Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 11.10 Activation energy from data -Study Notes - New Syllabus

11.10 Determination of Activation Energy using Calculations and Graphical Methods

The activation energy of a reaction is the minimum energy required for reactant particles to undergo a successful collision. It can be determined experimentally using rate data at different temperatures, either through direct calculation or by applying a graphical method based on the Arrhenius equation.

Arrhenius Equation

The Arrhenius equation relates the rate constant of a reaction to temperature and activation energy.

\( \mathrm{k = A e^{-\frac{E_a}{RT}}} \)

Where:

• \( \mathrm{k} \) = rate constant
• \( \mathrm{A} \) = Arrhenius constant (frequency factor)
• \( \mathrm{E_a} \) = activation energy
• \( \mathrm{R} \) = gas constant
• \( \mathrm{T} \) = temperature in Kelvin

Linear Form of the Arrhenius Equation

Taking natural logarithms:

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

This equation has the form \( \mathrm{y = mx + c} \), where:

• \( \mathrm{y = \ln k} \)
• \( \mathrm{x = \frac{1}{T}} \)
• Gradient \( \mathrm{= -\frac{E_a}{R}} \)
• Intercept \( \mathrm{= \ln A} \)

Therefore, a graph of \( \mathrm{\ln k} \) against \( \mathrm{\frac{1}{T}} \) gives a straight line, allowing activation energy to be determined from the gradient.

Graphical Method

• Plot \( \mathrm{\ln k} \) on the y-axis and \( \mathrm{\frac{1}{T}} \) on the x-axis.
• Draw the best-fit straight line.
• Determine the gradient of the line.
• Use:

\( \mathrm{gradient = -\frac{E_a}{R}} \Rightarrow E_a = -(\text{gradient}) \times R \)

Since the gradient is negative, the activation energy is always a positive value.

Calculation Method (Using Two Data Points)

When two sets of experimental data are given, activation energy can be calculated directly using:

\( \mathrm{\ln \left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} – \frac{1}{T_2}\right)} \)

This equation is derived from the Arrhenius equation and allows \( \mathrm{E_a} \) to be calculated without plotting a graph.

Key Points

• Temperature must always be in Kelvin.
• The graph of \( \mathrm{\ln k} \) vs \( \mathrm{1/T} \) is linear.
• The gradient is negative and proportional to activation energy.
• Activation energy is usually expressed in \( \mathrm{kJ\ mol^{-1}} \).