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9.3 - Gibbs Free Energy and Thermodynamic Favorability

Overview

Some processes release energy and some absorb it. Some increase entropy and some decrease it. To decide whether a process is thermodynamically favored, you need both, and Gibbs free energy, GG, combines them into one number. The sign of ΔG∘\Delta G^\circ answers the question. The rest of Unit 9 (equilibrium, coupled reactions, and electrochemical cells) builds on this topic.

Standard conditions and ΔG°

ΔG∘\Delta G^\circ is the Gibbs free energy change when every reactant and product is in its standard state:
  • pure substances (solids and liquids),
  • solutions at 1.0 M concentration,
  • gases at a pressure of 1.0 atm (or 1.0 bar).
The degree sign, °, means standard state. Temperature isn't part of the definition, but tables usually list values at 298 K (25 °C). ΔG∘\Delta G^\circ is reported in kJ/molrxn\mathrm{kJ/mol_{rxn}}, per mole of reaction as written.

What "thermodynamically favored" means

Sign of ΔG∘\Delta G^\circ
Meaning
ΔG∘<0\Delta G^\circ < 0
Thermodynamically favored: products are favored
ΔG∘>0\Delta G^\circ > 0
Not favored (the reverse process is favored)
Older books call a process with ΔG∘<0\Delta G^\circ < 0 spontaneous. The AP exam uses "thermodynamically favored" instead, because "spontaneous" sounds like "sudden" or "without any cause," and neither is true. A favored process can be extremely slow (Topic 9.4), and it may need something to get it started. Use "thermodynamically favored" in your answers. (What ΔG∘=0\Delta G^\circ = 0 means, and how ΔG∘\Delta G^\circ relates to the equilibrium constant, comes in Topic 9.5.)

Calculating ΔG° from free energies of formation

Like enthalpy, free energy can be calculated from tabulated values for each substance, the standard Gibbs free energies of formation, ΔGf∘\Delta G_f^\circ. This equation is on the AP reference sheet:
ΔGreaction∘=∑ΔGf, products∘−∑ΔGf, reactants∘\Delta G^\circ_{\text{reaction}} = \sum \Delta G^\circ_{f,\,\text{products}} - \sum \Delta G^\circ_{f,\,\text{reactants}}
It works just like ΔH∘\Delta H^\circ from formation enthalpies in Unit 6. Multiply by coefficients, subtract products minus reactants, and remember that ΔGf∘=0\Delta G_f^\circ = 0 for an element in its standard state (unlike S∘S^\circ).

Example 1: Burning methane

Calculate ΔG∘\Delta G^\circ at 298 K for the combustion of methane. Use ΔGf∘\Delta G_f^\circ values (in kJ/mol) of −50.7-50.7 for CH4(g)\mathrm{CH_4(g)}, −394.4-394.4 for CO2(g)\mathrm{CO_2(g)}, and −237.1-237.1 for H2O(l)\mathrm{H_2O(l)}.
CH4(g)+2O2(g)→CO2(g)+2H2O(l)\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}
Step 1: Products. Multiply the water value by 2.
∑ΔGf, products∘=−394.4+2(−237.1)=−868.6 kJ\sum \Delta G^\circ_{f,\,\text{products}} = -394.4 + 2(-237.1) = -868.6\ \mathrm{kJ}
Step 2: Reactants. O2(g)\mathrm{O_2(g)} is an element, so its value is 0.
∑ΔGf, reactants∘=−50.7+2(0)=−50.7 kJ\sum \Delta G^\circ_{f,\,\text{reactants}} = -50.7 + 2(0) = -50.7\ \mathrm{kJ}
Step 3: Subtract.
ΔG∘=−868.6−(−50.7)=−817.9 kJ/molrxn\Delta G^\circ = -868.6 - (-50.7) = -817.9\ \mathrm{kJ/mol_{rxn}}
Answer: ΔG∘=−817.9 kJ/molrxn\Delta G^\circ = -817.9\ \mathrm{kJ/mol_{rxn}}. It's negative, so the combustion of methane is thermodynamically favored. (A gas stove still needs a spark to light. Topic 9.4 explains why.)

Enthalpy and entropy together

Two tendencies decide whether a process is favored:
  • Enthalpy: releasing energy (ΔH∘<0\Delta H^\circ < 0, exothermic) favors a process.
  • Entropy: increasing dispersal (ΔS∘>0\Delta S^\circ > 0) favors a process.
When they agree, the answer is easy. When they compete, you have to consider both. The CED names two everyday cases:
  • Water freezing. Freezing is exothermic (the molecules lock into a hydrogen-bonded crystal and release energy, which favors it) but ΔS<0\Delta S < 0 (the molecules lose freedom, which works against it). Below 0 °C enthalpy wins and water freezes. Above 0 °C entropy wins and ice melts.
  • Sodium nitrate dissolving. Dissolving NaNO3\mathrm{NaNO_3} is endothermic (the water gets colder, which works against it) but ΔS>0\Delta S > 0 (the ions spread through the solution, which favors it). The salt dissolves readily, so entropy wins. Topic 9.6 works through dissolving in detail.
If you know ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ at a given temperature, you can calculate ΔG∘\Delta G^\circ directly. This equation is also on the reference sheet:
ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ
  • ΔH∘\Delta H^\circ is the enthalpy part. A negative value pushes ΔG∘\Delta G^\circ down (more favored).
  • −TΔS∘-T\Delta S^\circ is the entropy part. A positive ΔS∘\Delta S^\circ makes this term negative (more favored).
  • TT multiplies ΔS∘\Delta S^\circ, so entropy matters more at high temperature.
Units warning: TT must be in kelvins. ΔS∘\Delta S^\circ is usually given in J/(K⋅molrxn)\mathrm{J/(K{\cdot}mol_{rxn})}, but ΔH∘\Delta H^\circ is in kJ/molrxn\mathrm{kJ/mol_{rxn}}. Divide ΔS∘\Delta S^\circ by 1000 to convert it to kilojoules before you multiply by TT. This is the most common error on this topic.
As a check, Example 1 gives the same answer this way: for methane combustion, ΔH∘=−890.3 kJ/molrxn\Delta H^\circ = -890.3\ \mathrm{kJ/mol_{rxn}} and ΔS∘=−243.0 J/(K⋅molrxn)\Delta S^\circ = -243.0\ \mathrm{J/(K{\cdot}mol_{rxn})}, so ΔG∘=−890.3−(298)(−0.2430)=−817.9 kJ/molrxn\Delta G^\circ = -890.3 - (298)(-0.2430) = -817.9\ \mathrm{kJ/mol_{rxn}}.

Example 2: When does limestone decompose?

For CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s) \rightarrow CaO(s) + CO_2(g)}, ΔH∘=+178.3 kJ/molrxn\Delta H^\circ = +178.3\ \mathrm{kJ/mol_{rxn}} and ΔS∘=+160.6 J/(K⋅molrxn)\Delta S^\circ = +160.6\ \mathrm{J/(K{\cdot}mol_{rxn})} (from Topic 9.2). (a) Calculate ΔG∘\Delta G^\circ at 298 K. (b) Above what temperature is the reaction thermodynamically favored?
(a) Step 1: Convert ΔS∘\Delta S^\circ to kilojoules.
ΔS∘=+160.6 J/(K⋅molrxn)×1 kJ1000 J=+0.1606 kJ/(K⋅molrxn)\Delta S^\circ = +160.6\ \mathrm{J/(K{\cdot}mol_{rxn})} \times \frac{1\ \mathrm{kJ}}{1000\ \mathrm{J}} = +0.1606\ \mathrm{kJ/(K{\cdot}mol_{rxn})}
Step 2: Substitute.
ΔG∘=178.3−(298 K)(0.1606 kJ/(K⋅molrxn))=178.3−47.9=+130.4 kJ/molrxn\Delta G^\circ = 178.3 - (298\ \mathrm{K})(0.1606\ \mathrm{kJ/(K{\cdot}mol_{rxn})}) = 178.3 - 47.9 = +130.4\ \mathrm{kJ/mol_{rxn}}
ΔG∘>0\Delta G^\circ > 0, so at room temperature the reaction is not favored. That's why limestone buildings don't fall apart into lime and CO2\mathrm{CO_2}. (The ΔGf∘\Delta G_f^\circ method gives the same −604.0+(−394.4)−(−1128.8)=+130.4 kJ/molrxn-604.0 + (-394.4) - (-1128.8) = +130.4\ \mathrm{kJ/mol_{rxn}}.)
(b) Both ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ are positive, so the reaction becomes favored at high temperature, once TΔS∘T\Delta S^\circ is larger than ΔH∘\Delta H^\circ. The switch happens where ΔG∘=0\Delta G^\circ = 0:
0=ΔH∘−TΔS∘⇒T=ΔH∘ΔS∘=178.3 kJ/molrxn0.1606 kJ/(K⋅molrxn)=1.110×103 K0 = \Delta H^\circ - T\Delta S^\circ \quad\Rightarrow\quad T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{178.3\ \mathrm{kJ/mol_{rxn}}}{0.1606\ \mathrm{kJ/(K{\cdot}mol_{rxn})}} = 1.110 \times 10^3\ \mathrm{K}
Answer: the reaction is favored above about 1110 K (about 837 °C). Lime kilns run hotter than this, around 900 °C. This assumes ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ stay about the same as the temperature changes, which the AP exam assumes too.

Example 3: Why water freezes below 0 °C

For H2O(l)→H2O(s)\mathrm{H_2O(l) \rightarrow H_2O(s)}, ΔH∘=−6.01 kJ/molrxn\Delta H^\circ = -6.01\ \mathrm{kJ/mol_{rxn}} and ΔS∘=−22.0 J/(K⋅molrxn)\Delta S^\circ = -22.0\ \mathrm{J/(K{\cdot}mol_{rxn})}, or −0.0220 kJ/(K⋅molrxn)-0.0220\ \mathrm{kJ/(K{\cdot}mol_{rxn})}. Is freezing favored at −10-10 °C (263 K)? At +10+10 °C (283 K)?
T=263 K:ΔG∘=−6.01−(263)(−0.0220)=−6.01+5.79=−0.22 kJ/molrxnT = 263\ \mathrm{K}: \quad \Delta G^\circ = -6.01 - (263)(-0.0220) = -6.01 + 5.79 = -0.22\ \mathrm{kJ/mol_{rxn}}T=283 K:ΔG∘=−6.01−(283)(−0.0220)=−6.01+6.23=+0.22 kJ/molrxnT = 283\ \mathrm{K}: \quad \Delta G^\circ = -6.01 - (283)(-0.0220) = -6.01 + 6.23 = +0.22\ \mathrm{kJ/mol_{rxn}}
Answer: freezing is favored at 263 K but not at 283 K. Freezing is enthalpy-driven: the negative ΔH∘\Delta H^\circ favors it, and the entropy term works against it. At 0 °C (273.15 K) the two parts cancel, since (273.15)(0.0220)=6.01(273.15)(0.0220) = 6.01, so ΔG∘=0\Delta G^\circ = 0. That's the melting point, where ice and water coexist.

Predicting favorability from the signs

Often you don't need numbers at all. The signs of ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ tell you when ΔG∘<0\Delta G^\circ < 0:
ΔH∘\Delta H^\circ
ΔS∘\Delta S^\circ
Favored (ΔG∘<0\Delta G^\circ < 0) at
Why
<0< 0
>0> 0
All temperatures
Both parts favor the process
>0> 0
<0< 0
No temperature
Both parts work against it
>0> 0
>0> 0
High temperature
Entropy wins once TΔS∘>ΔH∘T\Delta S^\circ > \Delta H^\circ
<0< 0
<0< 0
Low temperature
Enthalpy wins while ∣TΔS∘∣<∣ΔH∘∣|T\Delta S^\circ| < |\Delta H^\circ|
When ΔH∘<0\Delta H^\circ < 0 and ΔS∘>0\Delta S^\circ > 0, no calculation is needed: the process is favored. When ΔH∘>0\Delta H^\circ > 0 and ΔS∘<0\Delta S^\circ < 0, no calculation is needed either: it's not favored.
A graph shows why. With ΔH∘\Delta H^\circ and ΔS∘\Delta S^\circ treated as constants, ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ is a straight line: it starts at ΔH∘\Delta H^\circ when T=0T = 0, and its slope is −ΔS∘-\Delta S^\circ. The process is favored wherever the line is below the TT-axis (red). In the two mixed cases, the line crosses zero at T=ΔH∘/ΔS∘T = \Delta H^\circ / \Delta S^\circ.
9.3 - Gibbs Free Energy and Thermodynamic Favorability

Which factor drives the process?

The AP exam often asks whether a favored process is driven by enthalpy, entropy, or both. Use the signs:
Favored process with
Driven by
Examples
ΔH∘<0\Delta H^\circ < 0, ΔS∘<0\Delta S^\circ < 0
Enthalpy (entropy works against it)
Freezing, condensation, making NH3\mathrm{NH_3} at room temperature
ΔH∘>0\Delta H^\circ > 0, ΔS∘>0\Delta S^\circ > 0
Entropy (enthalpy works against it)
Dissolving NaNO3\mathrm{NaNO_3}, ice melting above 0 °C, limestone above 1110 K
ΔH∘<0\Delta H^\circ < 0, ΔS∘>0\Delta S^\circ > 0
Both
Many combustion reactions that make more gas
Don't just say "both" because both appear in ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ. If one of them has the unfavorable sign, it is working against the process. Connect the signs to evidence too: if a reaction mixture gets colder, it is endothermic, so if it still happens, entropy must be driving it. If a gas forms from solids or liquids, ΔS∘>0\Delta S^\circ > 0.

Example 4: Making ammonia (AP-style)

N2(g)+3H2(g)→2NH3(g)\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}
For this reaction, ΔH∘=−92.2 kJ/molrxn\Delta H^\circ = -92.2\ \mathrm{kJ/mol_{rxn}} and ΔS∘=−198.7 J/(K⋅molrxn)\Delta S^\circ = -198.7\ \mathrm{J/(K{\cdot}mol_{rxn})}. (a) Explain the sign of ΔS∘\Delta S^\circ in terms of the particles. (b) Calculate ΔG∘\Delta G^\circ at 298 K and state whether the reaction is thermodynamically favored. (c) Is the reaction driven by enthalpy, entropy, or both? Justify. (d) A student claims that because ΔH∘<0\Delta H^\circ < 0, the reaction is still thermodynamically favored at 725 K. Do you agree? Justify with a calculation.
(a) Four moles of gas (1 N2\mathrm{N_2} + 3 H2\mathrm{H_2}) become two moles of gas. With fewer gas particles moving independently, the matter is less dispersed, so ΔS∘<0\Delta S^\circ < 0.
(b) Convert ΔS∘\Delta S^\circ to −0.1987 kJ/(K⋅molrxn)-0.1987\ \mathrm{kJ/(K{\cdot}mol_{rxn})}, then substitute:
ΔG∘=−92.2−(298)(−0.1987)=−92.2+59.2=−33.0 kJ/molrxn\Delta G^\circ = -92.2 - (298)(-0.1987) = -92.2 + 59.2 = -33.0\ \mathrm{kJ/mol_{rxn}}
ΔG∘<0\Delta G^\circ < 0, so the reaction is thermodynamically favored at 298 K.
(c) Enthalpy. ΔH∘<0\Delta H^\circ < 0 favors the reaction. ΔS∘<0\Delta S^\circ < 0 works against it, because the entropy term −TΔS∘=+59.2 kJ/molrxn-T\Delta S^\circ = +59.2\ \mathrm{kJ/mol_{rxn}} is positive. The reaction is favored only because the enthalpy part is larger in size at this temperature.
(d) Disagree.
ΔG∘=−92.2−(725)(−0.1987)=−92.2+144.1=+51.9 kJ/molrxn\Delta G^\circ = -92.2 - (725)(-0.1987) = -92.2 + 144.1 = +51.9\ \mathrm{kJ/mol_{rxn}}
At 725 K, −TΔS∘-T\Delta S^\circ outweighs ΔH∘\Delta H^\circ, so ΔG∘>0\Delta G^\circ > 0 and the reaction is not favored. Since both signs are negative, the reaction is favored only at low temperature, below T=92.2/0.1987=464T = 92.2 / 0.1987 = 464 K (about 191 °C). Yet industrial ammonia plants run near 725 K (about 452 °C), with an iron catalyst and very high pressure, because at low temperatures the reaction is far too slow. Topic 9.4 is about exactly this gap between "favored" and "fast."

Common mistakes

  • Mixing J and kJ. Convert ΔS∘\Delta S^\circ from J to kJ (divide by 1000) before using ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ. Being off by a factor of 1000 can flip the sign of your answer.
  • Using °C instead of K. TT is always in kelvins.
  • Thinking "favored" means "fast." ΔG∘\Delta G^\circ says nothing about rate.
  • Saying "both" drive a reaction when one sign is unfavorable. Check each sign separately.
  • Treating elements' ΔGf∘\Delta G_f^\circ as nonzero, or their S∘S^\circ as zero. It's the other way around.
  • Dropping signs when subtracting. Write −TΔS∘-T\Delta S^\circ with parentheses, like −(298)(−0.1987)-(298)(-0.1987).

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