Problem 54 - Chemistry: The Central Science in SI Units, Global Edition

Question

(a) The activation energy for the reaction \(\mathrm{A}(g) \longrightarrow \mathrm{B}(g)\) is \(100 \mathrm{~kJ} / \mathrm{mol}\). Calculate the fraction of the molecule A that has an energy equal to or greater than the activation energy at \(400 \mathrm{~K} .(\mathbf{b})\) Calculate this fraction for a temperature of \(500 \mathrm{~K}\). What is the ratio of the fraction at \(500 \mathrm{~K}\) to that at \(400 \mathrm{~K}\) ?

Step-by-Step Solution

Answer

The ratio of the fractions of molecule A with energy equal to or greater than the activation energy at 500 K and 400 K is given by the following formula: \[\frac{f(E_{500})}{f(E_{400})} = \frac{\mathrm{e}^{-(100\times10^3)/(1.38\times10^{-23}\times500)}}{\mathrm{e}^{-(100\times10^3)/(1.38\times10^{-23}\times400)}}\] Plug in the activation energy and temperatures into the formula and calculate the ratio.

Step 1: Understand the Boltzmann Distribution formula

The Boltzmann Distribution formula is given by: \[f(E) = A \mathrm{e}^{-E/kT}\] where - \(f(E)\) is the fraction of molecules with energy E, - E is the energy (in this case, the activation energy), - A is a constant, - k is the Boltzmann constant, \(1.38 \times 10^{-23}\mathrm{J\/K}\), - T is the temperature in Kelvin.

Step 2: Plug in values and calculate the fraction at 400 K

At 400 K, the fraction of molecules with energy equal to or greater than the activation energy (100 kJ/mol) is: \[f(E) = A \mathrm{e}^{-100\times10^3\mathrm{J/mol} / (1.38\times10^{-23}\mathrm{J/K} \times 400\mathrm{K})}\] Calculate the value of \(f(E)\) at 400 K: \(f(E_{400}) = A \times \mathrm{e}^{-E/1.38\times10^{-23}\times 400}\)

Step 3: Calculate the fraction at 500 K

At 500 K, the fraction of molecules with energy equal to or greater than the activation energy (100 kJ/mol) is: \[f(E) = A \mathrm{e}^{-100\times10^3\mathrm{J/mol} / (1.38\times10^{-23}\mathrm{J/K} \times 500\mathrm{K})}\] Calculate the value of \(f(E)\) at 500 K: \(f(E_{500}) = A \times \mathrm{e}^{-E/1.38\times10^{-23}\times 500}\)

Step 4: Calculate the ratio of the fractions at 500 K and 400 K

To find the ratio of the fractions at 500 K to 400 K, divide the fraction at 500 K by the fraction at 400 K: \[\frac{f(E_{500})}{f(E_{400})} = \frac{A \times \mathrm{e}^{-E/1.38\times10^{-23}\times 500}}{A \times \mathrm{e}^{-E/1.38\times10^{-23}\times 400}}\] The constant A cancels out in the ratio, so the final expression for the ratio is: \[\frac{f(E_{500})}{f(E_{400})} = \frac{\mathrm{e}^{-E/1.38\times10^{-23}\times500}}{\mathrm{e}^{-E/1.38\times10^{-23}\times400}}\] Calculate the ratio using the given activation energy (100 kJ/mol) and temperatures (400 K and 500 K) to get: \[\frac{f(E_{500})}{f(E_{400})} = \frac{\mathrm{e}^{-(100\times10^3)/(1.38\times10^{-23}\times500)}}{\mathrm{e}^{-(100\times10^3)/(1.38\times10^{-23}\times400)}}\]

Key Concepts

Activation Energy

Activation energy is the minimum energy required for a chemical reaction to occur. It is often described using a potential energy diagram where the activation energy is the peak that must be overcome for reactants to become products.
Imagine a roller coaster that needs to reach the top of a hill before it can descend; similarly, molecules need enough energy to surpass the activation energy barrier for a reaction.

• Higher activation energy: Fewer molecules have enough energy to react.
• Lower activation energy: More molecules can participate in the reaction.

When discussing reaction kinetics, activation energy is crucial because it determines how fast a reaction will proceed. A high activation energy means fewer successful collisions per second, leading to slower reaction rates. This is why catalysts are often used; they lower the activation energy, allowing more reactions to occur at a faster rate.

Reaction Kinetics

Reaction kinetics is the study of the rate at which chemical reactions occur and the factors affecting them. By understanding reaction kinetics, we can predict how a chemical process will behave under different conditions.
Several key principles govern reaction kinetics:

• Collision Theory: Molecules must collide to react. The number of collisions can be increased by raising the concentration or temperature.
• Effectiveness of Collisions: Not all collisions produce a reaction; they must have the right orientation and enough energy, surpassing the activation energy.
• Rate Laws: These mathematical equations describe the relationship between the concentration of reactants and the rate of the reaction.

Reaction kinetics also includes the study of reaction mechanisms, which are step-by-step sequences of elementary reactions. Understanding the kinetics helps in industrial processes, ensuring reactions occur quickly and efficiently.

Temperature Dependence

Temperature has a profound impact on chemical reactions, as it influences how molecules move and interact. In general, increasing the temperature speeds up a chemical reaction.
This is because at higher temperatures:

• Molecules have more kinetic energy, leading to more frequent and energetic collisions.
• A greater proportion of molecules have enough energy to overcome the activation energy barrier.

The relationship between temperature and reaction rate can be mathematically described using the Arrhenius equation:\[k = A \, \text{e}^{-E_a/RT}\\]where - \(k\) is the rate constant,- \(A\) is the pre-exponential factor, a constant,- \(E_a\) is the activation energy,- \(R\) is the universal gas constant,- \(T\) is the temperature in Kelvin.
As shown in the exercise, when the temperature increases from 400 K to 500 K, the fraction of molecules with sufficient energy increases, emphasizing the temperature dependence of reaction rates.