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 is approximately \(70.39^{\circ} \mathrm{C}\).

Step-by-step solution

Understand the Arrhenius Equation

The Arrhenius equation describes the temperature dependence of reaction rates. The 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, - \(R\) is the gas constant, and - \(T\) is the temperature in Kelvin.

Calculate the Temperature in Kelvin

We've been given the initial temperature in Celsius, but we need the temperature in Kelvin in order to use the Arrhenius equation. To convert Celsius to Kelvin, we will add 273.15: \(T_1 = 22^{\circ} \mathrm{C} + 273.15 = 295.15 \mathrm{K}\)

Use the Increased Rate Constant Factor

The rate constant has increased by a factor of 7. Therefore, we can represent the rate constant at the higher temperature as: \(k_2 = 7k_1\)

Set Up the Arrhenius Equation

We can now set up the Arrhenius equation at the higher temperature as: - For the initial temperature, \(k_1 = Ae^{-\frac{E_a}{R T_1}}\) - For the higher temperature, \(k_2 = Ae^{-\frac{E_a}{R T_2}}\)

Divide the Two Equations

Divide the second equation by the first equation: \(\frac{k_2}{k_1} = \frac{Ae^{-\frac{E_a}{R T_2}}}{Ae^{-\frac{E_a}{R T_1}}}\) Since \(k_2 = 7k_1\), substitute: \(7 = \frac{Ae^{-\frac{E_a}{R T_2}}}{Ae^{-\frac{E_a}{R T_1}}}\) The pre-exponential factors (A) cancel out, and we are left with: \(7 = \frac{e^{-\frac{E_a}{R T_2}}}{e^{-\frac{E_a}{R T_1}}}\)

Solve for the Higher Temperature

Rearrange the equation to solve for \(T_2\). Take the natural logarithm of both sides: \(\ln{7} = -\frac{E_a}{R T_2} + \frac{E_a}{R T_1}\) Now, make \(T_2\) the subject of the equation: \(T_2 = \frac{E_a(RT_1)}{RT_1\ln{7} + E_a}\) We know that \(E_a = 54.0 \mathrm{~kJ/mol} = 54000 \mathrm{J/mol}\), \(R = 8.314 \mathrm{J/mol*K}\), and \(T_1 = 295.15 \mathrm{K}\). Substitute these values into the equation: \(T_2 = \frac{54000(8.314)(295.15)}{(8.314)(295.15)\ln{7} + 54000}\)

Calculate the Higher Temperature

Compute the value of \(T_2\): \(T_2 \approx 343.54 \mathrm{K}\) Now convert the temperature back to Celsius: \(T_2 = 343.54 \mathrm{K} - 273.15 = 70.39^{\circ} \mathrm{C}\) The higher temperature is approximately \(70.39^{\circ} \mathrm{C}\).

Key concepts

Activation Energy

Activation energy (\(E_a\)) is the minimum energy that reactants must possess for a chemical reaction to occur. Think of it as the 'energy barrier' that reactants need to overcome to transform into products. In our everyday life, this concept is similar to needing a certain push to start a task—the push in chemistry is the activation energy.

In the given exercise, the reaction's activation energy is stated to be 54.0 kJ/mol. This value is crucial when examining how temperature affects reaction rates. By understanding the role of activation energy, we can predict and calculate the rate at which chemical reactions occur under different thermal conditions.

Rate Constant

The rate constant (\(k\)) is a proportionality factor that links the rate of a reaction to the concentration of the reactants. It is affected by various factors, with temperature being one of the most significant. In the exercise, an increase in temperature from 22°C to a higher temperature causes the rate constant to increase sevenfold, indicating that the rate at which the reaction occurs has also increased significantly.

Importance of Rate Constant

• A higher rate constant implies a faster reaction.
• The rate constant provides insight into the kinetics of a reaction.
• It allows for the calculation of reaction rates under varying conditions.

Recognizing the importance of the rate constant helps us appreciate how tiny changes in temperature can greatly influence the speed of chemical reactions.

Temperature and Reaction Rate

Temperature and the reaction rate are closely interlinked. Generally, as temperature increases, so does the reaction rate, up to a point. This is because higher temperatures provide more energy to the reactants, increasing the chances of collisions with enough energy to surpass the activation energy barrier.

Moreover, higher temperatures can also increase the number of effective collisions per unit time. This relationship is a cornerstone of chemical kinetics and is quantitatively expressed through the Arrhenius equation. The exercise exemplifies this relationship, with a temperature rise leading to an increase in the rate constant—and, consequently, the reaction rate—by a factor of seven.

Arrhenius Equation Problem Solving

To solve an Arrhenius equation problem, we need to understand its components and how to manipulate the equation. The exercise provided is a perfect example. We see that an accurate understanding of logarithms and exponential functions is vital to solve for the unknown temperature (\(T_2\)).

Key Steps in Problem Solving:

• Converting the given temperatures to Kelvin.
• Understanding how the rate constant changes with temperature.
• Applying the Arrhenius equation to set up a ratio of the rate constants at two different temperatures.
• Using algebraic manipulation, such as taking logarithms, to solve for the unknown.

Employing these steps, this problem demonstrates how to determine a new temperature at which a reaction's rate constant increases by a specified factor.