A + 2B `rarr C, the rate equation for this reaction is given as
Rate = k[A] [B].
If the concentration of A is kept the same but that of B is doubled what will happen to the rate itelf?

Options

A. halved
B. the same
C. doubled
D. quadrupled

Explanation

To solve the problem, we need to analyze the effect of changing the concentration of reactant B on the rate of the reaction given by the rate equation:

**Rate = k[A][B]**

Where:
- [A] is the concentration of reactant A,
- [B] is the concentration of reactant B,
- k is the rate constant.

### Step-by-Step Solution:

1. **Identify the initial rate expression**:
The rate of the reaction is given by:
\[ \text{Rate} = k[A][B] \]

2. **Define the initial conditions**:
Let’s denote the initial concentration of A as \([A]\) and the initial concentration of B as \([B]\). The initial rate of the reaction can be expressed as:
\[ R = k[A][B] \]

3. **Change the concentration of B**:
According to the problem, the concentration of A is kept constant, while the concentration of B is doubled. Therefore, the new concentration of B becomes:
\[ [B]_{\text{new}} = 2[B] \]

4. **Write the new rate expression**:
The new rate of the reaction, denoted as \(R'\), can be expressed as:
\[ R' = k[A][B]_{\text{new}} = k[A](2[B]) = 2k[A][B] \]

5. **Relate the new rate to the initial rate**:
Now we can relate the new rate \(R'\) to the initial rate \(R\):
\[ R' = 2R \]

6. **Conclusion**:
This shows that when the concentration of B is doubled while keeping the concentration of A constant, the rate of the reaction also doubles. Therefore, the final answer is:
\[ \text{The rate is doubled.} \]

### Final Answer:
The rate of the reaction will be doubled.