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 increased 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 \(2A + B \rightarrow \text{Products}\) and derive the order of the reaction with respect to the reactants, which will help us find the unit of the rate constant \(k\).

### Step-by-Step Solution:

1. **Understanding the Reaction**:
The reaction is given as \(2A + B \rightarrow \text{Products}\). We need to determine the order of the reaction with respect to \(A\) and \(B\).

2. **Analyzing the Effect of Concentration of \(B\)**:
- When the concentration of \(B\) is doubled, the half-life of the reaction does not change.
- This implies that the half-life is independent of the concentration of \(B\), which suggests that the order of the reaction with respect to \(B\) is zero (0).
- Therefore, we can conclude that the reaction rate does not depend on \(B\).

3. **Analyzing the Effect of Concentration of \(A\)**:
- When the concentration of \(A\) is doubled, the rate of the reaction increases by two times.
- This indicates that the order of the reaction with respect to \(A\) is one (1).

4. **Writing the Rate Law**:
- Based on the orders we have determined, we can write the rate law for the reaction:
\[ \text{Rate} = k [A]^1 [B]^0 = k [A] \]
- Here, \(k\) is the rate constant, \([A]\) is the concentration of \(A\), and \([B]\) does not appear in the rate law since its order is zero.

5. **Determining the Overall Order of the Reaction**:
- The overall order of the reaction is the sum of the orders with respect to each reactant:
\[ \text{Overall Order} = 1 + 0 = 1 \]

6. **Finding the Unit of the Rate Constant \(k\)**:
- The unit of rate for a reaction is typically expressed as mol L\(^{-1}\) s\(^{-1}\).
- For a first-order reaction (which we have determined), the unit of the rate constant \(k\) can be derived from the rate law:
\[ \text{Rate} = k [A] \implies k = \frac{\text{Rate}}{[A]} = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1} \]
- Therefore, the unit of the rate constant \(k\) for this reaction is:
\[ \text{Unit of } k = \text{L}^{-1} \text{mol}^{-1} \text{s}^{-1} \]

### Final Answer:
The unit of the rate constant \(k\) for the reaction \(2A + B \rightarrow \text{Products}\) is:
\[ \text{L}^{-1} \text{mol}^{-1} \text{s}^{-1} \]