For the reaction `2N_2O_5 to 4NO_2 +O_2` rate of reaction and rate constant are `1.04 xx 10^(-4)and 3.4 xx 10^(-5)sec^(-1)` respectively. The concentration of `N_2O_5` at that time will be :

Options

A. 1.732
B. 3
C. 3.06
D. `3.4 xx 10^(-5)`

Answer

To find the concentration of \( N_2O_5 \) for the reaction \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \), we can use the relationship between the rate of reaction, the rate constant, and the concentration of the reactants.

### Step-by-Step Solution:

1. **Write the Rate Law Expression**:
The rate of the reaction can be expressed as:
\[ r = k [N_2O_5]^n \]
where \( r \) is the rate of the reaction, \( k \) is the rate constant, and \( n \) is the stoichiometric coefficient of \( N_2O_5 \) in the balanced equation. For this reaction, \( n = 2 \) because the balanced equation shows that 2 moles of \( N_2O_5 \) are consumed.

2. **Substitute the Known Values**:
We know:
- Rate of reaction \( r = 1.04 \times 10^{-4} \, \text{mol/L/s} \)
- Rate constant \( k = 3.4 \times 10^{-5} \, \text{s}^{-1} \)

Therefore, we can rewrite the rate law as:
\[ r = k [N_2O_5]^2 \]

3. **Rearrange to Solve for Concentration**:
Rearranging the equation to isolate \( [N_2O_5] \):
\[ [N_2O_5]^2 = \frac{r}{k} \]
Taking the square root of both sides gives:
\[ [N_2O_5] = \sqrt{\frac{r}{k}} \]

4. **Substitute the Values into the Equation**:
Substitute the values of \( r \) and \( k \):
\[ [N_2O_5] = \sqrt{\frac{1.04 \times 10^{-4}}{3.4 \times 10^{-5}}} \]

5. **Calculate the Value**:
First, calculate the fraction:
\[ \frac{1.04 \times 10^{-4}}{3.4 \times 10^{-5}} = 3.0588 \]
Now take the square root:
\[ [N_2O_5] = \sqrt{3.0588} \approx 1.748 \]

6. **Final Result**:
The concentration of \( N_2O_5 \) is approximately \( 1.748 \, \text{mol/L} \).

### Conclusion:
The concentration of \( N_2O_5 \) at that time will be approximately \( 1.748 \, \text{mol/L} \).