Question

Consider the hypothetical reaction \(\mathrm{A}_{2}(g)+\mathrm{B}_{2}(g) \longrightarrow\) \(2 \mathrm{AB}(g),\) where the rate law is: $$-\frac{\Delta\left[\mathrm{A}_{2}\right]}{\Delta t}=k\left[\mathrm{A}_{2}\right]\left[\mathrm{B}_{2}\right]$$ The value of the rate constant at \(302^{\circ} \mathrm{C}\) is \(2.45 \times 10^{-4} \mathrm{L} / \mathrm{mol}\). s, and at \(508^{\circ} \mathrm{C}\) the rate constant is \(0.891 \mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\). What is the activation energy for this reaction? What is the value of the rate constant for this reaction at \(375^{\circ} \mathrm{C} ?\)

Short answer

The activation energy for this reaction is approximately 1.364 x 10^5 J/mol. The value of the rate constant for this reaction at 375 °C is approximately 1.056 x 10^-3 L/mol.s.

Step-by-step solution

Step 1. Write down the Arrhenius equation

The Arrhenius equation is given as: \(k = Ae^{\frac{-E_a}{RT}}\) where: - k is the rate constant - A is the pre-exponential factor - \(E_a\) is the activation energy - R is the gas constant (8.314 J/mol.K) - T is the temperature in Kelvin

Step 2. Convert Celsius temperature to Kelvin and use ln on Arrhenius equation

For our ease, we will convert the Celsius temperatures given to Kelvin: - T1 = 302 °C + 273.15 = 575.15 K - T2 = 508 °C + 273.15 = 781.15 K Now, let's take a natural logarithm on both sides of the Arrhenius equation: \(ln(k) = ln(A) - \frac{E_a}{RT}\)

Step 3. Setup two equations and eliminate the unknown pre-exponential factor

Now, we substitute the two given rate constant values at two different temperatures: For T1: \(ln(2.45 \times 10^{-4}) = ln(A) - \frac{E_a}{8.314 \times 575.15}\) -----(1) For T2: \(ln(0.891) = ln(A) - \frac{E_a}{8.314 \times 781.15}\) -----(2) Now, we will eliminate A from the equations. Subtract equation (1) from equation (2): \(ln\left(\frac{0.891}{2.45 \times 10^{-4}}\right) = \frac{E_a}{8.314}\left(\frac{1}{575.15} - \frac{1}{781.15}\right)\)

Step 4. Solve for the activation energy

Rearrange and solve for \(E_a\): \(E_a = 8.314 \times ln\left(\frac{0.891}{2.45 \times 10^{-4}}\right) \times \frac{1}{\left(\frac{1}{575.15} - \frac{1}{781.15}\right)}\) \(E_a \approx 1.364 \times 10^5 \text{ J/mol}\) So the activation energy for this reaction is approximately 1.364 x 10^5 J/mol.

Step 5. Calculate the value of the rate constant at 375 °C

First, let's convert the desired temperature to Kelvin: - T3 = 375 °C + 273.15 = 648.15 K Now, we can use any of the original equations (such as equation 1) and plug in the activation energy to calculate the pre-exponential factor A. From equation (1): \(ln(A) = ln(2.45 \times 10^{-4}) + \frac{1.364 \times 10^5}{8.314 \times 575.15}\) Now calculate A: \(A \approx 5.544 \times 10^6 \text{ L/mol.s}\) Now use the Arrhenius equation to find the rate constant at 375 °C (T3): \(k_3 = 5.544 \times 10^6 \cdot e^{\frac{-1.364 \times 10^5}{8.314 \times 648.15}}\) \(k_3 \approx 1.056 \times 10^{-3} \text{ L/mol.s}\) The value of the rate constant for this reaction at 375 °C is approximately 1.056 x 10^-3 L/mol.s.

Key concepts

Arrhenius Equation

Understanding the Arrhenius equation is key to grasping why some chemical reactions occur faster at higher temperatures. It links the rate constant of a reaction to the temperature and activation energy, providing invaluable insights into the behavior of chemical reactions under various conditions.

The Arrhenius equation is mathematically expressed as:
\[ k = Ae^{\frac{-E_a}{RT}} \]
where \( k \) is the rate constant, \( A \) represents the pre-exponential factor (often related to the frequency of collisions), \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the absolute temperature in Kelvin.

By taking the natural logarithm of this equation, we get a linear relationship that allows us to extract meaningful values such as the activation energy—a crucial hurdle that molecules must overcome to react—by investigating the reaction rates at different temperatures.

Rate Constant

The rate constant, symbolized by \( k \), is quite literally the 'constant' factor of proportionality in the rate law of a chemical reaction, connecting the reactant concentrations to the reaction rate. A unique value for every reaction, it changes with temperature and is influenced by the presence of a catalyst.

In our problem, we saw how the values of the rate constant at different temperatures allowed us to calculate the activation energy via the Arrhenius equation. This signifies a direct relationship between the rate constant and temperature: as the temperature increases, so does the rate constant, generally resulting in a faster reaction. This temperature dependency is mathematically captured by the Arrhenius equation, underscoring the centrality of the rate constant in chemical kinetics.

Understanding the rate constant's role paves the way to control reaction speeds—a factor critical in industries ranging from pharmaceuticals to materials science.

Chemical Kinetics

Chemical kinetics is the study of reaction rates and the factors that affect them. This field of chemistry is vital for developing new reactions and optimizing existing ones, as it governs the speed at which products are formed from reactants.

In the textbook exercise, the rate law equation \[ -\frac{\Delta[\mathrm{A}_{2}]}{\Delta t}=k[\mathrm{A}_{2}][\mathrm{B}_{2}] \]
is an example of kinetic analysis. It tells us that the rate at which \( \mathrm{A}_{2} \) disappears is proportional to its concentration and that of \( \mathrm{B}_{2} \). To further elucidate the mechanics of the reaction, we apply the Arrhenius equation and explore the role of activation energy in rate determination.

Through such exercises, we learn not only how to predict and control reaction speeds but also how to design reactions for desired outcomes, be it in making a new medicine or reducing pollution in car exhausts. Chemical kinetics, encompassing rate laws, reaction mechanisms, and the Arrhenius equation, offers a robust framework for such endeavors.