JEE Main · 2025 · Shift-ImediumCK-145

For a reaction, N2O5(g) 2NO2(g) + 12O2(g) in a constant volume container, no products were present initially. The final…

Chemical Kinetics · Class 12 · JEE Main Previous Year Question

Question

For a reaction, N2O5(g)→2NO2(g)+12O2(g)\text{N}_2\text{O}_{5(g)} \rightarrow 2\text{NO}_{2(g)} + \dfrac{1}{2}\text{O}_{2(g)} in a constant volume container, no products were present initially. The final pressure of the system when 50% of reaction gets completed is:

(a) 7/27/2 times of initial pressure (b) 55 times of initial pressure (c) 5/25/2 times of initial pressure (d) 7/47/4 times of initial pressure

Options
  1. a

    7/27/2 times of initial pressure

  2. b

    55 times of initial pressure

  3. c

    5/25/2 times of initial pressure

  4. d

    7/47/4 times of initial pressure

    ✓
Correct Answerd

7/47/4 times of initial pressure

Detailed Solution

Reaction: N2O5→2NO2+12O2\text{N}_2\text{O}_5 \rightarrow 2\text{NO}_2 + \tfrac{1}{2}\text{O}_2

At constant volume, pressure is proportional to moles.

Let initial pressure of N2O5=P0\text{N}_2\text{O}_5 = P_0.

| Species | t=0t = 0 | At 50% completion | |---|---|---| | N2O5\text{N}_2\text{O}_5 | P0P_0 | P0−P02=P02P_0 - \tfrac{P_0}{2} = \tfrac{P_0}{2} | | NO2\text{NO}_2 | 00 | 2×P02=P02 \times \tfrac{P_0}{2} = P_0 | | O2\text{O}_2 | 00 | 12×P02=P04\tfrac{1}{2} \times \tfrac{P_0}{2} = \tfrac{P_0}{4} |

Total pressure:

Ptotal=P02+P0+P04=2P0+4P0+P04=7P04P_{\text{total}} = \frac{P_0}{2} + P_0 + \frac{P_0}{4} = \frac{2P_0 + 4P_0 + P_0}{4} = \frac{7P_0}{4}

Ptotal=74×P0\boxed{P_{\text{total}} = \frac{7}{4} \times P_0}

Answer: (d) 7/4 times of initial pressure

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