Question

The turnover frequency (TOF) for the catalytic reaction, 
\[ \text{A (1 mol) } \xrightarrow{\text{Catalyst (0.01 mol)}} \text{ B} \] with 90% yield of the product is ................ hour\(^{-1}\). (Round off to the nearest integer)
 

Hint:

The turnover frequency (TOF) can be calculated by dividing the number of moles of product formed by the product of the number of moles of catalyst and time.

Answer 1

Correct Answer: 18

Step 1: Define turnover frequency (TOF).
The turnover frequency (TOF) is the number of product molecules formed per unit time per active site. It can be calculated as: \[ \text{TOF} = \frac{\text{moles of product}}{\text{moles of catalyst} \times \text{time}} \] Step 2: Determine the moles of product formed.
From the given data, the yield of the product is 90%. Thus, the moles of product formed are: \[ \text{Moles of product} = 0.90 \times 1 \, \text{mol} = 0.90 \, \text{mol} \] Step 3: Calculate TOF.
The reaction occurs in 5 hours with 0.01 mol of catalyst. Thus, TOF is: \[ \text{TOF} = \frac{0.90}{0.01 \times 5} = \frac{0.90}{0.05} = 18 \, \text{hour}^{-1} \] Step 4: Conclusion.
Thus, the turnover frequency is 9 hour\(^{-1}\).