Chapter 13 - Sections 13.1-13.7 - Exercises - Problems by Topic - Page 641: 59

$17 s^{-1}$

Given, $E_{a}= 56.8 kJ/mol=56800 J/mol$ $A= 1.5\times10^{11}/s$ T= (25+273.15)K= 298.15 K Recall that $k=Ae^{-\frac{E_{a}}{RT}}$ Then, $k= 1.5\times10^{11}s^{-1}\times e^{-\frac{56800J/mol}{8.314J mol^{-1}K^{-1}\times298.15K}}$ $=17 s^{-1}$