\(2N_2O_5 (g) \longrightarrow 2N_2O_4 (g) + O_2\) is a 1st order reaction. It occurs at constant volume. Calculate the rate constant if total pressure is \(0.8\, \ atm\) initially and total pressure is \(1\, \ atm\) after \(100\, \ s\)

\(6.9 \times 10^{-3}\, \ s^{-1}\)

The correct answer is option 3. \(6.9 \times 10^{-3}\, \ s^{-1}\).

To solve this problem, we need to use the relationship between the rate constant \( k \) for a first-order reaction and the change in total pressure over time.

Given,

Initial total pressure, \( P_{\text{initial}} = 0.8 \, \text{atm} \)

Total pressure after \( t = 100 \, \text{s} \), \( P_{\text{final}} = 1 \, \text{atm} \)

The reaction is: \( 2N_2O_5(g) \longrightarrow 2N_2O_4(g) + O_2(g) \)

For the given reaction, the change in the number of moles of gas is:

\(\Delta n = (2 + 1) - 2 = 1 \text{ mole of gas} \)

If \( x \) is the extent of reaction (i.e., the amount of \( N_2O_5 \) decomposed), the total pressure can be expressed as:

Initially, the pressure due to \( N_2O_5 \) alone is \( P_{\text{initial}} \).

As the reaction proceeds, the pressure changes due to the formation of \( N_2O_4 \) and \( O_2 \).

The total pressure after the reaction has occurred for time \( t \) is:

\(P_{\text{final}} = P_{\text{initial}} + x \)

Given:

Initially, \( P_{\text{total}} = 0.8 \, \text{atm} \)

After \( 100 \, \text{s} \), \( P_{\text{total}} = 1 \, \text{atm} \)

Thus, the change in pressure due to the reaction \( x = 1 \, \text{atm} - 0.8 \, \text{atm} = 0.2 \, \text{atm} \).

For a first-order reaction, the rate constant \( k \) can be calculated using the formula:

\( k = \frac{2.303}{t} \log\left(\frac{P_{\text{final}} - P_{\text{initial}} + P_{\text{initial}}}{P_{\text{initial}}}\right) \)

Substituting the values:

\(k = \frac{2.303}{100 \, \text{s}} \log\left(\frac{1 \, \text{atm} - 0.2 \, \text{atm} + 0.8 \, \text{atm}}{0.8 \, \text{atm}}\right) \)

\(k = \frac{2.303}{100} \log\left(\frac{1.6 \, \text{atm}}{0.8 \, \text{atm}}\right) \)

\( k = \frac{2.303}{100} \log(2) \)

We know that:

\( \log(2) \approx 0.3010 \)

So:

\( k = \frac{2.303 \times 0.3010}{100} \)

\(k \approx \frac{0.693}{100} \, \text{s}^{-1} \)

\(k \approx 6.93 \times 10^{-3} \, \text{s}^{-1} \)

Thus, the correct answer is option 3: \( 6.9 \times 10^{-3} \, \text{s}^{-1} \).