Problem 51 - Chemistry The Molecular Science

Question

These data were obtained for the rate constant for reaction of an unknown compound with water: \begin{tabular}{cc} \hline\(T\left({ }^{\circ} \mathrm{C}\right)\) & \(\mathrm{k}\left(\mathrm{s}^{-1}\right)\) \\ \hline 56.2 & \(1.04 \times 10^{-5}\) \\ 78.2 & \(1.45 \times 10^{-4}\) \\ \hline \end{tabular} (a) Calculate the activation energy and frequency factor for this reaction. (b) Estimate the rate constant of the reaction at a temperature of \(100.0^{\circ} \mathrm{C}\).

Step-by-Step Solution

Answer

The activation energy is approximately 80 kJ/mol, and the frequency factor is \(3.15 \times 10^7\). At 100°C, the rate constant is \(2.35 \times 10^{-3} s^{-1}\).

Step 1: Convert Temperatures to Kelvin

Start by converting the given temperatures in Celsius to Kelvin using the formula \(T(K) = T(°C) + 273.15\). For the temperatures given: \(T_1 = 56.2 + 273.15 = 329.35 \, K\) \(T_2 = 78.2 + 273.15 = 351.35 \, K\)

Step 2: Use the Arrhenius Equation

Recall the Arrhenius equation: \(k = A e^{-\frac{E_a}{RT}}\) where \(k\) is the rate constant, \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the universal gas constant \(8.314 \, J/mol \, K\), and \(T\) is the temperature in Kelvin. Here, \(k_1 = 1.04 \times 10^{-5} \, s^{-1}\) and \(k_2 = 1.45 \times 10^{-4} \, s^{-1}\).

Step 3: Arrhenius Equation Rearrangement

Rearrange the Arrhenius equation to find the activation energy by taking the natural logarithm: \(\ln(k) = \ln(A) - \frac{E_a}{R} \cdot \frac{1}{T}\). Use it for both given rate constants.

Step 4: Simplify to Calculate Activation Energy

Subtract the equations from Step 3 to eliminate \(\ln(A)\): \(\ln(k_2) - \ln(k_1) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\) This simplifies to \(\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)\).

Step 5: Calculate \(E_a\)

Substitute values: \(\ln\left(\frac{1.45 \times 10^{-4}}{1.04 \times 10^{-5}}\right) = \frac{E_a}{8.314} \left(\frac{1}{329.35} - \frac{1}{351.35}\right)\)Solve for \(E_a\): \(E_a \approx 80000 \, J/mol\) or \(80 \, kJ/mol\).

Step 6: Calculate the Frequency Factor \(A\)

Use the activation energy to solve for \(A\). Select one temperature and rate constant pair, e.g., \(T_1 = 329.35 \, K\) and \(k_1 = 1.04 \times 10^{-5} \, s^{-1}\): \(k_1 = A e^{-\frac{80000}{8.314 \times 329.35}}\) Solve for \(A\): \(A \approx 3.15 \times 10^7 \, s^{-1}\).

Step 7: Estimate Rate Constant at 100°C

Convert \(100°C\) to Kelvin: \(T_3 = 100 + 273.15 = 373.15 \, K\). Use the Arrhenius equation to solve for \(k_3\): \(k_3 = 3.15 \times 10^7 e^{-\frac{80000}{8.314 \times 373.15}}\).

Step 8: Solve for \(k_3\)

Calculate \(e^{-\frac{80000}{3103.79}}\) and then \(k_3\): \(k_3 \approx 2.35 \times 10^{-3} \, s^{-1}\).

Key Concepts

Arrhenius Equation

The Arrhenius Equation is a cornerstone in chemical kinetics, providing a mathematical relationship between the rate constant (k) of a reaction and its temperature. It is expressed as:\[ k = A e^{-\frac{E_a}{RT}} \]where:

• \( k \) is the reaction rate constant, which measures the speed of a chemical reaction.
• \( A \) is the frequency factor, representing the number of times molecules collide with the correct orientation to react per unit time.
• \( E_a \) refers to the activation energy, acting as a barrier that reactants need to overcome for a reaction to proceed.
• \( R \) is the universal gas constant, valued at \( 8.314 \, J/mol \, K \).
• \( T \) is the absolute temperature in Kelvin.

The equation elegantly shows that the rate constant increases with increasing temperature, highlighting how heat can accelerate chemical reactions by providing energy to surpass the activation energy barrier. It also exemplifies that the frequency factor \( A \) helps determine the likelihood of reactants successfully colliding and reacting.

Activation Energy

Activation energy \( E_a \) is a vital concept in chemical kinetics and is defined as the minimum energy required for reactants to transform into products. This energy barrier must be overcome for a reaction to occur, ensuring that molecules have enough energy to break and reform bonds.

In our given problem, we calculated the activation energy using the Arrhenius Equation's rearranged version for two different temperatures and their corresponding rate constants. The steps involved natural logarithms to isolate \( E_a \) and were as follows:

• Compute the natural logarithm of each rate constant to utilize the equation \( \ln(k) = \ln(A) - \frac{E_a}{R} \cdot \frac{1}{T} \).
• Subtract these two equations to derive an expression that eliminates \( \ln(A) \), thereby allowing us to focus on \( E_a \).
• Input known values and solve for \( E_a \), which for our example resulted in approximately \( 80000 \, J/mol \).

This activation energy barrier plays a crucial role in determining how quickly a reaction proceeds; lower activation energies lead to faster reactions under the same conditions.

Frequency Factor

The frequency factor, often denoted as \( A \), is a key component in the Arrhenius Equation. It describes the number of times reactant molecules collide with the correct orientation and sufficient energy within a specific time frame. The orientation and energy are critical because not every collision leads to a reaction.

To calculate \( A \), once we have \( E_a \), we can rearrange the Arrhenius Equation for one of the given temperatures and solve for \( A \). This entails:

• Selecting one pair of temperature and its corresponding rate constant.
• Plugging in these known values into the equation \( k = A e^{-\frac{E_a}{RT}} \).
• Solving for \( A \) gives us insight into the pre-exponential factor, revealing how often particles collide and attempt to react.

In the provided example, the calculated frequency factor was approximately \( 3.15 \times 10^7 \, s^{-1} \). This high number indicates frequent successful collisions under the given conditions, even though not all collisions lead to reaction.

Reaction Rate Constant

The reaction rate constant \( k \) is an essential parameter in kinetics, indicating the speed of a reaction at a given temperature. It is influenced by both the activation energy and the frequency factor. The rate constant changes with temperature, as predicted by the Arrhenius Equation.

To calculate the rate constant at a new temperature, as we did using \( 100°C \), the following steps apply:

• First, convert the temperature to Kelvin to use in the equation.
• Use the expression \( k = A e^{-\frac{E_a}{RT}} \), substituting the values for \( A \), \( E_a \), and the converted temperature.
• Compute the new \( k \) value, indicating the probable speed of the reaction at this higher temperature.

In the exercise, temperature increase raised the rate constant to \( 2.35 \times 10^{-3} \, s^{-1} \), showing that the reaction became significantly faster at \( 100°C \). This aligns with kinetic principles that suggest reactions hasten with rising temperatures due to more molecule collisions aided by higher thermal energy.