The balance between entropy and enthalpy determines the feasibility of a…

The idea

Gibbs free energy combines the two competing factors in one number: ∆G = ∆H − T∆S. The enthalpy term favours change that releases energy, while the entropy term, weighted by the absolute temperature, favours change that increases dispersal. Because the temperature multiplies ∆S, the entropy contribution grows as the system is heated, which is why heating can make an endothermic reaction go. Four combinations follow. Negative ∆H with positive ∆S is feasible at every temperature; positive ∆H with negative ∆S at none; negative ∆H with negative ∆S only at low temperature; positive ∆H with positive ∆S only at high temperature. The unit trap is built in, since ∆H is tabulated in kJ mol⁻¹ and ∆S in J K⁻¹ mol⁻¹.

What earns the marks

• Quoting ∆G = ∆H − T∆S with T as the absolute temperature in kelvin.
• Converting ∆S from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹, or ∆H into joules, before substituting.
• Correct substitution with signs preserved.
• Interpreting the size and sign of the T∆S term against ∆H to decide the outcome.

Common mistakes and how to put them right

• Substituting ∆H in kilojoules and ∆S in joules without converting, which makes T∆S a thousand times too large.

• Using a temperature in degrees Celsius instead of kelvin.

• Adding T∆S rather than subtracting it.

• Assuming that a negative ∆H alone guarantees a negative ∆G.

• Treating ∆G as fixed for a reaction, when it depends on the temperature.

Exam tips

• Write the units next to every substituted value; the conversion is worth a mark by itself.
• State the temperature in kelvin explicitly, even when the question gives 25 °C.
• Before calculating, predict the sign from the signs of ∆H and ∆S so an arithmetic slip is obvious.