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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:
   $$E_a=-8.314\cdot 290\cdot \ln \left(\frac{5.0\times {10}^{-4}}{3.0\times {10}^{10}}\right)\approx 75,280\text{J/mol}.$$