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GeneralCLASS 12CBSE
answered 22 May 2026

2N₂O₅(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

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GENERAL · CLASS 12