Question

Review | Constants Periodic Table Suppose that a catalyst lowers the activation barrier of a reaction from 124 kJ/mol to 54 kJ/mol. Part A By what factor would you expect the reaction rate to increase at 25°C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.) Express your answer using two significant figures. EVO AD O P ? Submit Request Answer

Answer 1

ln K1 = ln A - Ea1/RT

ln K2 = ln A - Ea2/RT

Subtract 1st one from 2nd:

ln K2 - ln K1 = Ea1/RT - Ea2/RT

ln K2/K1 = (Ea1-Ea2)/RT

Here:

Ea1-Ea2 = 124 KJ/mol - 54 KJ/mol = 70 KJ/mol = 70000 J/mol

T= 25.0 oC

= (25.0+273) K

= 298 K

use:

ln K2/K1 = (Ea1-Ea2)/RT

ln K2/K1 = (70000.0)/(8.314*298)

ln K2/K1 = 28.25

K2/K1 = e^(28.25)

K2/K1 = 1.9*10^12

Answer: 1.9*10^12