R2.2.13 Arrhenius factor (Higher Level Only)

The Arrhenius Factor and Activation Energy: Understanding Reaction Rates

The Role of the Arrhenius Factor in Collision Theory

In collision theory, three critical conditions must be met for a chemical reaction to occur:

Collision Frequency: Reactant particles must collide with one another.
Proper Orientation: The particles must align correctly during the collision to allow bonds to break and form.
Sufficient Energy: The particles must possess enough kinetic energy to overcome the activation energy barrier ( $E_{a}$ ).

The Arrhenius factor, $A$ , quantifies the first two conditions: how often particles collide and how frequently those collisions occur with the proper orientation.

Example

For small, simple molecules, $A$ tends to be large because most collisions have favorable orientations. In contrast, $A$ is smaller for large, complex molecules, where fewer collisions occur in the correct orientation.

Analogy

Think of $A$ as the number of green lights you encounter on a road trip. The more green lights (favorable collisions) you hit, the faster you reach your destination (reaction completion).

The Arrhenius Equation: A Mathematical Model of Reaction Rates

The Arrhenius equation provides a quantitative relationship between the rate constant ( $k$ ) of a reaction, the temperature ( $T$ ), and the activation energy ( $E_{a}$ ):

$k = A e^{- \frac{E_{a}}{R T}}$

Where:

$k$ : rate constant (units depend on the reaction order)
$A$ : Arrhenius factor, or frequency factor
$E_{a}$ : activation energy (in joules per mole, ${J mol}^{- 1}$ )
$R$ : universal gas constant ( $8.31 {J K}^{- 1} {mol}^{- 1}$ )
$T$ : absolute temperature (in kelvins, $K$ )

Key Insights from the Arrhenius Equation:

Temperature Dependence:
- As $T$ increases, the exponential term $e^{- \frac{E_{a}}{R T}}$ grows larger, leading to a higher $k$ .
- This explains why reactions proceed faster at elevated temperatures.
Impact of $E_{a}$ :
- A higher $E_{a}$ reduces the number of particles with sufficient energy to overcome the activation barrier, resulting in a smaller $k$ .
Importance of $A$ :
- The value of $A$ reflects the likelihood of favorable collisions, making it an essential factor in determining reaction rates.

Tip

Small increases in temperature can cause significant changes in reaction rates due to the sensitivity of the exponential term $e^{- \frac{E_{a}}{R T}}$ to temperature.

Determining Activation Energy and the Arrhenius Factor

Experimental data can be used to determine $E_{a}$ and $A$ by employing the logarithmic form of the Arrhenius equation:

$\ln k = - \frac{E_{a}}{R} \cdot \frac{1}{T} + \ln A$

This equation takes the form of a straight line ( $y = m x + c$ ), where:

$y = \ln k$
$x = \frac{1}{T}$
Slope ( $m$ ) = $- \frac{E_{a}}{R}$
Intercept ( $c$ ) = $\ln A$

Experimental Procedure:

Measure $k$ at Different Temperatures: Perform the reaction at several temperatures and record the rate constant ( $k$ ) for each.
Plot $\ln k$ vs. $\frac{1}{T}$ : Create a graph of $\ln k$ (y-axis) against $\frac{1}{T}$ (x-axis).
Determine $E_{a}$ : Calculate the slope of the line ( $- \frac{E_{a}}{R}$ ) and rearrange to find $E_{a}$ .
Determine $A$ : Use the y-intercept ( $\ln A$ ) to calculate $A$ .

Example question

Determining $E_{a}$ and $A$

The rate constants ( $k$ ) for a reaction are measured at two temperatures:

At $T_{1} = 300 K$ , $k_{1} = 1.2 \times 10^{- 3} s^{- 1}$
At $T_{2} = 320 K$ , $k_{2} = 4.5 \times 10^{- 3} s^{- 1}$

Calculate the activation energy ( $E_{a}$ ) and the Arrhenius factor ( $A$ ).

Solution

Step 1: Use the Two-Point Form of the Arrhenius Equation

The logarithmic form can be rearranged to:

$\ln (\frac{k_{2}}{k_{1}}) = \frac{E_{a}}{R} (\frac{1}{T_{1}} - \frac{1}{T_{2}})$

Substitute the known values:
$\ln (\frac{4.5 \times 10^{- 3}}{1.2 \times 10^{- 3}}) = \frac{E_{a}}{8.31} (\frac{1}{300} - \frac{1}{320})$

Evaluate each term:
$\ln (3.75) = \frac{E_{a}}{8.31} (0.00333 - 0.00313)$
$1.32 = \frac{E_{a}}{8.31} \cdot 0.00020$

Solve for $E_{a}$ :
$E_{a} = \frac{1.32}{0.00020} \cdot 8.31 = 54900 {J mol}^{- 1} or 54.9 {kJ mol}^{- 1}$

Step 2: Calculate $A$

Using the logarithmic form of the Arrhenius equation at $T_{1} = 300 K$ :
$\ln k_{1} = - \frac{E_{a}}{R} \cdot \frac{1}{T_{1}} + \ln A$

Substitute known values:
$\ln (1.2 \times 10^{- 3}) = - \frac{54, 900}{8.31} \cdot \frac{1}{300} + \ln A$
$- 6.73 = - 22.0 + \ln A$

Solve for $\ln A$ :
$\ln A = - 6.73 + 22.0 = 15.27$

Exponentiate to find $A$ :
$A = e^{15.27} = 4.3 \times 10^{6} s^{- 1}$

Final Answer:

Activation energy ( $E_{a}$ ): $54.9 {kJ mol}^{- 1}$
Arrhenius factor ( $A$ ): $4.3 \times 10^{6} s^{- 1}$

Note

This worked example shows how experimental data can be analyzed to calculate $E_{a}$ and $A$ , providing insight into the reaction's energy landscape and collision dynamics.

Reflection and Practice

Self review

Why does the rate constant increase exponentially with temperature?
How does the value of $A$ differ for simple vs. complex molecules?
The rate constant for a reaction doubles when the temperature increases from 298 K to 308 K. Calculate the activation energy ( $E_{a}$ ).

Theory of Knowledge

To what extent does the Arrhenius equation bridge the gap between theoretical models and experimental observations?
How does this interplay shape our understanding of chemical kinetics?

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