GeneralCLASS 12CBSE
answered 22 May 20262N₂O₅(g) → 4NO₂(g) + O₂(g). Initial P = 50 mmHg; P at 30 min = 87.5 mmHg. Find P at 60 min.
A.VERIFIED ANSWERfact-checked by tutors
Step-1: Let N₂O₅ decrease by 2p. Total pressure = 50 + 3p:
50 + 3p = 87.5 ⇒ p = 12.5 mmHg
P(N₂O₅) at 30 min = 50 − 2(12.5) = 25 mmHg
Step-2: Rate constant (first order):
k = (1/30) × ln(50/25) = ln(2)/30 = 0.02310 min⁻¹
Step-3: P(N₂O₅) at 60 min (= two half-lives = 25% of initial):
P(N₂O₅) at 60 min = 50/4 = 12.5 mmHg
Step-4: Total pressure at 60 min:
Decrease = 50 − 12.5 = 37.5 = 2p ⇒ p = 18.75
P_total = 50 + 3(18.75) = 106.25 mmHg
Answer: Total pressure after 60 minutes = 106.25 mm Hg