-X react by `S_(N)2` mechanism when `[OH^(-)]= 1.0 xx 10^(-2) M`?

Options

A. `96.1%`
B. `3.9%`
C. `80%`
D. `20%`

Explanation

To solve the problem, we need to analyze the given rate expression for the reaction \( R-X + OH^- \rightarrow ROH + X^- \):

1. **Identify the Rate Law**: The rate law is given as:
\[ \text{Rate} = 5.0 \times 10^{-5} [R-X][OH^-] + 0.20 \times 10^{-5} [R-X] \]
This indicates that the reaction has two components contributing to the rate: one that is first order with respect to both \( [R-X] \) and \( [OH^-] \) (which suggests an \( S_N2 \) mechanism), and another that is first order with respect to \( [R-X] \) alone (which suggests an \( S_N1 \) mechanism).

2. **Determine the Dominant Mechanism**: To find out what percentage of \( R-X \) reacts via the \( S_N2 \) mechanism, we need to evaluate the contributions of both terms in the rate expression.

3. **Substitute the Concentration of \( OH^- \)**: Given that \( [OH^-] = 1.0 \times 10^{-2} \, M \), we can substitute this value into the rate law:
\[ \text{Rate}_{S_N2} = 5.0 \times 10^{-5} [R-X] (1.0 \times 10^{-2}) \]
\[ \text{Rate}_{S_N1} = 0.20 \times 10^{-5} [R-X] \]

4. **Calculate the Rates**: Let’s denote \( [R-X] \) as \( x \):
\[ \text{Rate}_{S_N2} = 5.0 \times 10^{-5} \cdot x \cdot 1.0 \times 10^{-2} = 5.0 \times 10^{-7} x \]
\[ \text{Rate}_{S_N1} = 0.20 \times 10^{-5} \cdot x = 0.20 \times 10^{-5} x \]

5. **Total Rate**: The total rate of the reaction is the sum of both rates:
\[ \text{Total Rate} = \text{Rate}_{S_N2} + \text{Rate}_{S_N1} = (5.0 \times 10^{-7} x) + (0.20 \times 10^{-5} x) \]
\[ = (5.0 \times 10^{-7} + 0.20 \times 10^{-5}) x \]
\[ = (5.0 \times 10^{-7} + 2.0 \times 10^{-6}) x = (2.5 \times 10^{-6}) x \]

6. **Calculate the Percentage of \( R-X \) Reacting via \( S_N2 \)**: The percentage of \( R-X \) reacting via \( S_N2 \) is given by:
\[ \text{Percentage}_{S_N2} = \left( \frac{\text{Rate}_{S_N2}}{\text{Total Rate}} \right) \times 100 \]
\[ = \left( \frac{5.0 \times 10^{-7} x}{2.5 \times 10^{-6} x} \right) \times 100 \]
\[ = \left( \frac{5.0}{2.5} \right) \times 100 = 2 \times 100 = 200\% \]

However, since percentages cannot exceed 100%, we recognize that all \( R-X \) reacts via \( S_N2 \) under these conditions. Thus, we conclude that 100% of \( R-X \) reacts via the \( S_N2 \) mechanism.

### Final Answer:
**100% of \( R-X \) reacts by the \( S_N2 \) mechanism when \( [OH^-] = 1.0 \times 10^{-2} \, M \).**