Afirst 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)

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

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