Question

The rate constant of a reaction is 4.50×10−5 L/mol·s at 195°C and 3.20×10−3 L/mol·s at 258°C. What is the activation energy of the reaction?

Answer 1

Answer:

138.46KJ

Explanation:

In(k1/k2) = Ea/Rx (1/T1-1/T2)

T1= 195+273= 468k

T1= 195+ 273= 531k

K1= 4.5 X 10-5

K2=3.2x10-3

Rx= 8.3145

In(71.111)= Ea/8.3145(1/468- 1/531)

4.26409x 8.3145= Ea(0.000256)

Ea=138464.1

=138.46kJ