For an elementary chemical reaction, `A_(2) underset(k_(-1))overset(k_(1))(hArr) 2A`, the expression for `(d[A])/(dt)` is

Options

A. `k_(1) [A_(2)]-k_(1)[A]^(2)`
B. `2k_(1) [A_(2)]-k_(1) [A]^(2)`
C. `2k_(1)[A_(2)]-2k_(1) [A]^(2)`
D. `k_(1) [A_(2)]+k_(-1)[A]^(2)`

Explanation

To derive the expression for \(\frac{d[A]}{dt}\) for the elementary reaction \(A_2 \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} 2A\), we will follow these steps:

### Step 1: Write the forward and backward reactions
The forward reaction is:
\[ A_2 \xrightarrow{k_1} 2A \]
The backward reaction is:
\[ 2A \xrightarrow{k_{-1}} A_2 \]

### Step 2: Express the rates of the reactions
For the forward reaction, the rate of formation of \(A\) is given by:
\[ \text{Rate}_{\text{forward}} = k_1 [A_2] \]

Since 2 moles of \(A\) are produced for every mole of \(A_2\) that reacts, we express the change in concentration of \(A\) as:
\[ \frac{d[A]}{dt} = 2k_1 [A_2] \]

For the backward reaction, the rate of disappearance of \(A\) is given by:
\[ \text{Rate}_{\text{backward}} = k_{-1} [A]^2 \]

Since 2 moles of \(A\) are consumed to produce 1 mole of \(A_2\), we express the change in concentration of \(A\) as:
\[ -\frac{d[A]}{dt} = k_{-1} [A]^2 \]

### Step 3: Combine the rates
To find the overall rate of change of \(A\), we combine the rates from the forward and backward reactions:
\[ \frac{d[A]}{dt} = \text{Rate}_{\text{forward}} - \text{Rate}_{\text{backward}} \]
Substituting the expressions we derived:
\[ \frac{d[A]}{dt} = 2k_1 [A_2] - k_{-1} [A]^2 \]

### Final Expression
Thus, the expression for \(\frac{d[A]}{dt}\) is:
\[ \frac{d[A]}{dt} = 2k_1 [A_2] - k_{-1} [A]^2 \]

---