Chemistry - Chemical Kinetics Question with Solution | TestHub
ChemistryChemical KineticsOrder and Molecularity of ReactionsMedium2 minQB
ChemistryMediumsingle choice
A second-order rate constant is 0.0300 L mol⁻¹ s⁻¹. Express it in mL molecule⁻¹ min⁻¹ using = 6.00 × 10²³ mol⁻¹.
Options:
Answer:
B
Solution:
Multiply by 1000 mL L⁻¹ and 60 s min⁻¹, then divide by because one mole contains molecules. The result is 0.0300 × 1000 × 60/(6.00 × 10²³) = 3.00 × 10⁻²¹ mL molecule⁻¹ min⁻¹. Dividing by 60 instead of multiplying would reverse the time conversion.
Stream:JEE_ADVSubject:ChemistryTopic:Chemical KineticsSubtopic:Order and Molecularity of Reactions
⏱ 2mℹ️ Source: QB
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