SCIENCE · INVESTIGATE AND EXPLAIN

Build a dimensionless quotient and separate direction from speed

An equilibrium quotient belongs to a written reaction, a temperature and a declared activity convention.

Learning outcome: Vary A and the prescribed common gamma separately. Compute each normalized activity before Q; locate the direction boundary and explain why Q−K is not a rate.

Starting recap

Refresh dimensionless activities, component balances, admissible roots, normalized partial pressures and independent versus shared measurement errors.

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Try

For A+B⇌C, concentrations .20,.20,.10 M and common γ=1 give Q=2.50; compared with K=4 this favors forward change.

Open the experiment

Predict

Will a common activity coefficient cancel when the reaction changes the total stoichiometric number of dissolved particles?

Your prediction and reason

Experiment

Vary A and the prescribed common gamma separately. Compute each normalized activity before Q; locate the direction boundary and explain why Q−K is not a rate.

Normalize activities before constructing a quotient

Worked teaching model · synthetic values · Fictional A+B⇌C, [B]=.20 M and [C]=.10 M. All three concentration-based activities are γc/c°, c°=1 M; a common prescribed γ is an illustrative nonideal state, not a concentration law. Compare dimensionless Q with supplied K=4 at fixed temperature.

A concentration: 0.2 mol/L Common concentration activity coefficient: 1 ratio

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Q at selected gamma: solid

Axes rescale. A quotient comparison, inventory intersection and relaxation trajectory have different horizontal meanings; inspect labels rather than treating every curve as elapsed time.

All calculated plot coordinates

Q at selected gamma
fraction of A concentration range .10 to .50 M	dimensionless Q
0	5
0.01666667	4.6875
0.03333333	4.411765
0.05	4.166667
0.06666667	3.947368
0.08333333	3.75
0.1	3.571429
0.1166667	3.409091
0.1333333	3.26087
0.15	3.125
0.1666667	3
0.1833333	2.884615
0.2	2.777778
0.2166667	2.678571
0.2333333	2.586207
0.25	2.5
0.2666667	2.419355
0.2833333	2.34375
0.3	2.272727
0.3166667	2.205882
0.3333333	2.142857
0.35	2.083333
0.3666667	2.027027
0.3833333	1.973684
0.4	1.923077
0.4166667	1.875
0.4333333	1.829268
0.45	1.785714
0.4666667	1.744186
0.4833333	1.704545
0.5	1.666667
0.5166667	1.630435
0.5333333	1.595745
0.55	1.5625
0.5666667	1.530612
0.5833333	1.5
0.6	1.470588
0.6166667	1.442308
0.6333333	1.415094
0.65	1.388889
0.6666667	1.363636
0.6833333	1.339286
0.7	1.315789
0.7166667	1.293103
0.7333333	1.271186
0.75	1.25
0.7666667	1.229508
0.7833333	1.209677
0.8	1.190476
0.8166667	1.171875
0.8333333	1.153846
0.85	1.136364
0.8666667	1.119403
0.8833333	1.102941
0.9	1.086957
0.9166667	1.071429
0.9333333	1.056338
0.95	1.041667
0.9666667	1.027397
0.9833333	1.013514
1	1

Calculated under the declared model
Quantity	Value	Unit
A activity	0.2	dimensionless
B activity	0.2	dimensionless
C activity	0.1	dimensionless
Dimensionless reaction quotient	2.5	dimensionless
Q minus K	-1.5	dimensionless

Forward association is favored toward equilibrium.

Because this reaction changes the number of dissolved particles, even a common activity coefficient does not cancel: Q=.10/(γ[A]×.20) with concentrations normalized by 1 M. Q versus K identifies a thermodynamic direction, not a reaction speed or the size of the final change.

These calculated packets are not performed chemical measurements. Keep feasible inventories, activity conventions, fitting observations and independent checks distinct.

Forward association is favored toward equilibrium.

Observation to attach to a capture

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

Understand

Start with activities, not bare unit-bearing numbers

For the fictional association A+B⇌C, define a_i=γ_i c_i/c° with c°=1 mol/L. The dimensionless quotient is a_C/(a_A a_B). At fixed temperature its equilibrium value is K. A numerical concentration in millimoles per liter cannot be substituted unchanged where a normalized molar concentration is intended.

Keep the common coefficient that does not cancel

At A=.20, B=.20 and C=.10 M, ideal activities give Q=.10/(.20×.20)=2.50. If all three prescribed coefficients are .50, the activities are .10,.10,.05 and Q=5.00. This is not contradictory: one factor remains in the denominator because association reduces the total stoichiometric particle count.

The common gamma control illustrates a state-dependent correction. It is not a universal law saying all real solutes have equal activity coefficients, nor a complete thermodynamically consistent concentration dependence.

Predict the permitted direction

With supplied K=4, Q=2.50 favors forward association, Q=5.00 favors reverse dissociation and Q=4 is equilibrium. These comparisons say nothing about how quickly an inhibited transformation occurs. Nor do they determine the final amount without component balances.

Transfer and check

Keep B=.20 and C=.10 M but choose A=.40 M and gamma=.625 outside the displayed grid. Predict Q before calculating: it is .10/(.625×.40×.20)=2.00, so forward change is favored. Explain why doubling A alone halves Q but does not halve K. Then identify the missing observation needed to infer an actual relaxation time.

Source and evidence boundary

Conceptual reference checked 30 October 2026: OpenStax Chemistry 2e. These original hypothetical packets use explicitly defined activities, inventories and response assumptions. No material is mixed, heated, pressurized or measured. Numerical compatibility does not certify practical technique or uniquely establish a real chemical mechanism. Retain your prediction, feasible inventory, quotient check and changed-case interpretation.

Apply

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

Which response or standard-state assumption must be held fixed when comparing inferred constants?

Your explanation and limitations

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Previous: Test an intermediate profile without singular equal-rate arithmetic Next: Transform equilibrium constants with exact reaction statements

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