In a reaction `2A + B rarr A_(2)B`, the reactant `A` will disappear at

Options

A. Half the rate that `B` will decrease
B. The same rate that `B` will decrease
C. Twice the rate that `B` will decrease
D. The same rate that `A_(2)B` will form

Explanation

To determine how the reactant A will disappear in the reaction \(2A + B \rightarrow A_2B\), we can analyze the stoichiometry of the reaction and the rates of change of the reactants and products.

### Step-by-Step Solution:

1. **Write the Reaction:**
The given reaction is:
\[ 2A + B \rightarrow A_2B \]

2. **Define the Rate of Reaction:**
The rate of reaction can be expressed in terms of the change in concentration of the reactants and products. For this reaction, we can write:
\[ -\frac{1}{2} \frac{d[A]}{dt} = -\frac{d[B]}{dt} = \frac{d[A_2B]}{dt} \]

3. **Relate the Rates:**
From the stoichiometry of the reaction:
- For every 2 moles of A that react, 1 mole of B reacts.
- Therefore, the rate at which A disappears is twice the rate at which B disappears.

From the equation:
\[ -\frac{d[A]}{dt} = 2 \left(-\frac{d[B]}{dt}\right) \]

4. **Conclusion on the Disappearance of A:**
This means that the rate of disappearance of A is twice the rate of disappearance of B. Hence, if we denote the rate of disappearance of B as \(r_B\), then the rate of disappearance of A, \(r_A\), is:
\[ r_A = 2r_B \]

5. **Final Answer:**
Therefore, the correct option regarding how the reactant A will disappear is that A will disappear at twice the rate at which B disappears.