The rate of the reaction
`3A + 2B rarr` Products
is given by the rate expresison: Rate `= k[A][B]^(2)`
If `A` is taken in excess, the order of the reaction would be

Options

A. `3`
B. `2`
C. `1`
D. `5`

Explanation

To determine the order of the reaction given the rate expression and the condition that reactant A is taken in excess, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Rate Expression**:
The rate of the reaction is given as:
\[ \text{Rate} = k[A][B]^2 \]
Here, \(k\) is the rate constant, \([A]\) is the concentration of reactant A, and \([B]\) is the concentration of reactant B.

2. **Understand the Concept of Excess Reactant**:
When one reactant is in excess, its concentration does not change significantly during the reaction. Therefore, it can be treated as a constant for the purpose of calculating the rate.

3. **Modify the Rate Expression**:
Since A is in excess, we can consider its concentration \([A]\) to be constant. Let’s denote this constant concentration as \(k'\), where:
\[ k' = k[A] \]
Thus, the rate expression can be rewritten as:
\[ \text{Rate} = k'[B]^2 \]

4. **Determine the Order of the Reaction**:
The order of a reaction is determined by the sum of the powers of the concentration terms in the rate expression. In the modified rate expression \( \text{Rate} = k'[B]^2 \), the only variable concentration is \([B]\) raised to the power of 2.

5. **Conclusion**:
Therefore, the order of the reaction with respect to reactant B is 2. Since A is in excess and treated as a constant, it does not contribute to the order of the reaction.

### Final Answer:
The order of the reaction is **2**.

---