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Chapter 16: Q16.60P (page 731)

At 25鈭�C , what is the fraction of collisions with energy equal to or greater than an activation energy of 100. kJ/mol?

Short Answer

Expert verified

The fraction of collisions is equal to2.96脳10-18.

Step by step solution

01

Arrhenius equation

The Arrhenius equationis written like this:

k=Ae鈥�Ea/RTk=Af

Where k is the rate constant, A is the frequency factor, Ea is the reaction's activation energy at a certain temperature T, and R is the gas constant .The equation f equals e-Ea/RTand represents the proportion of collisions with a given energy.

02

Determine the fraction of collisions

Determine the fraction of collisions by substituting the values:

f=e-Ea/RT=e100kJ/mol8.314J/mol.k273+25K脳1000j1kJ=e-40.36=2.96脳10-18

The fraction of collisions is equal to 2.96脳10-18.

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