Question

The half life for N2O5 decomposition reaction is 2.4hours at STP. Starting with 10.8 grams of N2O5, how much oxygen will be obtained after a period of 9.6 hours?​

Answer 1

• Oxygen that will be obtained after a period of 9.6 hours is 0.07 liter.

Given-

• Half life of N₂O₅ = 2.4 hours
• Initial mass of N₂O₅ = 10.8 grams
• Moles of N₂O₅ = 10.8/108 = 0.1 moles

Decomposition reaction of N₂O₅ is-

N₂O₅ → 2 NO₂ + 1/2 O₂

Half life of N₂O₅ = 2.4 hours

T 1/2 = ln 2 / k = 2.4 hours

⇒ k = ln2 / 2.4

Now, at T = 9.6 hours

k = ln (0.1/N)/9.6 = ln 2 / 9.6

By solving the above equation we get N = 0.1 / 16

Now number of liter of O₂ = N × 22.4 / 2 = 0.07 Liter.

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