In the reaction `2A + B to A_2B`, if the concentration of A is doubled and of B is halved, then the rate of the reaction will:

Options

A. Increase by four times
B. Decrease by two times
C. Increase by two times
D. Remain the same

Answer

To solve the problem, we need to analyze how the changes in concentrations of reactants affect the rate of the reaction. The reaction given is:

\[ 2A + B \rightarrow A_2B \]

### Step 1: Write the Rate Law Expression
The rate of a reaction can be expressed using the rate law, which depends on the concentrations of the reactants raised to the power of their coefficients in the balanced equation. For this reaction, the rate law can be written as:

\[ R = k [A]^2 [B]^1 \]

where:
- \( R \) is the rate of the reaction,
- \( k \) is the rate constant,
- \([A]\) is the concentration of reactant A,
- \([B]\) is the concentration of reactant B.

### Step 2: Determine Initial Concentrations
Let’s denote the initial concentrations of A and B as \([A]\) and \([B]\).

### Step 3: Apply Changes to Concentrations
According to the problem:
- The concentration of A is doubled: \([A] \rightarrow 2[A]\)
- The concentration of B is halved: \([B] \rightarrow \frac{[B]}{2}\)

### Step 4: Write the New Rate Expression
Now, substituting the new concentrations into the rate law, we get the new rate \( R' \):

\[ R' = k (2[A])^2 \left(\frac{[B]}{2}\right)^1 \]

### Step 5: Simplify the New Rate Expression
Now, let's simplify \( R' \):

\[ R' = k \cdot (2^2 [A]^2) \cdot \left(\frac{[B]}{2}\right) \]
\[ R' = k \cdot 4[A]^2 \cdot \frac{[B]}{2} \]
\[ R' = k \cdot 2[A]^2 [B] \]

### Step 6: Relate New Rate to Initial Rate
Now, we can relate \( R' \) to the initial rate \( R \):

\[ R = k [A]^2 [B] \]

Thus, we can express \( R' \) in terms of \( R \):

\[ R' = 2 \cdot R \]

### Conclusion
The new rate of the reaction \( R' \) is twice the initial rate \( R \). Therefore, the rate of the reaction will **increase by 2 times**.