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

Given the rate constant: \[ K_{N_2O_5} = 2 \times 4.606 \times 10^{-2} \text{ S}^{-1} \] \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] At time \( t = 0 \): \[ P_i = 0.6 \quad \text{atm} \] At time \( t = 100 \) seconds: \[ P_f = 0.6 - P \quad \text{atm} \] \[ P = \text{amount of decomposition} \] We now apply the rate law for a first-order reaction: \[ 2 \times 4.606 \times 10^{-2} = \frac{2.303}{100} \log \left( \frac{0.6}{0.6 - P} \right) \] Simplifying: \[ 4 \log_{10} \left( \frac{0.6}{0.6 - P} \right) = 10^4 \] \[ \frac{0.6}{0.6 - P} = 10^4 \] \[ P = (6000 - 0.6) \times 10^{-4} = 5999.4 \times 10^{-4} = 0.59994 \quad \text{atm} \] Therefore, the total pressure is: \[ P_{\text{total}} = 0.6 + \frac{P}{2} = 0.6 + \frac{0.29997}{2} = 0.89997 \quad \text{atm} \] Thus, the total pressure is approximately: \[ P_{\text{total}} = 899.97 \times 10^{-3} \quad \text{atm} \] The correct value of \( x \) is \( 900 \) (rounded to the nearest integer). 

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: \( \boxed{897} \)