Question

If the rate constant for a particular reaction doubles between 20 °C and 40 °C, calculate the value of the activation energy for the reaction.

Answer 1

Arrhenius equation

lnk = Ae^-Ea/RT

Where,

A = Frequency factor

Ea = Activation energy

R = gas constant , 8.314(J/K mol)

T = Temperature

Therefore,

At 20°C

k = Ae^-Ea/(8.314(J/K mol)×293.15K) -----eq1

At 40°C

2k = Ae^-Ea/(8.314(J/K mol)× 313.15K) -----eq2

If divide eq2 by eq1

2k/k = e^-Ea/(2603.5J/mol)/e^-Ea/(2437.2J/mol)

ln2 = -Ea/2603.5J/mol - (-Ea/2437.2J/mol)

2.303log2= Ea(- 1/2603.5(J/mol) +1/2437.2(J/mol))

0.0000262(J/mol)^-1Ea = 0.6933

Ea = 0.6933/0.0000262(J/mol)^-1

= 26461J/mol

= 26.46kJ/mol

Therefore,

The activation energy of the reaction is 26.46kJ/mol