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Overview of the Arrhenius Equation

The Arrhenius equation is a foundational concept in chemical kinetics. It connects the rate constant (\( k \)) with activation energy (\( E_a \)), temperature (\( T \)), and the pre-exponential factor (\( A \)).

The equation is expressed as:

\( k = A \cdot e^{\frac{-E_a}{R \cdot T}} \)

Key terms explained:

  • \( k \): Rate constant, indicating the reaction’s speed.
  • \( A \): Pre-exponential factor, reflecting the frequency of effective collisions.
  • \( E_a \): Activation energy, the minimum energy required for a reaction to proceed.
  • \( R \): Gas constant (\( 8.314 \, \text{J/mol·K} \)).
  • \( T \): Absolute temperature in Kelvin.

This equation emphasizes how higher temperatures or lower activation energy increases reaction rates.

Other Relationships

    Logarithmic Form: \( \ln(k) = \ln(A) – \frac{E_a}{RT} \)

    This linearized form helps determine \( E_a \) and \( A \) using an Arrhenius plot.

    Temperature Dependence: \( \frac{k_2}{k_1} = e^{\frac{E_a}{R} \left(\frac{1}{T_1} – \frac{1}{T_2}\right)} \)

    This equation relates rate constants at two temperatures.

The Rate Constant and Its Role in Reaction Dynamics

The rate constant (\( k \)) is a pivotal parameter in chemical kinetics, dictating the speed of a reaction. It depends on the reaction’s nature and temperature. The relationship between the rate constant and reactant concentrations is described by the rate equation:

\( \text{Rate} = k \cdot [A]^m \cdot [B]^n \)

Where:

  • \( [A] \) and \( [B] \): Reactant concentrations.
  • \( m \) and \( n \): Reaction orders with respect to \( A \) and \( B \).
  • \( k \): Rate constant.

Units in the Arrhenius Equation

Each term in the Arrhenius equation requires specific units for consistency:

  • Rate constant (\( k \)): Depends on reaction order. For a first-order reaction, units are \( \text{s}^{-1} \).
  • Activation energy (\( E_a \)): Typically in \( \text{J/mol} \) or \( \text{kJ/mol} \).
  • Temperature (\( T \)): Always in Kelvin.
  • Pre-exponential factor (\( A \)): Shares units with \( k \).

Ensuring correct unit conversions is crucial for accurate calculations. Use our calculator to check accuracy of your calculations.

Arrhenius Plot

The Arrhenius equation can be linearized for graphical analysis:

\( \ln(k) = \ln(A) – \frac{E_a}{R} \cdot \frac{1}{T} \)

Plotting \( \ln(k) \) against \( \frac{1}{T} \) yields a straight line, where:

  • Slope: \( -\frac{E_a}{R} \), revealing the activation energy.
  • Intercept: \( \ln(A) \), representing the pre-exponential factor.

This method facilitates experimental determination of \( E_a \) and \( A \).

The Effect of Catalysts

Catalysts enhance reaction rates by lowering \( E_a \). While the Arrhenius equation does not explicitly include catalysts, their effect is observed in the reduced activation energy, leading to a higher \( k \) for the same \( T \).

How to Solve Problems Using the Arrhenius Equation

  1. Identify the Given Values:

    Start by noting the provided quantities, such as \( A \), \( E_a \), \( T \), or \( k \).

  2. Select the Appropriate Formula:

    Decide which form of the Arrhenius equation to use:

    • Use \( k = A e^{-\frac{E_a}{RT}} \) if \( k \) is to be calculated directly.
    • Use \( \ln(k) = \ln(A) – \frac{E_a}{RT} \) for an Arrhenius plot or logarithmic analysis.
    • Use \( \frac{k_2}{k_1} = e^{\frac{E_a}{R} \left(\frac{1}{T_1} – \frac{1}{T_2}\right)} \) for comparing rates at different temperatures.
  3. Substitute the Known Values:

    Ensure all units are consistent:

    • \( E_a \): Convert to Joules (\( 1 \, \text{kJ} = 1000 \, \text{J} \)) if given in kilojoules.
    • \( T \): Use Kelvin (\( \text{K} = \text{°C} + 273.15 \)).
    • \( R \): Use \( 8.314 \, \text{J mol}^{-1} \text{K}^{-1} \).

  4. Perform Calculations:

    Solve step-by-step using a calculator or computational tool. For logarithmic equations, use:

    • \( \ln(x) \) for natural log
    • \( e^x \) for the exponential function

  5. Interpret the Results:

    Verify that the units are correct and the value is reasonable for the given context (e.g., a positive rate constant).

Calculation Examples

Example 1: Determining the Rate Constant (\( k \))

Given:

  • \( A = 1.0 \times 10^{13} \, \text{s}^{-1} \)
  • \( E_a = 75,000 \, \text{J/mol} \)
  • \( T = 298 \, \text{K} \)

Solution:

\( k = A \cdot e^{\frac{-E_a}{R \cdot T}} = 1.0 \times 10^{13} \cdot e^{\frac{-75,000}{8.314 \cdot 298}} \approx 1.74 \times 10^{-1} \, \text{s}^{-1} \)

Example 2: Calculating Activation Energy (\( E_a \))

Given:

  • \( k = 2.0 \times 10^{-5} \, \text{s}^{-1} \)
  • \( A = 1.0 \times 10^{13} \, \text{s}^{-1} \)
  • \( T = 298 \, \text{K} \)

Solution:

\( E_a = -R \cdot T \cdot \ln\left(\frac{k}{A}\right) \)

Substitute values to find \( E_a \).

Questions for Practice

  1. What is the relationship between the rate constant (\( k \)) and activation energy (\( E_a \))?
    The rate constant (\( k \)) is inversely related to the activation energy (\( E_a \)). Higher activation energy results in a lower \( k \), meaning a slower reaction rate at a given temperature. This relationship is quantified by the Arrhenius equation \( k = A \cdot e^{\frac{-E_a}{R \cdot T}} \).
  2. Explain how the Arrhenius equation accounts for temperature changes in reaction rates.
    The Arrhenius equation shows that as temperature increases, the exponential factor \( e^{\frac{-E_a}{R \cdot T}} \) becomes larger, leading to an increased rate constant (\( k \)). This reflects the greater number of molecules with sufficient energy to overcome the activation energy barrier.
  3. Describe the significance of the pre-exponential factor (\( A \)) in the equation.
    The pre-exponential factor (\( A \)) represents the frequency of collisions and the proper orientation of reactants for a successful reaction. It determines the upper limit of the rate constant (\( k \)) and depends on the specific characteristics of the reaction.
  4. How can an Arrhenius plot be used to calculate activation energy?
    An Arrhenius plot graphs \( \ln(k) \) versus \( \frac{1}{T} \). The slope of the line is equal to \( -\frac{E_a}{R} \). By determining the slope experimentally, the activation energy (\( E_a \)) can be calculated.
  5. Derive the units of \( k \) for a second-order reaction.
    For a second-order reaction, the rate equation is \( \text{Rate} = k \cdot [A]^2 \). The unit of the rate is \( \text{mol/L·s} \) and the unit of concentration is \( \text{mol/L} \). Thus, the units of \( k \) are derived as:
    \[ \text{Units of } k = \frac{\text{mol/L·s}}{(\text{mol/L})^2} = \text{L/mol·s}. \]
  6. Calculate \( k \) for a reaction with \( A = 2.0 \times 10^{12} \, \text{s}^{-1} \), \( E_a = 85,000 \, \text{J/mol} \), and \( T = 310 \, \text{K} \).
    Using the Arrhenius equation:
    \[ k = A \cdot e^{\frac{-E_a}{R \cdot T}}, \] where \( R = 8.314 \, \text{J/mol·K} \):
    \[ k = 2.0 \times 10^{12} \cdot e^{\frac{-85,000}{8.314 \cdot 310}} \approx 1.13 \, \text{s}^{-1}. \]
  7. Determine \( E_a \) if \( k = 5.0 \times 10^{-4} \, \text{s}^{-1} \), \( A = 3.0 \times 10^{10} \, \text{s}^{-1} \), and \( T = 290 \, \text{K} \).
    Rearrange the Arrhenius equation to solve for \( E_a \):
    \[ E_a = -R \cdot T \cdot \ln\left(\frac{k}{A}\right), \] substituting the given values:
    \[ E_a = -8.314 \cdot 290 \cdot \ln\left(\frac{5.0 \times 10^{-4}}{3.0 \times 10^{10}}\right) \approx 75,280 \, \text{J/mol}. \]