Topic 12 · page 9 of 17

Thermodynamic and kinetic stability

Thermodynamic stability asks whether a change is feasible. Kinetic stability asks whether it is fast enough to see. A change can be feasible and still not happen, because of a high activation energy.

9 min read12.11

Read first: The feasibility temperature

Two kinds of stable 12.11

Everything so far was about direction. ΔStotal\Delta S_\text{total} positive means a reaction can happen. It says nothing about whether it will happen in a time you can observe. Speed depends on the activation energy, the energy that colliding particles need before they react.

Magnesium burning has ΔStotal\Delta S_\text{total} = +3824 J K⁻¹ mol⁻¹ at 298 K (page 6), yet a coil of magnesium ribbon sits in the lab unchanged. Mixtures of methane and air, or of hydrogen and oxygen, sit unchanged until a spark. Diamond is not the stable form of carbon, yet it lasts for ever.

A reaction profile. Enthalpy is on the vertical axis and reaction pathway on the horizontal. Reactants are on a higher level on the left and products on a lower level on the right, with a delta H arrow pointing down. In between is a tall hump, with an arrow from the reactants to its top labelled E a, large. A note says the reaction is feasible but slow.
A reaction that is feasible but kinetically stable: the products are favoured, but the large activation energy means almost no collisions have enough energy to react.
Thermodynamic stability Kinetic stability
The question Is the change feasible? Is the change fast?
Decided by the sign of ΔStotal\Delta S_\text{total} the size of the activation energy
Affected by a catalyst? no yes
"Stable" means ΔStotal\Delta S_\text{total} is negative: the change will not happen the activation energy is high: the change is too slow

Worked exampleDiamond and graphite

Diamond turning into graphite has ΔH\Delta H ≈ −2 kJ mol⁻¹. The entropies are 2.4 J K⁻¹ mol⁻¹ for diamond and 5.7 for graphite.

  1. ΔSsystem\Delta S_\text{system} = 5.7 − 2.4 = +3.3 J K⁻¹ mol⁻¹
  2. ΔSsurroundings\Delta S_\text{surroundings} = 2000 ÷ 298 = +6.7 J K⁻¹ mol⁻¹
  3. ΔStotal\Delta S_\text{total} ≈ +10 J K⁻¹ mol⁻¹, which is positive. Diamond is thermodynamically unstable relative to graphite.
  4. Converting diamond to graphite means breaking the rigid network of strong C–C bonds. The activation energy is enormous, so at room temperature the change does not occur. Diamond is kinetically stable.

A catalyst lowers the activation energy and so changes the rate. It does not change ΔH\Delta H, ΔSsystem\Delta S_\text{system} or ΔStotal\Delta S_\text{total}, so it cannot make an unfeasible reaction go.

Check yourself

  1. Define thermodynamic stability and kinetic stability.
  2. Magnesium reacts with oxygen with ΔStotal\Delta S_\text{total} = +3824 J K⁻¹ mol⁻¹, yet magnesium ribbon does not burn in air unless it is heated. Explain.
  3. Substance X is thermodynamically stable with respect to decomposition: ΔStotal\Delta S_\text{total} is negative for its decomposition. Can a catalyst make it decompose? Explain.
  4. A reaction has ΔStotal\Delta S_\text{total} = +200 J K⁻¹ mol⁻¹ and shows no change after a week. Give the most likely reason.
  5. Why does adding a catalyst not change the value of ΔStotal\Delta S_\text{total} for a reaction?

Answers cover these until you have tried the questions

  1. Thermodynamic stability: a substance is thermodynamically stable if the change is not feasible, that is, ΔStotal\Delta S_\text{total} for the change is negative. Kinetic stability: a substance is kinetically stable if the change is feasible but too slow to observe, because the activation energy is high.
  2. The reaction is thermodynamically feasible but magnesium is kinetically stable towards oxygen: the activation energy is high, so very few collisions have enough energy to react until the ribbon is heated.
  3. No. A catalyst only lowers the activation energy and increases the rate. It does not change ΔStotal\Delta S_\text{total}. If ΔStotal\Delta S_\text{total} is negative, the decomposition is not feasible at that temperature whatever the catalyst.
  4. The reaction is feasible but has a high activation energy, so it is too slow: the reactants are kinetically stable.
  5. A catalyst provides a different route with a lower activation energy, but the reactants and products are the same, so ΔH\Delta H and ΔSsystem\Delta S_\text{system} are unchanged, and so is ΔStotal\Delta S_\text{total}. It changes only the rate.