For the reaction : `2N_(2)O_(5) rarr 4NO_(2) +O_(2)`, the rate of reaction in terms of `O_(2)` is `d[O_(2)]`/dt. In terms of `N_(2)O_(5)`, it will be :

Options

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

Explanation

To find the rate of reaction in terms of \( N_2O_5 \) for the reaction:

\[ 2N_{2}O_{5} \rightarrow 4NO_{2} + O_{2} \]

we will follow these steps:

### Step 1: Write the rate expression
The rate of a reaction can be expressed in terms of the change in concentration of reactants and products. For the given reaction, we can express the rate in terms of the disappearance of the reactant \( N_2O_5 \) and the appearance of the products \( NO_2 \) and \( O_2 \).

### Step 2: Identify the stoichiometric coefficients
In the balanced equation, the stoichiometric coefficients are:
- For \( N_2O_5 \): 2
- For \( NO_2 \): 4
- For \( O_2 \): 1

### Step 3: Write the rate of disappearance of \( N_2O_5 \)
Since \( N_2O_5 \) is a reactant that is disappearing, we will denote its rate of change as a negative quantity. The rate of disappearance of \( N_2O_5 \) can be expressed as:

\[ -\frac{d[N_2O_5]}{dt} \]

### Step 4: Relate the rates using stoichiometry
According to the stoichiometric coefficients, the relationship between the rate of disappearance of \( N_2O_5 \) and the rates of appearance of the products can be established. The rate of reaction in terms of \( N_2O_5 \) is related to the rate of formation of \( O_2 \) and \( NO_2 \) by their respective coefficients:

\[ -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{1}{2} \frac{d[O_2]}{dt} \]

### Step 5: Write the final expression for the rate in terms of \( N_2O_5 \)
From the relationship established, we can express the rate of reaction in terms of \( N_2O_5 \):

\[ \text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} \]

### Final Answer:
Thus, the rate of reaction in terms of \( N_2O_5 \) is:

\[ -\frac{1}{2} \frac{d[N_2O_5]}{dt} \]

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