Question

In a study of the reaction

2 N₂O(g) ⇔ 2 N₂(g) + O₂(g)

quantities of all three gases were injected into a reaction vessel. The N₂O consisted entirely of isotopically labeled 15N₂O. Analysis of the reaction mixture after 1 day revealed the presence of compounds with molar masses 28, 29, 30, 32, 44, 45, and 46 g/mol. Identify the compounds and account for their presence.

Answer 1

The compounds present are N₂O, N₂O + O₂, 2 N₂O, 2 N₂O + N₂, 2 N₂ + O₂, N₂O + 2 N₂, and 2 N₂. They are formed during the reaction 2 N₂O(g) ⇔ 2 N₂(g) + O₂(g).

Step-by-step explanation:

The compounds with molar masses 28, 29, 30, 32, 44, 45, and 46 g/mol can be identified as:

• 28 g/mol: N₂O (nitrous oxide)
• 29 g/mol: N₂O + O₂
• 30 g/mol: 2 N₂O
• 32 g/mol: 2 N₂O + N₂
• 44 g/mol: 2 N₂ + O₂
• 45 g/mol: N₂O + 2 N₂
• 46 g/mol: 2 N₂

These compounds are present due to the reaction 2 N₂O(g) ⇔ 2 N₂(g) + O₂(g). The different molar masses observed indicate the different combinations and proportions of N₂O, N₂, and O₂ that formed during the reaction.