SCIENCE · INVESTIGATE AND EXPLAIN

Distinguish a catalytic time scale from an equilibrium ratio

Dynamic equilibrium has equal opposing rates, not zero molecular transformation.

Learning outcome: Compare multiplier settings from two starting compositions. Verify the common endpoint, matching nonzero equilibrium rates and different approach speeds.

Starting recap

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Try

With kf=.40 and kr=.10 s⁻¹, total1 M gives Aeq=.20 M and a relaxation time2 s. Multiplying both constants by3 keeps Aeq but changes the time to2/3 s.

Open the experiment

Predict

Would changing only the forward constant represent a catalyst for the same thermodynamic equilibrium?

Your prediction and reason

Experiment

Compare multiplier settings from two starting compositions. Verify the common endpoint, matching nonzero equilibrium rates and different approach speeds.

Change equilibration speed without changing its endpoint

Worked teaching model · synthetic values · Fictional elementary isomer pair A⇌B, total1.00 M, kf=.40s⁻¹ and kr=.10s⁻¹ times the same multiplier. K=kf/kr=4; Aeq=.20 M. This bounded catalytic model changes both rates consistently, not the equilibrium constant or conserved inventory.

Common catalyst rate multiplier: 1 ratio Initial A fraction: 0.8 fraction

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A: solid
B: long dashes

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

A
time (s)	concentration (mol/L)
0	0.8
0.1666667	0.7520266
0.3333333	0.707889
0.5	0.6672805
0.6666667	0.6299188
0.8333333	0.5955444
1	0.5639184
1.166667	0.5348211
1.333333	0.5080503
1.5	0.4834199
1.666667	0.4607589
1.833333	0.4399098
2	0.4207277
2.166667	0.4030793
2.333333	0.3868419
2.5	0.3719029
2.666667	0.3581583
2.833333	0.3455126
3	0.3338781
3.166667	0.3231738
3.333333	0.3133254
3.5	0.3042644
3.666667	0.2959278
3.833333	0.2882579
4	0.2812012
4.166667	0.2747087
4.333333	0.2687353
4.5	0.2632395
4.666667	0.2581832
4.833333	0.2535311
5	0.249251
5.166667	0.2453131
5.333333	0.2416901
5.5	0.2383567
5.666667	0.2352899
5.833333	0.2324683
6	0.2298722
6.166667	0.2274838
6.333333	0.2252863
6.5	0.2232645
6.666667	0.2214044
6.833333	0.219693
7	0.2181184
7.166667	0.2166698
7.333333	0.2153369
7.5	0.2141106
7.666667	0.2129824
7.833333	0.2119444
8	0.2109894
8.166667	0.2101107
8.333333	0.2093023
8.5	0.2085585
8.666667	0.2078742
8.833333	0.2072446
9	0.2066654
9.166667	0.2061325
9.333333	0.2056421
9.5	0.205191
9.666667	0.204776
9.833333	0.2043941
10	0.2040428

B
time (s)	concentration (mol/L)
0	0.2
0.1666667	0.2479734
0.3333333	0.292111
0.5	0.3327195
0.6666667	0.3700812
0.8333333	0.4044556
1	0.4360816
1.166667	0.4651789
1.333333	0.4919497
1.5	0.5165801
1.666667	0.5392411
1.833333	0.5600902
2	0.5792723
2.166667	0.5969207
2.333333	0.6131581
2.5	0.6280971
2.666667	0.6418417
2.833333	0.6544874
3	0.6661219
3.166667	0.6768262
3.333333	0.6866746
3.5	0.6957356
3.666667	0.7040722
3.833333	0.7117421
4	0.7187988
4.166667	0.7252913
4.333333	0.7312647
4.5	0.7367605
4.666667	0.7418168
4.833333	0.7464689
5	0.750749
5.166667	0.7546869
5.333333	0.7583099
5.5	0.7616433
5.666667	0.7647101
5.833333	0.7675317
6	0.7701278
6.166667	0.7725162
6.333333	0.7747137
6.5	0.7767355
6.666667	0.7785956
6.833333	0.780307
7	0.7818816
7.166667	0.7833302
7.333333	0.7846631
7.5	0.7858894
7.666667	0.7870176
7.833333	0.7880556
8	0.7890106
8.166667	0.7898893
8.333333	0.7906977
8.5	0.7914415
8.666667	0.7921258
8.833333	0.7927554
9	0.7933346
9.166667	0.7938675
9.333333	0.7943579
9.5	0.794809
9.666667	0.795224
9.833333	0.7956059
10	0.7959572

Calculated under the declared model
Quantity	Value	Unit
Equilibrium A	0.2	mol/L
Equilibrium B	0.8	mol/L
Relaxation time constant	2	s
Forward equilibrium rate	0.08	mol/(L s)
Reverse equilibrium rate	0.08	mol/(L s)
A after 2 seconds	0.4207277	mol/L

At equilibrium the opposing rates agree; the elementary transformations have not stopped.

Multiplying both rate constants preserves kf/kr while shortening the relaxation time. Changing only the forward rate would change this model equilibrium and would not represent a catalyst for the same thermodynamic reaction. These first-order rates are an explicit mechanism model, not inferred from the net equation alone.

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

At equilibrium the opposing rates agree; the elementary transformations have not stopped.

Observation to attach to a capture

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

Understand

Specify the elementary model

For a fictional first-order isomer pair A⇌B at total concentration1 M, the rate equation is dA/dt=−kf A+kr B. Substituting B=1−A yields a linear relaxation toward Aeq=kr/(kf+kr). This form is a supplied elementary mechanism, not a universal inference from a net reaction arrow.

At kf=.40 and kr=.10 s⁻¹, Aeq=.20 and Beq=.80 M. The forward and reverse equilibrium rates both equal .080 M/s. Equal rates maintain stable macroscopic concentrations even though transformations continue.

Derive the approach time

The solution is A(t)=Aeq+(A0−Aeq)exp[−(kf+kr)t]. The time constant is1/(kf+kr)=2 s. Multiplying both constants by the same catalytic factor changes this time without changing their ratio K=4 or the endpoint.

A model that accelerates only the forward step while retaining the reverse constant would change its predicted equilibrium. It cannot represent a catalyst for the same thermodynamic reaction at unchanged conditions. A real catalytic mechanism may contain additional intermediates, but detailed consistency still constrains its equilibrium result.

Transfer and check

Take multiplier2 and A0=.80 M. After2 s, A=.20+.60exp(−2)≈.281201 M and B≈.718799 M. At equilibrium the opposing rates are each .160 M/s. Explain why a doubled equilibrium turnover is compatible with no net concentration change, and why a fast approach alone does not measure the value of K.

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 mechanistic assumption permits using K=kf/kr here, and why is it not automatic for an arbitrary net equation?

Your explanation and limitations

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Previous: Check phase existence before imposing saturation Next: Bound an inferred constant and test a changed inventory

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