The rate law of the reaction `A+2Brarr` Product is given by `(d("Product"))/(dt)= k[A]^(2)[B]`. A is taken in large excess, the order of the reqaction will be

Options

A. 3
B. 1
C. 2
D. 0

Explanation

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

### Step-by-Step Solution:

1. **Understand the Given Rate Law**:
The rate law for the reaction is given as:
\[ \frac{d(\text{Product})}{dt} = k[A]^2[B] \]

2. **Identify the Condition of Excess Reactant**:
Since A is taken in large excess, its concentration will remain relatively constant throughout the reaction. This means that changes in the concentration of A will not significantly affect the rate of reaction.

3. **Simplify the Rate Expression**:
When A is in large excess, we can treat its concentration as a constant. Let's denote this constant concentration of A as \( [A]_0 \). The rate law can then be simplified:
\[ \frac{d(\text{Product})}{dt} = k[A]^2[B] \approx k' [B] \]
where \( k' = k[A]_0^2 \) is a new rate constant that incorporates the constant concentration of A.

4. **Determine the Order of the Reaction**:
In the simplified rate expression \( \frac{d(\text{Product})}{dt} = k' [B] \), the reaction rate now depends only on the concentration of B. The order of the reaction with respect to B is 1, since the concentration of B is raised to the first power.

5. **Conclude the Overall Order of the Reaction**:
Since the reaction is first order with respect to B and zero order with respect to A (due to its excess), the overall order of the reaction is:
\[ \text{Overall Order} = 0 + 1 = 1 \]

### Final Answer:
The order of the reaction when A is taken in large excess is **1**.