Question

A first order reaction has a rate constant of 0.543 at 25 c and 6.47 at 47

c. Calculate the value of the activation energy in kilojoules (enter answer to one decimal place)

Answer 1

The activation energy Ea can be related to rate constant (k) at temperature (T) through the equation:

ln(k2/k1) = Ea/R[1/T1 - 1/T2]

where :

k1 is the rate constant at temperature T1

k2 is the rate constant at temperature T2

R = gas constant = 8.314 J/K-mol

Given data:

k1 = 0.543 s-1; T1 = 25 C = 25+273 = 298 K

k2 = 6.47 s-1; T = 47 C = 47+273 = 320 K

ln(6.47/0.543) = Ea/8.314 [1/298 - 1/320]

2.478 = 2.774 *10^-5 Ea

Ea = 0.8934*10^5 J = 89.3 kJ