Question

A certain reaction quadruples in rate when the temperatures is increased from 25 degree celsius to 35 degree celsius. what is the activation energy for this reaction in kj/mole?

Answer 1

Let k? and k? be the rate constant at T? and T? respectively. ]

The rate constants are given by Arrhenius equation:
k? = A?e^[ -Ea/(R?T?) ]
k? = A?e^[ -Ea/(R?T?) ]

Consider the ratio of the rate constants:
k?/k? = ( A?e^[ -Ea/(R?T?) ] ) / ( A?e^[ -Ea/(R?T?) ] )
= e^[ -Ea/(R?T?) ] / e^[ -Ea/(R?T?) ]
= e^[ -Ea/(R?T?) - ( - Ea/(R?T?) ]
= e^[ (Ea/R) ? (1/T? - 1/T?) ]

Solve for activation energy:
k?/k?= e^[ (Ea/R) ? (1/T? - 1/T?) ]
<=>
ln( k?/k? )= (Ea/R) ? (1/T? - 1/T?)
=>
Ea = R ? ln( k?/k? ) / (1/T? - 1/T?)


So the activation energy of this reaction is:
Ea = 8.314472J/molK ? ln(3) / (1/(25+273.15)K - 1/(35+273.15)K)
= 49,503J/mol
= 49.503kJ/mol