Question

The activation energy of a certain reaction is \(65.7 \mathrm{~kJ} / \mathrm{mol}\) How many times faster will the reaction occur at \(50^{\circ} \mathrm{C}\) than at \(0^{\circ} \mathrm{C}\) ?

Short answer

The reaction will occur approximately 8.64 times faster at 50°C compared to 0°C.

Step-by-step solution

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. To do this, we add 273.15 to each temperature: 0°C = 273.15 K, 50°C = 273.15 + 50 = 323.15 K.

Calculate \(\frac{k1}{k2}\)

We will be solving for the rate constants at both temperatures, then finding the ratio of these rate constants to determine how many times faster the reaction will be. Let \(k1\) be the rate constant at 0°C (273.15 K) and \(k2\) be the rate constant at 50°C (323.15 K). Lets find the ratio \(\frac{k1}{k2}\) using the Arrhenius equation. Dividing the Arrhenius equation for \(k1\) by the equation for \(k2\), we get: \[\frac{k1}{k2} = \frac{A\text{exp}\left(-\frac{E_a}{8.314 \times 273.15}\right)}{A\text{exp}\left(-\frac{E_a}{8.314 \times 323.15}\right)}\]

Simplify the equation

We can now simplify the equation to solve for the ratio between the rate constants. Observe that the pre-exponential factor, \(A\), cancels out: \[\frac{k1}{k2} = \frac{\text{exp}\left(-\frac{E_a}{8.314 \times 273.15}\right)}{\text{exp}\left(-\frac{E_a}{8.314 \times 323.15}\right)}\] Next, we can combine the exponential terms: \[\frac{k1}{k2} = \text{exp}\left(\frac{E_a}{8.314}\times\left(\frac{1}{273.15} -\frac{1}{323.15}\right)\right)\]

Substitute given values and calculate

Now, substitute the given value for the activation energy (65,700 J/mol) into the equation, and calculate the ratio: \[\frac{k1}{k2} = \text{exp}\left(\frac{65700}{8.314}\times\left(\frac{1}{273.15} -\frac{1}{323.15}\right)\right)\] \[\frac{k1}{k2} \approx 8.64\]

Final answer

The reaction will occur approximately 8.64 times faster at 50°C compared to 0°C.

Key concepts

Activation Energy

Activation energy is a crucial concept in chemical kinetics. It refers to the minimum amount of energy required for a chemical reaction to occur. Think of it like a hill that reactants must climb over to transform into products. Without sufficient energy, the reactants can't make it over the hill, and the reaction won't happen.

In our lesson, we found that the activation energy of the reaction is 65.7 kJ/mol. A higher activation energy means reactions proceed more slowly because fewer molecules have enough energy to get over the energy barrier. Conversely, a lower activation energy means the reaction can happen more quickly.

Understanding activation energy helps in controlling how fast reactions occur, which is essential in industrial processes and even in everyday life, like cooking.

Arrhenius Equation

The Arrhenius equation is a formula that shows the relationship between the temperature and the rate constant of a reaction. It provides a way to calculate how fast a reaction is based on temperature and activation energy. This equation is given by:

• \[ k = A \times e^{-\frac{E_a}{RT}} \]

where:

• \( k \) is the rate constant
• \( A \) is the pre-exponential factor
• \( E_a \) is the activation energy
• \( R \) is the universal gas constant (8.314 J/mol K)
• \( T \) is the temperature in Kelvin

With this equation, by knowing the activation energy and the temperatures involved, we can calculate the change in the rate constant.

In the original exercise, we used the Arrhenius equation to determine how many times faster the reaction proceeds by comparing the rate constants at two different temperatures. This approach helps illustrate how temperature affects chemical processes, making it a pivotal tool for both chemists and engineers.

Rate Constant

The rate constant, denoted as \( k \), is a significant measurement in the study of chemical kinetics. It is a proportionality constant in the rate equation of a reaction that provides the relationship between the concentration of reactants and the rate of the reaction.

The value of \( k \) is influenced by factors like temperature and activation energy. Higher values of \( k \) generally indicate a faster reaction rate.

In our example, we compared the rate constants \( k_1 \) and \( k_2 \) at two different temperatures using the Arrhenius equation. This allowed us to see how the rate of reaction changes with temperature changes.

Understanding the rate constant helps in predicting how changing conditions will affect the speed of a chemical process, offering invaluable insights for controlling reactions in both laboratory and industry settings.