The reaction `N_2O_5` (in `C Cl_4` solution) `to 2NO_2(1) + 0.5O_2(g)` is of first order in `N_2O_5` with rate constant `6.2 xx 10^(-1) sec^(-1)`. What is the value of rate of reaction when `[N_2O_2] = 1.25 mol. L^(-1)` ?

Options

A. `7.75 xx 10^(-1) mol.l^(-1)s^(-1)`
B. `6.35 xx 10^(-3) mol.l^(-1)s^(-1)`
C. `5.15 xx 10^(-5) mol.l^(-1) s^(-1)`
D. `3.85 xx 10^(-1) mol.l^(-1) s^(-1)`

Answer

To solve the problem, we need to determine the rate of the reaction given the concentration of \( N_2O_5 \) and the rate constant. The reaction is first order with respect to \( N_2O_5 \).

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data:**
The reaction is:
\[ N_2O_5 \xrightarrow{} 2NO_2 + 0.5O_2 \]
Given data includes:
- Rate constant \( k = 6.2 \times 10^{-1} \, \text{s}^{-1} \)
- Concentration of \( N_2O_5 = 1.25 \, \text{mol/L} \)

2. **Write the Rate Law Expression:**
For a first-order reaction, the rate law is expressed as:
\[ \text{Rate} = k [N_2O_5] \]
where \( [N_2O_5] \) is the concentration of \( N_2O_5 \).

3. **Substitute the Given Values into the Rate Law:**
Now, substitute the values of \( k \) and \( [N_2O_5] \):
\[ \text{Rate} = (6.2 \times 10^{-1} \, \text{s}^{-1}) \times (1.25 \, \text{mol/L}) \]

4. **Calculate the Rate:**
Perform the multiplication:
\[ \text{Rate} = 6.2 \times 1.25 \times 10^{-1} \, \text{mol/L/s} \]
\[ \text{Rate} = 7.75 \times 10^{-1} \, \text{mol/L/s} \]

5. **Final Answer:**
Thus, the rate of the reaction when the concentration of \( N_2O_5 \) is \( 1.25 \, \text{mol/L} \) is:
\[ \text{Rate} = 0.775 \, \text{mol/L/s} \]