Question

If the rate constant increases from 0.40 M –1 s –1 at 25°C to 0.80 M...

If the rate constant increases from 0.40 M –1 s –1 at 25°C to 0.80 M –1 s –1 at 35°C, what is the activation energy in kJ/mol for this reaction?

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Answer #1

Answer :

To calculate activation energy (Ea)

Given,

Rate constant at 25°C (K1 ) = 0.40 M-1 s-1

Rate constant at 35°C (K2) = 0.80 M-1 s-1.

T1 = 25°C = 298 K

T2 = 35°C = 308 K

To solve this problem we can directly use two point form of Arrhenius equation

ln ( K2/ K1) = - Ea / R [ ( 1/T2) - ( 1/T1) ]

Step by step calculation is given in the image attached

Ea (activation energy) = 52.86 kJ /mol

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