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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What you'll be able to do: Describe what is happening at the particle level at equilibrium and interpret the size of K.
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.
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.
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.
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)
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.
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 state where the forward and reverse reactions proceed at equal rates, so concentrations remain constant.
The fixed ratio of product to reactant terms, each raised to its stoichiometric coefficient, at a given temperature.
Concentration, pressure and catalysts change the position of equilibrium but not the value of K.
A catalyst changes how fast equilibrium is reached, never where it lies.
A system reaches the same K whether it starts from pure reactants or pure products.
K = ([C]ᶜ [D]ᵈ) / ([A]ᵃ [B]ᵇ)
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.
0 of 2 steps revealed.
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.
0 of 3 steps revealed.
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.
0 of 3 steps revealed.
Why it's wrong: Equilibrium equalises rates, not amounts.
Check instead: Use K to judge which side is favored.
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.
Why it's wrong: It accelerates both directions equally.
Check instead: Change temperature if you want to change K.
Why it's wrong: Different starting amounts settle to different concentrations but the same ratio.
Check instead: Remember K is fixed by temperature alone.
No practice questions are available for this topic yet. You can still practice the whole unit.
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.
This lesson is original Chem Help content. No external sources were adapted.