Temperature and rate: Arrhenius equation and activation energy
Rate rises with temperature: k = A e^(−Ea/RT). A 10 K rise roughly doubles k, because many more molecules now have the activation energy.
Why temperature speeds up reactions
For most reactions, a 10 K rise in temperature makes the rate constant nearly double. Arrhenius explained this with the equation k = A e^(−Ea/RT). Reactant molecules must first form an unstable activated complex, for example when H₂ and I₂ meet to form 2HI. To reach it, they need a minimum extra energy called the activation energy (Ea). Only molecules with energy equal to or greater than Ea can react. The term e^(−Ea/RT) is the fraction of molecules that have this energy. Raising T makes this fraction much larger, so k rises sharply. Taking logs: ln k = −Ea/RT + ln A.
How to draw the three graphs
- Energy profile: y-axis Potential energy, x-axis Reaction coordinate. Draw a level for reactants, a hump, then a lower level for products (for an exothermic reaction). Label the top of the hump activated complex, the height from reactants to the top Ea, and the gap between reactants and products ΔH.
- Maxwell-Boltzmann distribution: y-axis fraction of molecules, x-axis kinetic energy. Draw a curve that rises to a peak (the most probable kinetic energy) and tails off. Draw a second curve for T + 10 that is flatter and shifted to higher energy. Draw a vertical line at Ea and shade the area to its right under each curve: the shaded area at T + 10 is about twice as large.
- Arrhenius plot: y-axis ln k, x-axis 1/T. Draw a straight line sloping down. Label "slope = −Ea/R" and "intercept = ln A".
Arrhenius equation at two temperatures
- 1At T₁: ln k₁ = −Ea/RT₁ + ln A. At T₂: ln k₂ = −Ea/RT₂ + ln A.
- 2Subtract: ln k₂ − ln k₁ = (Ea/R)(1/T₁ − 1/T₂), since A is the same at both temperatures.
- 3Change to log: log(k₂/k₁) = [Ea/(2.303R)] × (1/T₁ − 1/T₂).
- 4★So, log(k₂/k₁) = [Ea/(2.303R)] × [(T₂ − T₁)/(T₁T₂)].
★ key step
Worked example: activation energy when the rate doubles
The rate constant of a reaction doubles when the temperature is raised from 300 K to 310 K. Calculate Ea. (R = 8.314 J K⁻¹ mol⁻¹, log 2 = 0.3010)
- Given: k₂/k₁ = 2, T₁ = 300 K, T₂ = 310 K.
- Formula: Ea = 2.303R × log(k₂/k₁) × T₁T₂/(T₂ − T₁).
- 2.303 × 8.314 = 19.15 J K⁻¹ mol⁻¹ (rounded).
- T₁T₂/(T₂ − T₁) = (300 × 310)/10 = 93,000/10 = 9300 K.
- Ea = 19.15 × 0.3010 × 9300 = 5.764 × 9300 J mol⁻¹.
- Ea = 53,600 J mol⁻¹ (to three significant figures) = 53.6 kJ mol⁻¹.
Ea ≈ 53.6 kJ mol⁻¹
Activity: sodium thiosulphate and hydrochloric acid
Reaction: Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + SO₂(g) + S(s) + H₂O(l). The fine sulphur makes the solution milky. You need 0.1 M Na₂S₂O₃, 1 M HCl, five conical flasks, a paper with a cross drawn on it, a thermometer, a water bath and a stopwatch.
- 1Concentration: put 10, 20, 30, 40 and 50 mL of Na₂S₂O₃ solution in five flasks. Add water to make each 50 mL, so only the concentration changes.
- 2Place one flask on the cross. Add 10 mL of 1 M HCl, start the stopwatch and look down through the flask.
- 3Stop the watch when the cross just disappears. Record the time t. Repeat for the other flasks.
- 4★Observation: the more concentrated the thiosulphate, the shorter the time. Rate ∝ 1/t, so the rate rises with concentration.
- 5Temperature: take the same volumes (say 10 mL thiosulphate + 40 mL water) in each flask, warm them in the water bath to about 30, 40, 50 and 60 °C, add the acid and time as before.
- 6Observation: the time roughly halves for each 10 °C rise, so the rate roughly doubles. Plot 1/t against temperature.
★ key step
A pressure cooker cooks rice and dal faster because water boils at a higher temperature inside it, so the cooking reactions speed up. A refrigerator does the opposite and slows down spoiling.
3 marks: “The rate constant of a reaction doubles when the temperature rises from 310 K to 320 K. Calculate Ea. (R = 8.314 J K⁻¹ mol⁻¹, log 2 = 0.3010)” Ea = 2.303R log 2 × T₁T₂/(T₂ − T₁) = 19.15 × 0.3010 × (310 × 320)/10 = 5.764 × 9920 ≈ 57,200 J mol⁻¹ = 57.2 kJ mol⁻¹.
2 marks: “Draw a labelled energy profile for an exothermic reaction and mark the activation energy.” Potential energy against reaction coordinate: reactant level, hump labelled activated complex, lower product level; Ea from reactants to top; ΔH from reactants to products.
2 marks: “Write the Arrhenius equation and explain how a plot of ln k against 1/T gives Ea.” k = A e^(−Ea/RT), so ln k = −Ea/RT + ln A. The plot is a straight line of slope −Ea/R; Ea = −slope × R.
Ten degrees up, rate doubles up. Arrhenius plot: ln k against 1/T, slope minus Ea over R.