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Chapter 13: Problem 78

At high altitudes, water boils at a temperature lower than $100.0^{\circ} \mathrm{C}$ due to the lower air pressure. A rule of thumb states that the time to hard-boil an egg doubles for every \(10.0^{\circ} \mathrm{C}\) drop in temperature. What activation energy does this rule imply for the chemical reactions that occur when the egg is cooked?

Short Answer

Expert verified
Based on the given rule of thumb, derive an equation for the activation energy (Ea) of the chemical reactions that occur when the egg is cooked. The equation is: \(E_{a} (kJ/mol) = \frac{-ln(2)R(T(T-10))}{\frac{10}{T(T-10)}- \frac{1}{T}} * \frac{1}{1000}\) Where T is the temperature in Kelvin and R is the gas constant (8.314 J/mol K).

Step by step solution

01

Write down the equation for the rate constant ratio

We know that the time doubles for every 10.0°C drop in temperature. Therefore, we can infer that the rate constant (since time is inversely proportional to the rate in a chemical reaction) would be halved. Let \(k_{1}\) and \(k_{2}\) be the rate constants at temperatures \(T_{1}\) and \(T_{2}\) respectively, we can write: \(\frac{k_{2}}{k_{1}} = \frac{1}{2}\)
02

Convert temperature to Kelvin

We are given that a drop of 10.0°C should be considered. We need to convert this temperature difference in Kelvin. In the Kelvin scale, a difference of 1°C is equal to a difference of 1 K. Therefore, the temperature difference is 10.0 K. Let's assign these temperatures as \(T_{1} = T\) and \(T_{2} = T - 10.0\).
03

Express rate constant ratio in terms of Arrhenius equation

Now, we can use the Arrhenius equation to express the rate constant ratio in terms of the activation energy and temperature: \(\frac{k_{2}}{k_{1}} = \frac{Ae^{\frac{-E_{a}}{R(T - 10)}}}{Ae^{\frac{-E_{a}}{RT}}}\) \(A\)s cancel out and we have: \(\frac{1}{2} = \frac{e^{\frac{-E_{a}}{R(T - 10)}}}{e^{\frac{-E_{a}}{RT}}}\)
04

Solve for the activation energy, Ea

Now, let's solve for the activation energy \(E_{a}\): \(\frac{1}{2} = e^{\frac{E_{a}(T - T + 10)}{RT(T - 10)} - \frac{E_{a}}{RT}}\) Taking the natural logarithm: \(- ln(2) = \frac{E_{a} \cdot 10}{R(T(T-10))} - \frac{E_{a}}{RT}\) Now, let's simplify and isolate the \(E_{a}\) terms: \(E_{a} = \frac{-ln(2)R(T(T-10))}{\frac{10}{T(T-10)}- \frac{1}{T}}\) One returns the answer in \(kJ/mol\), so the final equation: \(E_{a} = \frac{-ln(2)R(T(T-10))}{\frac{10}{T(T-10)}- \frac{1}{T}} * \frac{1}{1000}\)
05

Answer

The activation energy \(E_{a}\) for the chemical reactions that occur when the egg is cooked, according to the given rule of thumb, can be represented by the equation: \(E_{a} (kJ/mol) = \frac{-ln(2)R(T(T-10))}{\frac{10}{T(T-10)}- \frac{1}{T}} * \frac{1}{1000}\) where T is the temperature in Kelvin and R is the gas constant (8.314 J/mol K).

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