Entropy and Enthalpy of Mixing

Pour two liquids together, and something odd happens for free. No heat escapes, no volume changes. Yet the process is spontaneous and irreversible. That "something for nothing" is entropy of mixing. It is one of the cleanest examples of a purely statistical driving force in chemistry. Understanding why mixing is spontaneous even with zero enthalpy change explains several things. It explains why oil and water refuse to mix. It explains why gases always mix completely. It also explains why "like dissolves like" is really a statement about enthalpy competing with entropy. It also sets up the entire framework of ideal solutions. This article derives the entropy, enthalpy, and volume of mixing for an ideal solution from the Gibbs energy of mixing. You will see the statistical origin of the entropy term. You will see why an ideal mixture has zero enthalpy and volume change. Then you will see two worked numerical examples and the traps that catch students on exams.

Key takeaways

Mixing two ideal substances always increases entropy, even with no energy change: $\Delta_{mix}S > 0$ for any composition.
For an ideal solution, $\Delta_{mix}H = 0$ and $\Delta_{mix}V = 0$ : molecules do not "notice" whether their neighbor is like or unlike.
The entropy of mixing comes purely from the increased number of ways to arrange particles among positions: $\Delta_{mix}S = -nR\sum_i x_i \ln x_i$ .
Mixing is spontaneous solely because $\Delta_{mix}G = \Delta_{mix}H - T\Delta_{mix}S$ is dominated by the $-T\Delta_{mix}S$ term.
Real solutions deviate from this ideal picture through nonzero enthalpy and volume changes, captured later by excess functions.
Key idea: ideal mixing is driven entirely by the combinatorial increase in entropy, not by any energetic preference between molecules.

The core insight

When mixing is ideal, intermolecular interactions between A and B are indistinguishable in strength from A–A and B–B interactions, so no energy is released or absorbed and no volume changes; the entire thermodynamic drive to mix comes from the vastly larger number of spatial arrangements available to the mixed state, which is a purely entropic, combinatorial effect quantified by $\Delta_{mix}S = -nR\sum_i x_i \ln x_i$ .

The physical picture

Imagine two liquids, A and B, each made of molecules that are chemically different but physically nearly identical in size and in the strength of their intermolecular forces. Before mixing, A molecules occupy one container and B molecules occupy another. After removing the partition, molecules move freely and can occupy either region. Nothing about the energetics has changed: an A molecule sitting next to another A feels the same attraction as it would next to a B, because the interactions are assumed equal by definition of "ideal." What has changed is the number of distinguishable spatial arrangements available to the system. This is the same logic used to derive the entropy of mixing two ideal gases, and it is imported unchanged into the liquid solution picture, with the added assumption that molecules of A and B are similar enough in size to occupy an interchangeable lattice of sites. Statistically, spreading $n_A$ A-particles and $n_B$ B-particles over $n_A + n_B$ sites vastly increases the number of accessible microstates compared to keeping them segregated. Since entropy measures $k \ln W$ for the number of microstates $W$ , more accessible arrangements means higher entropy. Because ideal mixing involves no change in intermolecular energy, $\Delta_{mix}H = 0$ . Because the molecules are assumed to be the same size and pack no more or less efficiently together than apart, $\Delta_{mix}V = 0$ . The entire Gibbs energy of mixing, which you have already seen is negative and drives spontaneity, is here built entirely from the entropy term:

\Delta_{mix}G = \Delta_{mix}H - T\Delta_{mix}S = 0 - T\Delta_{mix}S

Figure

Lattice model: segregated versus mixed arrangements of A (blue) and B (red) particles

The huge jump in $W$ for even this tiny 24-particle toy system, from 1 arrangement to over 2.7 million, illustrates why $\Delta_{mix}S$ is always positive: mixing multiplies the number of accessible microstates.

The core relations

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

where: $n$ is total moles, $x_A, x_B$ are mole fractions, and this is derived from chemical potentials in the Gibbs energy of mixing.

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

where: sum runs over all components, $x_i$ mole fraction of component $i$ , always positive since $0 < x_i < 1$ .

\Delta_{mix}H = 0 \qquad \Delta_{mix}V = 0

where: valid for an ideal solution, where A–A, B–B, and A–B interactions are energetically and sterically equivalent.

\Delta_{mix}S_{max} = nR\ln 2 \quad (\text{at } x_A = x_B = 1/2)

where: the entropy of mixing is maximized at equal mole fractions for a two-component mixture.

Derivation AFrom the Gibbs energy of mixing to entropy, enthalpy, and volume of mixing

Goal. Starting from the already-established ideal Gibbs energy of mixing, extract $\Delta_{mix}S$ , $\Delta_{mix}H$ , and $\Delta_{mix}V$ using standard thermodynamic relations.

Step 1. State the starting point. For an ideal solution of $n_A$ moles of A and $n_B$ moles of B, the Gibbs energy of mixing at constant $T$ and $p$ is

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

This result follows from writing the chemical potential of each ideal component as $\mu_i = \mu_i^* + RT\ln x_i$ and summing $G_{mixture} - G_{pure}$ over both components, a step assumed known from the Gibbs energy of mixing.

Step 2. Differentiate with respect to temperature to get entropy. Thermodynamics gives the general relation

\left(\frac{\partial G}{\partial T}\right)_p = -S

Apply this to $\Delta_{mix}G$ . Because $x_A$ and $x_B$ are mole fractions, fixed once the amounts $n_A$ and $n_B$ are fixed, they do not depend on $T$ . The only temperature dependence in $\Delta_{mix}G$ is the explicit prefactor $T$ . Differentiating term by term,

\Delta_{mix}S = -\left(\frac{\partial \Delta_{mix}G}{\partial T}\right)_p = -nR\left(x_A \ln x_A + x_B \ln x_B\right)

which is positive because each $\ln x_i$ is negative (since $0<x_i<1$ ) and the whole expression carries a leading minus sign.

Step 3. Differentiate with respect to pressure to get volume. The general relation is

\left(\frac{\partial G}{\partial p}\right)_T = V

Since $\Delta_{mix}G = nRT(x_A\ln x_A + x_B \ln x_B)$ contains no explicit pressure dependence at all — mole fractions and the gas constant term are pressure-independent for an ideal, incompressible mixing model — the derivative is identically zero:

\Delta_{mix}V = \left(\frac{\partial \Delta_{mix}G}{\partial p}\right)_T = 0

Step 4. Recover the enthalpy of mixing from the Gibbs–Helmholtz-type identity. Since $\Delta_{mix}G = \Delta_{mix}H - T\Delta_{mix}S$ , rearrange for the enthalpy:

\Delta_{mix}H = \Delta_{mix}G + T\Delta_{mix}S

Substitute the two results already found:

\Delta_{mix}H = nRT(x_A\ln x_A + x_B\ln x_B) + T\left[-nR(x_A\ln x_A + x_B\ln x_B)\right]

The two terms are equal in magnitude and opposite in sign, so they cancel exactly:

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

This confirms self-consistently that an ideal solution mixes with no heat effect and no volume change — the entire $-T\Delta_{mix}S$ term is responsible for the negative $\Delta_{mix}G$ that makes mixing spontaneous.

Worked examples

Example 1 — Foundational: entropy of mixing for an equimolar binary mixture. Calculate $\Delta_{mix}S$ , $\Delta_{mix}H$ , and $\Delta_{mix}G$ when 1.00 mol of liquid benzene is mixed with 1.00 mol of liquid toluene at 298 K, treating the mixture as ideal. Total moles: $n = 2.00$ mol. Mole fractions: $x_A = x_B = 0.500$ .

\Delta_{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})(0.500\ln 0.500 + 0.500\ln 0.500)

Since $\ln 0.500 = -0.6931$ ,

\Delta_{mix}S = -(2.00)(8.314)\left[(0.500)(-0.6931) + (0.500)(-0.6931)\right] = -(2.00)(8.314)(-0.6931)

\Delta_{mix}S = 11.53\ \text{J K}^{-1}

Because benzene and toluene are structurally very similar (both aromatic, similar size and polarizability), the ideal assumption is excellent here, so:

\Delta_{mix}H \approx 0 \ \text{J}, \qquad \Delta_{mix}V \approx 0 \ \text{cm}^3

\Delta_{mix}G = \Delta_{mix}H - T\Delta_{mix}S = 0 - (298\ \text{K})(11.53\ \text{J K}^{-1}) = -3.44\times 10^3\ \text{J} = -3.44\ \text{kJ}

Sanity check: $\Delta_{mix}G$ is negative, as it must be for a spontaneous process, and its magnitude (a few kJ) is typical for mixing entropy effects at room temperature — much smaller than a covalent bond energy, consistent with this being a purely statistical, not chemical, effect.

Example 2 — Applied: unequal mole fractions and maximum entropy composition. A process chemist mixes 3.00 mol of hexane (A) with 1.00 mol of heptane (B), treated as an ideal solution, at 350 K. Find $\Delta_{mix}S$ , $\Delta_{mix}G$ , and compare this to the entropy of mixing an equimolar mixture of the same total 4.00 mol. Also state the mole fraction at which $\Delta_{mix}S$ per mole is maximized. Mole fractions: $x_A = 3.00/4.00 = 0.750$ , $x_B = 1.00/4.00 = 0.250$ . Total $n = 4.00$ mol.

\ln(0.750) = -0.2877, \qquad \ln(0.250) = -1.3863

\Delta_{mix}S = -nR(x_A \ln x_A + x_B \ln x_B) = -(4.00)(8.314)\left[(0.750)(-0.2877)+(0.250)(-1.3863)\right]

= -(4.00)(8.314)\left[-0.2158 - 0.3466\right] = -(4.00)(8.314)(-0.5623)

\Delta_{mix}S = 18.70\ \text{J K}^{-1}

\Delta_{mix}G = 0 - (350\ \text{K})(18.70\ \text{J K}^{-1}) = -6.55\times10^3\ \text{J} = -6.55\ \text{kJ}

Now compare with an equimolar 2.00 mol + 2.00 mol split of the same 4.00 mol total, where $x_A = x_B = 0.500$ :

\Delta_{mix}S_{equimolar} = -(4.00)(8.314)(2)(0.500)(-0.6931) = 23.05\ \text{J K}^{-1}

The equimolar case gives a larger entropy of mixing (23.05 J/K versus 18.70 J/K for the same total moles), confirming that $\Delta_{mix}S$ per mole is maximized at $x_A = x_B = 0.5$ for a two-component mixture — the point of maximum "surprise" or configurational freedom. This is the origin of the $nR\ln 2 = (4.00)(8.314)(0.6931) = 23.05$ J/K value obtained above, matching exactly.

Limiting cases & common traps

Situation	Behavior	Why
$x_i \to 0$ or $x_i \to 1$	$\Delta_{mix}S \to 0$	Nothing to mix; only one component present
$x_A = x_B = 0.5$ (two components)	$\Delta_{mix}S$ maximal, $= nR\ln 2$	Maximum configurational uncertainty
Ideal solution assumed	$\Delta_{mix}H = 0$ , $\Delta_{mix}V = 0$	A–A, B–B, A–B interactions equal in strength
Real solution (e.g. ethanol + water)	$\Delta_{mix}H \neq 0$ , $\Delta_{mix}V \neq 0$	Hydrogen bonding, size mismatch — needs excess functions
Mixing two ideal gases	Same $\Delta_{mix}S$ formula applies	Statistical argument is identical regardless of phase

The chemist behind it

The statistical picture of entropy underlying $\Delta_{mix}S = -nR\sum_i x_i \ln x_i$ traces back to Ludwig Boltzmann's identification of entropy with the number of microscopic arrangements, $S = k\ln W$ . Its application to solutions and mixtures was formalized and popularized through the statistical thermodynamics developed in the early twentieth century, connecting the macroscopic Gibbs energy of mixing derived from chemical potentials to a microscopic counting argument over lattice sites — the same combinatorial logic used later in the Flory–Huggins treatment of polymer solutions, where size mismatch between components produces deviations from this simple ideal formula.

Origins · 1 minLudwig Boltzmann and how this was discovered

Ludwig Boltzmann

1844–1906 · Established that entropy counts microscopic arrangements, the statistical foundation of

\Delta_{mix}S

Ask the tutor: How do excess functions modify this ideal picture when hexane is mixed with ethanol instead of another hexane-like liquid?

Why does mixing increase entropy if no bonds are broken or formed?

Entropy is not only about chemical bonds; it also measures the number of ways particles can be spatially arranged. Before mixing, A and B molecules are confined to separate regions, so there is essentially one type of arrangement. After mixing, each molecule can occupy any position in the shared volume, multiplying the number of accessible microstates enormously. Since $S = k\ln W$ , more accessible arrangements directly means higher entropy, even though the total internal energy is unchanged because no new intermolecular interactions of different strength have formed.

Is Δmix H really always exactly zero in real life?

No — exactly zero only holds for the idealized model where A–A, B–B, and A–B interactions are equal. Real mixtures like ethanol and water, or hexane and octanol, have nonzero enthalpies of mixing because hydrogen bonding or dispersion forces differ between like and unlike molecules. These deviations are captured by the excess enthalpy $H^E$ , a topic that follows directly from this ideal baseline in the study of real solutions and excess functions.

Does the ideal gas entropy of mixing formula apply to liquids too?

Yes, remarkably the same formula, $\Delta_{mix}S = -nR\sum_i x_i \ln x_i$ , applies to ideal liquid solutions as well as ideal gas mixtures. The underlying combinatorial argument — counting arrangements of particles among available positions — does not depend on whether those positions are gas-phase volume elements or liquid lattice sites, as long as the components are similar enough in size and interaction strength to be treated as ideal.

Why is Δmix V zero for an ideal solution?

Zero volume of mixing means the total volume of the mixture equals the sum of the volumes of the pure components before mixing. This holds when molecules of A and B are similar enough in size and shape that they pack together exactly as efficiently as they pack with their own kind. Real solutions often show small positive or negative $\Delta_{mix}V$ , reflecting size mismatch or specific interactions like hydrogen bonding, which is why volume changes are used experimentally to probe non-ideality.

How does this connect to Raoult's law?

Raoult's law describes the vapor pressure lowering of an ideal solution and follows from the same chemical potential expression, $\mu_i = \mu_i^* + RT\ln x_i$ , that generates $\Delta_{mix}G$ here. Because the entropy of mixing derived in this article is built from that identical expression, ideal solutions obeying Raoult's law automatically show zero enthalpy and volume of mixing, and this consistency is exactly what defines an ideal solution in the first place.