Analyze the generalized rate data:
`RX + M^(ɵ) rarr` Product
`|{:("Experiment",[RX] "Substrate",[M^(ɵ)] "Attaking species","Rate"),(I,0.10 M,0.10 M,1.2 xx 10^(-4)),(II,0.20 M,0.10 M,2.4 xx 10^(-4)),(III,0.10 M,0.20 M,2.4 xx 10^(-4)),(IV,0.20 M,0.20 M,4.8 xx 10^(-4)):}|`
The value of rate constant for the give experiment data is

Options

A. `1.2 xx 10^(-2)`
B. `1.2 xx 10^(-4)`
C. `1.2 xx 10^(-3)`
D. `2.4 xx 10^(-3)`

Explanation

To solve the problem, we need to analyze the given experimental data and determine the rate constant (k) for the reaction \( RX + M^{-} \rightarrow \text{Product} \). We will follow these steps:

### Step 1: Write the Rate Law
The rate law for the reaction can be expressed as:
\[ \text{Rate} = k [RX]^x [M^{-}]^y \]
where \( x \) and \( y \) are the orders of the reaction with respect to \( RX \) and \( M^{-} \), respectively.

### Step 2: Determine the Order with Respect to \( RX \)
We will compare the rates of Experiment I and Experiment II to find the value of \( x \).

From the data:
- Experiment I: \([RX] = 0.10 \, M\), \([M^{-}] = 0.10 \, M\), Rate = \(1.2 \times 10^{-4}\)
- Experiment II: \([RX] = 0.20 \, M\), \([M^{-}] = 0.10 \, M\), Rate = \(2.4 \times 10^{-4}\)

Using the rate law:
\[ \frac{\text{Rate}_I}{\text{Rate}_{II}} = \frac{k [0.10]^x [0.10]^y}{k [0.20]^x [0.10]^y} \]
This simplifies to:
\[ \frac{1.2 \times 10^{-4}}{2.4 \times 10^{-4}} = \frac{[0.10]^x}{[0.20]^x} \]
\[ \frac{1}{2} = \left(\frac{0.10}{0.20}\right)^x \]
\[ \frac{1}{2} = \left(\frac{1}{2}\right)^x \]
From this, we find that \( x = 1 \).

### Step 3: Determine the Order with Respect to \( M^{-} \)
Next, we will compare the rates of Experiment I and Experiment III to find the value of \( y \).

From the data:
- Experiment I: Rate = \(1.2 \times 10^{-4}\)
- Experiment III: \([RX] = 0.10 \, M\), \([M^{-}] = 0.20 \, M\), Rate = \(2.4 \times 10^{-4}\)

Using the rate law:
\[ \frac{\text{Rate}_I}{\text{Rate}_{III}} = \frac{k [0.10]^x [0.10]^y}{k [0.10]^x [0.20]^y} \]
This simplifies to:
\[ \frac{1.2 \times 10^{-4}}{2.4 \times 10^{-4}} = \frac{[0.10]^y}{[0.20]^y} \]
\[ \frac{1}{2} = \left(\frac{0.10}{0.20}\right)^y \]
\[ \frac{1}{2} = \left(\frac{1}{2}\right)^y \]
From this, we find that \( y = 1 \).

### Step 4: Write the Final Rate Law
Now that we have determined the orders, we can write the rate law:
\[ \text{Rate} = k [RX]^1 [M^{-}]^1 = k [RX] [M^{-}] \]

### Step 5: Calculate the Rate Constant \( k \)
We can use any of the experiments to find \( k \). Let's use Experiment I:
\[ 1.2 \times 10^{-4} = k (0.10)(0.10) \]
\[ 1.2 \times 10^{-4} = k (0.01) \]
\[ k = \frac{1.2 \times 10^{-4}}{0.01} = 1.2 \times 10^{-2} \, \text{L mol}^{-1} \text{s}^{-1} \]

### Final Answer
The value of the rate constant \( k \) is:
\[ \boxed{1.2 \times 10^{-2} \, \text{L mol}^{-1} \text{s}^{-1}} \]