The decomposition of `N_2O_5` in `C Cl_4` solution was studied. `N_2O_5 → 2NO_2 + 1/2O_2`. The rate constant of the reaction is 6.2 x `10^(-4) sec^(-1)`. Calculate the rate when the concentration of `N_2O_5` is 1.25 molar.

Options

A. 6.45 x `10^-4`
B. 7.45 x `10^-4`
C. 6.75 x `10^-4`
D. 7.75 x `10^-4`

Answer

To calculate the rate of the decomposition of \( N_2O_5 \), we can use the rate law for the reaction. The reaction is given as:

\[ N_2O_5 \rightarrow 2NO_2 + \frac{1}{2}O_2 \]

### Step 1: Write the Rate Law Expression
For the reaction, the rate law can be expressed as:

\[ \text{Rate} = k [N_2O_5]^n \]

where:
- \( k \) is the rate constant,
- \( [N_2O_5] \) is the concentration of \( N_2O_5 \),
- \( n \) is the order of the reaction with respect to \( N_2O_5 \).

Since the decomposition of \( N_2O_5 \) is a first-order reaction, we have \( n = 1 \). Therefore, the rate law simplifies to:

\[ \text{Rate} = k [N_2O_5] \]

### Step 2: Substitute the Values
We know from the problem statement that:
- The rate constant \( k = 6.2 \times 10^{-4} \, \text{sec}^{-1} \)
- The concentration of \( N_2O_5 = 1.25 \, \text{mol/L} \)

Now, substituting these values into the rate law expression:

\[ \text{Rate} = (6.2 \times 10^{-4} \, \text{sec}^{-1}) \times (1.25 \, \text{mol/L}) \]

### Step 3: Perform the Calculation
Now we perform the multiplication:

\[ \text{Rate} = 6.2 \times 1.25 \times 10^{-4} \, \text{sec}^{-1} \]

Calculating \( 6.2 \times 1.25 \):

\[ 6.2 \times 1.25 = 7.75 \]

Thus, we have:

\[ \text{Rate} = 7.75 \times 10^{-4} \, \text{sec}^{-1} \]

### Final Answer
The rate of the reaction when the concentration of \( N_2O_5 \) is 1.25 molar is:

\[ \text{Rate} = 7.75 \times 10^{-4} \, \text{mol/L/sec} \]

---