Hydrolyiss of an alkyl halide `(RX)` by dilute alkali `[OH]^(ɵ)` takes place ismultaneously by `SN^(2)` and `SN^(1)` pathways. A plot of `-(1)/([RX]) (d[R-X])/(dt)` vs `[OH]^(ɵ)` is a straight line of the slope equal to `2 xx 10^(3) mol^(-1) L h^(-1)` and intercept equal to `1 xx 10^(2) h^(-1)`. Calculate the initial rate `("mole" L^(-1) min^(-1))` of consumption of `RX` when the reaction is carried out taking `mol L^(-1)` of `RX` and `0.1 mol L^(-1)` of `[OH]^(ɵ)` ions.

Explanation

To solve the problem, we need to calculate the initial rate of consumption of the alkyl halide \( RX \) when the concentrations of \( RX \) and \( OH^- \) ions are given. The reaction proceeds via both \( SN^2 \) and \( SN^1 \) pathways.

### Step-by-Step Solution:

1. **Write the Rate Expressions**:
- For the \( SN^2 \) pathway:
\[ \text{Rate}_{SN^2} = k_2 [RX][OH^-] \]
- For the \( SN^1 \) pathway:
\[ \text{Rate}_{SN^1} = k_1 [RX] \]

2. **Combine the Rate Expressions**:
The overall rate of consumption of \( RX \) can be expressed as:
\[ -\frac{d[R-X]}{dt} = k_2 [RX][OH^-] + k_1 [RX] \]
This can be factored as:
\[ -\frac{d[R-X]}{dt} = [RX] \left( k_2 [OH^-] + k_1 \right) \]

3. **Divide by [RX]**:
To express the rate in terms of \( \frac{1}{[RX]} \):
\[ -\frac{1}{[RX]} \frac{d[R-X]}{dt} = k_2 [OH^-] + k_1 \]

4. **Identify the Slope and Intercept**:
From the problem statement, we know:
- Slope \( m = 2 \times 10^3 \, \text{mol}^{-1} \text{L} \text{h}^{-1} \)
- Intercept \( c = 1 \times 10^2 \, \text{h}^{-1} \)

Therefore, we can write:
\[ -\frac{1}{[RX]} \frac{d[R-X]}{dt} = 2 \times 10^3 [OH^-] + 1 \times 10^2 \]

5. **Substitute the Known Concentrations**:
Given:
- \( [RX] = 1.0 \, \text{mol L}^{-1} \)
- \( [OH^-] = 0.1 \, \text{mol L}^{-1} \)

Substitute these values into the equation:
\[ -\frac{1}{1.0} \frac{d[R-X]}{dt} = 2 \times 10^3 \times 0.1 + 1 \times 10^2 \]

6. **Calculate the Rate**:
\[ -\frac{d[R-X]}{dt} = 2 \times 10^3 \times 0.1 + 1 \times 10^2 \]
\[ -\frac{d[R-X]}{dt} = 200 + 100 = 300 \, \text{mol L}^{-1} \text{h}^{-1} \]

7. **Convert to Minutes**:
Since the question asks for the rate in \( \text{mol L}^{-1} \text{min}^{-1} \):
\[ -\frac{d[R-X]}{dt} = \frac{300 \, \text{mol L}^{-1} \text{h}^{-1}}{60 \, \text{min h}^{-1}} = 5 \, \text{mol L}^{-1} \text{min}^{-1} \]

### Final Answer:
The initial rate of consumption of \( RX \) is \( 5 \, \text{mol L}^{-1} \text{min}^{-1} \).