The rate of reaction `A+2Bto` Products is given by `-(d[A])/(dt)=k[A][B]^(2)`.If B is present in large excess, the order of reaction is

Options

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

Answer

To solve the problem, we need to analyze the given rate law and the conditions provided.

### Step-by-Step Solution:

1. **Understand the Rate Law**: The rate of the reaction is given by the equation:
\[ -\frac{d[A]}{dt} = k[A][B]^2 \]
where \( k \) is the rate constant, \([A]\) is the concentration of reactant A, and \([B]\) is the concentration of reactant B.

2. **Identify the Condition**: The problem states that B is present in large excess. This means that the concentration of B does not change significantly during the reaction.

3. **Simplify the Rate Law**: Since B is in large excess, we can treat its concentration as a constant. Let’s denote the concentration of B as \( [B] = C \) (where \( C \) is a constant). Therefore, we can rewrite the rate law as:
\[ -\frac{d[A]}{dt} = k[A][B]^2 = k[A]C^2 \]
Here, \( kC^2 \) is a new constant (let's call it \( k' \)), so we can express the rate as:
\[ -\frac{d[A]}{dt} = k'[A] \]

4. **Determine the Order of Reaction**: The rate law now is:
\[ -\frac{d[A]}{dt} = k'[A] \]
This indicates that the reaction is first order with respect to A, because the rate depends linearly on the concentration of A.

5. **Calculate the Overall Order**: The overall order of the reaction is the sum of the orders with respect to each reactant. Since we have:
- First order with respect to A: 1
- Zero order with respect to B (because its concentration is constant and does not affect the rate): 0

Therefore, the overall order of the reaction is:
\[ 1 + 0 = 1 \]

### Final Answer:
The order of the reaction is **1**.