How to Find Activation Energy with Temperature and Rate Constant: A Comprehensive Guide

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Activation energy is a crucial concept in the field of chemical kinetics. It helps us understand how the rate of a chemical reaction is affected by temperature. By quantifying the activation energy, we can determine the energy barrier that a reactant must overcome for a reaction to occur. In this blog post, we will explore how to find activation energy using temperature and the rate constant.

The Arrhenius Equation: The Key to Finding Activation Energy

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Explanation of the Arrhenius Equation

The Arrhenius equation, named after the Swedish scientist Svante Arrhenius, connects the rate constant of a reaction with the temperature and activation energy. It is expressed as:

$k = Ae^{-\frac{E_a}{RT}}$

Here, $k$ represents the rate constant, $A$ is the pre-exponential factor, $E_a$ is the activation energy, $R$ is the gas constant, and $T$ is the temperature in Kelvin.

The Relationship between Activation Energy, Temperature, and the Rate Constant

According to the Arrhenius equation, the rate constant of a reaction increases exponentially with increasing temperature. This relationship demonstrates that as the temperature rises, more reactant molecules possess sufficient energy to overcome the activation energy barrier, resulting in a faster reaction rate.

The activation energy serves as a measure of the minimum energy required for a chemical reaction to occur. It represents the energy difference between the reactants’ energy and the transition state energy. A higher activation energy implies a slower reaction rate, as fewer reactant molecules can overcome the energy barrier.

Importance of the Arrhenius Equation in Calculating Activation Energy

The Arrhenius equation plays a crucial role in calculating activation energy because it provides a mathematical relationship between temperature, the rate constant, and activation energy. By experimentally determining the rate constant at different temperatures, we can apply the Arrhenius equation to compute the activation energy.

Step-by-Step Guide on How to Calculate Activation Energy

Gathering Necessary Data: Temperature and Rate Constant

To calculate the activation energy, we need to collect two essential pieces of information: the rate constant at different temperatures and the corresponding temperatures. The rate constant can be obtained through experimental measurement or theoretical calculations, while the temperatures are recorded during the experiments.

Applying the Arrhenius Equation

Once we have the rate constant and the corresponding temperatures, we can apply the Arrhenius equation to find the activation energy. Rearranging the equation, we get:

$\ln{\left(\frac{k_2}{k_1}\right)} = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)$

In this equation, $k_1$ and $k_2$ represent the rate constants at temperatures $T_1$ and $T_2$ respectively. $R$ is the gas constant.

Worked-out Example: Calculating Activation Energy

Let’s work through an example to illustrate how to calculate activation energy using the Arrhenius equation. Suppose we have the following data:

$k_1 = 0.05 \, \text{min}^{-1}$ at $T_1 = 300 \, \text{K}$
$k_2 = 0.12 \, \text{min}^{-1}$ at $T_2 = 350 \, \text{K}$
$R = 8.314 \, \text{J}\cdot\text{mol}^{-1}\cdot\text{K}^{-1}$

Plugging the values into the rearranged Arrhenius equation, we have:

$\ln{\left(\frac{0.12}{0.05}\right)} = -\frac{E_a}{8.314}\left(\frac{1}{350} - \frac{1}{300}\right)$

Simplifying the equation, we find:

$E_a = -\frac{8.314 \times 300 \times 350}{0.05 \times (350 - 300)} \times \ln{\left(\frac{0.12}{0.05}\right)}$

Evaluating this expression, we obtain the value for activation energy.

Common Mistakes and Misconceptions in Calculating Activation Energy

Incorrect Data Collection

One common mistake when calculating activation energy is gathering inaccurate or imprecise data. It is crucial to ensure that the rate constant and temperature values are measured or recorded correctly. Any errors in data collection can significantly impact the accuracy of the calculated activation energy.

Misapplication of the Arrhenius Equation

Another common error is misapplying the Arrhenius equation. It is important to remember that this equation is only valid within a certain temperature range and for reactions that follow the assumptions of the Arrhenius theory. Using the equation outside its applicable range or for reactions that deviate from the assumptions can lead to erroneous results.

Misinterpretation of Results

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Lastly, misinterpreting the results can also occur. Activation energy is a measure of the energy barrier, not the total energy change in a reaction. It is essential to understand the significance of activation energy in relation to the reaction rate and the overall energy profile of the chemical reaction.

Calculating activation energy using temperature and the rate constant is a valuable tool in understanding the kinetics of chemical reactions. The Arrhenius equation provides a mathematical framework for this calculation, allowing us to quantify the energy barrier that reactants must overcome for a reaction to proceed. By avoiding common mistakes and properly interpreting the results, we can derive meaningful insights into the reaction mechanisms and improve our understanding of various chemical processes.

Numerical Problems on How to find activation energy with temperature and rate constant

Problem 1:

The rate constant for a reaction at a certain temperature is given as $k_1 = 5.2 \times 10^{-3} \text{ s}^{-1}$ . At a higher temperature, the rate constant is found to be $k_2 = 1.2 \times 10^{-2}\text{ s}^{-1}$ . Calculate the activation energy of the reaction.

Solution:

The Arrhenius equation relates the rate constant ( $k$ ) to the activation energy ( $E_a$ ) and the temperature ( $T$ ) through the equation:

$k = A\cdot e^{-\frac{E_a}{RT}}$

Where:
– $k$ is the rate constant,
– $A$ is the pre-exponential factor,
– $E_a$ is the activation energy,
– $R$ is the gas constant (8.314 J/(mol·K)),
– $T$ is the temperature in Kelvin.

To find the activation energy, we can rearrange the Arrhenius equation as follows:

$\frac{k_2}{k_1} = \frac{A\cdot e^{-\frac{E_a}{RT_2}}}{A\cdot e^{-\frac{E_a}{RT_1}}}$

Simplifying the equation gives:

$\frac{k_2}{k_1} = e^{-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}$

Taking the natural logarithm of both sides:

$\ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Now we can solve for $E_a$ :

$E_a = -R \cdot \left(\frac{1}{T_2}-\frac{1}{T_1}\right)^{-1} \cdot \ln\left(\frac{k_2}{k_1}\right)$

Substituting the given values:

$E_a = -\left(8.314 \text{ J/(mol·K)}\right) \cdot \left(\frac{1}{T_2}-\frac{1}{T_1}\right)^{-1} \cdot \ln\left(\frac{1.2 \times 10^{-2} \text{ s}^{-1}}{5.2 \times 10^{-3} \text{ s}^{-1}}\right)$

Problem 2:

The rate constant for a reaction at a temperature of 300 K is $k_1 = 4.8 \times 10^{-4} \text{ s}^{-1}$ . If the reaction rate is observed to double when the temperature is increased to 400 K, calculate the activation energy.

Solution:

Using the same approach as in Problem 1, we can determine the activation energy using the Arrhenius equation:

$E_a = -\left(8.314 \text{ J/(mol·K)}\right) \cdot \left(\frac{1}{T_2}-\frac{1}{T_1}\right)^{-1} \cdot \ln\left(\frac{k_2}{k_1}\right)$

Substituting the given values:

$E_a = -\left(8.314 \text{ J/(mol·K)}\right) \cdot \left(\frac{1}{400 \text{ K}}-\frac{1}{300 \text{ K}}\right)^{-1} \cdot \ln\left(\frac{2k_1}{k_1}\right)$

Problem 3:

The rate constant for a reaction at a certain temperature is $k_1 = 2.5 \times 10^{-5} \text{ s}^{-1}$ . When the temperature is increased by 10°C, the rate constant becomes $k_2 = 4.5 \times 10^{-4} \text{ s}^{-1}$ . Calculate the activation energy of the reaction.

Solution:

Similar to the previous problems, we can use the Arrhenius equation to find the activation energy:

$E_a = -\left(8.314 \text{ J/(mol·K)}\right) \cdot \left(\frac{1}{T_2}-\frac{1}{T_1}\right)^{-1} \cdot \ln\left(\frac{k_2}{k_1}\right)$

Substituting the given values:

$E_a = -\left(8.314 \text{ J/(mol·K)}\right) \cdot \left(\frac{1}{T_2}-\frac{1}{T_1}\right)^{-1} \cdot \ln\left(\frac{4.5 \times 10^{-4} \text{ s}^{-1}}{2.5 \times 10^{-5} \text{ s}^{-1}}\right)$

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