Question

The precipitation of egg albumin in water at \(100^{\circ} \mathrm{C}\) has an activation energy of \(52.0 \mathrm{~kJ} / \mathrm{mol} .\) By what percent does the rate of precipitation decrease if the water is at \(92^{\circ} \mathrm{C} ?\)

Short answer

Answer: The rate of precipitation of egg albumin in water decreases by approximately \(33.24 \%\) when the temperature is decreased from \(100^{\circ} \mathrm{C}\) to \(92^{\circ} \mathrm{C}\).

Step-by-step solution

Convert temperatures to Kelvin

Before starting with calculations, we need to convert the given temperatures to Kelvin. To do this, we add \(273.15\) to the Celsius temperature. \(T_1 = 100^{\circ} \mathrm{C} + 273.15 = 373.15\mathrm{K}\) \(T_2 = 92^{\circ} \mathrm{C} + 273.15 = 365.15\mathrm{K}\)

Use the Arrhenius equation

The Arrhenius equation is given by: \(k(T) = Ae^{-\frac{E_a}{RT}}\) where \(k(T)\) is the rate constant at temperature \(T\), \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. To find the ratio of the rate constants at the two temperatures, we can take their ratio: \(\frac{k(T_1)}{k(T_2)} = \frac{Ae^{-\frac{E_a}{RT_1}}}{Ae^{-\frac{E_a}{RT_2}}}\) Notice that \(A\) cancels out, and we are left with: \(\frac{k(T_1)}{k(T_2)} = \frac{e^{-\frac{E_a}{RT_1}}}{e^{-\frac{E_a}{RT_2}}}\)

Calculate the ratio of rate constants

Now, we can plug in the values for the activation energy, temperatures, and gas constant to find the ratio of the rate constants: \(E_a = 52.0 \times 10^3 \mathrm{J/mol}\) \(R = 8.314 \mathrm{J/(mol\cdot K)}\) \(\frac{k(T_1)}{k(T_2)} = \frac{e^{-\frac{52.0\times10^3}{8.314 \cdot 373.15}}}{e^{-\frac{52.0\times10^3}{8.314 \cdot 365.15}}}\approx 1.498\)

Calculate the percent decrease in the rate of precipitation

Now that we have the ratio of the rate constants, we can calculate the percent decrease in the rate of precipitation: Percent decrease = \(\frac{k(T_1) - k(T_2)}{k(T_1)} \times 100\) Since \(\frac{k(T_1)}{k(T_2)} = 1.498\), we can rewrite the equation as: Percent decrease = \(\frac{1.498k(T_2) - k(T_2)}{1.498k(T_2)} \times 100\) Percent decrease = \(\frac{1.498k(T_2) - k(T_2)}{1.498k(T_2)} \times 100\) Percent decrease = \(\frac{0.498k(T_2)}{1.498k(T_2)} \times 100 \approx 33.24 \%\) Hence, the rate of precipitation of egg albumin in water decreases by approximately \(33.24 \%\) when the temperature of the water is decreased from \(100^{\circ} \mathrm{C}\) to \(92^{\circ} \mathrm{C}\).