Edexcel International A Level (IAL) Chemistry (YCH11) - Unit 4 - 11.13 Core Practical: activation energy-Study Notes - New Syllabus

11.13 Core Practical 10: Determination of Activation Energy of a Reaction

The activation energy of a reaction can be determined experimentally by measuring the rate constant at different temperatures and applying the Arrhenius equation. This practical allows the relationship between temperature and rate to be quantified, showing how increasing temperature increases the fraction of particles with sufficient energy to react.

Principle

The rate constant \( \mathrm{k} \) increases with temperature, and the activation energy can be calculated using the Arrhenius equation.

Arrhenius Equation

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

Linear Form

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

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

Experimental Method

• Carry out the reaction at several different temperatures.
• Measure the rate of reaction or determine the rate constant \( \mathrm{k} \) at each temperature.
• Convert temperatures to Kelvin.
• Calculate \( \mathrm{\ln k} \) and \( \mathrm{\frac{1}{T}} \) for each data point.
• Plot a graph of \( \mathrm{\ln k} \) against \( \mathrm{\frac{1}{T}} \).
• Determine the gradient of the straight line.

Determination of Activation Energy

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

The negative gradient reflects that increasing temperature leads to an increase in rate constant. Activation energy is always positive.

Alternative Method (Two Data Points)

If only two temperature values are given, activation energy can be calculated using:

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

Key Considerations

• Temperature must be measured accurately and converted to Kelvin.
• The same method must be used to determine rate at each temperature.
• A consistent reaction system must be maintained.
• Errors in measuring rate will affect the calculated activation energy.

Interpretation

• A higher activation energy means fewer particles have sufficient energy to react.
• Increasing temperature increases the fraction of particles with energy ≥ \( \mathrm{E_a} \).
• This explains the increase in rate with temperature.