Question

A reaction is followed and found to have a rate constant of 3.36 × 104 m-1s-1 at 344 k and a rate constant of 7.69 m-1s-1 at 219 k. determine the activation energy for this reaction. a reaction is followed and found to have a rate constant of 3.36 × 104 m-1s-1 at 344 k and a rate constant of 7.69 m-1s-1 at 219 k. determine the activation energy for this reaction. 12.5 kj/mol 11.5 kj/mol 23.8 kj/mol 58.2 kj/mol 42.0 kj/mol

Answer 1

The relation between rate constants at different temperatures, temperature and activation energy is known as Arrhenius equation

Arrhenius equation

$K=Ae^-{\frac{Ea}{RT} }$

For two temperatures

$Ea=R(\frac{ln\frac{k1}{k1} }{(\frac{1}{T1})-(\frac{1}{T2})})$

Where

Ea = ? = activation energy

k1 = 3.36 × 10⁴

T1=344 k

k2=7.69

T2=219K

R= gas constant = 8.314 J /molK

Putting values

$Ea= (8.314)(\frac{ln(\frac{7.69}{33600})}{(\frac{1}{344})(\frac{1}{219})}$

Ea = (-69.69)/(-0.00166) = 41981.93 J/mol

Or

Activation energy is 42.0 kJ /mol