Chemistry · Lesson

Dynamic equilibrium and reversible reactions

Describe equilibrium as matching opposing changes. Distinguish equal reaction rates from equal concentrations and from other reasons a composition can appear stationary.

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By the end, you can

Explain dynamic equilibrium under unchanged conditions.
Compare opposing rates without assuming equal concentrations.
Distinguish equilibrium from a steady flow or kinetically arrested state.

Macroscopic constancy can contain continuing change

For a reversible chemical reaction at equilibrium, the forward and reverse changes balance under the stated conditions. Molecules can continue to react while the macroscopic composition has no net change. The reversible arrow ⇌ describes the two directions; drawing it does not establish that a particular sample has already reached equilibrium.

The rates match on a compatible reaction basis. Concentrations need not be equal. Equilibrium describes a balance in opposing chemical change, rather than requiring the mixture to divide its material equally between every species.

Different concentrations can sustain equal rates

An original hypothetical elementary isomerization A ⇌ B uses k_f = 2.00 s⁻¹ and k_r = 0.500 s⁻¹ at fixed volume and conditions. A and B have the same conserved elemental composition. Its rate laws are r_f = k_f[A] and r_r = k_r[B]; these laws follow the stipulated elementary model, not just the overall arrow.

At [A] = 0.0200 mol L⁻¹ and [B] = 0.0800 mol L⁻¹, both rates are 0.0400 mol L⁻¹ s⁻¹. Four times as much B as A compensates for the smaller reverse rate constant. The total concentration of these isomers is 0.100 mol L⁻¹.

A model can approach equilibrium without stopping its steps

For the same fictional model beginning with [A]₀ = 0.100 mol L⁻¹ and [B]₀ = 0, the calculated solution is [A](t) = 0.0200 + 0.0800 exp(−2.50t), with numerical concentrations in mol L⁻¹ and t in seconds. [B](t) = 0.100 − [A](t). Treat the chosen model parameters as exact calculation inputs.

The plotted rate samples approach the same 0.0400 mol L⁻¹ s⁻¹ limit. At any finite time in this mathematical approach, a small net difference remains; the curves do not suddenly become exactly equal at an arbitrary plotted endpoint. Their continuing nonzero limit makes the dynamic balance visible.

Opposing rates approach the same nonzero limit

Reaction rate / mol L⁻¹ s⁻¹

Time / s

Calculated forward rate
Calculated reverse rate

Underlying data

Time / s	Reaction rate / mol L⁻¹ s⁻¹
0	0.2 · Calculated forward rate
0.2	0.13704490555402135 · Calculated forward rate
0.4	0.09886071058743078 · Calculated forward rate
0.6	0.07570082562374877 · Calculated forward rate
0.8	0.06165364531785804 · Calculated forward rate
1	0.053133599779823806 · Calculated forward rate
1.2	0.04796593093885823 · Calculated forward rate
1.4	0.04483158134757096 · Calculated forward rate
1.6	0.04293050222219747 · Calculated forward rate
1.8	0.041777439446118766 · Calculated forward rate
2	0.041078071519853675 · Calculated forward rate
2.2	0.040653883430154254 · Calculated forward rate
2.4	0.040396600348266615 · Calculated forward rate
2.6	0.04024055027087641 · Calculated forward rate
2.8	0.04014590111448872 · Calculated forward rate
3	0.04008849349922365 · Calculated forward rate
3.2	0.040053674020464405 · Calculated forward rate
3.4	0.040032554939041703 · Calculated forward rate
3.6	0.04001974556865387 · Calculated forward rate
3.8	0.04001197629278203 · Calculated forward rate
4	0.040007263988762 · Calculated forward rate
0	0 · Calculated reverse rate
0.2	0.015738773611494665 · Calculated reverse rate
0.4	0.02528482235314231 · Calculated reverse rate
0.6	0.03107479359406281 · Calculated reverse rate
0.8	0.03458658867053549 · Calculated reverse rate
1	0.036716600055044055 · Calculated reverse rate
1.2	0.038008517265285445 · Calculated reverse rate
1.4	0.038792104663107264 · Calculated reverse rate
1.6	0.03926737444445064 · Calculated reverse rate
1.8	0.03955564013847031 · Calculated reverse rate
2	0.03973048212003658 · Calculated reverse rate
2.2	0.03983652914246144 · Calculated reverse rate
2.4	0.03990084991293335 · Calculated reverse rate
2.6	0.0399398624322809 · Calculated reverse rate
2.8	0.03996352472137782 · Calculated reverse rate
3	0.039977876625194086 · Calculated reverse rate
3.2	0.039986581494883905 · Calculated reverse rate
3.4	0.03999186126523958 · Calculated reverse rate
3.6	0.039995063607836535 · Calculated reverse rate
3.8	0.039997005926804496 · Calculated reverse rate
4	0.0399981840028095 · Calculated reverse rate

Original hypothetical elementary A ⇌ B model: exact chosen k_f = 2.00 s⁻¹, k_r = 0.500 s⁻¹, fixed volume, [A]₀ = 0.100 mol L⁻¹ and [B]₀ = 0. Markers every 0.2 s are calculated from the stated exponential solution. Both approach 0.0400 mol L⁻¹ s⁻¹ as t grows; no finite-time exact equilibrium or measured data is asserted.

A stationary observation needs interpretation

An unchanged concentration alone is insufficient evidence for equilibrium. A flowing vessel can maintain a steady concentration because supply and removal balance its reaction. A process can also be so slow that a measurement detects little change during its observation window.

Combine composition observations with the system boundary, conditions and evidence for the relevant opposing processes. The equilibrium position and the time required to approach it are different questions. Changing the conditions can establish a different equilibrium state.

Worked examples

Check equal rates at unequal concentrations

Use the fictional elementary model at [A] = 0.0200 mol L⁻¹ and [B] = 0.0800 mol L⁻¹. Calculate its opposing rates.

Forward: r_f = 2.00 × 0.0200 = 0.0400 mol L⁻¹ s⁻¹.
Reverse: r_r = 0.500 × 0.0800 = 0.0400 mol L⁻¹ s⁻¹.
The concentrations differ by a factor of 4, while the rates match.

The model is in dynamic balance at these stipulated concentrations; neither opposing rate is zero.

Try it yourself

Must reactant and product concentrations be equal at equilibrium?
Show the answer and reasoning
No.
Matching opposing rates can occur at different concentrations.
Do the model’s elementary steps stop when equilibrium is reached?
Show the answer and reasoning
No.
Their nonzero opposing rates balance.
Does a ⇌ symbol prove a newly prepared mixture is already at equilibrium?
Show the answer and reasoning
No.
It specifies reversible directions; the current state still needs assessment.
Can a flowing steady state have unchanged composition without being equilibrium?
Show the answer and reasoning
Yes.
Supply, removal and reaction can balance while net chemical processes continue.

What to carry forward

Equal opposing rates do not require equal concentrations.
A reversible equation and a current equilibrium state are distinct.
The approach rate and equilibrium position answer different questions.
Interpret stationary measurements using system boundaries and conditions.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.