Reactivity 1.4 · Entropy and spontaneity

Supported HL focus. First assessment 2025; current brief acquired; full Chemistry guide not acquired. Remaining guide, assessment and practical requirements retain their recorded holds.

Prerequisites: read the stated quantities and units, use arithmetic and the model conditions below. Each lesson develops its own method before independent transfer.

These are original or explicitly fictional teaching examples, not actual measurements or completed assessed learner investigations.

What would explain this observation?

• A reaction can be endothermic and still be thermodynamically favourable. Enthalpy alone does not determine the direction favoured at a given temperature.
• Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.

Build the model

• Entropy is associated with energy dispersal and the number of accessible microscopic arrangements. Gibbs energy combines enthalpy, entropy and absolute temperature for a process at constant temperature and pressure.
• entropy: A state property related to energy dispersal and accessible arrangements; Gibbs energy: A thermodynamic quantity combining enthalpy and entropy contributions.

Choose evidence that can test it

• Use ΔG = ΔH - TΔS with consistent energy units. A negative Gibbs energy change indicates thermodynamic favourability for the stated conditions, not a fast rate. An activation barrier can make a favourable process slow.
• State whether values are standard-state quantities and record temperature in kelvin. Convert entropy from joules per kelvin per mole into kilojoules per kelvin per mole when enthalpy is in kilojoules per mole.

Work from known quantities

• State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
• Known: ΔH = +20 kJ per mole, ΔS = +100 J per kelvin per mole and T = 300 K. Convert ΔS = 0.100 kJ per kelvin per mole. ΔG = ΔH - TΔS = 20 - 300×0.100 = -10 kJ per mole. The process is favourable under the stated approximation.

Example:

ΔH=30 kJ per mole, ΔS=0.10 kJ per kelvin per mole and T=400 K. Find ΔG. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.

Check the conclusion and its limits

• Thermodynamic favourability does not establish reaction rate. Celsius cannot replace kelvin in TΔS, and a unit conversion error can change the result by a factor of 1,000.
• Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.

Warn:

A thermodynamically favourable reaction must be fast. This claim is false: Thermodynamic favourability does not establish reaction rate. Celsius cannot replace kelvin in TΔS, and a unit conversion error can change the result by a factor of 1,000.

Key:

Entropy, Gibbs energy and feasibility: Use ΔG = ΔH - TΔS with consistent energy units. A negative Gibbs energy change indicates thermodynamic favourability for the stated conditions, not a fast rate. An activation barrier can make a favourable process slow.