Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsRate of Reactions/Rate LawMedium2 minQB

ChemistryMediummultiple choice

For two reactions:

I: A $⟶$ B, Rate constant $k_{1} = 1 se c^{- 1}$

II: C $⟶$ D, Rate constant $k_{2} = 1 M^{- 1} se c^{- 1}$

Starting with an initial concentration of 1 M for each reactant, the time taken to reach 0.5 M is the same. Identify the correct options.

Answer:

B, C, D

Solution:

$r_{1} = k_{1} [A]; r_{2} = k_{2} [C]^{2}$

$t_{1/2} = \frac{ln 2}{k _{1}} = \frac{1}{k _{2} \times 1} \Rightarrow k_{2} = \frac{k _{1}}{ln 2}$

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

⏱ 2mℹ️ Source: QB

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