All subjects · Chemistry · WCH14 (Unit 4)

Activation energy, the Arrhenius equation and catalysts (Core Practical 10)

The rate constant rises steeply with temperature. The Arrhenius equation turns rate data at several temperatures into a value for the activation energy, which is Core Practical 10.

The Arrhenius equation (spec 11.10)

k = A e-Ea/RT, and in logarithmic form: ln k = ln A - Ea / (R T).

The equation is given in the exam if you need it. A plot of ln k against 1/T is a straight line of gradient -Ea/R, so Ea = -gradient x R.

Using two temperatures

Subtracting the equation at two temperatures gives ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). This is a quick check, but a graph using several temperatures is more reliable because it averages out random errors.

Core Practical 10 (spec 11.13)

Time a reaction to a fixed end point at several temperatures, for example the reaction of sodium thiosulfate with acid, timed until a mark under the flask is hidden by sulfur. Because the same amount of reaction happens each time, k is proportional to 1/t, so ln(1/t) against 1/T has the same gradient, -Ea/R. Warm each solution separately to the same temperature in a water bath before mixing, measure the temperature of the mixture, and convert °C to K.

Heterogeneous catalysts (spec 11.11)

A homogeneous catalyst is in the same phase as the reactants; a heterogeneous catalyst is in a different phase, usually a solid with gaseous reactants. Gas molecules are adsorbed onto the surface, bonds within them weaken, they react, and the products desorb. This route has a lower activation energy. Examples: iron in the Haber process, nickel in the hydrogenation of alkenes. Catalysts are used finely divided or on a support to give a large surface area, and can be poisoned by impurities that bind strongly to the surface.

Worked example

A graph of ln k against 1/T has a gradient of -6000 K. Find Ea.

  1. Gradient = -Ea / R
  2. Ea = 6000 x 8.31 = 49 860 J mol-1

Answer: Ea = +49.9 kJ mol-1

Key idea: ln k = ln A - Ea/RT. Plot ln k (or ln(1/t)) against 1/T: gradient = -Ea/R. Heterogeneous catalysts work by adsorption, bond weakening and desorption at a surface.

Check yourself

  1. A plot of ln k against 1/T for a reaction has a gradient of -9200 K. Calculate the activation energy in kJ mol-1. (R = 8.31 J mol-1 K-1)
    Show answer

    +76.5 kJ mol-1. Ea = -gradient x R = 9200 x 8.31 = 76 452 J mol-1 = +76.5 kJ mol-1.

  2. The rate constant of a reaction is 1.0 x 10-3 s-1 at 300 K and 4.0 x 10-3 s-1 at 320 K. Calculate Ea in kJ mol-1 using ln(k2/k1) = (Ea/R)(1/T1 - 1/T2). (R = 8.31 J mol-1 K-1)
    Show answer

    +55.3 kJ mol-1. ln 4.0 = 1.386; 1/300 - 1/320 = 2.083 x 10-4 K-1; Ea = 8.31 x 1.386 / 2.083 x 10-4 = 55297 J mol-1 = +55.3 kJ mol-1.

  3. Which graph gives a straight line from which Ea can be found?
    • ln k against 1/T
    • k against T
    • ln k against T
    • 1/k against T
    Show answer

    ln k against 1/T. ln k = ln A - (Ea/R)(1/T) has the form y = c + mx with x = 1/T.

  4. In Core Practical 10, why must temperatures be converted to kelvin before plotting?
    • Kelvin values are smaller
    • It makes the gradient positive
    • Water boils at 100 K
    • The Arrhenius equation uses absolute temperature
    Show answer

    The Arrhenius equation uses absolute temperature. T in the equation is the absolute temperature; 1/T in °C would give a meaningless graph.

  5. In Core Practical 10, why can ln(1/t) be plotted instead of ln k?
    • 1/t equals the activation energy
    • Each run reaches the same fixed end point, so k is proportional to 1/t
    • The reaction is zero order
    • Time does not depend on temperature
    Show answer

    Each run reaches the same fixed end point, so k is proportional to 1/t. k = constant x (1/t), so ln k = ln(1/t) + constant: same gradient.

  6. What is the correct order of events at the surface of a heterogeneous catalyst?
    • Desorption, reaction, adsorption
    • Dissolving, precipitation, filtration
    • Reaction, adsorption, desorption
    • Adsorption, bonds weaken and react, desorption
    Show answer

    Adsorption, bonds weaken and react, desorption. Reactants stick to the surface, react with a lower activation energy, then products leave.

  7. Why are solid catalysts in industry often used as a thin layer on a support or as fine powder?
    • To raise the activation energy
    • To give a large surface area for adsorption
    • To change the position of equilibrium
    • To stop the catalyst from being adsorbed
    Show answer

    To give a large surface area for adsorption. More surface means more sites where molecules can be adsorbed and react.

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Last updated 7 November 2026. Spotted a mistake? Tell us.