Question

A certain reaction has an activation energy of 64.0 kJ mol and a frequency factor of A, 4.40x1012 L mol What is the rate constant K of this reaction at 26.0 C? Express your answer with the appropriate units. Indicate the multiplication of units explicitly either with a multiplication dot (asterisk) or a dash.

Answer 1

1)

Given:

T = 26.0 oC

=(26.0+273)K

= 299.0 K

A = 4.4*10^12 L.mol-1.s-1

Ea = 64.0 KJ/mol

= 64000.0 J/mol

R = 8.314 J/mol.K

use:

K = A*e^(-Ea/RT)

= 4.4*10^12*e^(-64000.0/(8.314*299.0))

= 4.4*10^12*e^(-25.7453)

= 4.4*10^12*6.591*10^-12

= 29.0 L.mol-1.s-1

Answer: 29.0 L.mol-1.s-1

2)

Given:

T1 = 352 K

T2 = 426 K

K1 = 109 L.mol-1.s-1

K2 = 185 L.mol-1.s-1

use:

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

ln(185/109) = ( Ea/8.314)*(1/352 - 1/426)

0.529 = (Ea/8.314)*(4.935*10^-4)

Ea = 8912 J/mol

Ea = 8.91 KJ/mol

Answer: 8.91 KJ/mol