The rate constant for the reaction, `2N_2O_5 to 4NO_2 + O_2" is "4.0 xx 10^(-5)sec^(-1)`. If the rate is `3.4 xx 10^(-5)mol.l^(-1)s^(-1)`. Then the concentration of `N_2O_5`(in mol `litre^(-1)`) is :

Options

A. 1.4
B. 1.2
C. 0.04
D. 0.85

Answer

To solve the problem, we need to find the concentration of \( N_2O_5 \) given the rate constant and the rate of the reaction. The reaction is:

\[ 2N_2O_5 \rightarrow 4NO_2 + O_2 \]

### Step-by-Step Solution:

1. **Identify Given Data:**
- Rate constant \( k = 4.0 \times 10^{-5} \, \text{s}^{-1} \)
- Rate of the reaction \( r = 3.4 \times 10^{-5} \, \text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1} \)

2. **Write the Rate Law Expression:**
The rate of a reaction can be expressed as:
\[ r = k [N_2O_5]^n \]
Here, \( n \) is the order of the reaction with respect to \( N_2O_5 \). Since we are not given that this is an elementary reaction, we will treat it as a general case.

3. **Assume the Order of Reaction:**
For the reaction \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \), we can assume the reaction is second order with respect to \( N_2O_5 \) (as per stoichiometry). Therefore:
\[ r = k [N_2O_5]^2 \]

4. **Rearranging the Rate Law to Find Concentration:**
Rearranging the equation to solve for the concentration of \( N_2O_5 \):
\[ [N_2O_5]^2 = \frac{r}{k} \]
\[ [N_2O_5] = \sqrt{\frac{r}{k}} \]

5. **Substituting the Values:**
Substitute the values of \( r \) and \( k \):
\[ [N_2O_5] = \sqrt{\frac{3.4 \times 10^{-5}}{4.0 \times 10^{-5}}} \]

6. **Calculating the Concentration:**
First, calculate the fraction:
\[ \frac{3.4 \times 10^{-5}}{4.0 \times 10^{-5}} = 0.85 \]
Now take the square root:
\[ [N_2O_5] = \sqrt{0.85} \approx 0.922 \, \text{mol} \cdot \text{L}^{-1} \]

7. **Final Result:**
The concentration of \( N_2O_5 \) is approximately \( 0.922 \, \text{mol} \cdot \text{L}^{-1} \).