Q16.60P - Chemistry: Molecular Nature Of Matter And Change

Question

Question: At ${\mathbf{25}}^{\mathbf{0}}\mathit{C}$ , what is the fraction of collisions with energy equal to or greater than an activation energy of 100. kJ/mol?

Step-by-Step Solution

Answer

The fraction of collisions is equal to $2.96\times {10}^{-18}$ .

Step 1: Arrhenius equation

The Arrhenius equation is written like this:

 $$\begin{gathered} \mathit{k}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{-}{\mathbf{E}}_{\mathbf{a}}\mathbf{/}\mathbf{R}\mathbf{T}} \\ \mathit{k}\mathbf{=}\mathit{A}\mathit{f} \end{gathered}$$

Where k is the rate constant, A is the frequency factor, ${\mathit{E}}_{\mathbf{a}}$  is the reaction's activation energy at a certain temperature T, and R is the gas constant .The equation f equals ${\mathit{e}}^{\mathbf{-}\mathbf{E}\mathbf{a}\mathbf{/}\mathbf{R}\mathbf{T}}$ and represents the proportion of collisions with a given energy.

Step 2: Determine the fraction of collisions

Determine the fraction of collisions by substituting the values:

$$\begin{gathered} \backslash \mathrm{begingathered}f=e^{-Ea/RT} \\ =e^{\frac{100kJ/mol}{\left(8.314J/mol.k\right)\left(273+25\right)K}\times \frac{1000J}{1kJ}} \\ =e^{-40.36} \\ =2.96\times {10}^{-18} \end{gathered}$$

The fraction of collisions is equal to $2.96\times {10}^{-18}$ .