Problem 92 - Chemistry: An Atoms-Focused Approach

Question

Chlorine atoms react with methane, forming HC1 and \(\mathrm{CH}_{3} .\) The rate constant for the reaction is \(6.0 \times 10^{7} M^{-1} s^{-1}\) at \(298 \mathrm{K} .\) When the experiment was repeated at three other temperatures, the following data were collected: $$\begin{array}{ll}T(\mathrm{K}) & k\left(M^{-1} \mathrm{s}^{-1}\right) \\\303 & 6.5 \times 10^{7} \\\\\hline 308 & 7.0 \times 10^{7} \\\\\hline 313 & 7.5 \times 10^{7} \\\\\hline\end{array}$$ a. Calculate the values of the activation energy and the frequency factor for the reaction. b. What is the value of the rate constant in the lower stratosphere where \(T=218 \mathrm{K} ?\)

Step-by-Step Solution

Answer

Question: Determine the activation energy (Ea) and frequency factor (A) for the reaction given the following rate constants and temperatures: k1 = 6.0 * 10^7 M^-1 s^-1 at T1 = 298 K, k2 = 6.5 * 10^7 M^-1 s^-1 at T2 = 303 K. Calculate the rate constant at 218 K using the Arrhenius equation. Answer: The activation energy (Ea) for the reaction is approximately 2.10 * 10^4 J mol^-1, the frequency factor (A) is approximately 1.52 * 10^12 M^-1 s^-1, and the rate constant at 218 K is approximately 3.0 * 10^5 M^-1 s^-1.

Step 1: Determine the activation energy (Ea)

We have four data points of temperature and rate constants. We can use any two of them to determine the activation energy. We'll use the first two data points for this purpose: k1 = 6.0 * 10^7 M^-1 s^-1 at T1 = 298 K k2 = 6.5 * 10^7 M^-1 s^-1 at T2 = 303 K Using the Arrhenius equation, we will set up a system of equations: k1 / k2 = e^(Ea / (R * T1) - Ea / (R * T2)) Solve for Ea: Ea = R * (T1 * T2) * ln(k1 / k2) / (T2 - T1) Ea = (8.314 J mol^-1 K^-1) * (298K * 303K) * ln((6.0 * 10^7) / (6.5 * 10^7)) / (303K - 298K) Ea = 2.10 * 10^4 J mol^-1

Step 2: Determine the frequency factor (A)

Now that we have calculated the activation energy, we can determine the frequency factor (A). We can use any of the data points for this. We'll use the first data point: k1 = 6.0 * 10^7 M^-1 s^-1 at T1 = 298 K Rearrange the Arrhenius equation to solve for A: A = k1 / e^(-Ea / (R * T1)) A = (6.0 * 10^7 M^-1 s^-1) / e^( -2.10 * 10^4 J mol^-1 / (8.314 J mol^-1 K^-1 * 298 K)) A = 1.52 * 10^12 M^-1 s^-1

Step 3: Calculate the rate constant at T = 218 K

Now that we have calculated the activation energy (Ea) and the frequency factor (A), we can determine the rate constant (k) at the lower stratosphere temperature of T = 218 K using the Arrhenius equation: k = A * e^(-Ea / (R * T)) k = (1.52 * 10^12 M^-1 s^-1) * e^[-(2.1 * 10^4 J mol^-1) / (8.314 J mol^-1 K^-1 * 218 K)] k = 3.0 * 10^5 M^-1 s^-1 The value of the rate constant in the lower stratosphere where T = 218 K is approximately 3.0 * 10^5 M^-1 s^-1.

Key Concepts

Activation Energy

Activation energy, denoted as \( E_a \), is the minimum energy required for a chemical reaction to occur. It can be thought of as a barrier that reactant particles must overcome to transform into products. The Arrhenius equation highlights the relationship between activation energy and the rate constant: \[k = A \, \exp\left(-\frac{E_a}{RT}\right)\] where \( k \) is the rate constant, \( A \) is the frequency factor, \( R \) is the universal gas constant \( (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \), and \( T \) is the temperature in Kelvin.

• **Higher activation energy means a slower reaction**, as fewer molecules have sufficient energy to react.
• **Lower activation energy accelerates reactions**, enabling more molecules to attain the necessary energy to overcome the barrier.

We can calculate the activation energy using experimental data by rearranging the Arrhenius equation, typically using the natural logarithm to solve for \( E_a \).

Reaction Rate Constant

The reaction rate constant, \( k \), is a crucial element in chemical kinetics that describes the speed of a reaction. It varies with temperature and is calculated using the Arrhenius equation:\[k = A \, \exp\left(-\frac{E_a}{RT}\right)\] The rate constant is often used to gauge how fast reactants are converted into products.

• At higher temperatures, \( k \) usually increases, indicating faster reactions.
• At lower temperatures, \( k \) tends to decrease, meaning slower reaction rates.

Knowing \( k \) helps predict how quickly a reaction will proceed under certain conditions, making it vital in both industrial and laboratory settings. The value of \( k \) often changes significantly with small temperature shifts, given the exponential term in the Arrhenius equation.

Temperature Dependence

The temperature dependence of reaction rates is a significant aspect of chemical kinetics. According to the Arrhenius equation, the rate constant \( k \) depends exponentially on the temperature \( T \). This relationship can be expressed as:\[k = A \, \exp\left(-\frac{E_a}{RT}\right)\] As temperature increases, kinetic energy of molecules increases, resulting in:

• More collisions between reactant molecules.
• More energy available for molecules to overcome the activation energy barrier.

This renders reactions faster at higher temperatures. The exponential nature of the Arrhenius equation implies that even modest increases in temperature can lead to significant increases in the reaction rate, making temperature control critical in many chemical processes.

Frequency Factor

The frequency factor, denoted as \( A \), is a component of the Arrhenius equation that signifies the frequency of collisions between reactant molecules with correct orientation leading to product formation. It can be viewed in the rewritten Arrhenius equation:\[k = A \, \exp\left(-\frac{E_a}{RT}\right)\] The frequency factor comprises two major elements:

• The **collision frequency**, how often molecules collide.
• The **steric factor**, the fraction of collisions where molecules are in the correct orientation.

Though often regarded as a constant for a given reaction, \( A \) can actually be influenced by factors such as pressure and orientation. **Influencing \( A \) can enhance reaction rates**, especially in scenarios where the activation energy is already optimized, illustrating why controlling conditions like pressure is essential in industrial chemistry.