Question

A certain reaction has an activation energy of 54.0 \(\mathrm{kJ} / \mathrm{mol}\) . As the temperature is increased from \(22^{\circ} \mathrm{C}\) to a higher temperature, the rate constant increases by a factor of 7.00 . Calculate the higher temperature.

Short answer

The higher temperature of the reaction is approximately 62.25 °C. This is calculated using the Arrhenius equation and the given activation energy, initial temperature, and rate constant increase factor.

Step-by-step solution

Identify the Given Variables

The given variables are: - Activation energy (Ea) = 54 kJ/mol - Initial temperature (T1) = 22°C = 295.15 K (convert to Kelvin by adding 273.15) - Rate constant increases by a factor = 7

Write Down the Arrhenius Equation

The Arrhenius equation is expressed as: \[ k = Ae^{-\frac{Ea}{RT}}\] Where: - k is the rate constant, - A is the pre-exponential or frequency factor, - Ea is the activation energy, - R is the gas constant, and - T is the temperature in Kelvin.

Write Down the Arrhenius Equation for Both Temperatures

Let \(k_1\) be the initial rate constant at temperature \(T_1\) and \(k_2\) be the final rate constant at the higher temperature \(T_2\). Then we can write the Arrhenius equation for both temperatures: \[ k_1 = Ae^{-\frac{Ea}{RT_1}}\] and \[ k_2 = Ae^{-\frac{Ea}{RT_2}}\]

Relate the Rate Constants and Solve for T2

The rate constant increases by a factor of 7 i.e., \(k_2 = 7k_1\). Divide the second Arrhenius equation by the first to eliminate A: \[\frac{k_2}{k_1} = \frac{Ae^{-\frac{Ea}{RT_2}}}{Ae^{-\frac{Ea}{RT_1}}}\] Replace \(k_2\) with \(7k_1\) and simplify the equation: \[7 = e^{\frac{Ea}{R}(\frac{1}{T_1} - \frac{1}{T_2})}\] Now, we need to solve for \(T_2\).

Convert the Activation Energy to J/mol

In order to use the gas constant R in J/(mol·K), we need to convert the activation energy from kJ/mol to J/mol: \(Ea = 54\,kJ/mol \times 1000\,J/kJ = 54000\,J/mol\)

Insert the Given Variables and Solve for T2

Insert the values of Ea, T1, R, and rate constant increase factor into the equation to solve for T2: \[7 = e^{\frac{54000\,J/mol}{8.314\,J/(mol\cdot K)}(\frac{1}{295.15\,K} - \frac{1}{T_2})}\] To find \(T_2\), you can follow these steps: 1. Take the natural logarithm (ln) of both sides. 2. Solve for \(\frac{1}{T_2}\). 3. Take the reciprocal of the result to find \(T_2\). After solving, you should find the value of the higher temperature, \(T_2\approx 335.4 K\).

Convert the Higher Temperature to Celsius

Finally, convert the higher temperature from Kelvin to Celsius: Higher temperature = 335.4 K - 273.15 = 62.25 °C

Key concepts

Arrhenius Equation

The Arrhenius Equation is a crucial concept in understanding how the temperature influences the rate of a chemical reaction. It is expressed as \( k = Ae^{-\frac{Ea}{RT}} \), where:

• \( k \) represents the rate constant, which is a measure of the speed of the reaction.
• \( A \) is the pre-exponential factor, often associated with the frequency of collisions that lead to reactions.
• \( Ea \) is the activation energy, the energy barrier that must be overcome for a reaction to occur.
• \( R \) stands for the universal gas constant, which has a value of 8.314 J/(mol·K).
• \( T \) is the absolute temperature in Kelvin.

Utilizing this equation, we can determine how changes in temperature affect the reaction rate. By affecting \( k \), it helps chemists predict how fast a chemical process will proceed under varying conditions.

Temperature Dependence

Temperature Dependence in chemical kinetics refers to how the rate of a reaction changes with temperature. According to the Arrhenius Equation, even a small increase in temperature can lead to a significant increase in the rate constant \( k \).
This is because temperature changes affect:

• The number of molecules with enough energy to surpass the activation energy barrier. An increase in temperature means more molecules reach this energy threshold, leading to more successful collisions per unit time.
• The kinetic energy of the molecules. Higher temperatures typically mean molecules move faster, increasing the likelihood of collisions that can lead to reactions.

In the original exercise, increasing the temperature from 22°C to a higher temperature increased the rate constant by a factor of 7. This shows that the reaction becomes considerably faster with a small increase in temperature. Understanding temperature dependence is essential for controlling reaction rates in industrial and laboratory settings.

Rate Constant

The Rate Constant \( k \) is a fundamental part of chemical kinetics, indicating the speed of a chemical reaction at a given temperature. It is influenced by:

• Temperature, as evident from the Arrhenius Equation. An increase in temperature usually increases \( k \).
• Activation energy, with lower \( Ea \) resulting in higher \( k \) values.
• The Arrhenius pre-exponential factor \( A \), reflecting the number and orientation of collisions.

In practical terms, the rate constant quantifies how quickly reactants are converted into products. A high rate constant means a fast reaction, whereas a low rate constant indicates a slower process. In the original problem, the factor by which \( k \) increases (by 7 times) helps calculate the new temperature, demonstrating how \( k \) serves as a bridge between reaction kinetics and temperature.

Chemical Kinetics

Chemical Kinetics is the study of reaction rates and the factors that affect them. It is concerned with understanding how changes in conditions, like temperature, influence the speed of a reaction.
Key components of chemical kinetics include:

• The reaction rate, which describes the change in concentration of reactants or products over time.
• The activation energy \( Ea \), which acts as an energy barrier for the reaction to proceed.
• The rate law, which expresses the relationship between the rate of a chemical reaction and the concentration of its reactants.
• The rate constant \( k \), indicating how quickly a reaction occurs under specific conditions.

Chemical kinetics allows chemists to predict how fast a reaction will occur and to design processes that optimize the rate. Through exercises like the original problem, students learn to connect theoretical concepts with practical implications, enhancing their understanding of how reactions behave and how they can be controlled.