A reaction : A 2 + B → Products, involves the following mechanism : A 2 ⇌ 2 A (fast) ( A ˉ being the intermediate) A + B ⟶ k 2 → Products (slow). The rate law consistent to this mechanism is :

From slow step : rate = k 2 [ A ] [ B ] From fast step : K e = [ A 2 ] [ A ] 2 or [ A ] = K e 1/2 [ A ] 1/2 From (i) and (ii) rate = k 2 k e 1/2 [ A 2 ] 1/2 [ B ] = k [ A 2 ] 1/2 [ B ]

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Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsRate of Reactions/Rate LawMedium2 minPYQ_2016

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A reaction : $A_{2} + B \to$ Products, involves the following mechanism : $A_{2} ⇌ 2 A$ (fast) ( $\overset{ˉ}{A}$ being the intermediate) $A + B ⟶_{k_{2}} \to$ Products (slow). The rate law consistent to this mechanism is :

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Answer:

Solution:

From slow step : rate $= k_{2} [A] [B]$ From fast step : $K_{e} = \frac{[ A ] ^{2}}{[ A _{2} ]} or [A] = K_{e}^{1/2} [A]^{1/2}$ From (i) and (ii) $rate = k_{2} k_{e}^{1/2} [A_{2}]^{1/2} [B] = k [A_{2}]^{1/2} [B]$

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

⏱ 2mℹ️ Source: PYQ_2016

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