After midterm · Session 5 of 6 · 180 minutes
30 September 2026
Reconcile independent thermodynamic routes to K and then solve composition.
Before class: Review formation Gibbs energies, reaction enthalpy/entropy and integration of heat capacity.
| In class | Minutes |
|---|---|
| Recall and prediction | 10 |
| Concepts and derivation | 45 |
| Worked example | 30 |
| Break | 10 |
| Instructor lab demonstration | 25 |
| Guided student exploration | 35 |
| Discussion and interpretation | 15 |
| Exit question and independent task | 10 |
n_i=n_{i,0}+\nu_i\xi \Delta_rG=\sum_i\nu_i\mu_i=\Delta_rG^\circ+RT\ln Q
Stoichiometric coefficients are negative for reactants and positive for products. At an interior equilibrium, ΔrG=0.
\ln K=-\frac{\Delta_rG^\circ}{RT},\qquad Q=\prod_i a_i^{\nu_i}
K is tied to T and the fixed reaction/standard-state convention. Q describes the supplied state.
Q=K is an equilibrium condition, not an identity for every measured composition.
| Phase | Activity used |
|---|---|
| Gas | φᵢ yᵢ P/p° |
| Solute, molarity standard | γᵢ cᵢ/c° |
| Liquid solution | γᵢ xᵢ |
| Present pure solid/liquid | 1 |
Formation data and activities must use matching phases and standards.
Independent inputs can disagree; the lab preserves the disagreement.
Lab 12 synthetic A₂ ⇌ 2A at 350 K uses ΔrG°=−4.000 kJ/mol.
\ln K=\frac{4000}{R(350)}=1.37454,\qquad K\approx3.95326
With ΔrH°=50.000 kJ/mol and ΔrS°=54000/350 J mol⁻¹ K⁻¹, the H−TS route gives the same value.
\frac{d\ln K}{dT}=\frac{\Delta_rH^\circ(T)}{RT^2}
With constant ΔrH° over the stated interval, \ln\frac{K(T)}{K(T_r)}=\frac{\Delta_rH_r^\circ}{R}\left(\frac1{T_r}-\frac1T\right)
An endothermic reaction has increasing K with T in this approximation.
\Delta_rH^\circ(T)=\Delta_rH_r^\circ+\Delta_rC_p^\circ(T-T_r) \Delta_rS^\circ(T)=\Delta_rS_r^\circ+\Delta_rC_p^\circ\ln(T/T_r)
The integrated ln K adds \frac{\Delta_rC_p^\circ}{R}\left[\ln(T/T_r)+T_r/T-1\right]
to the constant-enthalpy expression.
Changing only Kref while holding reference H and S fixed creates two independent anchors that may disagree.
Point ΔG° and formation-energy data are unavailable away from their supplied T unless an explicit temperature model exists.
Keeping ln K avoids false zeros or infinities from exponent underflow/overflow.
For a neutral ideal-gas reaction, Q(\xi)=\prod_i\left[y_i(\xi)P/p^\circ\right]^{\nu_i}
Solve lnQ(ξ)−lnK=0 within nonnegative mole amounts. Pressure changes Q and equilibrium composition, not standard K at fixed T.
Adding inert gas at fixed total pressure differs from adding it at fixed volume.

Choose one temperature change and one pressure change.
Keep a table of lnK, Q, reaction direction and equilibrium extent. Explain which quantities change for each intervention.
Independently check one atom balance and one H−TS or van ’t Hoff calculation.
Reversing the reaction changes lnK to −lnK. Multiplying every coefficient by c changes lnK to c lnK.
Formation sums, reaction H, S, Cp and direct ΔG must transform on the same basis.
A numerical K without a written reaction is incomplete information.
At the same temperature, pressure increases and equilibrium conversion changes.
Does this mean the equilibrium constant changed?
What additional information is needed before using one ΔG° value at a second temperature?
Compare constant-property and heat-capacity routes; report ln K, Q and one extent/material-balance check.
Retain the calculator export, your worksheet, a comparison plot/table and one independent check. State an assumption that limits your conclusion.
Use the core labs on the learning path. Optional extensions are additional work.
Module reference deck · Lab sources and equations
IUPAC: standard electromotive force and standard equilibrium constant.
Synthetic worked examples illustrate calculations; they are not evidence of real-system accuracy.