At `250^(@)C` , the half life for the decomposition of `N_(2)O_(5)` is `5.7` hour and is independent of initial pressure of `N_(2)O_(5)`. The specific rate constant is

Options

A. `(0.693)/(7)`
B. `0.693xx5.7`
C. `(5.7)/(0.693)`
D. None of the above

Answer

To find the specific rate constant (k) for the decomposition of \( N_2O_5 \), we will use the relationship between half-life and the rate constant for a first-order reaction. Here are the steps to solve the problem:

### Step 1: Identify the Reaction Order
The problem states that the half-life of \( N_2O_5 \) is independent of the initial pressure, which indicates that the reaction is a first-order reaction.

### Step 2: Write the Half-Life Formula for First-Order Reactions
For a first-order reaction, the half-life (\( t_{1/2} \)) is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
where \( k \) is the specific rate constant.

### Step 3: Rearrange the Formula to Solve for \( k \)
We can rearrange the formula to solve for \( k \):
\[ k = \frac{0.693}{t_{1/2}} \]

### Step 4: Substitute the Given Half-Life Value
We are given that the half-life (\( t_{1/2} \)) is 5.7 hours. Substituting this value into the equation gives:
\[ k = \frac{0.693}{5.7} \]

### Step 5: Calculate the Value of \( k \)
Now, we perform the calculation:
\[ k = \frac{0.693}{5.7} \approx 0.121 \]

### Conclusion
The specific rate constant \( k \) for the decomposition of \( N_2O_5 \) at \( 250^\circ C \) is approximately \( 0.121 \, \text{h}^{-1} \).

---