Question

If the temperature in Problem 16.60 is increased to ${\mathbf{50}}^{\mathbf{\circ }}\mathbf{C}$, by what factor does the fraction of collisions with energy equal to or greater than the activation energy change?

Short answer

If the temperature in Problem 16.60 is increased ${50}^{\circ }\mathrm{C}$, the fraction of collisions increases by a factor of 22.7.

Step-by-step solution

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 at 50∘C

Substitute 100kJ/mol for Ea and 8.314 J/mol.K for R to get the proportion of collisions at 50C:

$$\begin{gathered} \mathrm{f}={\text{e}}^{-\mathrm{Ea}/\mathrm{RT}} \\ {\text{=e}}^{\frac{100\mathrm{kJ}/\mathrm{mol}}{\left(8.314\mathrm{J}/\mathrm{mol}.\mathrm{k}\right)\left(273+50\right)\text{K}}\text{×}\frac{1000\mathrm{j}}{1\mathrm{kJ}}} \\ {\text{=e}}^{-37.24} \\ =6.72\times {10}^{-17} \end{gathered}$$

Determine how the proportion of collisions increases as the temperature rises

Determine the increase in the fraction of collisions with the increase in temperature:

$$\begin{gathered} \mathrm{Increase}\mathrm{in}\mathrm{f}=\frac{6.72\times {10}^{-17}}{2.96\times {10}^{-18}} \\ =22.7 \end{gathered}$$

The fraction of collisions increases by a factor of 22.7.