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Chemical Kinetics question

2019 · Q56
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Chemical Kinetics question

2019 · Q56

IIT JAMChemistryChemical KineticsNumerical+2 / −0
The turnover frequency (TOF) for the catalytic reaction, A (1 mol) →5 hoursCatalyst (0.01 mol)\xrightarrow[5 \text{ hours}]{\text{Catalyst (0.01 mol)}}Catalyst (0.01 mol)5 hours​ B with 90% yield of the product is ‾\underline{\hspace{2cm}}​ hour −1^{-1}−1. (Round off to the nearest integer)
Numerical answer
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Correct answer: 18

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