Worked example 1
For N₂O₄(g) ⇌ 2 NO₂(g), K = 4.6 × 10⁻³ at 25 C. Describe the equilibrium mixture.
Try it first: Compare K with 1 before doing anything else.
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Chemical Equilibrium & SolubilityEquilibriumContent level: Core 18 min
Dynamic Equilibrium and the Equilibrium Constant
What you'll be able to do: Describe what is happening at the particle level at equilibrium and interpret the size of K.
Introduction
A reaction that reaches equilibrium has not stopped. Forward and reverse changes are still happening, just at matching rates, so the amounts you can measure stop changing.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Explain equilibrium as equal forward and reverse rates
- Interpret large, small and intermediate values of K
- State which changes do and do not alter the value of K
- Distinguish the position of equilibrium from the constant K
Lesson
Equilibrium is dynamic
When a reversible reaction starts, only the forward reaction can occur. As product builds up the reverse reaction speeds up and the forward reaction slows down. Eventually the two rates become equal. From that moment the measurable concentrations stop changing, but individual molecules are still converting in both directions.
Equal rates, not equal amounts
A common trap is to read equilibrium as meaning equal concentrations of reactants and products. It means equal rates. The actual amounts at equilibrium depend on the reaction and can be wildly lopsided.
Equilibrium means equal rates, not equal concentrations.
The equilibrium constant
For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant compares product terms to reactant terms, each raised to its coefficient. The value is constant at a fixed temperature no matter what starting amounts you used.
K = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ)
Reading the size of K
K much greater than 1 means the equilibrium mixture is mostly product and the reaction is said to lie to the right. K much less than 1 means mostly reactant. K near 1 means significant amounts of both are present.
K > 10³ is product-favored, K < 10⁻³ is reactant-favored, and anything in between means both are present in measurable amounts.
What changes K
Only temperature changes the value of K. Adding or removing a substance, changing volume or pressure, or adding a catalyst shifts the system to a new position of equilibrium, but the ratio it settles back to is the same K. A catalyst speeds up the forward and reverse reactions equally, so equilibrium arrives sooner with exactly the same composition.
A label can reveal continuing exchange
At equilibrium the forward and reverse rates are equal, so bulk concentrations remain steady. A labeled atom introduced into one side can still appear in products as the reactions continue. The label changes what is traceable, not the principle: a constant macroscopic composition can conceal molecular exchange in both directions.
Equal rates maintain the composition while molecules keep reacting.
Key ideas
Definition
Dynamic equilibrium
A state where the forward and reverse reactions proceed at equal rates, so concentrations remain constant.
Definition
Equilibrium constant K
The fixed ratio of product to reactant terms, each raised to its stoichiometric coefficient, at a given temperature.
Rule
Only temperature changes K
Concentration, pressure and catalysts change the position of equilibrium but not the value of K.
Key concept
Catalysts and equilibrium
A catalyst changes how fast equilibrium is reached, never where it lies.
Key concept
Direction independence
A system reaches the same K whether it starts from pure reactants or pure products.
Equation
Equilibrium constant
K = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ)
- [X] = equilibrium concentration in mol/L
- a, b, c, d = stoichiometric coefficients
Worked examples
Worked example 2
A sealed flask of H₂ and I₂ reaches equilibrium with HI. A student says the reaction has stopped. Correct the statement.
Try it first: Ask what would happen to a labelled iodine atom after equilibrium is reached.
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Worked example 3
Adding a catalyst to an equilibrium mixture: what happens to the time to reach equilibrium and to K?
Try it first: Recall which activation energies a catalyst lowers.
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Worked example 4
At equilibrium A ⇌ B, a small amount of isotopically labeled A is introduced without materially disturbing the mixture. Will labeled B eventually appear?
Try it first: Ask whether the forward reaction stops at equilibrium.
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Common mistakes
Reading equilibrium as equal concentrations of reactants and products.
Why it's wrong: Equilibrium equalises rates, not amounts.
Check instead: Use K to judge which side is favored.
Saying the reaction stops at equilibrium.
Why it's wrong: Molecules keep converting in both directions; only the net change is zero.
Check instead: Describe it as no net change rather than no reaction.
Claiming a catalyst increases the yield at equilibrium.
Why it's wrong: It accelerates both directions equally.
Check instead: Change temperature if you want to change K.
Thinking K depends on the starting concentrations.
Why it's wrong: Different starting amounts settle to different concentrations but the same ratio.
Check instead: Remember K is fixed by temperature alone.
After the lesson · Question bank
Independent practice
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What you should now know
At equilibrium the forward and reverse reactions occur at equal rates, so concentrations stay constant while molecules keep reacting. This is why equilibrium is called dynamic rather than static. The equilibrium constant K is a fixed number at a given temperature that compares products to reactants. A large K means products dominate at equilibrium, a small K means reactants do, and K near 1 means both are present in comparable amounts. K depends only on temperature: changing concentrations, pressures, or adding a catalyst shifts the position of equilibrium but never changes the value of K.
- Equilibrium is dynamic: equal rates, no net change
- K compares products to reactants at a fixed temperature
- Large K is product-favored, small K is reactant-favored
- Only temperature changes the value of K
- A catalyst changes the speed, never the position
Sources and further reading
This lesson is original Chem Help content. No external sources were adapted.