The rate of the reaction 2N₂O₅ → 4NO₂ + O₂ can be written in three ways: −d[N₂O₅]/dt = k[N₂O₅] d[NO₂]/dt = k′[N₂O₅] d[O₂]/dt = k″[N₂O₅] The relationship between k and k′ and between k and k″ are

Question

Dr. Abubacker · Accepted Answer

[Solution from source] Rate = −½ d[N₂O₅]/dt = ¼ d[NO₂]/dt = d[O₂]/dt. So d[NO₂]/dt = 2 × (−d[N₂O₅]/dt), giving k′ = 2k, and d[O₂]/dt = ½ × (−d[N₂O₅]/dt), giving k″ = k/2. Setting all three constants equal ignores the stoichiometric coefficients. Answer (4).

Q. The rate of the reaction 2N₂O₅ → 4NO₂ + O₂ can be written in three ways: −d[N₂O₅]/dt = k[N₂O₅] d[NO₂]/dt = k′[N₂O₅] d[O₂]/dt = k″[N₂O₅] The relationship between k and k′ and between k and k″ are

Explanation

Reviewed by Dr. Abubacker

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