Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A compound A dissociate by two parallel first order paths at certain temperature

  A g   k 1 ( min -1 )   2 B g k 1 = 6 . 9 3 × 1 0 - 3  min -1

A g   k 2 ( min -1 )   C g k 2 = 6 . 9 3 × 1 0 - 3  min -1

If reaction started with pure 'A' with 1 mole of A in 1 litre closed container with initial pressure 2 atm. What is the pressure (in atm) developed in container after 50 minutes from start of experiment ?
ChemistryChemical KineticsJEE Main
Options:
  • A 1.25
  • B 0.75
  • C 1.50
  • D 2.50
Solution:
2839 Upvotes Verified Answer
The correct answer is: 2.50
A g 2 B g ; k 1 = 6 . 9 3 × 1 0 - 3  min -1 A g C g ; k 2 = 6 . 9 3 × 1 0 - 3  min -1

k = ( k 1 + k 2 ) overall velocity constant = 6.93 x 2 x 10-3 min-1

t 1 / 2 = 0 . 6 9 3 × 1 0 0 0 6 . 9 3 × 2 = 5 0  min

i.e., after = 50 min

P A = 1 atm.

Since k1 = k2

P B = 0.5 x 2 = 1 atm (due to stoichio metric coefficient)

P C = 0.5 x 1 = 0.5 atm.

Total pressure = 2.5 atm.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.