Question

If the temperature in Problem 16.60 is increased to ${\mathbf{50}}^{\mathbf{0}}\mathit{C}$, 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 ${50}^{0}C$, the fraction of collisions increases by a factor of 22.7.

Step-by-step solution

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 at 500C

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

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

Step 3: Determine how the proportion of collisions increases as the temperature rises

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

$$\begin{gathered} Increaseinf=\frac{{6.72\times 10}^{-17}}{{2.96\times 10}^{-18}} \\ =22.7 \end{gathered}$$

The fraction of collisions increases by a factor of 22.7.