For a reaction , `2N_(2)O_(5)rarr4NO_(2)+O_(2),` the rate is directly proportional to `[N_(2)O_(5)].` At `45^(@)C, 90%` of the `N_(2)O_(5)` react in 3600 s. The value of the rate constant is

Options

A. `3.2xx10^(-4)s^(-1)`
B. `6.4xx10^(-4)s^(-1)`
C. `8.5xx10^(-4)s^(-1)`
D. `12.8xx10^(-4)s^(-1)`

Explanation

To find the rate constant \( k \) for the reaction \( 2N_2O_5 \rightarrow 4NO_2 + O_2 \), we can follow these steps:

### Step 1: Understand the reaction and the given data
The reaction shows that 2 moles of \( N_2O_5 \) decompose to form 4 moles of \( NO_2 \) and 1 mole of \( O_2 \). We are given that 90% of \( N_2O_5 \) reacts in 3600 seconds at \( 45^\circ C \).

### Step 2: Determine the remaining concentration of \( N_2O_5 \)
If 90% of \( N_2O_5 \) has reacted, then 10% remains. If we denote the initial concentration of \( N_2O_5 \) as \( [N_2O_5]_0 \), the concentration after 3600 seconds will be:
\[ [N_2O_5] = 0.1 [N_2O_5]_0 \]

### Step 3: Use the integrated rate law for first-order reactions
Since the rate is directly proportional to \( [N_2O_5] \), we can treat this as a first-order reaction. The integrated rate law for a first-order reaction is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{[N_2O_5]_0}{[N_2O_5]} \right) \]

### Step 4: Substitute the values into the equation
Substituting the values we have:
- \( t = 3600 \) seconds
- \( [N_2O_5]_0 = [N_2O_5] + 0.9 [N_2O_5]_0 \) implies \( [N_2O_5] = 0.1 [N_2O_5]_0 \)

Now, substituting into the equation:
\[ k = \frac{2.303}{3600} \log \left( \frac{[N_2O_5]_0}{0.1 [N_2O_5]_0} \right) \]
This simplifies to:
\[ k = \frac{2.303}{3600} \log(10) \]
Since \( \log(10) = 1 \):
\[ k = \frac{2.303}{3600} \]

### Step 5: Calculate the value of \( k \)
Now, we can calculate \( k \):
\[ k = \frac{2.303}{3600} \approx 6.4 \times 10^{-4} \, \text{s}^{-1} \]

### Final Answer
Thus, the value of the rate constant \( k \) is approximately \( 6.4 \times 10^{-4} \, \text{s}^{-1} \).

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