The rate constant for the reaction, `2N_(2)O_(5) rarr 4NO_(2) + O_(2)` is `3.0 xx 10^(-5) s^(-1)`. If the rate is `2.40 xx 10^(-5) mol L^(-1) s^(-1)`, then the initial concentration of `N_(2)O_(5)` (in `mol L^(-1)`) is

Options

A. 0.8
B. 0.7
C. 1.2
D. 1

Explanation

To find the initial concentration of \( N_2O_5 \) for the reaction \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \), we can follow these steps:

### Step 1: Identify the given values
- Rate constant \( k = 3.0 \times 10^{-5} \, s^{-1} \)
- Rate of the reaction \( \text{Rate} = 2.40 \times 10^{-5} \, mol \, L^{-1} \, s^{-1} \)

### Step 2: Determine the order of the reaction
The unit of the rate constant \( k \) is \( s^{-1} \), which indicates that the reaction is first order with respect to \( N_2O_5 \).

### Step 3: Write the rate law expression
For a first-order reaction, the rate law can be expressed as:
\[ \text{Rate} = k [N_2O_5] \]
Where \( [N_2O_5] \) is the concentration of \( N_2O_5 \).

### Step 4: Rearrange the rate law to find the concentration
We can rearrange the equation to solve for the concentration of \( N_2O_5 \):
\[ [N_2O_5] = \frac{\text{Rate}}{k} \]

### Step 5: Substitute the known values
Now, substitute the given values into the equation:
\[ [N_2O_5] = \frac{2.40 \times 10^{-5} \, mol \, L^{-1} \, s^{-1}}{3.0 \times 10^{-5} \, s^{-1}} \]

### Step 6: Calculate the concentration
Perform the calculation:
\[ [N_2O_5] = \frac{2.40}{3.0} \times 10^{-5} \, mol \, L^{-1} \, s^{-1} \cdot s^{-1} = 0.8 \, mol \, L^{-1} \]

### Conclusion
The initial concentration of \( N_2O_5 \) is \( 0.8 \, mol \, L^{-1} \).

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