For the following first order reaction : `N_2O_5` (in `CCl_4` solvent) `to 2NO_2` (solvent) `+1/2 O_2(g)` the velocity constant is `6.2 times 10^-4 s^-1`. The rate of the reaction at `[N_2O_5]`=1.25 moles/litre will be:

Options

A. `7.75 times 10^-4 molesl^-1sec
B. `6.35 times 10^-3 moles//l-sec`
C. `5.15 times 10^-5 moles//l-sec`
D. `3.85 times 10^-4 Moles//I-sec`

Answer

To solve the problem, we need to calculate the rate of a first-order reaction given the rate constant and the concentration of the reactant. The reaction is:

\[ N_2O_5 \, (in \, CCl_4) \rightarrow 2NO_2 \, (solvent) + \frac{1}{2} O_2(g) \]

### Step-by-step Solution:

1. **Identify the Rate Law for a First-Order Reaction**:
The rate of a first-order reaction is given by the equation:
\[ \text{Rate} = k \times [A] \]
where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant.

2. **Write Down the Given Values**:
- Rate constant \( k = 6.2 \times 10^{-4} \, s^{-1} \)
- Concentration of \( N_2O_5 \) \( [N_2O_5] = 1.25 \, \text{moles/litre} \)

3. **Substitute the Values into the Rate Law**:
Now, substitute the values of \( k \) and \( [N_2O_5] \) into the rate equation:
\[ \text{Rate} = (6.2 \times 10^{-4} \, s^{-1}) \times (1.25 \, \text{moles/litre}) \]

4. **Perform the Calculation**:
\[ \text{Rate} = 6.2 \times 10^{-4} \times 1.25 \]
\[ \text{Rate} = 7.75 \times 10^{-4} \, \text{moles/litre/s} \]

5. **Final Answer**:
The rate of the reaction at the given concentration is:
\[ \text{Rate} = 7.75 \times 10^{-4} \, \text{moles/litre/s} \]