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Chapter 12: Problem 64

A first-order reaction has rate constants of \(4.6 \times 10^{-2} \mathrm{~s}^{-1}\) and \(8.1 \times 10^{-2} \mathrm{~s}^{-1}\) at \(0^{\circ} \mathrm{C}\) and \(20 .{ }^{\circ} \mathrm{C}\), respectively. What is the value of the activation energy?

Short Answer

Expert verified
The value of the activation energy for the given first-order reaction is approximately \(47.9 \text{ kJ/mol}\).

Step by step solution

01

Write down the Arrhenius equation

The Arrhenius equation is given by: \[k = Ae^{\frac{-E_a}{RT}}\] Where: - \(k\) is the rate constant - \(A\) is the pre-exponential factor - \(E_a\) is the activation energy (the value we want to find) - \(R\) is the gas constant, which is equal to \( 8.314 \mathrm{~J} \cdot \mathrm{mol}^{-1} \cdot \mathrm{K}^{-1} \) - \(T\) is the temperature in Kelvin
02

Convert temperatures to Kelvin

Since the given temperatures are in Celsius, we need to convert them to Kelvin before using them in the Arrhenius equation. The conversion formula is: \[K = ^{\circ}C + 273.15\] For \(0^{\circ}C\): \[T_1 = 0 + 273.15 = 273.15 K\] For \(20^{\circ}C\): \[T_2 = 20 + 273.15 = 293.15 K\]
03

Write the Arrhenius equation for both temperatures

We have rate constants at both temperatures, so we can write the Arrhenius equation for \(T_1\) and \(T_2\): For \(T_1\): \[k_1 = Ae^{\frac{-E_a}{RT_1}}\] For \(T_2\): \[k_2 = Ae^{\frac{-E_a}{RT_2}}\]
04

Derive the activation energy from the two rate constant equations

Divide the equation for \(T_2\) by the equation for \(T_1\) to eliminate the pre-exponential factor, A: \[\frac{k_2}{k_1} = \frac{Ae^{\frac{-E_a}{RT_2}}}{Ae^{\frac{-E_a}{RT_1}}}\] Now, we can solve for the activation energy, \(E_a\): \[\frac{k_2}{k_1} = e^{\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)}\] Take the natural logarithm of both sides: \[\ln{\frac{k_2}{k_1}} = \frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\] Rearrange the equation to solve for \(E_a\): \[E_a = R\left(\frac{1}{T_1}-\frac{1}{T_2}\right)^{-1} \ln{\frac{k_2}{k_1}}\]
05

Insert the given values and calculate the activation energy

Now we can insert the given values for the rate constants and temperatures, as well as the value for the gas constant R: \[E_a = 8.314\left(\frac{1}{273.15}-\frac{1}{293.15}\right)^{-1} \ln{\frac{8.1 \times 10^{-2}}{4.6 \times 10^{-2}}}\] Calculate the activation energy: \[E_a \approx 47,852 \text{ J/mol}\] The value of the activation energy is approximately \(47.9 \text{ kJ/mol}\).

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