`2N_(2)O_(5)to4NO_(2)+O_(2)`. The rate of reaction in terms of `N_(2)O_(5)` will be

Options

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

Explanation

To express the rate of the reaction \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \) in terms of \( N_2O_5 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction given is:
\[ 2N_2O_5 \rightarrow 4NO_2 + O_2 \]

2. **Understand Rate of Reaction**:
The rate of a reaction can be expressed in terms of the change in concentration of reactants and products over time. For reactants, we denote the rate of disappearance, and for products, we denote the rate of appearance.

3. **Write Rate Expressions**:
The rate of disappearance of \( N_2O_5 \) can be expressed as:
\[ -\frac{d[N_2O_5]}{dt} \]
The negative sign indicates that the concentration of \( N_2O_5 \) is decreasing.

4. **Use Stoichiometry**:
According to the stoichiometry of the reaction:
- For every 2 moles of \( N_2O_5 \) that react, 4 moles of \( NO_2 \) are produced and 1 mole of \( O_2 \) is produced.
- Therefore, we can relate the rates of change of concentrations using their stoichiometric coefficients:
\[ -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{1}{1} \frac{d[O_2]}{dt} \]

5. **Express Rate in Terms of \( N_2O_5 \)**:
To express the rate in terms of \( N_2O_5 \), we can rearrange the equation:
\[ \frac{d[NO_2]}{dt} = -2 \frac{d[N_2O_5]}{dt} \]
\[ \frac{d[O_2]}{dt} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} \]

6. **Final Rate Expression**:
The rate of the reaction in terms of \( N_2O_5 \) is:
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} \]