For the reaction : `2N_(2)O_(5)rarr4NO2_(g)+O_(2)(g)` if the concentration of `NO_(2)` increases by `5.2xx10^(-3)M` in 100 sec, then the rate of reaction is :

Options

A. `1.3xx10^(-5)" M s"^(-1)`
B. `0.5xx10^(-4)" M s"^(-1)`
C. `7.6xx10^(-4)" M s"^(-1)`
D. `2xx10^(-4)" M s"^(-1)`

Explanation

To find the rate of the reaction given the increase in concentration of \( NO_2 \), 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 the Rate of Reaction**:
The rate of reaction can be expressed in terms of the change in concentration of the reactants and products. For the given reaction, the rate can be expressed as:
\[ \text{Rate} = -\frac{1}{2} \frac{d[N_2O_5]}{dt} = \frac{1}{4} \frac{d[NO_2]}{dt} = \frac{1}{2} \frac{d[O_2]}{dt} \]

3. **Given Information**:
We are given that the concentration of \( NO_2 \) increases by \( 5.2 \times 10^{-3} \, M \) in \( 100 \, s \).

4. **Calculate the Rate of Change of \( NO_2 \)**:
The rate of change of concentration of \( NO_2 \) can be calculated as:
\[ \frac{d[NO_2]}{dt} = \frac{5.2 \times 10^{-3} \, M}{100 \, s} \]

5. **Perform the Calculation**:
\[ \frac{d[NO_2]}{dt} = 5.2 \times 10^{-5} \, M/s \]

6. **Relate to the Rate of Reaction**:
Since the rate of reaction is related to the change in concentration of \( NO_2 \) by the stoichiometric coefficient, we have:
\[ \text{Rate} = \frac{1}{4} \frac{d[NO_2]}{dt} \]

7. **Substitute the Value**:
\[ \text{Rate} = \frac{1}{4} \times 5.2 \times 10^{-5} \, M/s = 1.3 \times 10^{-5} \, M/s \]

8. **Final Answer**:
The rate of the reaction is:
\[ 1.3 \times 10^{-5} \, M/s \]