Consider the reaction, $$2A + B \rightarrow$$ 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

Step 1: Determine the Reaction Order with respect to $$B$$

• The problem states: "When concentration of $$B$$ alone was doubled, the half-life did not change."
• For a reaction where the half-life ($$t_{1/2}$$) is independent of the initial concentration of that reactant, the reaction is first-order with respect to that specific component.
• Therefore, the order with respect to $B$ ($$y$$) = $$1$$.

Step 2: Determine the Reaction Order with respect to $$A$$

• The problem states: "When the concentration of $A$ alone was doubled, the rate increased by two times."
• Since doubling the concentration ($$2^1$$) causes a direct doubling of the rate ($$2^1$$), the rate is directly proportional to the concentration of $$A$$.
• Therefore, the order with respect to $$A$$ ($$x$$) = $$1$$.

Step 3: Calculate the Overall Order of the Reaction

The rate law can be written as:

$$\text{Rate} = k[A]^1[B]^1$$

$$\text{Overall Order } (n) = x + y = 1 + 1 = 2 \text{ (Second-order reaction)}$$

Step 4: Find the Unit of the Rate Constant ($$k$$)

The general formula for the units of a rate constant is:

$$\text{Unit of } k = (\text{mol L}^{-1})^{1-n} \text{ s}^{-1}$$

Substituting $$n = 2$$ into the formula:

$$\text{Unit of } k = (\text{mol L}^{-1})^{1-2} \text{ s}^{-1} = (\text{mol L}^{-1})^{-1} \text{ s}^{-1} = \text{L mol}^{-1} \text{ s}^{-1}$$

Conclusion:

The reaction is overall second-order, meaning the rate constant is expressed in liters per mole-second.

Answer: Option A — $$\text{L mol}^{-1}\text{s}^{-1}$$