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:
Show Hint
- \( \frac{7}{2} \) times of initial pressure
- 5 times of initial pressure
- \( \frac{5}{2} \) times of initial pressure
- \( \frac{7}{4} \) times of initial pressure
The Correct Option is D
Approach Solution - 1
To determine the final pressure of the system when 50% of the reaction is completed, we start by analyzing the given reaction:
\({N}_2{O}_5(g) \rightarrow 2{NO}_2(g) + \frac{1}{2} {O}_2(g)\)
- Initially, we have only \({N}_2{O}_5\) in the container with initial pressure P0.
- Let's assume the initial moles of \({N}_2{O}_5\) are 1. No products are present initially.
- When 50% of the reaction is completed, half of \({N}_2{O}_5\) is decomposed, so moles of \({N}_2{O}_5 = 0.5\).
- From the stoichiometry of the reaction, when \(0.5\) moles of \({N}_2{O}_5\) decompose, the moles of products formed are:
- \({NO}_2: 2 \times 0.5 = 1\)
- \({O}_2: \frac{1}{2} \times 0.5 = 0.25\)
- The total moles in the system when 50% reaction is completed: \(0.5 \text{ (remaining } {N}_2{O}_5) + 1 \text{ (formed } {NO}_2) + 0.25 \text{ (formed } {O}_2) = 1.75\)
- 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.
- The final pressure can be calculated as:
- \(\frac{\text{Final Moles}}{\text{Initial Moles}} \times P_0 = \frac{1.75}{1} \times P_0 = 1.75 \times P_0\)
- 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
Approach Solution -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.
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