Chemical Potential: Worked Problems

Introduction

Chemical-potential problems often combine several formulas that look similar but apply to different species and reference states. An ideal gas uses a partial-pressure ratio, an ideal liquid solvent uses mole fraction, a real solution uses activity, and a reaction uses a stoichiometric sum of chemical potentials. This workshop works through examples and shows how to choose the right expression before reaching for a calculator.

Core explanation

Start by identifying the physical change. For passive transfer of one species at fixed T and p, dG = (μ target − μ source)dn. For an ideal-gas component, μ i = μ i° + RT ln(p i/p°), with p i its partial pressure. For an ideal liquid solvent, μ A = μ A + RT ln x A. A real solution replaces x A with activity a A under a consistent standard. These equations share a logarithm but their standards are different, so one cannot equate numerical activities drawn from unrelated conventions without transforming their μ° values.

For ideal-gas mixing of distinct species initially separate at the same T and p, Δ mixG = RTΣ i n i ln x i, Δ mixS = −RΣ i n i ln x i and Δ mixH = 0. A negative Δ mixG is expected because each 0 < x i < 1. For a dilute nonelectrolyte solution separated from pure solvent by a solute-blocking membrane, Π ≈ c particle RT. The concentration must be in mol m⁻³ when R is in J mol⁻¹ K⁻¹ and pressure is desired in pascals.

For a reaction, Δ rG = Δ rG° + RT ln Q = RT ln(Q/K) at a fixed temperature when Δ rG° = −RT ln K. The quotient Q is a product of dimensionless activities with stoichiometric exponents. Q < K means forward reaction lowers G locally; Q > K means reverse reaction does. If temperature changes, use the van 't Hoff relation with Δ rH° and state the constant-enthalpy approximation before applying its integrated form.

The most common calculation mistakes are reference and unit mistakes. A logarithm needs a dimensionless ratio, not a pressure carrying units. A gas-mixture component needs its partial pressure, not total pressure. Osmotic pressure uses solution volume and dissolved-particle concentration, not solvent mass as in molality-based freezing-point formulas. A reaction quotient must use the balanced reaction's stoichiometric coefficients. Before trusting a number, predict its sign and rough scale from the physical situation.

Step-by-step reasoning

Read the process description and identify whether it concerns one component's transfer, mixing, membrane equilibrium or a reaction. Write the relevant μ or ΔG equation symbolically. Choose and state standard states, calculate mole fractions or partial pressures, and convert concentrations to compatible units. Insert numbers only after the argument of each logarithm is visibly dimensionless. End with a sign check and a sentence interpreting the result under the stated approximations.

Visual explanation

Draw a decision map with four branches from “What changes?”: gas composition leads to p i and RT ln(p i/p°); liquid solvent dilution leads to a A and RT ln a A; membrane balance leads to Π; reaction extent leads to RT ln(Q/K). Put a unit checkpoint before the numerical answer on every branch.

Real-world analogy

Different journeys can all be measured in kilometres, yet a map, a timetable and an altitude chart answer different questions. Chemical-potential equations similarly share energy-per-mole units while describing different constraints. Choosing the correct thermodynamic map matters more than memorising one impressive-looking logarithm.

Real-world example

In a gas–liquid process, an engineer may use a gas fugacity to describe a high-pressure vapour, solvent activity to describe a nonideal liquid and equality of one species' μ across phases to determine equilibrium. No single raw concentration represents all three effects. The simplified cases below show the core arithmetic before those real-system corrections are included.

Why?

Chemical potential is a partial molar Gibbs energy, so every process can be analysed as a sum of amount changes times μ. The different practical formulas are ways to evaluate μ for particular models. Organising the work around the process and its constraints prevents a valid formula from being applied to the wrong state.

Common misconception

One might use Δ rG° to decide direction without knowing current Q, or use pure solvent vapour pressure as if it were a gas component's partial pressure in a mixture. Another frequent error is to put mol L⁻¹ directly into Π = cRT with SI R and report pascals, missing a factor of 1000. These are model-selection and unit errors rather than subtle algebra mistakes.

Worked example

Three linked cases at 298 K illustrate the choices. First, mix 1.00 mol each of two distinct ideal gases, initially separate at the same p. Their final x values are 0.5, so Δ mixG = 2RT ln 0.5 ≈ −3.44 kJ and Δ mixS = 2R ln 2 ≈ 11.5 J K⁻¹. Second, a 0.010 mol L⁻¹ ideal nonelectrolyte solution has c = 10 mol m⁻³ and Π ≈ cRT = (10)(8.314)(298) ≈ 24.8 kPa. Third, a reaction has K = 5 and current Q = 1. Then Δ rG = RT ln(1/5) ≈ −3.99 kJ per mole of reaction extent, so forward change is favourable. The three values cannot be compared as though they were the same physical quantity: one is total mixing G, one pressure, and one Gibbs energy per reaction extent.

Quick check

1. An ideal gas mixture is at 2.0 bar total with species A mole fraction 0.25. What pressure enters μ A? Answer: Its partial pressure is p A = 0.25 × 2.0 = 0.50 bar. Use μ A = μ A° + RT ln(0.50 bar/p°), not a logarithm of the 2.0-bar total pressure.

Exam focus

Write assumptions near each formula: ideal gas, ideal solvent, dilute nonelectrolyte or activity-corrected real mixture. Check mole fraction against amount, molarity against volume and molality against solvent mass. Keep Gibbs-energy and pressure units distinct. For reaction direction, compare Q with K before computing the logarithm.

Advanced insight

These problems are joined by equality or minimisation of G under a constraint. Mixing explores more accessible states, osmosis balances solvent μ across a membrane, and reaction equilibrium sets the G slope along a stoichiometric path to zero. When ideal formulas fail, activities and fugacities change the numerical inputs while preserving this thermodynamic structure.

Summary

Chemical-potential calculations begin with the permitted change and a consistent standard state. Gas components use partial pressure, liquid components use activity, osmosis balances solvent μ and reactions use RT ln(Q/K). Sign, units and model assumptions are essential checks. The formulas are related through Gibbs energy but describe different constrained processes.

Practice questions

1. A real solvent has a A = 0.90 at 300 K. Find μ A − μ A on a pure-liquid standard scale. Answer: RT ln 0.90 = (8.314)(300)(−0.10536) ≈ −263 J mol⁻¹. The solvent's chemical potential is lower than in its pure reference state. 2. A dilute ideal solution has c particle = 25 mol m⁻³ at 300 K. Estimate Π. Answer: Π ≈ cRT = (25)(8.314)(300) ≈ 6.24 × 10⁴ Pa, or 62.4 kPa. 3. At a fixed temperature K = 0.20 but a mixture has Q = 0.050. Predict reaction direction and the sign of Δ rG. Answer: Q < K, so ln(Q/K) = ln 0.25 < 0. Forward reaction lowers G and Δ rG is negative for the reaction as written.