For the reaction `2N_(2)O_(5 )rarrNO_(2)+O_(2)` rate of reaction is :

Options

A. `(1)/(2)(d)/(dt)[N_(2)O_(5)]`
B. `2(d)/(dt)[N_(2)O_(5)]`
C. `(1)/(4)(d)/(dt)[NO_(2)]`
D. `4(d)/(dt)[NO_(2)]`

Explanation

To solve the problem regarding the rate of the reaction \(2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2}\), we need to follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ 2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2} \]

### Step 2: Identify the stoichiometric coefficients
From the balanced equation, we can identify the stoichiometric coefficients:
- For \(N_{2}O_{5}\), the coefficient is 2.
- For \(NO_{2}\), the coefficient is 4.
- For \(O_{2}\), the coefficient is 1.

### Step 3: Write the rate of reaction expression
The rate of reaction can be expressed in terms of the change in concentration of the reactants and products. The general form is:
\[ \text{Rate} = -\frac{1}{\text{stoichiometric coefficient}} \frac{d[\text{Reactant}]}{dt} = \frac{1}{\text{stoichiometric coefficient}} \frac{d[\text{Product}]}{dt} \]

For the reactant \(N_{2}O_{5}\):
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_{2}O_{5}]}{dt} \]

For the product \(NO_{2}\):
\[ \text{Rate} = \frac{1}{4} \frac{d[NO_{2}]}{dt} \]

For the product \(O_{2}\):
\[ \text{Rate} = \frac{1}{1} \frac{d[O_{2}]}{dt} \]

### Step 4: Combine the expressions
Since all expressions represent the same rate, we can equate them:
\[ -\frac{1}{2} \frac{d[N_{2}O_{5}]}{dt} = \frac{1}{4} \frac{d[NO_{2}]}{dt} = \frac{d[O_{2}]}{dt} \]

### Step 5: Identify the correct option
Now, we can analyze the options provided in the question:
1. \(-\frac{1}{2} \frac{d[N_{2}O_{5}]}{dt}\) (Incorrect, as it lacks the negative sign)
2. \(2 \frac{d[N_{2}O_{5}]}{dt}\) (Incorrect, wrong coefficient)
3. \(\frac{1}{4} \frac{d[NO_{2}]}{dt}\) (Correct, matches the derived expression)
4. \(4 \frac{d[NO_{2}]}{dt}\) (Incorrect, wrong coefficient)

Thus, the correct answer is option 3: \(\frac{1}{4} \frac{d[NO_{2}]}{dt}\).