For the reaction : 2 N 2 O 5 ( g ) ⟶ 4 NO 2 ( g ) + O 2 ( g ) if the concentration of NO 2 increases by 5.2 × 1 0 − 3 M in 100 sec , then the rate of reaction is:

The rate of reaction is 1/4 × ( Δ [ NO 2 ] /Δ t ) = 1/4 × ( 5.2 × 1 0 − 3 M /100 s ) = 1.3 × 1 0 − 5 M s − 1 .

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ChemistryChemical KineticsRate of Reactions/Rate LawMedium2 minQB

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For the reaction : $2 N_{2} O_{5} (g) ⟶ 4 NO_{2} (g) + O_{2} (g)$ if the concentration of $NO_{2}$ increases by $5.2 \times 1 0^{- 3} M$ in 100 sec , then the rate of reaction is:

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Answer:

Solution:

The rate of reaction is $1/4 \times (Δ [NO_{2}] /Δ t) = 1/4 \times (5.2 \times 1 0^{- 3} M /100 s) = 1.3 \times 1 0^{- 5} M s^{- 1}$ .

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

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