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. `0.5xx10^(-4)Ms^(-1)`
C. `2xx10^(-3)Ms^(-1)`
D. `2.5xx10^(-5)Ms^(-1)`

Answer

To solve the problem, we need to determine the rate of the reaction based on the increase in concentration of \( NO_2 \). The reaction given is:

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

### 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. **Write the expression for the rate of reaction**:
The rate of the reaction can be expressed in terms of the change in concentration of the products and reactants. For the reaction, we can write:
\[ \text{Rate} = -\frac{1}{2} \frac{\Delta [N_2O_5]}{\Delta t} = \frac{1}{4} \frac{\Delta [NO_2]}{\Delta t} = \frac{1}{2} \frac{\Delta [O_2]}{\Delta t} \]
Here, the negative sign indicates that the concentration of \( N_2O_5 \) is decreasing.

3. **Calculate the rate using the change in concentration of \( NO_2 \)**:
Since we know the change in concentration of \( NO_2 \) and the time interval, we can calculate the rate:
\[ \text{Rate} = \frac{1}{4} \cdot \frac{\Delta [NO_2]}{\Delta t} \]
Substituting the known values:
\[ \Delta [NO_2] = 5.2 \times 10^{-3} \, M, \quad \Delta t = 100 \, s \]
\[ \text{Rate} = \frac{1}{4} \cdot \frac{5.2 \times 10^{-3} \, M}{100 \, s} \]

4. **Perform the calculation**:
\[ \text{Rate} = \frac{1}{4} \cdot \left( 5.2 \times 10^{-5} \, M/s \right) = 1.3 \times 10^{-5} \, M/s \]

5. **Final result**:
The rate of the reaction is:
\[ \text{Rate} = 1.3 \times 10^{-5} \, \text{mol/s} \]