Question

At 47 °C, what is the fraction of collisions with energy equal or greater than an activation energy of 88.20 kJ/mol?

Answer 1

The fraction of collision is
`4.00*10^(-15)`

Step-by-step explanation:

Given that,

Temperature = 47°C

Activation energy = 88.20 KJ/mol

From Arrhenius equation,

`k=Ae^{-(E_(a))/(RT)}`

Here,
`e^{-(E_(a))/(RT)}`=fraction of collision

We need to calculate the fraction of collisions

Using formula of fraction of collisions

`f=e^{-(E_(a))/(RT)}`

Where f = fraction of collision

E = activation energy

R = gas constant

T = temperature

Put the value into the formula

`f=e^{-(88.20)/(8.314*10^(-3)*(47+273))}`

`f=4.00*10^(-15)`

Hence, The fraction of collision is
`4.00*10^(-15)`