Problem 104 - Chemistry The Central Science In Si Units

Question

The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: \begin{tabular}{ll} \hline Temperature \((\mathrm{K})\) & Rate Constant \(\left(\mathrm{s}^{-1}\right)\) \\ \hline 300 & \(3.2 \times 10^{-11}\) \\ 320 & \(1.0 \times 10^{-9}\) \\ 340 & \(3.0 \times 10^{-8}\) \\ 355 & \(2.4 \times 10^{-7}\) \\ \hline \end{tabular} From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\).

Step-by-Step Solution

Answer

The activation energy is approximately 171.8 kJ/mol.

Step 1: Understand the Arrhenius Equation

The Arrhenius equation describes how the rate constant \( k \) of a chemical reaction depends on temperature \( T \) and is given by: \[ k = A e^{-\frac{E_a}{RT}} \] where \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the ideal gas constant \( 8.314 \text{ J mol}^{-1} \text{ K}^{-1} \), and \( T \) is the temperature in Kelvin.

Step 2: Linearize the Arrhenius Equation

Taking the natural logarithm of both sides of the Arrhenius equation gives: \[ \ln k = \ln A - \frac{E_a}{R} \frac{1}{T} \] This equation is in the form of \( y = mx + b \), where \( y = \ln k \), \( x = \frac{1}{T} \), \( m = -\frac{E_a}{R} \), and \( b = \ln A \). We can plot \( \ln k \) versus \( \frac{1}{T} \) to find the slope \( m = -\frac{E_a}{R} \).

Step 3: Calculate \( \ln k \) for Each Temperature

Calculate the natural logarithm of the given rate constants: - For 300 K: \( \ln(3.2 \times 10^{-11}) \approx -24.58 \)- For 320 K: \( \ln(1.0 \times 10^{-9}) \approx -20.72 \)- For 340 K: \( \ln(3.0 \times 10^{-8}) \approx -17.62 \)- For 355 K: \( \ln(2.4 \times 10^{-7}) \approx -15.24 \)

Step 4: Calculate \( \frac{1}{T} \) for Each Temperature

Calculate \( \frac{1}{T} \) for each temperature in the dataset:- For 300 K: \( \frac{1}{300} = 0.00333 \)- For 320 K: \( \frac{1}{320} = 0.00313 \)- For 340 K: \( \frac{1}{340} = 0.00294 \)- For 355 K: \( \frac{1}{355} = 0.00282 \)

Step 5: Plot and Determine the Slope

You would plot \( \ln k \) versus \( \frac{1}{T} \) using the calculated values. Using linear regression or a graph, determine the slope of the line \( m = -\frac{E_a}{R} \). For these specific data points, the slope \( m \) is calculated to be approximately \( -2.068 \times 10^4 \).

Step 6: Calculate Activation Energy

Use the slope to solve for \( E_a \): \[ E_a = -m \cdot R \]Given \( m = -2.068 \times 10^4 \), the activation energy \( E_a \) is: \[ E_a = (-2.068 \times 10^4 \text{ K} ) \times (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \approx 171800 \text{ J mol}^{-1} \]Convert this to \( \text{kJ mol}^{-1} \):\( E_a = 171.8 \text{ kJ mol}^{-1} \).

Key Concepts

Arrhenius Equation

The Arrhenius equation is fundamental when studying chemical kinetics. It provides a clear link between the rate constant of a reaction and the temperature at which the reaction occurs. This equation is expressed as: \[ k = A e^{-\frac{E_a}{RT}} \]

• k is the rate constant, which gives the speed of a reaction at a given temperature.
• A is the pre-exponential factor, also sometimes called the frequency factor, representing the number of effective collisions between molecules.
• e denotes the exponential function, which reflects how the rate decreases with increasing activation energy.
• E_a stands for the activation energy, the minimum energy required for a reaction to proceed.
• R is the ideal gas constant with a value of 8.314 J/mol·K.
• T represents the absolute temperature in Kelvin.

By analyzing this formula, we see that as the temperature increases, the fraction \(\frac{E_a}{RT}\) decreases, resulting in a larger value of exponential \( e^{-\frac{E_a}{RT}} \), hence a higher rate constant \( k \). This means reactions tend to occur faster at higher temperatures.

Temperature Dependence

The relationship between temperature and reaction rates is a key aspect of chemical kinetics. According to the Arrhenius equation, the rate constant \( k \) is significantly influenced by temperature changes. The relationship is exponential, indicating that even small temperature changes can lead to large variations in reaction rates.Understanding this temperature dependence helps in practical applications, such as:

• Predicting how reaction speed changes with seasonal temperature variations.
• Controlling reaction conditions in industrial processes explicitly.
• Designing chemical storage and usage strategies to avoid unwanted reactions.

By using the linear form of the Arrhenius equation, \( \ln k = \ln A - \frac{E_a}{R} \frac{1}{T} \), we can plot \( \ln k \) against \( \frac{1}{T} \). The plot yields a straight line, allowing us to determine the activation energy. This insight reveals how sensitive a reaction's rate is to temperature changes, offering an intuitive understanding of why heating or cooling can either speed up or slow down a reaction.

Rate Constant

The rate constant \( k \) plays a crucial role in understanding how fast a reaction occurs at a given temperature. From a mathematical perspective, \( k \) is a proportionality factor in the rate law, linking the reaction rate to the concentrations of reactants. For first-order reactions, as given in the exercise, the rate depends solely on the concentration of one reactant and \( k \) has the units of second inverse \( \text{s}^{-1} \).What influences the rate constant?

• The intrinsic nature of the reactants (strength of bonds and molecular configuration).
• Temperature, as discussed, has a profound effect due to its impact on kinetic energy and collision frequency.
• Presence of catalysts that can alter the reaction path to lower the activation energy \( E_a \), hence increasing \( k \).

By analyzing rate constants at different temperatures, scientists can gather information about the dynamics of a reaction, including its mechanism and energy profile. Knowing \( k \) at various temperatures aids in making predictions about the reaction's behavior under different conditions, which is vital for chemical engineering and development.