The activation energy for a particular reaction is 102 kJ/mol. If the rate constant is 1.35 × 10⁻⁴ s⁻¹ at 302 K, what is the rate constant at 273 K?

Answer 1

Arrhenius equation:

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

k1 = 1.35 × 10⁻⁴ s⁻¹

k2 = ?

Ea = 102 kJ/mol

R = 0.008314 kJ/molK

T1 = 302K

T2 = 273K

ln(k2/1.45x10^-4 )= -102/0.008314(1/302 - 1/273)

ln(k2/1.45x10^-4 ) = -12,268 (0.00331 - 0.00366)

ln(k2/1.45x10^-4 ) = -12,268 (-0.00035)

ln(k2/1.45x10^-4 ) = 4.29

k2/1.45x10^-4 = 72.97

k2 = 0.011

May be a math error because one would expect k2 (at 273K) to be lower, not greater than k1 (at 302K). Please do double check the math.