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MolSynapse / Question bank / Surface Chemistry / Catalysis: homogeneous, heterogeneous, enzyme catalysis

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3sc107

In a run, $0.0200 mmol$ of catalyst forms $7.50 mmol$ of product in $25 min$ . Select all correct statements.

AThe turnover number is 3.75.
BThe turnover number is 375.
CThe turnover frequency is $0.25 s^{- 1}$ .
DThe turnover frequency is $15 s^{- 1}$ .

Answer: B, C

TON

= 7.50/0.0200 = 375

. TOF

= 375/25 = 15 mi n^{- 1} = 0.25 s^{- 1}

. So B and C are correct, and 15 is per minute, not per second.

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