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

### Step 1: Write the Rate Expressions
For the reaction, we can express the rates of change of the concentrations of the reactants and products. The reaction can be broken down into two parts:
1. The forward reaction: \(A_2 \rightarrow 2A\) with a rate constant \(k_1\)
2. The reverse reaction: \(2A \rightarrow A_2\) with a rate constant \(k_{-1}\)

The rate of the forward reaction is given by:
\[ \text{Rate}_{\text{forward}} = k_1 [A_2] \]
The rate of the reverse reaction is given by:
\[ \text{Rate}_{\text{reverse}} = k_{-1} [A]^2 \]

### Step 2: Set Up the Rate of Change for Each Species
The rate of change of the concentration of \(A_2\) and \(A\) can be expressed as:
\[ -\frac{d[A_2]}{dt} = k_1 [A_2] \quad \text{(since } A_2 \text{ is decreasing)} \]
\[ \frac{d[A]}{dt} = 2k_1 [A_2] - 2k_{-1} [A]^2 \quad \text{(since } A \text{ is increasing)} \]

### Step 3: Relate the Changes in Concentration
From the stoichiometry of the reaction, we know that:
\[ \frac{d[A]}{dt} = -2 \frac{d[A_2]}{dt} \]
Substituting the expression for \(\frac{d[A_2]}{dt}\):
\[ \frac{d[A]}{dt} = -2 \left(-\frac{d[A_2]}{dt}\right) = 2k_1 [A_2] - 2k_{-1} [A]^2 \]

### Step 4: Combine and Rearrange
We can express \(\frac{d[A]}{dt}\) in terms of the concentrations and rate constants:
\[ \frac{d[A]}{dt} = 2k_1 [A_2] - 2k_{-1} [A]^2 \]

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