For a reaction, `2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g)` rate of reaction is:

Options

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

Answer

To solve the question regarding the rate of the reaction \(2N_{2}O_{5}(g) \rightarrow 4NO_{2}(g) + O_{2}(g)\), we need to express the rate of the reaction in terms of the change in concentration of the reactants and products.

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
\[ 2N_{2}O_{5}(g) \rightarrow 4NO_{2}(g) + O_{2}(g) \]

2. **Define the rate of reaction**:
The rate of a reaction can be expressed in terms of the change in concentration of the reactants and products. For the given reaction, we can express the rate as:
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_{2}O_{5}]}{dt} = \frac{1}{4} \frac{d[NO_{2}]}{dt} = \frac{1}{1} \frac{d[O_{2}]}{dt} \]

3. **Explanation of the coefficients**:
- The coefficient of \(N_{2}O_{5}\) is 2, hence the negative sign and the factor of \(\frac{1}{2}\) in front of \(\frac{d[N_{2}O_{5}]}{dt}\).
- The coefficient of \(NO_{2}\) is 4, hence the factor of \(\frac{1}{4}\) in front of \(\frac{d[NO_{2}]}{dt}\).
- The coefficient of \(O_{2}\) is 1, hence the factor of \(\frac{1}{1}\) in front of \(\frac{d[O_{2}]}{dt}\).

4. **Final expression for the rate**:
Therefore, the rate of reaction can be summarized as:
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_{2}O_{5}]}{dt} = \frac{1}{4} \frac{d[NO_{2}]}{dt} = \frac{1}{1} \frac{d[O_{2}]}{dt} \]

5. **Identify the correct option**:
From the options provided, the correct expression for the rate of reaction is:
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_{2}O_{5}]}{dt} = \frac{1}{4} \frac{d[NO_{2}]}{dt} = \frac{d[O_{2}]}{dt} \]
This corresponds to option B.