Question

For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \). 

 

Hint:

For a first-order reaction, the rate law can be used to calculate the change in concentration or pressure over time. The logarithmic relationship is key in determining the pressure or concentration at a specific time.

Answer 1

The decomposition reaction is: \[ 2N_2O_5(g) \rightarrow 2N_2O_4(g) + O_2(g) \]

1. Reaction Order Analysis:
The rate constant \(k = 4.606 \times 10^{-2} s^{-1}\) suggests a first-order reaction. For first-order kinetics: \[ ln \frac{P_0}{P_t} = kt \] where \(P_0\) is initial \(N_2O_5\) pressure and \(P_t\) is pressure at time t.

2. Initial Conditions:
At t=0, total pressure = 0.6 atm (pure \(N_2O_5\)), so \(P_0 = 0.6\) atm.

3. Pressure Calculation After 100s:
\[ ln \frac{0.6}{P_t} = (4.606 \times 10^{-2}) \times 100 = 4.606 \] \[ \frac{0.6}{P_t} = e^{4.606} \approx 100 \] \[ P_t = \frac{0.6}{100} = 0.006 \text{ atm} \]

4. Stoichiometric Analysis:
Let \(x\) be the decrease in \(N_2O_5\) pressure. From stoichiometry: \[ 2x = P_0 - P_t = 0.6 - 0.006 = 0.594 \text{ atm} \] \[ x = \frac{0.594}{2} = 0.297 \text{ atm} \]

5. Partial Pressures at 100s:
- \(N_2O_4\) pressure: \(0.594\) atm (2x) - \(O_2\) pressure: \(0.297\) atm (x) - Remaining \(N_2O_5\): \(0.006\) atm

6. Total Pressure Calculation:
\[ P_{total} = 0.006 + 0.594 + 0.297 = 0.897 \approx 0.9 \text{ atm} \]

Final Answer:
The total pressure after 100 seconds is $0.9$ atm.

Answer 2

Given:

• Initial total pressure at \( t = 0 \): 0.6 atm
• Rate constant \( k = 4.606 \times 10^{-2} \, \text{s}^{-1} \)
• Time \( t = 100 \, \text{s} \)
• Reaction: \( 2 \text{N}_2\text{O}_5(g) \rightarrow 2 \text{N}_2\text{O}_4(g) + \text{O}_2(g) \)

Let the pressure of O₂ formed at time \( t \) be \( p \).

Then the decrease in pressure of \( \text{N}_2\text{O}_5 \) is \( 2p \), and the increase in pressure of \( \text{N}_2\text{O}_4 \) is also \( 2p \).

Total pressure at time \( t \) is:

\[ P(t) = (0.6 - 2p) + 2p + p = 0.6 + p \]

Using first-order kinetics:

\[ k = \frac{1}{t} \ln\left(\frac{[N_2O_5]_0}{[N_2O_5]_t}\right) = \frac{1}{t} \ln\left(\frac{0.6}{0.6 - 2p}\right) \]

Substituting values:

\[ 4.606 \times 10^{-2} = \frac{1}{100} \ln\left(\frac{0.6}{0.6 - 2p}\right) \Rightarrow \ln\left(\frac{0.6}{0.6 - 2p}\right) = 4.606 \]

\[ \frac{0.6}{0.6 - 2p} = e^{4.606} \approx 100 \Rightarrow 0.6 - 2p = \frac{0.6}{100} = 0.006 \Rightarrow 2p = 0.594 \Rightarrow p = 0.297 \]

Total pressure at time \( t = 100 \, \text{s} \):

\[ x = 0.6 + p = 0.6 + 0.297 = 0.897 \, \text{atm} \Rightarrow x = 897 \times 10^{-3} \, \text{atm} \]

Final Answer: \( \boxed{897} \)