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

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.