Gibbs Energy of Mixing

Pour two liquids together and they mix without any push from outside. No stirring is required in principle, no heat needs to be added or removed. Why does this happen spontaneously? The answer lies in the Gibbs energy of mixing, $\Delta_{\text{mix}}G$ , which is always negative for an ideal mixture at constant temperature and pressure. This quantity matters because it is the thermodynamic reason solutions exist at all. It explains why gases interdiffuse, why alloys form, and why a drop of dye spreads through water rather than staying clumped. It connects the microscopic idea of dispersal to the macroscopic criterion for spontaneous change. This article derives $\Delta_{\text{mix}}G$ for an ideal mixture from the chemical potentials of the components. You will see where the negative sign comes from. Then you will see how it splits into entropy and enthalpy contributions. Finally you will see two fully worked numerical examples.

Key takeaways

$\Delta_{\text{mix}}G = nRT(x_A \ln x_A + x_B \ln x_B)$ for an ideal mixture, and this is always negative.
The negative sign arises because $\ln x_i < 0$ for any mole fraction between 0 and 1.
Ideal mixing has zero enthalpy change; the entire driving force is entropic.
$\Delta_{\text{mix}}G$ is most negative near $x = 0.5$ and vanishes at the pure-component limits.
The same algebra you met for mixing ideal gases carries over unchanged to ideal solutions.
Key idea: mixing lowers Gibbs energy because spreading molecules over a larger accessible volume of composition space increases entropy, at no enthalpic cost, and $\Delta G = \Delta H - T\Delta S$ then guarantees $\Delta_{\text{mix}}G < 0$ .

The core insight

For an ideal mixture, $\Delta_{\text{mix}}H = 0$ and $\Delta_{\text{mix}}S = -nR\sum_i x_i \ln x_i > 0$ , so $\Delta_{\text{mix}}G = nRT\sum_i x_i \ln x_i$ is negative at every composition except the pure states — mixing is spontaneous purely because it multiplies the number of ways molecules can be arranged, not because any bonds get stronger.

The physical picture

Imagine two liquids, A and B, separated by a partition inside a container held at constant $T$ and $p$ . Each liquid is a pure substance with a well-defined chemical potential, $\mu_A^*$ and $\mu_B^*$ . When the partition is removed, the molecules of A and B intermingle. If the mixture is ideal — meaning A–A, B–B, and A–B interactions are all energetically equivalent — then no heat is exchanged in the process. The molecules do not care whether their neighbor is an A or a B. What does change is the number of accessible microscopic arrangements. Before mixing, all the A molecules are confined to one region and all the B molecules to another. After mixing, each type of molecule can be found anywhere in the combined volume. This is a purely statistical, entropic effect. It is exactly the same argument used earlier for two ideal gases expanding into a shared container: nothing energetic drives the process, but the accessible number of configurations rises enormously, so the entropy rises, and Gibbs energy falls. Because $G = H - TS$ and $\Delta_{\text{mix}}H = 0$ for the ideal case, the sign of $\Delta_{\text{mix}}G$ is dictated entirely by $-T\Delta_{\text{mix}}S$ . Since mixing always increases the entropy of two distinguishable, initially separated components, $\Delta_{\text{mix}}G$ is always negative — mixing is always spontaneous for an ideal mixture, at any nonzero temperature and any intermediate composition.

Figure

Gibbs energy of mixing versus composition for one mole of an equimolar A–B system, showing the symmetric negative dip

Figure

Microstate picture: separated pure phases versus the mixed state, illustrating why W increases and S increases

The core relations

\Delta_{\text{mix}}G = nRT\sum_i x_i \ln x_i

where: $n$ is total moles, $R$ the gas constant, $T$ absolute temperature, $x_i$ the mole fraction of component $i$ ; the sum runs over all components.

\Delta_{\text{mix}}S = -nR\sum_i x_i \ln x_i

where: this is always positive since each $\ln x_i < 0$ , and it is the sole source of the negative $\Delta_{\text{mix}}G$ .

\Delta_{\text{mix}}H = 0

where: holds specifically for the ideal-mixture (or ideal-gas) approximation, where interactions between unlike and like molecules are energetically indistinguishable.

\mu_i(\text{mixture}) = \mu_i^{*} + RT\ln x_i

where: $\mu_i^{*}$ is the chemical potential of pure $i$ at the same $T$ and $p$ ; this is the defining relation for an ideal mixture and the starting point of the derivation below.

Derivation AGibbs Energy of Mixing for an Ideal Mixture

Goal: starting from the chemical potentials of pure and mixed components, obtain $\Delta_{\text{mix}}G = nRT\sum_i x_i\ln x_i$ and show it is necessarily negative.

Step 1. Gibbs energy before mixing. Consider $n_A$ moles of pure liquid A and $n_B$ moles of pure liquid B, held separately at temperature $T$ and pressure $p$ . The total Gibbs energy of the unmixed system is simply the sum of the two pure Gibbs energies,

G_{\text{initial}} = n_A \mu_A^{*} + n_B \mu_B^{*}

since for a pure substance $G = n\mu^{*}$ .

Step 2. Gibbs energy after mixing. For a mixture, the total Gibbs energy is the sum of $n_i\mu_i$ over all components, where $\mu_i$ is the chemical potential of $i$ in the mixture. For an ideal mixture this is defined (in exact analogy with the ideal-gas mixing result) by

\mu_i = \mu_i^{*} + RT\ln x_i.

This says a component's chemical potential is lowered relative to its pure value by an amount that depends on how dilute it is in the mixture — the more dilute, the more negative $\ln x_i$ , the lower $\mu_i$ . Substituting,

G_{\text{final}} = n_A\left(\mu_A^{*} + RT\ln x_A\right) + n_B\left(\mu_B^{*} + RT\ln x_B\right).

Step 3. Subtract to isolate the mixing term. The Gibbs energy of mixing is defined as the difference,

\Delta_{\text{mix}}G = G_{\text{final}} - G_{\text{initial}}.

The pure-component terms $n_A\mu_A^{*}$ and $n_B\mu_B^{*}$ appear identically in both expressions and cancel exactly, leaving

\Delta_{\text{mix}}G = n_ART\ln x_A + n_BRT\ln x_B.

Step 4. Express in terms of mole fraction and total moles. Writing $n_A = n x_A$ and $n_B = n x_B$ , where $n = n_A + n_B$ is the total number of moles,

\Delta_{\text{mix}}G = nRT\left(x_A\ln x_A + x_B\ln x_B\right).

Generalizing to any number of components gives the boxed relation stated above.

Step 5. Show the sign is always negative. Because $0 < x_i < 1$ for every component actually present in a mixture, $\ln x_i$ is strictly negative for each term. Since $R$ , $T$ , and $n$ are all positive quantities, every term in the sum $nRT\, x_i\ln x_i$ is negative, so the sum itself must be negative. There is no composition (short of a pure substance, where the process is not really mixing) at which $\Delta_{\text{mix}}G$ can be positive or zero.

Step 6. Split into entropy and enthalpy. Differentiating $\Delta_{\text{mix}}G$ with respect to $T$ at constant $p$ and composition, using $(\partial G/\partial T)_p = -S$ , gives

\Delta_{\text{mix}}S = -\left(\frac{\partial \Delta_{\text{mix}}G}{\partial T}\right)_{p,x} = -nR\sum_i x_i\ln x_i,

which is positive by the same sign argument. Because $\Delta_{\text{mix}}G$ has no explicit dependence on $T$ beyond the prefactor $T$ itself, applying $G = H - TS$ term by term shows

\Delta_{\text{mix}}H = \Delta_{\text{mix}}G + T\Delta_{\text{mix}}S = 0.

\boxed{\Delta_{\text{mix}}G = nRT\sum_i x_i \ln x_i \;<\; 0 \quad\text{for all } 0<x_i<1}

Worked examples

Example 1 — Equimolar mixing of two liquids at room temperature. Take 1.00 mol of liquid A and 1.00 mol of liquid B, assumed to form an ideal mixture, at $T = 298\ \text{K}$ . Total moles: $n = n_A + n_B = 2.00\ \text{mol}$ . Mole fractions: $x_A = x_B = 0.500$ . Compute the entropy of mixing first:

\Delta_{\text{mix}}S = -nR(x_A\ln x_A + x_B\ln x_B)

= -(2.00\ \text{mol})(8.314\ \text{J K}^{-1}\text{mol}^{-1})\big[(0.500)\ln(0.500) + (0.500)\ln(0.500)\big]

= -(2.00)(8.314)\left[(0.500)(-0.6931) \times 2\right]\ \text{J K}^{-1}

= -(2.00)(8.314)(-0.6931)\ \text{J K}^{-1} = 11.53\ \text{J K}^{-1}.

Now the Gibbs energy of mixing, using either the direct formula or $\Delta_{\text{mix}}G = -T\Delta_{\text{mix}}S$ (valid since $\Delta_{\text{mix}}H = 0$ ):

\Delta_{\text{mix}}G = nRT(x_A\ln x_A + x_B\ln x_B) = -T\Delta_{\text{mix}}S

= -(298\ \text{K})(11.53\ \text{J K}^{-1}) = -3.44\times 10^{3}\ \text{J} = -3.44\ \text{kJ}.

Sanity check: per mole of mixture, this is $-3.44\ \text{kJ}/2.00\ \text{mol} = -1.72\ \text{kJ mol}^{-1}$ , matching the minimum labeled on the composition plot above, where the dip is deepest at $x = 0.5$ . The negative value confirms mixing is spontaneous, and the modest magnitude — a few kJ per mole, far smaller than a typical bond energy of hundreds of kJ per mole — confirms this is an entropic, not an energetic, effect.

Example 2 — Non-equimolar mixing and the "most negative" composition. A chemist mixes 3.00 mol of ideal liquid A with 1.00 mol of ideal liquid B at $T = 350\ \text{K}$ . Find $\Delta_{\text{mix}}G$ , and compare it with the equimolar case at the same total moles and temperature to confirm the mixing is less favorable than 50:50 mixing. Total moles: $n = 4.00\ \text{mol}$ . Mole fractions: $x_A = 3.00/4.00 = 0.750$ , $x_B = 1.00/4.00 = 0.250$ .

x_A\ln x_A = (0.750)\ln(0.750) = (0.750)(-0.2877) = -0.2158

x_B\ln x_B = (0.250)\ln(0.250) = (0.250)(-1.3863) = -0.3466

\sum_i x_i\ln x_i = -0.2158 + (-0.3466) = -0.5624

\Delta_{\text{mix}}G = nRT\sum_i x_i\ln x_i = (4.00\ \text{mol})(8.314\ \text{J K}^{-1}\text{mol}^{-1})(350\ \text{K})(-0.5624)

= (4.00)(8.314)(350)(-0.5624)\ \text{J} = -6.55\times 10^{3}\ \text{J} = -6.55\ \text{kJ}.

Per mole of mixture: $-6.55\ \text{kJ}/4.00\ \text{mol} = -1.64\ \text{kJ mol}^{-1}$ . Compare with an equimolar 2:2 mixture (same 4.00 mol total, same $T$ ): $x_A = x_B = 0.500$ ,

\sum_i x_i\ln x_i = 2(0.500)(-0.6931) = -0.6931,

\Delta_{\text{mix}}G = (4.00)(8.314)(350)(-0.6931)\ \text{J} = -8.07\times10^{3}\ \text{J} = -8.07\ \text{kJ},

or $-2.02\ \text{kJ mol}^{-1}$ . As expected, the equimolar mixture is more stable per mole ( $-2.02$ vs. $-1.64\ \text{kJ mol}^{-1}$ ): mixing is still spontaneous in both cases, but the driving force is strongest at $x = 0.5$ , where the entropic dispersal of molecules is maximal, and weakest as either mole fraction approaches 0 or 1.

Limiting cases & common traps

Situation	Behavior of $\Delta_{\text{mix}}G$	Note
$x_i \to 0$ or $x_i \to 1$	$\Delta_{\text{mix}}G \to 0$	No mixing occurs at the pure-substance limit; nothing to mix.
$x_A = x_B = 0.5$ (two components)	Most negative value	Maximum entropic dispersal at equimolar composition.
Real (non-ideal) mixtures	$\Delta_{\text{mix}}H \neq 0$	Enthalpy of mixing can be positive or negative; ideal formula is only an approximation.
$T \to 0$	$\Delta_{\text{mix}}G \to 0$	Since $\Delta_{\text{mix}}G = -T\Delta_{\text{mix}}S$ for ideal mixing, it scales linearly with $T$ .
More than two components	Still negative	Sum extends over all $i$ ; each term remains negative.

The chemist behind it

The chemical-potential formalism used here — writing $\mu_i = \mu_i^{*} + RT\ln x_i$ and building macroscopic mixing thermodynamics directly from it — traces to Josiah Willard Gibbs, whose 1870s papers on the equilibrium of heterogeneous substances founded the entire framework of chemical potentials, phase equilibrium, and free energy used throughout physical chemistry. Gibbs worked largely in isolation at Yale, publishing in the relatively obscure Transactions of the Connecticut Academy of Sciences, and his ideas were only widely appreciated in chemistry after being popularized in Europe by Wilhelm Ostwax and later systematized by Gilbert N. Lewis, who introduced the modern notation of activities and fugacities built directly on Gibbs's chemical potential.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist whose chemical potential underlies every mixing and phase-equilibrium calculation in this article

Ask the tutor: What happens to the Gibbs energy of mixing when the two components are not ideal and have a nonzero enthalpy of mixing?

Why is ΔmixG negative even though nothing "energetically favorable" happens?

Gibbs energy combines enthalpy and entropy through $G = H - TS$ . For an ideal mixture, $\Delta_{\text{mix}}H = 0$ because A–A, B–B, and A–B interactions are equally strong, so there is no energetic reward for mixing. However, entropy always rises when previously separated, distinguishable molecules are allowed to intermingle, since there are vastly more ways to arrange the mixed state than the separated one. With $\Delta H = 0$ and $\Delta S > 0$ , the term $-T\Delta S$ is negative, so $\Delta G$ must be negative — mixing is spontaneous on entropic grounds alone.

How does this relate to mixing ideal gases?

The derivation is structurally identical. For ideal gases, mixing at constant total pressure gives $\Delta_{\text{mix}}G = nRT\sum_i x_i\ln x_i$ because each gas expands into the combined volume, lowering its partial pressure and hence its chemical potential by $RT\ln x_i$ . For an ideal solution, the same algebraic form emerges because each component's chemical potential in the mixture is defined the same way, $\mu_i = \mu_i^{*} + RT\ln x_i$ . The physics differs — volume expansion for gases, statistical dispersal on a lattice-like model for liquids — but the resulting equation is identical.

Does ΔmixG depend on pressure?

Not directly, in the ideal case. The formula $\Delta_{\text{mix}}G = nRT\sum_i x_i\ln x_i$ contains no explicit $p$ dependence, because the pure and mixed states are compared at the same $T$ and $p$ , and the ideal-mixture assumption means volume changes on mixing are zero. In a real, non-ideal mixture, there can be a small volume of mixing, which would introduce weak pressure sensitivity through the $p\,\Delta V$ contributions buried in the true $\mu_i^{*}$ and activity terms.

Can ΔmixG ever be positive for a real mixture?

Yes, but that means the mixture is behaving non-ideally, and full miscibility may fail. If $\Delta_{\text{mix}}H$ is sufficiently positive (unfavorable A–B interactions) it can outweigh $-T\Delta_{\text{mix}}S$ at some compositions, making $\Delta_{\text{mix}}G$ positive there — this is exactly the mechanism behind partial miscibility and upper critical solution temperatures. For a strictly ideal mixture, though, this can never happen; $\Delta_{\text{mix}}G$ is guaranteed negative everywhere between the pure limits.

Why does ΔmixG vanish at x = 0 and x = 1?

At these limits, one of the mole fractions equals 1 and the other equals 0. The term $x_i\ln x_i$ vanishes both when $x_i = 1$ (since $\ln 1 = 0$ ) and when $x_i \to 0$ (since $x\ln x \to 0$ in that limit, even though $\ln x \to -\infty$ , because $x$ goes to zero faster). Physically, these limits correspond to not actually mixing anything — you have pure A or pure B, and there is no dispersal to gain entropy from.