The rate constant for the reaction, `2N_(2)O_(5) to 4NO_(2)+O_(2)` is `3xx10^(-5)"sec"^(-1)` . If the rate is `2.40xx10^(-5) "mol litre"^(-1) "sec"^(-1)` Then the concentration of `N_(2)O_(5)` ( in mol `"litre"^(-1)` ) is

Options

A. `1.4`
B. `1.2`
C. `0.04`
D. `0.8`

Explanation

To find the concentration of \( N_2O_5 \) in the reaction \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \), we can use the rate equation for a first-order reaction. Here are the steps to solve the problem:

### Step-by-Step Solution

1. **Identify the given values:**
- Rate constant, \( k = 3 \times 10^{-5} \, \text{sec}^{-1} \)
- Rate of reaction, \( \text{Rate} = 2.40 \times 10^{-5} \, \text{mol L}^{-1} \text{sec}^{-1} \)

2. **Write the rate equation for a first-order reaction:**
The rate of a first-order reaction is given by:
\[ \text{Rate} = k \cdot [N_2O_5] \]
where \( [N_2O_5] \) is the concentration of the reactant.

3. **Rearrange the equation to solve for \( [N_2O_5] \):**
\[ [N_2O_5] = \frac{\text{Rate}}{k} \]

4. **Substitute the known values into the equation:**
\[ [N_2O_5] = \frac{2.40 \times 10^{-5} \, \text{mol L}^{-1} \text{sec}^{-1}}{3 \times 10^{-5} \, \text{sec}^{-1}} \]

5. **Calculate the concentration:**
\[ [N_2O_5] = \frac{2.40}{3} \, \text{mol L}^{-1} = 0.8 \, \text{mol L}^{-1} \]

6. **Final answer:**
The concentration of \( N_2O_5 \) is \( 0.8 \, \text{mol L}^{-1} \).