Question

Assume water boils at ${\mathbf{100}}^{\mathbf{0}}\mathit{C}$ in Houston (near sea level), and at ${\mathbf{90}}^{\mathbf{0}}\mathit{C}$ in Cripple Creek, Colorado (near 9500 ft). If it takes 4.8 min to cook an egg in Cripple Creek and 4.5 min in Houston, what is${\mathbf{E}}_{\mathbf{a}}$ for this process?

Short answer

The activation energy $E_a$is $7.3\times {10}^{3}J/mol$.

Step-by-step solution

Step 1: The Arrhenius equation

The Arrhenius equation is written like this:

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

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.

Step 2: Determine the activation energy

Write the equation for the two rate constants, $k1$and $k_2$at temperature $T_1$ and $T_2$respectively and obtain new expression:

$$\begin{gathered} \ln k_1=\ln A-E_a/R{T}_1 \\ \ln k_2=\ln A-E_a/R{T}_2 \\ \ln k_2-\ln k_1=-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \\ \ln \left(\frac{k_2}{k_1}\right)=-\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right) \end{gathered}$$

Substitute the values and determine the activation energy $E_a$

$$\begin{gathered} \ln \left(\frac{1egg/4.5\min }{1egg/4.8\min }\right)=-\frac{E_a}{8.314J/mol.K}\left(\frac{1}{\left(273+90.0\right)K}-\frac{1}{\left(273+100.0\right)K}\right) \\ {E}_a=\frac{\left(8.314J/mol.K\right)\ln \left(\frac{4.8}{4.5}\right)}{\left(\frac{1}{363.2K}-\frac{1}{373.2K}\right)} \\ =7.3\times {10}^{3}J/mol \end{gathered}$$

Therefore, the activation energy is $7.3\times {10}^{3}J/mol$.