Free Energy and Feasibility

Learning Objectives (OCR Spec 5.2.2)

By the end of this lesson you should be able to:

• Recall and use the equation ΔG = ΔH - TΔS
• State the condition for a reaction to be feasible: ΔG ≤ 0
• Calculate ΔG° for a reaction from standard enthalpy and entropy data
• Determine the temperature at which a reaction becomes feasible
• Explain why kinetic barriers can prevent thermodynamically feasible reactions from occurring

Bringing Enthalpy and Entropy Together

Lessons 1-5 focused on enthalpy; lesson 6 introduced entropy. Neither quantity alone predicts whether a reaction will happen - we need to combine them. The combined quantity is the Gibbs free energy, G:

G = H - TS

The change in Gibbs free energy for a reaction is:

ΔG = ΔH - TΔS

where:

• ΔG is the change in Gibbs free energy (kJ mol^-1)
• ΔH is the change in enthalpy (kJ mol^-1)
• T is the temperature (K)
• ΔS is the change in entropy (kJ K^-1 mol^-1 - careful with units!)

The Feasibility Criterion

A reaction is thermodynamically feasible (also called spontaneous) if:

ΔG ≤ 0

If ΔG > 0 the forward reaction is not feasible; the reverse reaction would be feasible.

"Feasible" means thermodynamically allowed, not necessarily fast. A reaction with ΔG < 0 may still have an enormous activation energy and proceed too slowly to observe (see the end of this lesson).

The Four Cases

The sign of ΔG depends on the individual signs of ΔH and ΔS and on temperature T:

ΔH	ΔS	-TΔS	ΔG	Feasibility
negative	positive	negative	always negative	feasible at ALL T
positive	negative	positive	always positive	NEVER feasible
negative	negative	positive	depends on T	feasible at LOW T
positive	positive	negative	depends on T	feasible at HIGH T

Exothermic reactions with positive entropy change are always feasible. Endothermic reactions with negative entropy change are never feasible. Reactions where ΔH and ΔS have the same sign become feasible or infeasible only at certain temperatures.

Worked Example 1: CaCO3 Decomposition

Calculate ΔG° for the decomposition of calcium carbonate at 298 K and determine whether it is feasible at that temperature.

CaCO3(s) -> CaO(s) + CO2(g)

Data:

• ΔH° = +178 kJ mol^-1
• ΔS° = +161 J K^-1 mol^-1

Step 1: Convert ΔS° to kJ: 161 / 1000 = 0.161 kJ K^-1 mol^-1

Step 2: Apply the equation: ΔG° = ΔH° - TΔS°

ΔG° = 178 - (298 x 0.161)

ΔG° = 178 - 48.0

ΔG° = +130 kJ mol^-1

Step 3: ΔG° > 0, so the reaction is NOT feasible at 298 K. CaCO3 is stable at room temperature.

Worked Example 2: Finding the Temperature for Feasibility

At what temperature does the decomposition of CaCO3 become feasible? Use the same data.

Condition: The reaction becomes feasible when ΔG° = 0 (the borderline).

0 = ΔH° - TΔS°

T = ΔH° / ΔS°

T = 178 / 0.161

T = 1106 K (approximately 833 °C)

So above about 833 °C the decomposition becomes feasible. This is close to the observed value - lime kilns operate above about 900 °C.

Worked Example 3: Formation of Ammonia

N2(g) + 3H2(g) -> 2NH3(g)

Data:

• ΔH° = -92 kJ mol^-1
• ΔS° = -201 J K^-1 mol^-1 = -0.201 kJ K^-1 mol^-1

At 298 K:

ΔG° = -92 - (298 x -0.201)

ΔG° = -92 - (-59.9)

ΔG° = -92 + 59.9

ΔG° = -32 kJ mol^-1

The reaction is feasible at 298 K. But it is kinetically slow - see below.

At what temperature does it cease to be feasible?

T = ΔH° / ΔS° = -92 / -0.201 = 458 K (about 185 °C)

Above 458 K the -TΔS term becomes dominant and ΔG > 0. This is why the Haber process operates at 450 °C - a compromise between kinetics (rate) and thermodynamics (yield). Above 458 K the equilibrium constant falls, so increasing temperature reduces the yield. The high pressure (200 atm) compensates and pushes the equilibrium towards products.

Worked Example 4: Dissolving Ammonium Nitrate

NH4NO3(s) -> NH4+(aq) + NO3-(aq)

Data:

• ΔH° = +26 kJ mol^-1
• ΔS° = +108 J K^-1 mol^-1 = +0.108 kJ K^-1 mol^-1

At 298 K:

ΔG° = 26 - (298 x 0.108)

ΔG° = 26 - 32.2

ΔG° = -6.2 kJ mol^-1

Negative, therefore feasible. The dissolving is endothermic (absorbs heat from surroundings) but driven entropically. This is exactly how some instant cold packs work.

Worked Example 5: Finding the Temperature at Which Water Boils

H2O(l) -> H2O(g)

Data:

• ΔH° = +44 kJ mol^-1
• ΔS° = +118 J K^-1 mol^-1 = +0.118 kJ K^-1 mol^-1

Vaporisation becomes feasible when ΔG° ≤ 0:

T = ΔH° / ΔS° = 44 / 0.118 = 373 K (100 °C)

Exactly the boiling point of water at 1 atm, as it should be.

Signs, Temperature, and the Feasibility Map