NTA Abhyas JEE Main2020ChemistryChemical KineticsPractice
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 ?
Options
- A1.25
- B0.75
- C1.50
- D2.50
Correct answer
D. 2.50
Step-by-step solution
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 k 1 = k 2 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.