Chemical Thermodynamics · Lesson 9 of 10

Gibbs Energy, Temperature, and Equilibrium

“Use enthalpy and entropy together to predict direction and relate standard free energy to equilibrium.”

Learning Objectives

• Define Gibbs energy and derive the constant-temperature change. • Connect Gibbs-energy decrease to the total-entropy criterion at constant temperature and pressure. • Explain all four sign combinations of enthalpy and entropy. • Distinguish the equilibrium reaction Gibbs energy from its standard value. • Calculate standard Gibbs energy or an equilibrium constant with consistent units.

A direction criterion using system properties

Total entropy is a powerful criterion, but continually calculating the surroundings can be inconvenient. For a system in contact with a large thermal reservoir at fixed temperature and pressure, its enthalpy change tells us the heat exchanged with that reservoir when only pressure–volume work is involved. Combining this with the system entropy produces a criterion using the system’s own properties.

Definition

Gibbs energy

The state function $G = H - T S$ , where H is enthalpy, T absolute temperature, and S entropy. Like H and S, G is extensive. It is also often called Gibbs free energy.

At a fixed temperature, subtracting the initial G from the final G gives $Δ G = Δ H - T Δ S$ . Both terms have energy units. If enthalpy is in kJ but entropy is in J/K, convert one before subtracting. An increase of entropy makes the negative $- T Δ S$ contribution more favourable to a decrease in G.

Gibbs energy change at constant temperatureLaTeX

Δ G = Δ H - T Δ S

T must be in kelvin. ΔS is the system entropy change, not total system-plus-surroundings entropy.

For the specified constant-pressure process, the reservoir receives heat $- Δ H_{sys}$ , so $Δ S_{surr} = - Δ H_{sys} / T$ . Add the system entropy and multiply by T: $T Δ S_{total} = T Δ S_{sys} - Δ H_{sys} = - Δ G_{sys}$ . Thus a positive total-entropy change corresponds to a decrease in system Gibbs energy.

Gibbs and total entropyLaTeX

Δ G_{sys} = - T Δ S_{total}

For a closed system at fixed temperature and pressure under the stated heat/work conditions, spontaneous direction corresponds to decreasing G; equilibrium is a constrained minimum of G.

Gibbs tendency at fixed T and p	Interpretation
$Δ_{r} G < 0$	Forward chemical change is favoured at the current composition
$Δ_{r} G > 0$	Reverse chemical change is favoured at the current composition
$Δ_{r} G = 0$	Chemical equilibrium: no net driving force

For a finite specified transformation, $- Δ G$ also gives the maximum non-expansion work that could be delivered reversibly at constant temperature and pressure. Electrical work from a cell is an example of such useful work. Real irreversible operation delivers less. A positive spontaneous driving force does not promise that every joule becomes useful work or that the reaction is fast.

Example — An unfavourable enthalpy can be overcome

Problem
A reaction has $Δ H = + 40.0 kJ mo l^{- 1}$ and $Δ S = + 150 J K^{- 1} mo l^{- 1}$ . Calculate $Δ G$ at 300 K.

1.Convert entropy to compatible units: $150 J K^{- 1} mo l^{- 1} = 0.150 kJ K^{- 1} mo l^{- 1}$ .
2.Calculate $T Δ S = (300) (0.150) = 45.0 kJ mo l^{- 1}$ .
3.Then $Δ G = 40.0 - 45.0 = - 5.0 kJ mo l^{- 1}$ .
4.The entropy contribution outweighs the positive enthalpy. The indicated forward change is thermodynamically favourable under the specified conditions.

How temperature changes the balance

Temperature multiplies the entropy term, so it can change which contribution dominates. If enthalpy and entropy changes vary little over a chosen interval, their signs give four useful patterns. “High” and “low” are relative to the particular reaction: room temperature may be high enough for one reaction and too low for another.

Enthalpy sign	Entropy sign	Gibbs sign and temperature implication
Negative	Positive	Negative at all positive temperatures within the assumptions
Negative	Negative	Negative at sufficiently low T; positive at sufficiently high T
Positive	Positive	Positive at low T; negative at sufficiently high T
Positive	Negative	Positive at all positive temperatures within the assumptions

For the two competing-sign situations, the crossover occurs at $Δ G = 0$ . Rearranging gives $T = Δ H /Δ S$ with compatible units. At that temperature the net thermodynamic driving force is zero. It is not correct to call the equality point itself a spontaneously forward change. The simple classification assumes constant enthalpy and entropy and does not ignore phase changes outside that range.

Example — A temperature threshold

Problem
For $Δ H = + 400 kJ mo l^{- 1}$ and $Δ S = + 0.200 kJ K^{- 1} mo l^{- 1}$ , above what temperature is the forward change favoured?

1.Set $Δ G = Δ H - T Δ S = 0$ .
2.The crossover is $T = 400/0.200 = 2000 K$ .
3.Since both quantities are positive, increasing temperature makes the negative entropy term more important.
4.Thus $Δ G < 0$ for $T > 2000 K$ under the assumption that both values remain constant. At 2000 K the criterion gives zero driving force.

A minimum does not mean that reactions stop

At chemical equilibrium forward and reverse molecular events continue at equal rates, so composition stays steady. Thermodynamic reversibility means a limiting process reversible by an infinitesimal change; a chemically reversible reaction simply can proceed in both directions. These ideas are related but are not interchangeable descriptions of every actual reaction path.

At fixed temperature and pressure, an allowed shift away from a stable equilibrium composition would raise G. The equilibrium composition therefore lies at a constrained minimum. The reaction Gibbs energy, written $Δ_{r} G$ , measures the local driving force for advancing the reaction at its current composition; it is zero at equilibrium. This is distinct from the standard reaction value, $Δ_{r} G^{\circ}$ , which compares standard-state chemical potentials.

Gibbs energy and reaction progress— The curve is schematic: progress in either direction towards the minimum lowers the system Gibbs energy.

Standard Gibbs energy and the equilibrium constant

An equilibrium constant describes how products and reactants are represented at equilibrium. Its thermodynamic form is dimensionless: gas pressures appear as ratios to standard pressure, not as a logarithm of a quantity carrying units. At equilibrium, the standard reaction Gibbs energy is related to K through the gas constant and temperature.

Standard Gibbs energy and equilibriumLaTeX

Δ_{r} G^{\circ} = - R T ln K = - 2.303 R T lo g_{10} K

R = 8.314 J mol−1 K−1 gives ΔrG° in J mol−1. K is the dimensionless equilibrium constant for the balanced reaction at temperature T.

If $K > 1$ , $ln K$ is positive and $Δ_{r} G^{\circ}$ is negative. If $K < 1$ , the standard value is positive; if $K = 1$ , it is zero. A large K indicates equilibrium strongly favours products in the appropriate activity expression. The actual direction from a particular mixture also depends on composition, so a positive standard value does not forbid every forward change.

Why the two Gibbs values differ

One way to express the composition distinction is $Δ_{r} G = Δ_{r} G^{\circ} + R T ln Q$ , where Q has the same form as K but uses the current composition. At equilibrium Q equals K and $Δ_{r} G = 0$ , giving the relation above. Q is introduced only to explain why the standard and actual values need not share a sign.

Example — The tiny equilibrium constant for ozone formation

Problem
For $\frac{3}{2} O_{2} (g) \to O_{3} (g)$ at 298 K, use $K = 2.47 \times 1 0^{- 29}$ to find $Δ_{r} G^{\circ}$ .

1.Compute the base-ten logarithm: $lo g_{10} K = lo g_{10} 2.47 - 29 \approx - 28.607$ .
2.Use $Δ_{r} G^{\circ} = - 2.303 (8.314) (298) (- 28.607) J mo l^{- 1}$ .
3.The result is approximately $+ 1.63 \times 1 0^{5} J mo l^{- 1} = + 163 kJ mo l^{- 1}$ .
4.The large positive standard value is consistent with an extremely small equilibrium constant for ozone formation from oxygen under these conditions.

Example — Finding K from standard Gibbs energy

Problem
At 298 K, the standard reaction Gibbs energy for $2N H_{3} (g) + C O_{2} (g) \to N H_{2} CON H_{2} (aq) + H_{2} O (l)$ is −13.6 kJ/mol. Find K.

1.First convert the energy: $- 13.6 kJ mo l^{- 1} = - 13600 J mo l^{- 1}$ .
2.Rearrange: $ln K = - Δ_{r} G^{\circ} / (R T) = 13600/ [(8.314) (298)] \approx 5.489$ .
3.Therefore $K = e^{5.489} \approx 2.42 \times 1 0^{2}$ .
4.The equilibrium constant exceeds one, consistent with a negative standard reaction Gibbs energy.

Example — Dissociation and partial pressures

Problem
At 333 K, an initially pure sample of $N_{2} O_{4} (g)$ is 50% dissociated at total pressure 1.00 bar. Calculate K and $Δ_{r} G^{\circ}$ for $N_{2} O_{4} (g) ⇌ 2N O_{2} (g)$ .

1.Start with one mole. At equilibrium the amounts are 0.50 mol of $N_{2} O_{4}$ and 1.00 mol of $N O_{2}$ , for a total of 1.50 mol.
2.The mole fractions are $1/3$ and $2/3$ . Their partial pressures are $1/3$ bar and $2/3$ bar.
3.With $p^{\circ} = 1 bar$ , $K = [(p_{N O_{2}} / p^{\circ})^{2}] / (p_{N_{2} O_{4}} / p^{\circ}) = (2/3)^{2} / (1/3) = 4/3$ .
4.Then $Δ_{r} G^{\circ} = - (8.314) (333) ln (4/3) = - 796 J mo l^{- 1} \approx - 0.80 kJ mo l^{- 1}$ .
5.If the total pressure is instead 1.00 atm, convert it to 1.01325 bar before using the same standard pressure: K is approximately 1.351 and the result is about −0.83 kJ/mol. Neither result is hundreds of kJ/mol.

Combining $Δ_{r} G^{\circ} = Δ_{r} H^{\circ} - T Δ_{r} S^{\circ}$ with the K relation links thermal data to equilibrium yield. Conversely, a measured K determines standard Gibbs energy at that measurement temperature. Predicting another temperature requires suitable thermal data or an explicit approximation; a single K measurement is not enough by itself.

Common mistake

A reaction at equilibrium has $Δ_{r} G = 0$ , not necessarily $Δ_{r} G^{\circ} = 0$ . Also, exothermicity alone does not guarantee a large K, because the entropy contribution can change the standard Gibbs energy substantially.

Quick check

Which conditions give the usual Gibbs-energy criterion for a closed chemical system?

Quick check

A reaction has negative enthalpy change and positive entropy change. What follows if both are treated as constant?

Quick check

What is the actual reaction Gibbs energy at equilibrium?

Quick check

If the dimensionless equilibrium constant is less than one, what is the sign of standard reaction Gibbs energy?

Quick check

Why must entropy units be converted before combining 20 kJ/mol with T times 50 J mol−1 K−1?

Quick check

A positive standard Gibbs energy is reported for a reaction. Which conclusion is correct?

Practice Problems

Derive $Δ G = - T Δ S_{total}$ for the specified constant-temperature, constant-pressure conditions.
Find $Δ G$ at 298 K when $Δ H = - 50.0 kJ mo l^{- 1}$ and $Δ S = - 100 J K^{- 1} mo l^{- 1}$ .
For $Δ H = - 60.0 kJ mo l^{- 1}$ and $Δ S = - 150 J K^{- 1} mo l^{- 1}$ , find the crossover temperature and state which side favours the forward change.
Find $Δ_{r} G^{\circ}$ for K = 10 at 300 K.
Calculate K at 298 K if $Δ_{r} G^{\circ} = + 5.00 kJ mo l^{- 1}$ .
Explain why an equilibrium mixture need not have equal reactant and product amounts, and why $Δ_{r} G^{\circ}$ need not be zero there.

Key Takeaways

At constant temperature, $Δ G = Δ H - T Δ S$ combines both system contributions. At fixed temperature and pressure the natural direction lowers G, and equilibrium is a constrained minimum. Temperature changes the weight of the entropy contribution. The standard relation $Δ_{r} G^{\circ} = - R T ln K$ uses a dimensionless K and must be distinguished from actual reaction driving force.

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Spontaneity, Entropy, and the Thermodynamic Laws

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Chapter Summary and Practice

A direction criterion using system properties

How temperature changes the balance

A minimum does not mean that reactions stop

Standard Gibbs energy and the equilibrium constant

Quiz

Practice Problems

Key Takeaways