The rate of formation of a dimer in a second order dimerisation reaction is `9.1 xx 10^(-6) mol L^(-1)s^(1)` at 0.01 mol `L^(-1)` monomer concentration. Calculate the rate consant for the reaction.

Options

A. `9.1 xx 10^(-2) L mol^(-1) s^(-1)`
B. `9.1 xx 10^(-6) mol^(-1) s^(-1)`
C. `3 xx 10^(-4) L mol^(-1) s^(-1)`
D. `27.3 xx 10^(-2) L mol^(-1) s^(-1)`

Explanation

To solve the problem, we need to find the rate constant \( k \) for a second-order dimerization reaction. The reaction can be represented as:

\[ 2A \rightarrow A_2 \]

### Step 1: Write the rate law for the reaction
For a second-order reaction, the rate law can be expressed as:

\[ \text{Rate} = k [A]^2 \]

where:
- \( k \) is the rate constant,
- \( [A] \) is the concentration of the monomer.

### Step 2: Substitute the given values into the rate law
We are given:
- Rate = \( 9.1 \times 10^{-6} \, \text{mol L}^{-1} \text{s}^{-1} \)
- \( [A] = 0.01 \, \text{mol L}^{-1} \)

Substituting these values into the rate law gives:

\[ 9.1 \times 10^{-6} = k (0.01)^2 \]

### Step 3: Calculate \( (0.01)^2 \)
Calculating \( (0.01)^2 \):

\[ (0.01)^2 = 0.0001 \, \text{mol}^2 \text{L}^{-2} \]

### Step 4: Rearrange the equation to solve for \( k \)
Now, we can rearrange the equation to solve for \( k \):

\[ k = \frac{9.1 \times 10^{-6}}{0.0001} \]

### Step 5: Perform the division
Now, we perform the division:

\[ k = \frac{9.1 \times 10^{-6}}{1 \times 10^{-4}} = 9.1 \times 10^{-2} \, \text{L mol}^{-1} \text{s}^{-1} \]

### Final Answer
Thus, the rate constant \( k \) for the reaction is:

\[ k = 9.1 \times 10^{-2} \, \text{L mol}^{-1} \text{s}^{-1} \]

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