Question

69.The water in a pressure cooker boils at a temperature greater than 100°C because it is under pressure. At this higher temperature, the chemical reactions associated with the cooking of food take place at a greater rate.

(a) Some food cooks fully in 5 min in a pressure cooker at 112°C and in 10 minutes in an open pot at 100°C. Calculate the average activation energy for the reactions associated with the cooking of this food.

(b) How long will the same food take to cook in an open pot of boiling water in Denver, where the average atmospheric pressure is 0.818 atm and the boiling point of water is 94.4°C?

Short answer

Average activation energy required for cooking is$6.90\mathrm{x}{10}^4{\mathrm{Jmol}}^{-1}$

Step-by-step solution

First order kinetics;

Consider that cooking of food follows first order kinetics, then the eqaution is as follows

$$\begin{gathered} \ln \left(\frac{{\mathrm{C}}_0}{\mathrm{C}}\right)=\mathrm{kt}.......\left(1\right) \\ \mathrm{Where}{\mathrm{C}}_0\mathrm{is}\mathrm{the}\mathrm{initial}\mathrm{concentration}\mathrm{of}\mathrm{food} \\ \mathrm{C}=\mathrm{Final}\mathrm{concentration}\mathrm{of}\mathrm{food} \\ \mathrm{k}=\mathrm{rate}\mathrm{constant}\mathrm{of}\mathrm{the}\mathrm{reaction} \\ \mathrm{t}=\mathrm{time}\mathrm{required}\mathrm{forcooking} \end{gathered}$$

T is the temparature of the reaction where the concentration of reactents is C₀

Concentration of reactents at  different timings

If at temparatute T₁, the concentration of reactents are C₀and final concentration at C ,kis the rate constant and t is the time required for the reaction.

$\ln \left(\frac{{\mathrm{C}}_0}{\mathrm{C}}\right)={\mathrm{kt}}_1.......\left(1\right)$

In the similar way T₂ is the temparature of reactents where C₀ and C is the initial and final concentration of the reactents.

$\ln \left(\frac{{\mathrm{C}}_0}{\mathrm{C}}\right)={\mathrm{k}}_1{\mathrm{t}}_1.......\left(2\right)$

$\ln \left(\frac{{\mathrm{C}}_0}{\mathrm{C}}\right)={\mathrm{k}}_2{\mathrm{t}}_2.......\left(3\right)$

Average activation energy;

Where t₁ = 5 min t₂ = 10 min ,substituting these values in the eqaution 2 and 3

$\ln \left(\frac{{\mathrm{C}}_0}{\mathrm{C}}\right)=\mathrm{k}\mathrm{x}5\min .......\left(4\right)$

And

$\ln \left(\frac{{\mathrm{C}}_0}{\mathrm{C}}\right)=\mathrm{k}\mathrm{x}10\min ....\left(5\right)$

Divide the equation 4 by 5

$$\begin{gathered} \frac{{\mathrm{k}}_1\mathrm{x}5}{{\mathrm{k}}_2\mathrm{x}10}=1 \\ \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{10}{5} \\ \mathrm{or}\frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=2........\left(6\right) \end{gathered}$$

From arrehinus eqaution

$$\begin{gathered} \ln \frac{{\mathrm{k}}_1}{{\mathrm{k}}_2}=\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right).......\left(7\right) \\ {\mathrm{k}}_1=\mathrm{Rate}\mathrm{constant}\mathrm{at}{\mathrm{T}}_1 \\ {\mathrm{k}}_2=\mathrm{Rate}\mathrm{constant}\mathrm{at}{\mathrm{T}}_2 \\ {\mathrm{E}}_{\mathrm{a}}=\mathrm{activation}\mathrm{energy} \end{gathered}$$

Temparature are given in degree celsius,but it should be converted to kelvin units

$$\begin{gathered} {\mathrm{T}}_1={112}^0\mathrm{C} \\ =\left(112+273\right)\mathrm{K} \\ {\mathrm{T}}_1=385\mathrm{K} \\ {\mathrm{T}}_2={100}^0\mathrm{C} \\ =\left(100+273\right) \\ {\mathrm{T}}_2=373\mathrm{K} \end{gathered}$$

Substitute these values in equation-7 ,the following value is obtained

$$\begin{gathered} \ln 2=\frac{{\mathrm{E}}_{\mathrm{a}}}{8.315{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}}\left(\frac{1}{373\mathrm{K}}-\frac{1}{385\mathrm{K}}\right) \\ 0.693471=\frac{{\mathrm{E}}_{\mathrm{a}}}{8.315{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}}\left(0.0026809-0.0025974\right) \\ 0.693147={\mathrm{E}}_{\mathrm{a}}\mathrm{x}0.000010049615{\mathrm{J}}^{-1} \\ {\mathrm{E}}_{\mathrm{a}}=\frac{0.6931471}{0.00001004}\mathrm{J}{\mathrm{mol}}^{-1} \\ {\mathrm{E}}_{\mathrm{a}}=6.90\mathrm{x}{10}^4{\mathrm{Jmol}}^{-1} \end{gathered}$$

Average activation energy required for cooking is $6.90\mathrm{x}{10}^4{\mathrm{Jmol}}^{-1}$

Arrehinus equation:

Arrehinus equation:

$$\begin{gathered} {\text{k}}_1=\mathrm{Aexp}\left({\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_1\right)........\left(4\right) \\ {\mathrm{k}}_3=\mathrm{Aexp}\left({\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_3\right).......\left(5\right) \end{gathered}$$

Substituting the values of k₁and k₃ in equation 3

$$\begin{gathered} \mathrm{Aexp}\left({\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_1\right)\mathrm{x}{\mathrm{t}}_1=\mathrm{Aexp}\left({\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_1\right)\mathrm{x}{\mathrm{t}}_3 \\ {\mathrm{t}}_3=\frac{\mathrm{Aexp}\left({\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_1\right)\mathrm{x}{\mathrm{t}}_1}{\mathrm{Aexp}\left({\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_1\right)}.........\left(6\right) \end{gathered}$$

Substituing the values of E_a,T,R,t₁ in the eqaution-6

$$\begin{gathered} {\mathrm{T}}_1={112}^0\mathrm{C}\backslash \mathrm{hfill} \\ =\left(112+273\right)\mathrm{K} \\ {\mathrm{T}}_1=385\mathrm{K} \\ {\mathrm{T}}_2=94.4^0\mathrm{C} \\ =\left(94.4+273\right)\mathrm{K} \\ {\mathrm{T}}_2=367.4\mathrm{K} \\ {\mathrm{t}}_1=5\min \end{gathered}$$

Substitute these values in eqaution-6

$$\begin{gathered} {\mathrm{t}}_3=\frac{\exp \left(-6.9\mathrm{x}{10}^4\mathrm{Jmol}-1/8.315\mathrm{Jmolx}385\mathrm{K}\right)\mathrm{x}5\min }{\mathrm{Aexp}\left(-6.9\mathrm{x}{10}^4\mathrm{Jmol}/8.315\mathrm{Jmol}\mathrm{x}367.4\mathrm{K}\right)}.. \\ =\frac{\exp \left(-21.5539\right)\mathrm{x}5\min }{\exp \left(-22.5864\right)} \\ =\frac{0.0000000004358\mathrm{x}5\min }{0.0000000001552} \\ {\mathrm{t}}_3=14\mathrm{minutes} \end{gathered}$$

Therefore the time taken for the food to cook is 14 minutes at 94.4⁰C.