Question

For a reaction, $$ {N}_2{O}_5(g) \rightarrow 2{NO}_2(g) + \frac{1}{2} {O}_2(g) $$ in a constant volume container, no products were present initially. The final pressure of the system when 50% of the reaction gets completed is: 

• A. \( \frac{7}{2} \) times of initial pressure
• B. 5 times of initial pressure
• C. \( \frac{5}{2} \) times of initial pressure
• D. \( \frac{7}{4} \) times of initial pressure

Hint:

For reactions occurring in a constant volume container, use stoichiometry to calculate the total pressure by considering the moles of reactants and products formed at each step.

Answer 1

The Correct Option is D

To determine the final pressure of the system when 50% of the reaction is completed, we start by analyzing the given reaction:

\({N}_2{O}_5(g) \rightarrow 2{NO}_2(g) + \frac{1}{2} {O}_2(g)\) 

1. Initially, we have only \({N}_2{O}_5\) in the container with initial pressure P₀.
2. Let's assume the initial moles of \({N}_2{O}_5\) are 1. No products are present initially.
3. When 50% of the reaction is completed, half of \({N}_2{O}_5\) is decomposed, so moles of \({N}_2{O}_5 = 0.5\).
4. From the stoichiometry of the reaction, when \(0.5\) moles of \({N}_2{O}_5\) decompose, the moles of products formed are:
   • \({NO}_2: 2 \times 0.5 = 1\)
   • \({O}_2: \frac{1}{2} \times 0.5 = 0.25\)
5. The total moles in the system when 50% reaction is completed: \(0.5 \text{ (remaining } {N}_2{O}_5) + 1 \text{ (formed } {NO}_2) + 0.25 \text{ (formed } {O}_2) = 1.75\)
6. The change in total moles from initial (1 mole of \({N}_2{O}_5\)) to the final state (1.75 moles) results in a pressure change because volume and temperature are constant.
7. The final pressure can be calculated as:
   • \(\frac{\text{Final Moles}}{\text{Initial Moles}} \times P_0 = \frac{1.75}{1} \times P_0 = 1.75 \times P_0\)
8. This is equivalent to the final pressure being \(\frac{7}{4}\) times the initial pressure, since \(1.75 = \frac{7}{4}\).

Hence, the final pressure of the system is \(\frac{7}{4}\) times the initial pressure, confirming that the correct answer is:

\( \frac{7}{4} \) times of initial pressure 
 

Answer 2

Given: The initial pressure of \( N_2O_5 \) is \( P_0 \).

At 50% reaction completion:

- \( \frac{1}{2} P_0 \) of \( N_2O_5 \) decomposes. - This produces: - \( P_0 \) of \( NO_2 \), - \( \frac{P_0}{4} \) of \( O_2 \).

Final Pressure Calculation:

The total final pressure is the sum of the partial pressures of the products and the remaining \( N_2O_5 \) at 50% reaction completion: \[ P_{\text{final}} = \frac{P_0}{2} + P_0 + \frac{P_0}{4} = \frac{7P_0}{4}. \]

Conclusion:

Hence, the final pressure is \( \frac{7}{4} \) of the initial pressure.