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

Question

At ${\mathbf{25}}^{\mathbf{\circ }}\mathbf{C}$ , 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 $2.96\times {10}^{-18}.$

Arrhenius equation

The Arrhenius equationis written like this:

$$\begin{gathered} \mathrm{k}={\mathrm{Ae}}^{‐{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}} \\ \mathrm{k}=\mathrm{Af} \end{gathered}$$

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

Determine the fraction of collisions

Determine the fraction of collisions by substituting the values:

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

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