Raoult's law and ideal solutions

Result

Raoult's law and ideal solutions

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Statement

A liquid mixture is an ideal solution if every component satisfies \(\mu_i(\text{l})=\mu_i^{*}(\text{l})+RT\ln x_i\) at every composition \(0\lt x_i\lt1\), where \(\mu_i^{*}(\text{l})\) is the chemical potential of pure liquid \(i\) at the same temperature and pressure and \(x_i\) is its mole fraction. For such a mixture in equilibrium with a vapour that behaves as an ideal gas mixture, the partial vapour pressure of each component is proportional to its mole fraction in the liquid, \(p_i=x_i\,p_i^{*}\), with \(p_i^{*}\) the saturated vapour pressure of the pure liquid at that temperature; by Dalton’s law the total pressure above the solution is \(p=\sum_i x_i\,p_i^{*}\), a linear function of composition. Equivalently \(\Delta H_{\text{mix}}=0\), \(\Delta V_{\text{mix}}=0\) and \(\Delta G_{\text{mix}}=nRT\sum_i x_i\ln x_i\): an ideal solution mixes with no enthalpy and no volume change, purely for the entropy already computed in entropy-of-mixing.

Why it matters

Raoult’s law is the bridge between the chemical potential formalism of this degree and the two most-used pieces of practical solution chemistry: distillation and the colligative properties. Once \(p_i=x_ip_i^{*}\) is available, the vapour above a mixture is generally richer in the more volatile component than the liquid it came from, and repeating that enrichment is exactly what a fractionating column does; in the opposite limit, a non-volatile solute lowers the solvent’s vapour pressure by a fraction equal to the solute mole fraction, which is the origin of boiling-point elevation, freezing-point depression and osmosis.

It also fixes the reference state for everything that follows. Real mixtures are described by activities \(a_i=\gamma_ix_i\), and the whole activity-coefficient apparatus is only meaningful because there is a well-defined ideal baseline to measure deviations from. Raoult’s law is that baseline, in the same way that ideal-gas-law is the baseline from which the van der Waals equation measures gas non-ideality.

Hypotheses

The liquid mixture is ideal: \(\mu_i(\text{l})=\mu_i^{*}(\text{l})+RT\ln x_i\) at all compositions.Microscopically this requires the \(A\)–\(A\), \(B\)–\(B\) and \(A\)–\(B\) interactions to be equal in strength and the molecules to be comparable in size and shape, so that swapping a neighbour costs nothing. Drop it and both the enthalpy and the entropy of mixing change: ethanol–water shows large positive deviations and an azeotrope (Fails without).

The vapour is an ideal gas mixture, so \(\mu_i(\text{g})=\mu_i^{\circ}+RT\ln(p_i/p^{\circ})\) and \(p=\sum_ip_i\).Only at pressures low enough that the vapour’s own intermolecular forces are negligible. At elevated pressure the correct statement is in terms of fugacities, \(f_i=x_if_i^{*}\), and the pressure form acquires a fugacity-coefficient correction.

Every component is volatile enough to have a defined \(p_i^{*}\) at the working temperature, and the temperature is below the critical temperature of each.For an involatile solute (a salt, a sugar, a polymer) \(p_i^{*}\) does not exist and the law is applied to the solvent only; the solute is then described by Henry’s law or not at all.

The Poynting correction is neglected: the pure liquid reference is taken at the total pressure \(p\) rather than at its own \(p_i^{*}\).Rigorously \(a_i=(p_i/p_i^{*})\exp[-V_{\text{m},i}(p-p_i^{*})/RT]\). For water at \(298\,\text{K}\) under \(1\,\text{bar}\), \(V_{\text{m}}(p-p^{*})/RT\approx1.81\times10^{-5}\times10^{5}/2478\approx7\times10^{-4}\) — a \(0.07\%\) effect, negligible at ambient pressure and not negligible at \(100\,\text{bar}\).

The solute does not associate or dissociate on dissolving.Mole fractions must be counted over the species actually present. A strong electrolyte such as \(\text{NaCl}\) contributes close to two particles per formula unit, roughly doubling the vapour-pressure lowering (Fails without).

Proof

The derivation is in two halves. Steps 1–4 are exact for an ideal vapour and produce a definition of the liquid-phase activity in terms of measurable pressures; step 5 imposes ideality of the liquid, and steps 6–8 read off the consequences. Steps 9–10 supply the molecular content of the ideality hypothesis and the size of the correction that was dropped.

\[ \mu_i(\text{l},T,p,\{x\}) \;=\; \mu_i(\text{g},T,p_i) \]

Phase equilibrium. At constant \(T\) and \(p\), \(\mathrm{d}G=\sum_i\mu_i\,\mathrm{d}n_i\); transferring \(\mathrm{d}n\) of \(i\) from liquid to vapour changes \(G\) by \((\mu_i(\text{g})-\mu_i(\text{l}))\,\mathrm{d}n\), which must vanish at equilibrium for every component independently (gibbs-free-energy). A

\[ \mu_i(\text{g},T,p_i) \;=\; \mu_i^{\circ}(\text{g},T) + RT\ln\frac{p_i}{p^{\circ}} \]

The ideal-gas chemical potential, obtained by integrating \((\partial\mu/\partial p)_T=V_{\text{m}}=RT/p\) from the standard pressure \(p^{\circ}\) to the partial pressure \(p_i\) (ideal-gas-law, gibbs-equilibrium-constant). In an ideal gas mixture each component behaves as though it alone occupied the volume, so only \(p_i\) appears. A

\[ \mu_i^{*}(\text{l},T) \;=\; \mu_i^{\circ}(\text{g},T) + RT\ln\frac{p_i^{*}}{p^{\circ}} \]

The same two statements applied to the pure liquid \(i\) at the same temperature, in equilibrium with its own saturated vapour at pressure \(p_i^{*}\). This is the step that turns the unknown \(\mu_i^{\circ}(\text{g})\) into something measurable. A

\[ \mu_i(\text{l}) \;=\; \mu_i^{*}(\text{l}) + RT\ln\frac{p_i}{p_i^{*}} \;\equiv\; \mu_i^{*}(\text{l}) + RT\ln a_i \]

Subtracting step 3 from steps 1–2 cancels \(\mu_i^{\circ}(\text{g})\) and \(p^{\circ}\) identically. Nothing about the liquid has been assumed: this defines the activity \(a_i=p_i/p_i^{*}\) of a component of any liquid mixture whose vapour is ideal, and it is how activities are measured. B

\[ \text{Ideality (Hypotheses)}:\qquad \mu_i(\text{l}) \;=\; \mu_i^{*}(\text{l}) + RT\ln x_i \]

The defining property of an ideal solution: the only effect of the other components is the entropic dilution term, exactly the form the chemical potential takes in an ideal gas mixture with \(x_i\) in place of \(p_i/p\). Its molecular justification is step 9. B

\[ RT\ln\frac{p_i}{p_i^{*}} = RT\ln x_i \quad\Longrightarrow\quad \frac{p_i}{p_i^{*}} = x_i \quad\Longrightarrow\quad \boxed{\,p_i = x_i\,p_i^{*}\,} \]

Equating steps 4 and 5, dividing by \(RT\neq0\) and exponentiating; \(\ln\) is injective on \((0,\infty)\), so the equality of logarithms gives equality of arguments. Ideality is therefore equivalent to \(a_i=x_i\), i.e. to \(\gamma_i=1\). A

\[ p=\sum_i p_i=\sum_i x_i p_i^{*};\qquad \text{binary: } p = p_B^{*} + \left(p_A^{*}-p_B^{*}\right)x_A \]

Dalton’s law for the ideal vapour, then \(x_B=1-x_A\) for a binary mixture. The total pressure is linear in the liquid mole fraction, interpolating between the two pure vapour pressures — the straight line that experimental \(p\)–\(x\) diagrams are compared against. A

\[ y_A=\frac{p_A}{p}=\frac{x_Ap_A^{*}}{x_Ap_A^{*}+(1-x_A)p_B^{*}} \quad\Longleftrightarrow\quad \frac{y_A}{y_B}=\alpha\,\frac{x_A}{x_B},\qquad \alpha\equiv\frac{p_A^{*}}{p_B^{*}} \]

Dividing the partial pressure by the total. Written as a ratio the composition dependence collapses into the single constant \(\alpha\), the relative volatility: if \(\alpha\gt1\) the vapour is enriched in \(A\) at every composition, which is why repeated vaporisation and condensation separates an ideal pair completely. B

\[ \Delta U_{\text{mix}} = n\,x_Ax_B\,W,\qquad W \equiv zN_{\text{A}}\left[w_{AB}-\tfrac{1}{2}\left(w_{AA}+w_{BB}\right)\right] \]

The lattice (Bragg–Williams) count: with \(z\) nearest neighbours per site, random mixing and pairwise interaction energies \(w_{AA},w_{BB},w_{AB}\), the number of \(A\)–\(B\) contacts per mole is \(zN_{\text{A}}x_Ax_B\) and every such contact replaces half an \(A\)–\(A\) and half a \(B\)–\(B\) contact. The interchange energy \(W\) vanishes exactly when \(w_{AB}\) equals the arithmetic mean of \(w_{AA}\) and \(w_{BB}\) — in particular whenever the three interactions are equal in strength, which is the microscopic content of step 5; \(W=0\) then forces \(\Delta H_{\text{mix}}=0\) and leaves the random-mixing entropy of entropy-of-mixing untouched, reproducing \(\mu_i=\mu_i^{*}+RT\ln x_i\). C

\[ \mu_i^{*}(\text{l},p) = \mu_i^{*}(\text{l},p_i^{*}) + \int_{p_i^{*}}^{p}V_{\text{m},i}\,\mathrm{d}p' \approx \mu_i^{*}(\text{l},p_i^{*}) + V_{\text{m},i}\left(p-p_i^{*}\right) \]

Step 3 evaluated the pure-liquid reference at \(p_i^{*}\) but step 5 uses it at the total pressure \(p\); the difference is the Poynting term, with \(V_{\text{m},i}\) essentially constant because liquids are nearly incompressible. Carrying it through gives \(p_i=x_ip_i^{*}\exp[V_{\text{m},i}(p-p_i^{*})/RT]\), an exponential of order \(10^{-4}\) at ambient pressure (Hypotheses) and the reason the law is quoted without it. C

Result

\[ p_i = x_i\,p_i^{*},\qquad p=\sum_i x_i\,p_i^{*},\qquad y_i=\frac{x_ip_i^{*}}{\sum_j x_jp_j^{*}} \]

\[ \Delta H_{\text{mix}}=0,\qquad \Delta V_{\text{mix}}=0,\qquad \Delta G_{\text{mix}}=nRT\sum_i x_i\ln x_i\lt0 \]

Reading. In an ideal solution a molecule of \(i\) cannot tell what its neighbours are; the only thing dilution does is reduce the fraction of surface sites from which \(i\) can escape, so the escaping tendency — and hence the partial vapour pressure — falls in exact proportion to \(x_i\). The vapour is then richer in whichever component is more volatile, by the fixed factor \(\alpha=p_A^{*}/p_B^{*}\).

Scope. Exact only for an ideal liquid mixture in equilibrium with an ideal vapour, with the Poynting correction dropped. For real mixtures it is a limiting law: \(p_i\to x_ip_i^{*}\) as \(x_i\to1\), i.e. every solvent obeys Raoult’s law in the dilute-solute limit, whatever the solute (Corollaries).

Corollaries & converses

Relative lowering of vapour pressure. For a single involatile solute \(B\) in solvent \(A\), \(p=x_Ap_A^{*}\) so \(\dfrac{p_A^{*}-p}{p_A^{*}}=1-x_A=x_B\). The fractional lowering equals the solute mole fraction and depends on how many solute particles there are, not what they are — the definition of a colligative property, and the root of boiling-point elevation, freezing-point depression and osmotic pressure.
Relative volatility is composition-independent. \(y_A/y_B=\alpha\,x_A/x_B\) with \(\alpha=p_A^{*}/p_B^{*}\) constant at fixed \(T\); an ideal pair therefore has no azeotrope unless \(\alpha=1\), and can in principle be separated to any purity by fractional distillation.
Raoult’s law is the solvent limit; Henry’s law is the solute limit. For a real binary, \(p_A\to x_Ap_A^{*}\) as \(x_A\to1\) while \(p_B\to K_Bx_B\) as \(x_B\to0\), with the Henry constant \(K_B\ne p_B^{*}\) in general. The two coincide, \(K_B=p_B^{*}\), exactly when the solution is ideal over the whole range.
Duhem–Margules. The Gibbs–Duhem relation \(x_A\,\mathrm{d}\ln p_A+x_B\,\mathrm{d}\ln p_B=0\) forces the two behaviours to occur together: if \(A\) obeys Raoult’s law over a composition interval then \(B\) obeys Henry’s law over the same interval (Problems).
Ideality is symmetric. If one component of a binary mixture obeys \(p_A=x_Ap_A^{*}\) at every composition, the same Gibbs–Duhem argument forces \(p_B=x_Bp_B^{*}\) at every composition: there is no such thing as a half-ideal binary solution.
Regular solutions. Keeping \(W\ne0\) in step 9 but retaining random mixing gives \(RT\ln\gamma_A=Wx_B^{2}\), so \(p_A=x_Ap_A^{*}\exp(Wx_B^{2}/RT)\): positive \(W\) (unlike neighbours repel) gives positive deviations, negative \(W\) gives negative deviations, and both reduce to Raoult’s law as \(x_B\to0\).

Fails without

Equal interaction energies dropped — positive deviations and minimum-boiling azeotropes: when unlike contacts are less favourable than the average of the like ones (\(W\gt0\)) mixing is endothermic, both partial pressures exceed \(x_ip_i^{*}\), the total pressure curve bulges above the Raoult line and it acquires a maximum — ethanol–cyclohexane is the clean case. Ethanol–water reaches the same pressure maximum by a subtler route worth stating carefully: its excess Gibbs energy is positive (about \(+1\,\text{kJ mol}^{-1}\) at equimolar, hence the positive deviations), but its enthalpy of mixing is exothermic and its volume change negative in the water-rich range, so the deviation there is carried by a large negative excess entropy — water orders around the ethyl group — and not by \(W\gt0\) at all. Positive deviation means \(G^{\text{E}}\gt0\); it does not by itself mean \(\Delta H_{\text{mix}}\gt0\). At the pressure maximum, however it arises, \(y_i=x_i\) and distillation stops enriching: ethanol–water boils unchanged at \(78.2\,{}^{\circ}\text{C}\) and about \(95.6\%\) ethanol by mass at \(1\,\text{atm}\), below the boiling point of either pure liquid. No amount of fractionation gets past it, which is why absolute ethanol is not made by simple distillation.
Equal interaction energies dropped the other way — negative deviations and maximum-boiling azeotropes: acetone and chloroform form a hydrogen bond with each other that neither forms with itself (\(W\lt0\)), so both partial pressures fall below the Raoult line and the boiling curve has a maximum near \(64\,{}^{\circ}\text{C}\) — above pure acetone (\(56.1\,{}^{\circ}\text{C}\)) and pure chloroform (\(61.2\,{}^{\circ}\text{C}\)). The measured \(\Delta H_{\text{mix}}\) is exothermic and \(\Delta V_{\text{mix}}\) is negative, both flatly contradicting the ideal Result.
Comparable molecular size dropped — polymer solutions: a polymer of \(N=10^{3}\) repeat units has, in dilute solution, a mole fraction about \(N\) times smaller than its volume fraction, so mole-fraction Raoult’s law predicts almost no solvent-pressure lowering where the measured lowering is large. The combinatorial entropy must be recounted per segment, giving \(\ln a_1=\ln\phi_1+(1-1/N)\phi_2+\chi\phi_2^{2}\) (flory-huggins) with volume fractions \(\phi\) in place of mole fractions.
Non-dissociation dropped — electrolytes: \(0.100\,\text{mol}\) of \(\text{NaCl}\) in water lowers the vapour pressure by very nearly twice what \(0.100\,\text{mol}\) of sucrose does, because the solute species are \(\text{Na}^{+}\) and \(\text{Cl}^{-}\) and it is the total particle mole fraction that enters. The observed factor is a few per cent below \(2\) at ordinary concentrations because long-range ion–ion attraction lowers the activity of each ion (debye-huckel), so even the corrected count is not the whole story.
Ideal vapour dropped — high pressure: above a few tens of bar the vapour’s own non-ideality makes \(p_i\) a poor proxy for escaping tendency; step 2 must be replaced by \(\mu_i(\text{g})=\mu_i^{\circ}+RT\ln(f_i/p^{\circ})\) and the law survives only in the fugacity form \(f_i=x_if_i^{*}\).

Common errors

“Use the mass fraction, or the molality, in \(p_i=x_ip_i^{*}\).” The law is written in mole fractions of the liquid phase; masses must be divided by molar masses first. The error is largest exactly where the molar masses differ most, which is the interesting case.
“\(x_i\) is the mole fraction in the vapour.” \(x_i\) is the liquid composition; the vapour composition is the derived quantity \(y_i=x_ip_i^{*}/p\), and \(y_i=x_i\) only when \(p_A^{*}=p_B^{*}\) or at an azeotrope.
“\(p=x_{\text{solvent}}\,p_{\text{solvent}}^{*}\) always.” That form assumes the solute contributes nothing to the vapour. If the second component is volatile its own \(x_Bp_B^{*}\) must be added, and the total pressure can then be higher, not lower, than \(p_A^{*}\).
“Salt dissolved in water behaves like sugar dissolved in water.” Mole fractions count particles in solution: \(\text{MgCl}_2\) contributes close to three, \(\text{NaCl}\) close to two, sucrose exactly one.
“Henry’s constant is the solute’s vapour pressure.” \(K_B=p_B^{*}\) holds only for an ideal solution; for a real one \(K_B\) is the slope of the tangent to \(p_B\) at \(x_B=0\) and can differ from \(p_B^{*}\) by orders of magnitude.
“\(\Delta G_{\text{mix}}\lt0\) means every pair of liquids is miscible.” That conclusion is a consequence of ideality, not of thermodynamics; a large positive \(W\) makes \(\Delta G_{\text{mix}}\) non-convex and the mixture separates into two phases.

Discussion

Raoult reached the law empirically in 1887, by weighing how much the vapour pressure and the freezing point of a solvent fell when known masses of many different solutes were dissolved in it. The striking regularity he found — that equal numbers of dissolved molecules produce equal lowering, regardless of chemical identity — was one of the strongest pieces of evidence available at the time that solutions contain discrete molecules at all, and it made vapour-pressure and freezing-point measurements the standard route to molar masses for the next half-century (Problems). The thermodynamic derivation given above came later, from Gibbs’s chemical potential, and it inverts the logic: instead of Raoult’s law being an experimental regularity, ideality of the liquid is the postulate and the law is a theorem.

Which real mixtures are close to ideal is exactly the question step 9 answers. Benzene–toluene, hexane–heptane, bromobenzene–chlorobenzene and mixtures of isotopologues are all near-ideal because the two species are similar in size, shape and polarity, so \(W\approx0\). Anything involving hydrogen bonding, strong dipoles or a large size mismatch is not. It is worth being clear that ideality is a statement about a pair, not about a substance: water is not “non-ideal”, but water–ethanol is.

The practical payoff is distillation. Because \(\alpha=p_A^{*}/p_B^{*}\) is composition-independent for an ideal pair, each vaporisation–condensation cycle multiplies the ratio \(x_A/x_B\) by the same factor \(\alpha\), and \(n\) theoretical plates multiply it by \(\alpha^{n}\); separating a pair with \(\alpha\) close to \(1\) needs many plates, which is why the number of plates is the natural measure of a column. An azeotrope is precisely a composition where \(\alpha_{\text{effective}}=1\), and the reason it exists is that \(\gamma_A\) and \(\gamma_B\) vary with composition — the one thing ideality forbids.

The modern reading of Raoult’s law is that it fixes a convention rather than a fact. Writing \(a_i=\gamma_ix_i\) with \(\gamma_i\to1\) as \(x_i\to1\) is the symmetric or Raoult convention, natural for solvents and for mixtures of like liquids; writing \(a_B=\gamma_B^{H}x_B\) with \(\gamma_B^{H}\to1\) as \(x_B\to0\) is the asymmetric or Henry convention, natural for a solute that has no pure-liquid state at the working conditions, such as dissolved \(\text{O}_2\). The same physical \(\mu_B\) is being described in both, so the two activity coefficients differ only by the constant ratio \(K_B/p_B^{*}\); tabulated equilibrium constants are meaningless until it is stated which convention their activities use.

Common misconceptions. That Raoult’s law is a law of nature that real solutions violate. It is the definition of a reference behaviour, and its real status is as a limiting law: every solvent obeys it as its mole fraction approaches \(1\), which is guaranteed by the Gibbs–Duhem relation, not by any assumption about the solute. That is why colligative formulae derived from it work well for dilute solutions of substances whose mixtures are wildly non-ideal at intermediate composition.

Worked examples

Example 1. Benzene and toluene form a near-ideal pair. At \(298\,\text{K}\) the standard tabulated saturated vapour pressures are \(p_{\text{ben}}^{*}=12.70\,\text{kPa}\) and \(p_{\text{tol}}^{*}=3.79\,\text{kPa}\). A solution is made from \(39.06\,\text{g}\) of benzene (\(M=78.11\,\text{g mol}^{-1}\)) and \(138.2\,\text{g}\) of toluene (\(M=92.14\,\text{g mol}^{-1}\)). Find the total vapour pressure and the composition of the vapour.

\[ n_i=\frac{m_i}{M_i}:\qquad n_{\text{ben}}=\frac{39.06}{78.11}=0.5001\,\text{mol},\qquad n_{\text{tol}}=\frac{138.2}{92.14}=1.500\,\text{mol} \]

Raoult’s law is in mole fractions, so the masses must be converted first (Common errors). A

\[ x_{\text{ben}}=\frac{n_{\text{ben}}}{n_{\text{ben}}+n_{\text{tol}}}=\frac{0.500}{2.000}=0.250,\qquad x_{\text{tol}}=1-x_{\text{ben}}=0.750 \]

Mole fractions of the liquid phase; they sum to \(1\) by construction. A

\[ p_{\text{ben}}=x_{\text{ben}}p_{\text{ben}}^{*}=0.250\times12.70=3.175\,\text{kPa},\qquad p_{\text{tol}}=x_{\text{tol}}p_{\text{tol}}^{*}=0.750\times3.79=2.843\,\text{kPa} \]

The Result applied to each component separately; each partial pressure is below the corresponding pure value, as it must be for \(x_i\lt1\). A

\[ p=p_{\text{ben}}+p_{\text{tol}}=3.175+2.843=6.018\approx6.02\,\text{kPa} \]

Dalton’s law (step 7 of the Proof). As a check, the linear form gives \(p=3.79+(12.70-3.79)(0.250)=3.79+2.228=6.02\,\text{kPa}\). A

\[ y_{\text{ben}}=\frac{p_{\text{ben}}}{p}=\frac{3.175}{6.018}=0.528,\qquad y_{\text{tol}}=0.472 \]

The vapour is \(52.8\%\) benzene although the liquid is only \(25.0\%\) benzene: the more volatile component is enriched by \(\alpha=12.70/3.79=3.35\), and indeed \(y/(1-y)=1.12=3.35\times(0.250/0.750)\) as step 8 requires. B

\[ p=6.02\,\text{kPa},\qquad y_{\text{ben}}=0.528,\qquad y_{\text{tol}}=0.472 \]

Reading. One vaporisation step has more than doubled the benzene mole fraction; condensing this vapour and re-evaporating it would multiply \(x_{\text{ben}}/x_{\text{tol}}\) by \(3.35\) again, which is the whole principle of a fractionating column.

Scope. Requires benzene–toluene to be ideal at \(298\,\text{K}\), which it is to within about \(1\%\) — the two molecules are similar in size and both non-polar aromatics, so \(W\approx0\) in step 9.

Example 2. \(45.0\,\text{g}\) of sucrose (\(\text{C}_{12}\text{H}_{22}\text{O}_{11}\), \(M=342.30\,\text{g mol}^{-1}\), involatile and non-dissociating) is dissolved in \(200.0\,\text{g}\) of water (\(M=18.015\,\text{g mol}^{-1}\)) at \(298\,\text{K}\), where pure water has \(p^{*}=3.169\,\text{kPa}\). Find the vapour pressure of the solution and the vapour-pressure lowering.

\[ p = x_{\text{w}}p^{*} \quad\Longrightarrow\quad \Delta p \equiv p^{*}-p = \left(1-x_{\text{w}}\right)p^{*} = x_{\text{suc}}\,p^{*} \]

Rearranged symbolically first: with only one volatile component, the total pressure is the solvent partial pressure, and the lowering is the solute mole fraction times the pure-solvent pressure (Corollaries). A

\[ n_{\text{suc}}=\frac{45.0}{342.30}=0.13146\,\text{mol},\qquad n_{\text{w}}=\frac{200.0}{18.015}=11.102\,\text{mol} \]

Amounts of each species. Sucrose does not dissociate, so \(0.13146\,\text{mol}\) of solute means \(0.13146\,\text{mol}\) of solute particles (Hypotheses). A

\[ x_{\text{suc}}=\frac{0.13146}{0.13146+11.102}=\frac{0.13146}{11.233}=0.011703,\qquad x_{\text{w}}=0.988297 \]

Note how small the solute mole fraction is even at this fairly concentrated \(22.5\,\text{g}\) per \(100\,\text{g}\) of water: water’s low molar mass means it dominates the particle count. A

\[ p = 0.988297\times3.169 = 3.1319 \approx 3.132\,\text{kPa} \]

Substituting into the Result. The pure-water value \(3.169\,\text{kPa}\) at \(298\,\text{K}\) is the standard tabulated saturated vapour pressure. A

\[ \Delta p = x_{\text{suc}}p^{*} = 0.011703\times3.169 = 0.0371\,\text{kPa} = 37.1\,\text{Pa} \]

Consistent with step 1 and with \(3.169-3.132\); the fractional lowering \(\Delta p/p^{*}=1.17\%\) equals \(x_{\text{suc}}\) exactly, independent of what the solute is — the colligative statement. B

\[ p = 3.132\,\text{kPa},\qquad \Delta p = 37.1\,\text{Pa},\qquad \frac{\Delta p}{p^{*}} = x_{\text{suc}} = 0.0117 \]

Reading. A heavy sugar load barely dents the vapour pressure, because colligative effects count particles and one sucrose molecule weighs nineteen times a water molecule. The same \(45.0\,\text{g}\) as \(\text{NaCl}\) would be \(0.770\,\text{mol}\) of formula units and close to \(1.54\,\text{mol}\) of ions, giving \(x_{\text{ions}}=1.540/(1.540+11.102)=0.1218\) and \(\Delta p\approx386\,\text{Pa}\) — about ten times as much lowering. (Ten, not the twelve the particle counts alone suggest: the mole fraction is not linear in the solute amount once the solute stops being dilute.)

Scope. Assumes the solution is ideal in the solvent, which for a \(1.2\,\text{mol}\%\) non-electrolyte is a good approximation, and that sucrose is completely involatile, which it is.

Problems

Hexane and heptane form an ideal solution. At \(298\,\text{K}\), \(p_{\text{hex}}^{*}=20.2\,\text{kPa}\) and \(p_{\text{hep}}^{*}=6.09\,\text{kPa}\). For a liquid with \(x_{\text{hex}}=0.400\), find the total vapour pressure and the vapour composition.
Solution
Apply \(p_i=x_ip_i^{*}\) to each component: \(p_{\text{hex}}=0.400\times20.2=8.08\,\text{kPa}\) and \(p_{\text{hep}}=0.600\times6.09=3.654\,\text{kPa}\).
Dalton’s law gives \(p=8.08+3.654=11.73\,\text{kPa}\), which the linear form confirms: \(p=6.09+(20.2-6.09)(0.400)=6.09+5.644=11.73\,\text{kPa}\).
Vapour composition: \(y_{\text{hex}}=p_{\text{hex}}/p=8.08/11.73=0.689\), so \(y_{\text{hep}}=0.311\). The relative volatility check: \(\alpha=20.2/6.09=3.317\) and \(\alpha x_{\text{hex}}/x_{\text{hep}}=3.317\times(0.400/0.600)=2.211\), which is \(y_{\text{hex}}/y_{\text{hep}}=0.6886/0.3114=2.211\) as step 8 requires.
For the same hexane–heptane system at \(298\,\text{K}\), what liquid composition is in equilibrium with a vapour containing \(90.0\,\text{mol}\%\) hexane, and what is the total pressure then?
Solution
Invert step 8 symbolically before substituting. From \(\dfrac{y_A}{1-y_A}=\alpha\dfrac{x_A}{1-x_A}\), solving for \(x_A\) gives \(x_A=\dfrac{y_A}{\alpha-(\alpha-1)y_A}\).
With \(\alpha=20.2/6.09=3.317\) and \(y_A=0.900\): the denominator is \(3.317-2.317\times0.900=3.317-2.085=1.232\), so \(x_{\text{hex}}=0.900/1.232=0.731\).
Then \(p=6.09+(20.2-6.09)(0.731)=6.09+10.31=16.40\,\text{kPa}\). Check: \(p_{\text{hex}}=0.731\times20.2=14.77\,\text{kPa}\) and \(14.77/16.40=0.900\) as required. A liquid that is \(73\%\) hexane is needed to produce a \(90\%\) hexane vapour — the enrichment per plate is real but limited, and reaching \(99\%\) would take several more stages.
Dissolving \(5.00\,\text{g}\) of an involatile, non-dissociating solid in \(100.0\,\text{g}\) of benzene (\(M=78.11\,\text{g mol}^{-1}\)) lowers the vapour pressure at \(298\,\text{K}\) from \(12.70\,\text{kPa}\) to \(12.42\,\text{kPa}\). Determine the molar mass of the solute.
Solution
Symbols first. From the Corollaries, \(\dfrac{\Delta p}{p^{*}}=x_B=\dfrac{n_B}{n_A+n_B}\), so \(n_B=n_A\dfrac{x_B}{1-x_B}\) and \(M_B=\dfrac{m_B}{n_B}\).
Numbers: \(x_B=(12.70-12.42)/12.70=0.28/12.70=0.02205\). The solvent amount is \(n_A=100.0/78.11=1.2802\,\text{mol}\), so \(n_B=1.2802\times\dfrac{0.02205}{0.97795}=1.2802\times0.02254=0.02886\,\text{mol}\).
Hence \(M_B=5.00/0.02886=173\,\text{g mol}^{-1}\).
The frequently taught shortcut \(x_B\approx n_B/n_A\) would give \(n_B=0.02823\,\text{mol}\) and \(M_B=177\,\text{g mol}^{-1}\), a \(2\%\) error — acceptable here, but it is only valid for \(x_B\ll1\) and should not be used at the concentration of Worked example 2.
For the benzene–toluene solution of Worked example 1 (\(0.500\,\text{mol}\) benzene, \(1.500\,\text{mol}\) toluene) at \(298\,\text{K}\), compute \(\Delta G_{\text{mix}}\), \(\Delta S_{\text{mix}}\) and \(\Delta H_{\text{mix}}\), and state what drives the mixing. Take \(R=8.314\,\text{J K}^{-1}\text{mol}^{-1}\).
Solution
From the Result, \(\Delta G_{\text{mix}}=nRT\sum_ix_i\ln x_i\) with \(n=2.000\,\text{mol}\), \(x_{\text{ben}}=0.250\), \(x_{\text{tol}}=0.750\).
The bracket: \(0.250\ln0.250+0.750\ln0.750=0.250(-1.3863)+0.750(-0.28768)=-0.34657-0.21576=-0.56234\).
The prefactor: \(nRT=2.000\times8.314\times298=4955\,\text{J}\). So \(\Delta G_{\text{mix}}=4955\times(-0.56234)=-2787\,\text{J}=-2.79\,\text{kJ}\).
Since \(\Delta H_{\text{mix}}=0\) for an ideal solution, \(\Delta G_{\text{mix}}=-T\Delta S_{\text{mix}}\) gives \(\Delta S_{\text{mix}}=2787/298=+9.35\,\text{J K}^{-1}\), which the direct formula \(-nR\sum x_i\ln x_i=-2.000\times8.314\times(-0.56234)=9.35\,\text{J K}^{-1}\) confirms (entropy-of-mixing).
Mixing is driven entirely by entropy: there is no enthalpy release at all, because in an ideal solution a benzene–toluene contact is energetically identical to the half benzene–benzene plus half toluene–toluene contact it replaces (\(W=0\) in step 9). The mixture is spontaneous only because there are vastly more ways to arrange the molecules mixed than separated.
Prove the Duhem–Margules statement: for a binary liquid mixture at fixed \(T\) and \(p\) whose vapour is ideal, if component \(A\) obeys \(p_A=x_Ap_A^{*}\) throughout some composition interval, then \(p_B=K_Bx_B\) with \(K_B\) constant throughout the same interval. Then say when \(K_B=p_B^{*}\).
Solution
Start from the Gibbs–Duhem relation at constant \(T,p\): \(x_A\,\mathrm{d}\mu_A+x_B\,\mathrm{d}\mu_B=0\).
By step 4 of the Proof, \(\mu_i=\mu_i^{*}+RT\ln(p_i/p_i^{*})\) for each component, so \(\mathrm{d}\mu_i=RT\,\mathrm{d}\ln p_i\) at fixed \(T\) (the pure-liquid terms are constants). Dividing by \(RT\neq0\): \(x_A\,\mathrm{d}\ln p_A+x_B\,\mathrm{d}\ln p_B=0\).
Suppose \(p_A=x_Ap_A^{*}\) on the interval. Then \(\ln p_A=\ln x_A+\text{const}\), so \(\mathrm{d}\ln p_A=\mathrm{d}\ln x_A=\mathrm{d}x_A/x_A\), and the first term is simply \(x_A\cdot\mathrm{d}x_A/x_A=\mathrm{d}x_A\).
Substituting: \(\mathrm{d}x_A+x_B\,\mathrm{d}\ln p_B=0\). Since \(x_B=1-x_A\) gives \(\mathrm{d}x_A=-\mathrm{d}x_B\), this is \(x_B\,\mathrm{d}\ln p_B=\mathrm{d}x_B\), i.e. \(\mathrm{d}\ln p_B=\mathrm{d}x_B/x_B=\mathrm{d}\ln x_B\).
Integrating across the interval, \(\ln p_B=\ln x_B+\text{const}\), so \(p_B=K_Bx_B\) with \(K_B\) a constant on that interval — Henry’s law. The constant of integration is fixed by any one measured point in the interval, not by the theory.
\(K_B=p_B^{*}\) requires the interval to extend to \(x_B=1\), where \(p_B\) must equal \(p_B^{*}\) by definition. If \(A\) obeys Raoult’s law only near \(x_A=1\) — the dilute-solute case, which is the general situation for a real mixture — then \(K_B\) is the slope of the tangent to the true \(p_B(x_B)\) curve at \(x_B=0\) and need not resemble \(p_B^{*}\) at all. A mixture ideal over the whole range has \(K_B=p_B^{*}\), which is the Corollaries’ statement that Raoult and Henry behaviour coincide exactly for an ideal solution.