SCIENCE · INVESTIGATE AND EXPLAIN

Use K and Q to decide the actual direction

Connect standard free energy to K and actual reaction free energy to Q/K.

Learning outcome: Connect standard free energy to K and actual reaction free energy to Q/K.

Starting recap

K and Q use dimensionless activities; K belongs to a specified reaction and temperature.

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Try

At 298.15 K a hypothetical reaction has K=100, but a prepared sample may have Q=10000.

Open the experiment

Predict

Does K greater than one force every prepared sample to react forward?

Your prediction and reason

Experiment

Hold logK at 2 and compare logQ 0, 2 and 4. Then reverse the sign of logK and explain the new standard-state preference.

Distinguish K, Q and actual direction

Worked teaching model · synthetic values · Dimensionless activity quotient and equilibrium constant at 298.15 K. ΔrG° = −RT ln K; ΔrG = RT ln(Q/K). This is a local direction criterion, not a finite reaction yield or rate.

log10 K: 2 decade log10 Q: 0 decade

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Conditional calculations, not measured observations
Quantity	Value	Unit
K	100	dimensionless
Q	1	dimensionless
ΔG°	-11.41602	kJ/mol reaction
ΔrG	-11.41602	kJ/mol reaction

Forward direction thermodynamically favored

K>1 describes equilibrium preference for the written reaction; even then a product-rich Q>K favors reverse change. Q=K, not Q=1, is the equilibrium test.

Predict before changing a control. Save two contrasting captures and explain what is held fixed. Displayed digits are calculation precision, not experimental uncertainty.

Forward direction thermodynamically favored

K>1 describes equilibrium preference for the written reaction; even then a product-rich Q>K favors reverse change. Q=K, not Q=1, is the equilibrium test.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Two different questions

Does a written reaction favor products at equilibrium? The standard Gibbs energy and K answer that question. Which way should a particular prepared mixture change? That requires its reaction quotient Q. Confusing these questions is a common source of sign errors.

For compatible standard states, ΔrG°=−RT ln K and ΔrG=ΔrG°+RT ln Q=RT ln(Q/K). Both logarithms require dimensionless quantities. Activities are ratios to reference states; simply placing a dimensional concentration inside a logarithm hides that normalization.

Work the three mixtures

At 298.15 K, let K=100. Then ΔrG°≈−11.416 kJ/mol reaction. If Q=1, the actual ΔrG equals that negative standard value. If Q=100, ΔrG=0. If Q=10000, Q/K=100 and ΔrG≈+11.416 kJ/mol: the reverse direction is favored even though K remains greater than one.

Change only Q in the lab and record the three cases. Explain the product-rich case without saying that K has changed. At fixed temperature the same reaction retains the same equilibrium constant under this model; preparation changes the quotient.

What a large constant does not guarantee

A large K does not prove that an experiment reaches equilibrium quickly, or that every reactant molecule disappears. It describes an equilibrium relation. Actual equilibrium amounts also depend on initial amounts, conservation laws, phase availability and the form of the quotient. A large value for a squared or cubed activity expression cannot be interpreted as an unqualified product mole fraction.

Reversing the reaction gives Kreverse=1/K and ΔrG°reverse=−ΔrG°forward. Multiplying the equation by two gives Knew=K² and ΔrG°new=2ΔrG°. The lab uses one fixed equation; distinguish a changed equation from a changed prepared mixture.

Changed-case exercise

For A⇌B with ideal activities, let K=0.01 and Q=0.0001. Calculate Q/K=0.01, so the actual forward ΔrG is negative even though the standard reaction favors A. State how B/A changes as equilibrium is approached. It rises toward 0.01; no rate or elapsed time can be inferred. This transfer checks whether you use the ratio rather than a memorized sign of ΔG°.

Source scope and continuation

Original teaching explanation, calculations and practice. Framework alignment: AP Chemistry topic 9.5 in the College Board course framework, effective Fall 2024, checked 30 October 2026. These pages do not reproduce restricted examination questions or establish AP authorization. The investigation is a bounded computational model, not a conducted laboratory.

Return to the preceding Unit 9 lesson when its reasoning is unfamiliar.

Continue to Explain why dissolution is a balance of competing effects.

Define “near zero” relative to the thermal scale

Write g = ΔrG°/(RT). Then K = exp(−g), so g = 0 gives K = 1 exactly. Small positive g gives K slightly below one; small negative g gives K slightly above one. A free energy is only small for this purpose relative to RT in matching units, not relative to an arbitrary number of kilojoules.

Changed-case calculation: Compare g = +0.02, −0.02, +5 and −5. Predict the side of one and relative size before calculating. Explain why changing the written reaction coefficients also changes g and K without changing the physical equilibrium mixture.

Compare the dimensionless scale

The constants are approximately 0.98020, 1.02020, 0.006738 and 148.413, respectively. The first pair is near one; the second pair is not. Doubling a reaction doubles g and squares K. Even K = 1 is an activity-expression statement, not a guarantee that all species concentrations are equal.

The relation and its standard-state limits are supported by OpenStax free energy, checked 31 October 2026. All comparison values are original.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Show your calculation or representation, explain the changed case, and separate the stated model from measured evidence. Connect standard free energy to K and actual reaction free energy to Q/K.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Separate a favorable direction from an accessible rate Next: Explain why dissolution is a balance of competing effects

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