Question

A reaction has a rate constant of 1.12×10−2 s−1 at 400.0 K and 0.690 s−1 at 450.0 K. Part 1: Determine the activation barrier for the reaction. Part 2: What is the value of the rate constant at 425 K?

Answer 1

1. 1)

   Given:

   T1 = 400 K

   T2 = 450 K

   K1 = 1.12*10^-2 s-1

   K2 = 0.69 s-1

   use:

   ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

   ln(0.69/1.12*10^-2) = ( Ea/8.314)*(1/400 - 1/450)

   4.1208 = (Ea/8.314)*(2.778*10^-4)

   Ea = 123337 J/mol

   Ea = 123 KJ/mol

   Answer: 123 KJ/mol

   2)

   Given:

   T1 = 400 K

   T2 = 425 K

   K1 = 1.12*10^-2 s-1

   Ea = 123337 J/mol

   use:

   ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

   ln(K2/1.12*10^-2) = (123337/8.314)*(1/400 - 1/425)

   ln(K2/1.12*10^-2) = 14835*(1.471*10^-4)

   K2 = 9.924*10^-2 s-1

   Answer: 9.92*10^-2 s-1