For the reaction ` 2 A + B to A_(2) B`, the rate law given is

Options

A. `k[2A][B]`
B. `k[A]^(2) [B]`
C. `k[A] [B]^(3)`
D. `k [A]^(2) [B]`

Explanation

To determine the rate law for the reaction \(2 A + B \rightarrow A_2B\), we can follow these steps:

### Step 1: Identify the Reaction and Its Stoichiometry
The given reaction is \(2 A + B \rightarrow A_2B\). The stoichiometry indicates that 2 moles of reactant A and 1 mole of reactant B are consumed to produce 1 mole of product \(A_2B\).

### Step 2: Understand the Concept of Rate Law
The rate law expresses the rate of a chemical reaction in terms of the concentration of the reactants. It is generally written as:
\[ \text{Rate} = k [A]^m [B]^n \]
where \(k\) is the rate constant, and \(m\) and \(n\) are the orders of the reaction with respect to reactants A and B, respectively.

### Step 3: Determine the Order of the Reaction
For elementary reactions (which this reaction is assumed to be), the order of the reaction can be directly inferred from the stoichiometry of the reactants.
- The coefficient of A is 2, so the order with respect to A is 2.
- The coefficient of B is 1, so the order with respect to B is 1.

### Step 4: Write the Rate Law Expression
Based on the stoichiometry and the orders determined:
\[ \text{Rate} = k [A]^2 [B]^1 \]

### Step 5: Finalize the Rate Law
Thus, the final rate law expression for the reaction \(2 A + B \rightarrow A_2B\) is:
\[ \text{Rate} = k [A]^2 [B] \]