Consider a reaction, `2A + B rarr` Products
When concentration of `B` alone was doubled, the half-life did not change. When the concentration of `A` alone was doubled, the rate incresed by two times. The unit of rate constant for this reaction is :

Options

A. `L "mol"^(-1) s^(-1)`
B. No unit
C. `"mol" L^(-1) s^(-1)`
D. `s^(-1)`

Explanation

To solve the problem, we need to analyze the given information about the reaction and determine the order of the reaction with respect to each reactant. Then, we can calculate the unit of the rate constant based on the overall order of the reaction.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is given as:
\[ 2A + B \rightarrow \text{Products} \]

2. **Analyze the Effect of Concentration of B**:
- When the concentration of \( B \) is doubled, the half-life does not change.
- This indicates that the reaction is **first order** with respect to \( B \) because the half-life of a first-order reaction is independent of the concentration of the reactant.

3. **Analyze the Effect of Concentration of A**:
- When the concentration of \( A \) is doubled, the rate of the reaction increases by two times.
- This suggests that the reaction is **first order** with respect to \( A \) as well.

4. **Determine the Rate Law**:
The rate law for the reaction can be expressed as:
\[ \text{Rate} = k[A]^x[B]^y \]
From our analysis:
- Since the reaction is first order with respect to \( A \) and first order with respect to \( B \), we have \( x = 1 \) and \( y = 1 \).
- Therefore, the rate law becomes:
\[ \text{Rate} = k[A]^1[B]^1 = k[A][B] \]

5. **Determine the Overall Order of the Reaction**:
- The overall order of the reaction is the sum of the orders with respect to each reactant:
\[ \text{Overall Order} = x + y = 1 + 1 = 2 \]

6. **Determine the Unit of the Rate Constant (k)**:
- The unit of rate is typically expressed as concentration per time:
\[ \text{Unit of Rate} = \text{mol L}^{-1} \text{s}^{-1} \]
- For a second-order reaction, the unit of the rate constant \( k \) can be derived from the rate law:
\[ \text{Rate} = k[A][B] \implies k = \frac{\text{Rate}}{[A][B]} \]
- Substituting the units:
\[ k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})(\text{mol L}^{-1})} = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^2 \text{L}^{-2}} = \text{L mol}^{-1} \text{s}^{-1} \]

### Final Answer:
The unit of the rate constant \( k \) for this reaction is:
\[ \text{L mol}^{-1} \text{s}^{-1} \]