Question

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

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

Step-by-step solution

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

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 \(20^{\circ}C\): \[T_2 = 20 + 273.15 = 293.15 K\]

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 \(T_2\): \[k_2 = Ae^{\frac{-E_a}{RT_2}}\]

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 \(E_a\): \[E_a = R\left(\frac{1}{T_1}-\frac{1}{T_2}\right)^{-1} \ln{\frac{k_2}{k_1}}\]

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 \(47.9 \text{ kJ/mol}\).

Key concepts

First-order reaction

In the realm of chemical reactions, a first-order reaction stands out due to its simplicity. A first-order reaction is one where the rate of reaction is directly proportional to the concentration of a single reactant. This implies that if the concentration of the reactant doubles, the reaction rate also doubles. Mathematically, the rate of a first-order reaction can be expressed as:\[Rate = k[A]\]where:

• \( k \) is the rate constant
• \( [A] \) is the concentration of the reactant

The proportionality makes first-order reactions relatively straightforward to analyze, particularly when studying their kinetics through experiments. These reactions are common in nature and are often used to model various processes, such as radioactive decay and some enzymatic reactions. Understanding the concept of a first-order reaction is essential for grasping how reaction rates change in response to concentration variations.

Arrhenius equation

The Arrhenius equation is a critical component in the study of chemical kinetics. This equation illustrates how the rate constant \( k \) of a reaction varies with temperature. It is expressed mathematically as: \[k = Ae^{\frac{-E_a}{RT}}\]Where:

• \( A \) is the pre-exponential factor, related to the frequency of collisions and the orientation of reactant molecules
• \( E_a \) is the activation energy, the minimum energy that reactant molecules need to initiate a reaction
• \( R \) is the universal gas constant (8.314 J·mol^-1·K^-1)
• \( T \) is the absolute temperature in Kelvin

The equation highlights the effect of temperature on reaction rates: as temperature increases, the rate constant \( k \) generally increases, assuming the activation energy remains constant. This relationship underscores the importance of thermal energy in overcoming the activation barrier, enabling reactions to proceed more rapidly at higher temperatures.

Rate constant

The rate constant, denoted as \( k \), is a pivotal term in the kinetics of chemical reactions. It serves as a proportionality factor in the rate equation that determines the speed of a reaction under specific conditions. The value of \( k \) is influenced by various factors like temperature and the nature of the reactants. For first-order reactions, \( k \) has units of time inverse (s^-1) and indicates how quickly a reaction proceeds. The Arrhenius equation allows us to see that the rate constant increases with temperature. Its value, at any given temperature, provides insight into how frequently reactant molecules successfully collide and convert into products. In the given problem, we are provided with rate constants at two different temperatures, which helps in calculating the activation energy using the Arrhenius equation. This constant is essential for quantitatively describing reaction rates and deducing the kinetic parameters of a reaction.

Temperature conversion

In chemical kinetics, temperature plays a crucial role in dictating reaction dynamics. However, temperatures need to be in Kelvin when using the Arrhenius equation for accurate calculations. This is because the Kelvin scale is an absolute temperature scale starting from absolute zero, where kinetic motion ceases. To convert from the Celsius scale to Kelvin, the formula is straightforward: \[K = ^{\circ}C + 273.15\]Using this conversion, we can accurately substitute temperature values into the Arrhenius equation. In our problem, converting 0ºC to Kelvin results in 273.15 K, while 20ºC converts to 293.15 K. This conversion is indispensable because it reflects the absolute thermal energy available to reactant molecules, which influences how effectively they can overcome the activation energy and proceed in a reaction.