Activation energy and catalysisEdexcel International A Level Chemistry: Revision notes
Section 1
Activation energy and temperature
The activation energy, Ea, is the minimum energy colliding particles need for a reaction to occur. At a given temperature the molecules have a range of energies. Raising the temperature increases the proportion of molecules with energy equal to or greater than Ea, so more collisions per second are successful. Even a 10 K rise can roughly double the rate, and the rate constant k increases with temperature.
A catalyst provides an alternative route with a lower activation energy, so a greater proportion of molecules can react at the same temperature.
Section 2
The Arrhenius equation
The Arrhenius equation relates the rate constant to temperature: k = A e^(−Ea/RT), where A is a constant, Ea is the activation energy in J mol⁻¹, R is 8.31 J K⁻¹ mol⁻¹ and T is the temperature in kelvin.
Taking natural logs gives ln k = ln A − Ea/RT. This has the form y = c + mx, so a graph of ln k against 1/T is a straight line with gradient = −Ea/R. Then Ea = −gradient × R.
The equation is given if needed, but you must be able to use it.
Convert °C to K before finding 1/T, and convert Ea from J mol⁻¹ to kJ mol⁻¹ at the end. A negative Ea means you forgot the minus sign in the gradient.
Section 3
Calculating Ea from two sets of data
Subtracting the Arrhenius equation at two temperatures gives ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).
Worked example. k = 1.5 × 10⁻³ s⁻¹ at 298 K and 6.0 × 10⁻³ s⁻¹ at 318 K.
ln(k₂/k₁) = ln 4.0 = 1.386. 1/298 − 1/318 = 2.11 × 10⁻⁴ K⁻¹.
Ea = 8.31 × 1.386 ÷ 2.11 × 10⁻⁴ = 54 600 J mol⁻¹ = 54.6 kJ mol⁻¹.
The same equation can predict k at a new temperature if Ea is known. A graph using several temperatures is more reliable than two points, because it averages out errors.
Keep the larger temperature as T₂ with the larger k, and check that your Ea is positive.
Section 4
Core Practical 10: finding the activation energy
A common method uses sodium thiosulfate and hydrochloric acid, which form a sulfur precipitate. The time t for a cross to disappear is measured at several temperatures.
- Use the same volumes and concentrations each time.
- Bring each solution to temperature in a water bath, mix them, and record the temperature and time.
- Use at least five temperatures over about 20 to 60 °C.
- Rate is proportional to 1/t.
- Plot ln(1/t) against 1/T (T in kelvin). The gradient is −Ea/R, so Ea = −gradient × 8.31.
Limitations include judging when the cross disappears and the temperature changing during the reaction.
Section 5
Heterogeneous catalysts in industry
In many industrial gas-phase reactions the catalyst is a solid (heterogeneous catalyst), for example iron in the Haber process. The reaction takes place on its surface:
- Reactant molecules are adsorbed onto active sites, which weakens bonds and holds them close together.
- The reaction occurs through a route with a lower activation energy.
- The products are desorbed, freeing the sites.
The catalyst is used as pellets or a finely divided solid to give a large surface area, and the rate increases with surface area. The catalyst lets the reaction run at a lower temperature, saving energy. It is easy to separate from the gaseous products.
Must Know
- Higher temperature gives a greater proportion of molecules with E ≥ Ea
- ln k = ln A − Ea/RT; plot ln k against 1/T; gradient = −Ea/R
- Use R = 8.31 J K⁻¹ mol⁻¹ and T in kelvin
- ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂)
- Core Practical 10: time t at several temperatures, plot ln(1/t) against 1/T
- A solid catalyst provides a surface for the reaction: adsorption, lower Ea, desorption
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Activation energy and catalysis
- The Arrhenius equation, ln k = ln A − Ea/RT, relates the rate constant k of a reaction to the absolute temperature T, where Ea is the activation energy and R is the gas constant, 8.31 J K⁻¹ mol⁻¹. A student plots ln k on the y-axis against 1/T on the x-axis for one reaction, with T in kelvin.The gradient of the line is −6.50 × 10³. Calculate the activation energy in kJ mol⁻¹.2 marks
- In an experiment to find the activation energy of a reaction, a student mixes sodium thiosulfate solution with dilute hydrochloric acid and measures the time t for a cross viewed through the flask to disappear behind the sulfur precipitate. The experiment is repeated at several temperatures using the same volumes and concentrations each time.Explain, using collision theory, why the time for the cross to disappear is shorter at higher temperatures.2 marks
- The rate constant of a first-order decomposition is 2.40 × 10⁻⁴ s⁻¹ at 300 K and 8.80 × 10⁻⁴ s⁻¹ at 320 K. Use R = 8.31 J K⁻¹ mol⁻¹.Calculate the activation energy of the reaction in kJ mol⁻¹, using ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂).3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).