Question

Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 12.5 kJ or greater. Report your answer as a percentage and to 2 sig figs. Do not include the "%" sign in your answer. Do not use scientific notation.

Answer 1

Answer:

0.023

Explanation:

The Arrhenius' equation states that:

$k = A*e^{\frac{-Ea}{RT}$

Where k is the velocity constant of the reaction, A is the constant of the collisions, Ea is the activation energy (the energy necessary to the molecules have so the reaction will happen), R is the gas constant (8.314 J/molK) and T is the temperature.

This equation is derivated of:

k = pZf

Where

p=fraction of collisions that occur with reactant molecules properly oriented

f=fraction of collisions having energy greater than the activation energy

Z=frequency of collisions

Thus, p*Z = A, and

f = $e^{\frac{-Ea}{RT} }$

So, if the energy of the molecules is 12.5 kJ/mol = 12500 J/mol, thus the fraction will be:

f = $e^{\frac{-12500}{8.314*400} }$

f = 0.023