The rate law of reaction `A+2Bto`product given by
`r=K(A)^(2)(B)^(1)`. If A is taken in excess the order of the reaction
will be :-

Options

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

Answer

To determine the order of the reaction when reactant A is taken in excess, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Rate Law**: The rate law of the reaction is given as:
\[ r = K[A]^2[B]^1 \]
where \( r \) is the rate of the reaction, \( K \) is the rate constant, and \( [A] \) and \( [B] \) are the concentrations of reactants A and B, respectively.

2. **Identify Excess Reactant**: The problem states that A is taken in excess. When a reactant is in excess, its concentration does not change significantly during the reaction. Therefore, we can treat its concentration as constant.

3. **Simplify the Rate Law**: Since A is in excess, we can denote its concentration as a constant value. Let’s say:
\[ [A] = C \quad \text{(a constant)} \]
Thus, the rate law can be simplified to:
\[ r = K[C]^2[B]^1 \]
Here, \( K[C]^2 \) can be treated as a new constant, say \( K' \):
\[ K' = K[C]^2 \]
Therefore, the rate law now looks like:
\[ r = K'[B]^1 \]

4. **Determine the Order of the Reaction**: The order of a reaction is the sum of the powers of the concentration terms in the rate law. In the simplified rate law \( r = K'[B]^1 \), the only concentration term is \( [B] \) raised to the power of 1. Thus, the order of the reaction is:
\[ \text{Order} = 1 \]

5. **Conclusion**: When A is taken in excess, the reaction behaves as a first-order reaction with respect to B. Therefore, the overall order of the reaction is 1.

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