Start free

Chemistry · Chemical Equilibrium

JEE Main 2025 — 24 January, Morning Shift — Question 38

For a reaction, N2O5( g)→2NO2( g)+12O2( g)\mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} in a constant volume container, no

products were present initially. The final pressure of the system when 50%50 \% of reaction gets

completed is

  1. Option A:

    7/27 / 2 times of initial pressure

  2. Option B:

    5 times of initial pressure

  3. Option C:

    5/25 / 2 times of initial pressure

  4. Option D:

    7/47 / 4 times of initial pressure

    Correct

Answer: D

Step-by-step solution

N2O5( g)⟶2NO2( g)+12O2( g)\quad \mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})} \longrightarrow 2 \mathrm{NO}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})}

t=0P0−−t=tP0−x2xx2\begin{array}{llll} \mathrm{t}=0 & \mathrm{P}_{0} & - & - \mathrm{t}=\mathrm{t} & \mathrm{P}_{0}-\mathrm{x} & 2 \mathrm{x} & \frac{\mathrm{x}}{2} \end{array}

x=P02x=\frac{P_{0}}{2}

Ptotal =P0−P02+P0+P04=74P0\mathrm{P}_{\text {total }}=\mathrm{P}_{0}-\frac{\mathrm{P}_{0}}{2}+\mathrm{P}_{0}+\frac{\mathrm{P}_{0}}{4}=\frac{7}{4} \mathrm{P}_{0}

Answer key and solution verified before publishing.

Practise Chemical Equilibrium

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2025
Paper
24 January, Morning Shift
Subject
Chemistry
Chapter
Chemical Equilibrium
Topic
Factors affecting equilibrium &Le Chatelier's Principle

More Chemical Equilibrium questions from this paper

For a reaction, N 2 O 5( g ) rightarrow 2 NO 2( g ) +1/2 O 2( g ) in… | JEE Main 2025 PYQ with Solution · DhiX AI