JEE Main2015ChemistryChemical KineticsActual
The reaction 2 N 2 O 5 g → 4 N O 2 g + O 2 ( g ) follows first order kinetics. The pressure of a vessel containing only N 2 O 5 was found to increase from 50 mm Hg to 87.5 mm Hg in 30 min. The pressure exerted by the gases after 60 min. Will be (Assume temperature remains constant) :
Options
- A106.25 mm Hg
- B125 mm Hg
- C116.25 mm Hg
- D150 mm Hg
Correct answer
A. 106.25 mm Hg
Step-by-step solution
2 N 2 O 5 g → 4 N O 2 g + O 2 g at t=0, P 0 0 0 at t=30 min. (P 0 -x) 2x x 2 at t=60 min. (P 0 -y) 2y y 2 ∵ P 0 = 50 m m H g ∵ a f t e r 30 m i n . ⟹ P 0 - x + 2 x + x 2 = 87.5 P 0 + 3 2 x = 87.5 3 2 x = 37.5 ; x = 25 m m H g P 0 = 50 ; P 30 m i n = 25 m m H g That mean 30 min. is Half life of reaction. P 0 → 30 m i n . P 0 2 → 30 m i n . P 0 4 at t= 60 min P N 2 O 5 = P 0 4 ⟹ 50 4 = 12.5 m m H g P 0 - y = 12.5 ⟹ 50 - y = 12.5 y = 37.5 m m H g Total P after 60 min. ⟹ P 0 - y + 2 y + y 2 ⟹ 50 × 3 2 × 37.5 = 106.25 m