For the reaction : `2N_(2)O_(5)rarr4NO_(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)Ms^(-1)`
B. `5xx10^(-4)Ms^(-1)`
C. `7.6xx10^(-4)Ms^(-1)`
D. `2xx10^(-3)Ms^(-1)`

Answer

To solve the problem, we need to determine the rate of the reaction based on the change in concentration of the product \( NO_2 \) over a given time period. The reaction given is:

\[ 2N_2O_5 \rightarrow 4NO(g) + O_2(g) \]

### Step-by-Step Solution:

1. **Identify the change in concentration of \( NO_2 \)**:
The concentration of \( NO_2 \) increases by \( 5.2 \times 10^{-3} \, M \) over a time period of \( 100 \, s \).

2. **Determine the rate of formation of \( NO_2 \)**:
The rate of formation of a product in a reaction can be expressed as:
\[ \text{Rate} = \frac{\Delta [NO_2]}{\Delta t} \]
where \( \Delta [NO_2] \) is the change in concentration of \( NO_2 \) and \( \Delta t \) is the time interval.

3. **Substituting the values**:
Here, \( \Delta [NO_2] = 5.2 \times 10^{-3} \, M \) and \( \Delta t = 100 \, s \).
\[ \text{Rate} = \frac{5.2 \times 10^{-3} \, M}{100 \, s} = 5.2 \times 10^{-5} \, M/s \]

4. **Relate the rate to the stoichiometry of the reaction**:
According to the stoichiometry of the reaction, the rate of the reaction can also be expressed in terms of the change in concentration of \( NO_2 \):
\[ \text{Rate} = \frac{1}{4} \frac{d[NO_2]}{dt} \]
Therefore, we can write:
\[ \text{Rate} = \frac{1}{4} \times 5.2 \times 10^{-5} \, M/s \]

5. **Calculate the final rate**:
\[ \text{Rate} = \frac{5.2 \times 10^{-5}}{4} = 1.3 \times 10^{-5} \, M/s \]

### Final Answer:
The rate of the reaction is \( 1.3 \times 10^{-5} \, M/s \).

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