2 NO ( g ) + O 2 ( g ) → 2 NO 2 ( g ) , dt d [ NO 2 ] = 0.052 mol dm − 3 s − 1 . Calculate − dt d [ O 2 ]

− dt d [ O 2 ] = 2 1 dt d [ NO 2 ] = 2 1 × 0.052 = 0.026 mol dm − 3 s − 1

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ChemistryChemical KineticsRate of Reactions/Rate LawEasy2 minPYQ_2022

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$2 NO (g) + O_{2} (g) \to 2 NO_{2} (g), \frac{d [ NO _{2} ]}{dt} = 0.052 mol dm^{- 3} s^{- 1}$ . Calculate $- \frac{d [ O _{2} ]}{dt}$

Answer:

Solution:

$- \frac{d [ O _{2} ]}{dt} = \frac{1}{2} \frac{d [ NO _{2} ]}{dt} = \frac{1}{2} \times 0.052 = 0.026 mol dm^{- 3} s^{- 1}$

Stream:MHT_CETSubject:ChemistryTopic:Chemical KineticsSubtopic:Rate of Reactions/Rate Law

⏱ 2mℹ️ Source: PYQ_2022

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