While studying the decompoistion of gaseous `N_(2)O_(5)`, it is observed that a plot of logarithm of its partial pressure versus time is linear. What kinetic parameters can be obtained form this observation?

Explanation

To solve the question regarding the decomposition of gaseous \( N_2O_5 \) and the implications of a linear plot of the logarithm of its partial pressure versus time, we can follow these steps:

### Step 1: Understand the relationship between logarithm of partial pressure and time
The problem states that a plot of the logarithm of the partial pressure of \( N_2O_5 \) versus time is linear. This suggests that the reaction follows first-order kinetics.

### Step 2: Recall the first-order reaction equation
For a first-order reaction, the relationship between the concentration (or partial pressure) of the reactant and time can be expressed as:
\[ \log \left( \frac{P}{P_0} \right) = -\frac{k}{2.303} t + \text{constant} \]
where:
- \( P \) is the partial pressure at time \( t \)
- \( P_0 \) is the initial partial pressure
- \( k \) is the rate constant
- \( t \) is time

### Step 3: Identify the kinetic parameters
From the linear plot of \( \log \left( \frac{P}{P_0} \right) \) versus time, we can identify two key kinetic parameters:
1. **Order of the Reaction**: Since the plot is linear, it confirms that the decomposition of \( N_2O_5 \) is a first-order reaction.
2. **Rate Constant (k)**: The slope of the line obtained from the plot is equal to \(-\frac{k}{2.303}\). Therefore, we can calculate the rate constant \( k \) from the slope of the line.

### Step 4: Conclusion
Thus, from the observation that the plot of the logarithm of the partial pressure versus time is linear, we can conclude that:
- The reaction follows first-order kinetics.
- The rate constant \( k \) can be determined from the slope of the plot.