Arrhenius equation and activation energyEdexcel A-Level Chemistry: Revision notes

The activation energy, Eₐ, is the minimum energy colliding particles need for a reaction to occur. Only particles with energy ≥ Eₐ react.

When the temperature rises, a greater proportion of molecules have energy ≥ Eₐ, so there are more successful collisions per second and the rate constant, k, increases. A catalyst provides an alternative route with a lower Eₐ, so a greater proportion of collisions are successful at the same temperature.

Small changes in temperature make a large change in k, because k depends exponentially on Eₐ/RT.

The Arrhenius equation links the rate constant to temperature:

$k = A e^{-E_a/RT}$

Taking natural logarithms gives

$\ln k = \ln A - \dfrac{E_a}{R} \times \dfrac{1}{T}$

where A is a constant (the pre-exponential factor), R is the gas constant, 8.31 J K⁻¹ mol⁻¹, and T is the temperature in kelvin. The equation is given in the Edexcel data booklet if needed, so the skill is using it.

This has the form y = mx + c, with y = ln k and x = 1/T, so the gradient = −Eₐ/R and the intercept is ln A.

1. Measure k (or a quantity proportional to k) at several temperatures.
2. Convert each temperature to kelvin (add 273) and calculate 1/T.
3. Calculate ln k for each.
4. Plot ln k (y-axis) against 1/T (x-axis); the line is straight with a negative gradient.
5. Calculate the gradient using a large triangle (or two points far apart on the line).
6. Eₐ = −gradient × R, which gives J mol⁻¹. Divide by 1000 for kJ mol⁻¹.

Worked example. k = 0.128 s⁻¹ at 300 K and 0.792 s⁻¹ at 330 K. 1/T = 3.33 × 10⁻³ and 3.03 × 10⁻³ K⁻¹; ln k = −2.06 and −0.233. Gradient = 1.83 ÷ (−3.0 × 10⁻⁴) = −6.0 × 10³ K. Eₐ = 6.0 × 10³ × 8.31 = 5.0 × 10⁴ J mol⁻¹ = 50 kJ mol⁻¹.

If concentrations are kept the same in every experiment, the rate is proportional to k. In practice, time taken for a visible change, t, is measured, and the rate is proportional to 1/t. So a graph of ln(1/t) against 1/T has the same gradient, −Eₐ/R, as a graph of ln k against 1/T. Only the intercept changes.

This works because the same amount of reaction has happened when the colour change or cross disappearance occurs.

A common method uses sodium thiosulfate and hydrochloric acid: Na₂S₂O₃ + 2HCl → 2NaCl + S + SO₂ + H₂O. Sulfur makes the mixture cloudy, hiding a cross under the flask.

• Keep volumes and concentrations constant; change only the temperature.
• Bring both solutions to the temperature in a water bath before mixing.
• Start the timer on mixing and stop it when the cross disappears (same observer, same cross).
• Use at least five temperatures over about 30 K, with repeats.
• Plot ln(1/t) against 1/T, and calculate Eₐ from the gradient.

Sources of error include the judgement of when the cross disappears, and temperature changes during the run. Reduce these by using the same observer and by using a large temperature range.