A solid catalyst site has an initial turnover frequency of $50 s^{- 1}$ . The catalyst deactivates by first-order decay of activity with a half-life of $20 min$ . The total number of turnovers per site before activity is lost completely (integrated over infinite time) is $N \times 1 0^{3}$ . Find $N$ to one decimal. Use $ln 2 = 0.693$ .

Numerical answer type, tolerance 0.1.

Answer: 86.6

Activity

a = e^{- k_{d} t}

with

k_{d} = 0.693/1200 s = 5.7750 e - 04 s^{- 1}

. Total turnovers

= \int_{0}^{\infty} 50 e^{- k_{d} t} d t = 50/ k_{d} = 86580

. This is 86.6

\times 1 0^{3}