Equilibrium and free energy
Separate equilibrium composition from approach speed and relate a dimensionless equilibrium constant to standard reaction Gibbs energy.
Start with: Concentration, exponentials and ratios. A reversible reaction can proceed in both directions.
01 · Commit to a prediction
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02 · Change an assumption
Before changing a control, say what should move and why. Start with the experiments below. Reset restores the starting model; it preserves your written responses.
Starting at kf = 0.3 and kr = 0.1, double both to 0.6 and 0.2. First predict which quantity stays fixed.
The ratio remains 3, so B equilibrium stays 0.75 mol/L. The relaxation time halves from 2.5 to 1.25 s. This deliberate paired change isolates overall speed at fixed ratio.
Reset. Change only initial B to 1 mol/L.
The trajectory approaches the same 0.75 mol/L from above. Initial composition changes the direction, not the equilibrium under these conditions.
0.05 to 1 · step 0.05
0.05 to 1 · step 0.05
0 to 1 · step 0.05
1 to 30 · step 1
Calculated model output. The table gives the same values. Displayed values are rounded; calculations keep full precision.
03 · Connect the mechanism
In this closed elementary A ⇌ B model, forward flux is kf[A] and reverse flux is kr[B]. At equilibrium their equality stops net change while both conversions continue. K specifies the ratio at equilibrium. The sum of rate constants sets how quickly this particular model approaches it. Favorable reaction Gibbs energy does not mean a reaction is fast.
C = [A]+[B] is the fixed total concentration, 1 mol/L. Both rate constants are positive, 0.05 to 1 s⁻¹ here. t is seconds. K is dimensionless: activities are concentrations divided by the same 1 mol/L reference in this ideal solution. R = 8.314462618 J/(mol·K), T = 298 K. ΔG° is the standard molar reaction Gibbs energy in J/mol. It is not the actual ΔG at every composition.
With kf = 0.3 s⁻¹ and kr = 0.1 s⁻¹, equilibrium has [B]/[A] = 3 and [A]+[B] = 1. Hence [B] = 0.75 and [A] = 0.25 mol/L. Both equilibrium fluxes equal 0.075 mol/(L·s). Starting from all A, after 10 s [B] = 0.75(1−e⁻⁴) ≈ 0.736263 mol/L. ΔG° = −RT ln 3 ≈ −2.722 kJ/mol.
A closed, ideal, well-mixed system with constant temperature and elementary first-order rates. Actual ΔG = ΔG° + RT ln Q, where Q = aB/aA, and is zero at equilibrium. At pure A or B, this logarithmic expression has a limiting divergence; the kinetic trajectory remains defined. Both constants are kept positive so K is finite. A catalyst consistent with the same thermodynamics speeds both directions without changing K. Temperature changes can alter K.
The annotated sources distinguish established results from this lesson’s original examples.
04 · Follow the structure
Opposing conversions can continue at equal rates with unequal concentrations.
Boundary: Complex reaction networks may not have a single exponential relaxation.
Equal crossing rates can stabilize unequal occupancies.
Boundary: Human choices are not ideal chemical activities; assigning a Gibbs energy to a crowd is unsupported.
05 · Retrieve without hints
Write an answer before opening its feedback. Later, return directly here without rereading above. Recognition, explanation and transfer are separate outcomes. No page action or answer reveal measures mastery.
Recognition and explanation
No. At equilibrium the net driving force and net reaction rate vanish while forward and reverse events continue.
Self-check: Distinguish molecular events from net change and ΔG from ΔG°.
Calculation
Each concentration is 0.5 mol/L. Each flux is 0.2 × 0.5 = 0.1 mol/(L·s). K = 1, so ΔG° = 0.
Self-check: Balance fluxes and check the total concentration.
Novel transfer
Changing only kf changes kf/kr and thus K in this elementary model. A catalyst for the same equilibrium cannot change K; reverse kinetics must remain thermodynamically consistent.
Self-check: Separate a new reaction model from faster approach to the same equilibrium.
On a later day, try again and record actual evidence in the curriculum. A later unaided explanation and a fresh transfer problem give stronger evidence than immediate familiarity. No reminder is scheduled.
Sources & scope
Original teaching examples. Reference links need a connection; the lesson itself does not. Built 2026-10-11. Learner understanding is not assessed.