Free Energy and Feasibility

Spec Mapping — OCR H432 Module 5.2.2 — Free energy and feasibility, covering the definition of Gibbs free-energy change $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ΔG∘=ΔH∘−TΔS∘, the feasibility criterion $\Delta G \leq 0$ΔG≤0, numerical calculations of $\Delta_r G^\circ$Δr​G∘ from standard enthalpy and entropy data (with unit-consistency between kJ for $H$H and J K$^{-1}$−1 mol$^{-1}$−1 for $S$S), determination of the critical temperature $T_{crit} = \Delta H^\circ/\Delta S^\circ$Tcrit​=ΔH∘/ΔS∘ at which a reaction shifts between feasible and non-feasible, qualitative analysis of the four sign cases for $\Delta H^\circ$ΔH∘ and $\Delta S^\circ$ΔS∘, and the kinetic-versus-thermodynamic distinction (diamond-to-graphite is feasible but kinetically prevented) (refer to the official OCR H432 specification document for exact wording).

Lessons 1–5 built the enthalpy story; Lesson 6 added entropy. Either function alone is insufficient: exothermic reactions with negative $\Delta S$ΔS can fail (Haber at high $T$T), and endothermic reactions with positive $\Delta S$ΔS can succeed (NH$_4$4​NO$_3$3​ dissolving). The two state functions combine via Gibbs's free-energy function ($\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ΔG∘=ΔH∘−TΔS∘, 1873), and the criterion $\Delta G \leq 0$ΔG≤0 tells you whether a process is thermodynamically allowed at any specified $T$T. This lesson builds the equation from first principles, walks through four sign cases, computes $\Delta_r G^\circ$Δr​G∘ for six worked examples including Group-2-carbonate decomposition, derives the critical temperature for feasibility switching, and closes with the thermodynamic-vs-kinetic distinction (the diamond-to-graphite paradox).

Key Equation: the Gibbs free-energy change for a reaction is $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ΔG∘=ΔH∘−TΔS∘ with $\Delta G^\circ$ΔG∘ and $\Delta H^\circ$ΔH∘ in kJ mol$^{-1}$−1, $T$T in K, and $\Delta S^\circ$ΔS∘ in J K$^{-1}$−1 mol$^{-1}$−1 (so the entropy term must be divided by 1000 before substitution). The reaction is thermodynamically feasible when $\Delta G^\circ \leq 0$ΔG∘≤0, and the temperature at which feasibility switches on or off is the critical temperature $T_{crit} = \Delta H^\circ/\Delta S^\circ$Tcrit​=ΔH∘/ΔS∘.

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Learning Objectives (OCR Spec 5.2.2)

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

• State and use the equation $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ΔG∘=ΔH∘−TΔS∘ with full unit discipline (kJ vs J)
• Apply the feasibility criterion $\Delta G \leq 0$ΔG≤0 to predict whether a reaction proceeds at a given temperature
• Identify the four sign cases ($\Delta H$ΔH vs $\Delta S$ΔS) and predict the corresponding feasibility behaviour with $T$T
• Calculate the critical temperature $T_{crit} = \Delta H^\circ/\Delta S^\circ$Tcrit​=ΔH∘/ΔS∘ at which a reaction's feasibility flips, and explain physically what happens above and below $T_{crit}$Tcrit​
• Apply Gibbs analysis to dissolution (NH$_4$4​NO$_3$3​ cold-pack), thermal decomposition (Group 2 carbonates), gas-phase synthesis (Haber), and phase change (water $\to$→ steam at 100 °C)
• Distinguish thermodynamic feasibility ($\Delta G < 0$ΔG<0) from kinetic accessibility (the activation-energy barrier) — diamond-to-graphite as the canonical example
• (Going-further) link $\Delta G^\circ$ΔG∘ to the equilibrium constant via $\Delta G^\circ = -RT \ln K$ΔG∘=−RTlnK

Building $\Delta G$ΔG from $\Delta H_{\text{system}}$ΔHsystem​ and $\Delta S_{\text{system}}$ΔSsystem​

The condition for spontaneity is $\Delta S_{\text{universe}} > 0$ΔSuniverse​>0, where $\Delta S_{\text{universe}} = \Delta S_{\text{system}} + \Delta S_{\text{surroundings}}$ΔSuniverse​=ΔSsystem​+ΔSsurroundings​. At constant pressure the heat the reaction dumps into the surroundings is $-\Delta H_{\text{system}}$−ΔHsystem​, absorbed at temperature $T$T:

$\Delta S_{\text{surroundings}} = -\frac{\Delta H_{\text{system}}}{T}$ΔSsurroundings​=−TΔHsystem​​

Substituting and multiplying through by $-T$−T defines the Gibbs free energy change:

$\boxed{\Delta G_{\text{system}} = \Delta H_{\text{system}} - T\Delta S_{\text{system}}}$ΔGsystem​=ΔHsystem​−TΔSsystem​​

The criterion $\Delta S_{\text{universe}} > 0$ΔSuniverse​>0 becomes $\Delta G_{\text{system}} < 0$ΔGsystem​<0 — feasibility is now expressed in terms of system properties alone, which is what an experimentalist can measure or look up.

Key Definitions:

• Gibbs free energy ($G$G) — a state function with units of kJ mol$^{-1}$−1, defined $G = H - TS$G=H−TS at constant temperature.
• Standard Gibbs free-energy change ($\Delta G^\circ$ΔG∘) — the value of $\Delta G$ΔG when reactants and products are in their standard states (298 K unless specified, 100 kPa, 1 mol dm$^{-3}$−3 for aqueous species, pure form for condensed phases).
• Thermodynamic feasibility — a process is feasible when $\Delta G \leq 0$ΔG≤0. Equivalent to $\Delta S_{\text{universe}} \geq 0$ΔSuniverse​≥0.
• Critical temperature ($T_{crit}$Tcrit​) — the value of $T$T at which $\Delta G = 0$ΔG=0 for a reaction. Above or below this temperature the reaction's feasibility switches.

Units, Units, Units — the Single Biggest Mark-Loser

Unit discipline is the #1 cause of mark-loss in 5.2.2. Memorise:

• $\Delta H^\circ$ΔH∘: kJ mol$^{-1}$−1.
• $T$T: kelvin (NOT °C — add 273).
• $\Delta S^\circ$ΔS∘: J K$^{-1}$−1 mol$^{-1}$−1 — joules, not kilojoules. Divide by 1000 before substituting.

Example: $\Delta S^\circ = +200$ΔS∘=+200 J K$^{-1}$−1 mol$^{-1}$−1 at 300 K gives $T\Delta S^\circ = 300 \times 0.200 = +60$TΔS∘=300×0.200=+60 kJ mol$^{-1}$−1. Not $300 \times 200 = +60{,}000$300×200=+60,000.

The Feasibility Criterion and the Four Cases

The sign of $\Delta G$ΔG depends on the signs of $\Delta H$ΔH and $\Delta S$ΔS and on $T$T. Four cases exhaust the possibilities:

Case	$\Delta H^\circ$ΔH∘	$\Delta S^\circ$ΔS∘	$-T\Delta S^\circ$−TΔS∘	$\Delta G^\circ$ΔG∘	Feasibility	Example
1	negative (exo)	positive (entropy gained)	negative	always negative	feasible at all $T$T	combustion of H$_2$2​ in pure O$_2$2​ on a catalyst
2	positive (endo)	negative (entropy lost)	positive	always positive	never feasible	spontaneous separation of mixed gases
3	negative (exo)	negative (entropy lost)	positive	depends on $T$T	feasible at low $T$T only	Haber synthesis N$_2$2​ + 3H$_2$2​ $\to$→ 2NH$_3$3​
4	positive (endo)	positive (entropy gained)	negative	depends on $T$T	feasible at high $T$T only	CaCO$_3$3​ $\to$→ CaO + CO$_2$2​

The first two cases are always in the same direction regardless of temperature: the signs combine to give $\Delta G$ΔG always negative or always positive. The last two are temperature-dependent: the feasibility switches at the critical temperature $T_{crit} = \Delta H^\circ/\Delta S^\circ$Tcrit​=ΔH∘/ΔS∘. Below $T_{crit}$Tcrit​ the $\Delta H^\circ$ΔH∘ term dominates; above $T_{crit}$Tcrit​ the $T\Delta S^\circ$TΔS∘ term dominates.

graph TD
  A["delta G = delta H - T delta S"] --> B{"delta H?"}
  B -->|negative| C{"delta S?"}
  B -->|positive| D{"delta S?"}
  C -->|positive| E["Feasible at ALL T<br/>(Case 1)"]
  C -->|negative| F["Feasible at LOW T only<br/>(Case 3) -- Haber"]
  D -->|negative| G["NEVER feasible<br/>(Case 2)"]
  D -->|positive| H["Feasible at HIGH T only<br/>(Case 4) -- carbonate decomp"]

Worked Example 1: Calcium-Carbonate Decomposition at 298 K

Calculate $\Delta_r G^\circ$Δr​G∘ for the thermal decomposition of calcium carbonate at 298 K and determine whether it is feasible at that temperature.

$\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$CaCO3​(s)→CaO(s)+CO2​(g)

Data: $\Delta_r H^\circ = +178$Δr​H∘=+178 kJ mol$^{-1}$−1, $\Delta_r S^\circ = +161$Δr​S∘=+161 J K$^{-1}$−1 mol$^{-1}$−1.

Step 1 — convert entropy units. $\Delta_r S^\circ = 161/1000 = 0.161$Δr​S∘=161/1000=0.161 kJ K$^{-1}$−1 mol$^{-1}$−1.

Step 2 — apply the Gibbs equation.

$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ = 178 - (298 \times 0.161) = 178 - 48.0 = +130\;\text{kJ mol}^{-1}$Δr​G∘=Δr​H∘−TΔr​S∘=178−(298×0.161)=178−48.0=+130kJ mol−1

Step 3 — feasibility. $\Delta_r G^\circ > 0$Δr​G∘>0, so the reaction is NOT feasible at 298 K. CaCO$_3$3​ is thermodynamically stable at room temperature — exactly why limestone outcrops persist for geological timescales (CO$_2$2​-loss to the atmosphere would require T $>$> T$_{crit}$crit​).

Worked Example 2: Critical Temperature for CaCO$_3$3​ Decomposition

At what temperature does the decomposition of CaCO$_3$3​ become feasible? Same data.

Condition for marginal feasibility: $\Delta_r G^\circ = 0$Δr​G∘=0 (the borderline between feasible and non-feasible).

$0 = \Delta_r H^\circ - T_{crit}\Delta_r S^\circ \quad\Rightarrow\quad T_{crit} = \frac{\Delta_r H^\circ}{\Delta_r S^\circ}$0=Δr​H∘−Tcrit​Δr​S∘⇒Tcrit​=Δr​S∘Δr​H∘​

$T_{crit} = \frac{178}{0.161} = 1106\;\text{K} \approx 833\;°\text{C}$Tcrit​=0.161178​=1106K≈833°C

Above about 833 °C the $T\Delta S^\circ$TΔS∘ term overtakes the $\Delta H^\circ$ΔH∘ term, $\Delta_r G^\circ$Δr​G∘ becomes negative, and decomposition becomes feasible. This is very close to the temperature at which commercial lime kilns are operated (typically 900–1000 °C, to ensure rapid kinetics on top of the marginal thermodynamics). The CaCO$_3$3​ $\to$→ CaO + CO$_2$2​ process is the chemistry of cement manufacture: about 8 % of global anthropogenic CO$_2$2​ emissions come from this single reaction in cement kilns, because the CO$_2$2​ liberated is geochemical (locked in the limestone for hundreds of millions of years) rather than fossil-fuel derived.

Worked Example 3: Ammonia Synthesis (Haber)

$\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$N2​(g)+3H2​(g)→2NH3​(g)

Data: $\Delta_r H^\circ = -92$Δr​H∘=−92 kJ mol$^{-1}$−1, $\Delta_r S^\circ = -201$Δr​S∘=−201 J K$^{-1}$−1 mol$^{-1}$−1 (= $-0.201$−0.201 kJ K$^{-1}$−1 mol$^{-1}$−1).

At 298 K:

$\Delta_r G^\circ = -92 - (298 \times -0.201) = -92 - (-59.9) = -92 + 59.9 = -32\;\text{kJ mol}^{-1}$Δr​G∘=−92−(298×−0.201)=−92−(−59.9)=−92+59.9=−32kJ mol−1

Feasible at 298 K ($\Delta G^\circ < 0$ΔG∘<0). But — and this is the critical exam-style discriminator — the reaction is kinetically intractable at 298 K because N$_2$2​ has a triple bond of bond enthalpy 945 kJ mol$^{-1}$−1, giving a huge activation energy. The Haber process therefore operates at 450 °C to make the reaction go (kinetics) — but at this temperature thermodynamics fights back, because $\Delta_r S^\circ$Δr​S∘ is negative.

Critical temperature (above which Haber becomes infeasible):

$T_{crit} = \frac{\Delta_r H^\circ}{\Delta_r S^\circ} = \frac{-92}{-0.201} = 458\;\text{K} \approx 185\;°\text{C}$Tcrit​=Δr​S∘Δr​H∘​=−0.201−92​=458K≈185°C

Above 458 K $\Delta_r G^\circ > 0$Δr​G∘>0 and the equilibrium yield of NH$_3$3​ falls. The Haber process at 450 °C therefore operates above $T_{crit}$Tcrit​ for thermodynamic feasibility — a deliberate compromise. The lost yield at high $T$T is recovered by (a) very high pressure (200 atm), which by Le Chatelier shifts the equilibrium back to the side with fewer moles of gas (the product side), and (b) recycling unreacted N$_2$2​ and H$_2$2​ through the reactor — a single pass converts only $\sim 15$∼15 % of the feed, but successive recycles drive the cumulative yield above 95 %. The full chemistry is a textbook example of how rate, equilibrium, and economic factors interact to set industrial operating conditions.

Worked Example 4: Dissolving Ammonium Nitrate (the Cold-Pack)

$\text{NH}_4\text{NO}_3\text{(s)} \rightarrow \text{NH}_4^+\text{(aq)} + \text{NO}_3^-\text{(aq)}$NH4​NO3​(s)→NH4+​(aq)+NO3−​(aq)

Data: $\Delta_r H^\circ = +26$Δr​H∘=+26 kJ mol$^{-1}$−1 (endothermic), $\Delta_r S^\circ = +108$Δr​S∘=+108 J K$^{-1}$−1 mol$^{-1}$−1 ($= +0.108$=+0.108 kJ K$^{-1}$−1 mol$^{-1}$−1).

At 298 K:

$\Delta_r G^\circ = 26 - (298 \times 0.108) = 26 - 32.2 = -6.2\;\text{kJ mol}^{-1}$Δr​G∘=26−(298×0.108)=26−32.2=−6.2kJ mol−1

$\Delta_r G^\circ < 0$Δr​G∘<0, so dissolving is feasible — even though it is endothermic. The dissolving absorbs heat from the surroundings (which is exactly the cold-pack effect: the pack chills your sports injury), but the entropy gain from dispersing two moles of ions into water drives the process anyway. This is Case 4 of the four-case table — an entropically-driven endothermic feasible reaction. Cold-packs work because of this single thermodynamic feature.

The lower bound on the operating temperature is found by demanding $\Delta G^\circ \leq 0$ΔG∘≤0, giving:

$T_{crit} = \frac{\Delta_r H^\circ}{\Delta_r S^\circ} = \frac{26}{0.108} = 241\;\text{K} \approx -32\;°\text{C}$Tcrit​=Δr​S∘Δr​H∘​=0.10826​=241K≈−32°C

So NH$_4$4​NO$_3$3​ dissolution remains feasible down to about $-32$−32 °C. Below this temperature the entropy gain is insufficient to overcome the endothermic enthalpy; in practice, of course, water freezes well before this temperature is reached, and the cold-pack physics shifts to ice-and-salt slurry chemistry.

Worked Example 5: Water Boils at 373 K (a Phase-Transition Sanity Check)

$\text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{O(g)}$H2​O(l)→H2​O(g)

Data: $\Delta_{\text{vap}} H^\circ = +44$Δvap​H∘=+44 kJ mol$^{-1}$−1, $\Delta_{\text{vap}} S^\circ = +118$Δvap​S∘=+118 J K$^{-1}$−1 mol$^{-1}$−1 ($= +0.118$=+0.118 kJ K$^{-1}$−1 mol$^{-1}$−1).

Vaporisation is feasible when $\Delta G^\circ \leq 0$ΔG∘≤0. At the marginal feasibility temperature:

$T_{crit} = \frac{\Delta_{\text{vap}} H^\circ}{\Delta_{\text{vap}} S^\circ} = \frac{44}{0.118} = 373\;\text{K} = 100\;°\text{C}$Tcrit​=Δvap​S∘Δvap​H∘​=0.11844​=373K=100°C