Equilibrium · Lesson 2 of 14
How Chemical Equilibrium Is Established
“Follow reacting particles and concentration graphs to understand why chemical equilibrium remains active.”
• Explain how the forward and reverse rates approach equality. • Interpret concentration–time graphs without assuming equal concentrations. • Use water-transfer and isotope evidence to explain dynamic behaviour. • State the conditions under which equilibrium can be approached from either direction.
A reverse reaction becomes possible as products accumulate
Imagine a closed vessel initially containing only reactants A and B. They form products C and D. Once products exist, collisions between them can regenerate the reactants. A reversible reaction is represented by a double arrow because both chemical directions are possible under the conditions being considered. The net change depends on the difference between their rates.
At first, there may be little or no reverse reaction because little product is present. As products accumulate and reactants are consumed, the forward rate commonly falls and the reverse rate rises. Eventually they are equal. The concentrations then remain constant because the amount consumed per unit time is replenished at the same rate. Constancy does not mean that the reactants have all disappeared.
A dynamic state in which the forward and reverse reaction rates are equal and the macroscopic composition remains constant under unchanged conditions.
Some equilibria leave mostly reactants; others leave mostly products; still others contain substantial amounts of both. These are different extents of reaction, not different definitions of equilibrium. Industrial conditions are chosen partly to favour the useful product, but a good final yield and a fast reaction are separate requirements.
Problem
A reaction forms 0.030 mol of product each minute while the reverse reaction consumes 0.020 mol each minute. Is it at equilibrium?
- 1.Compare the two simultaneous rates: .
- 2.The net production is .
- 3.The composition is still changing. Equilibrium requires equal forward and reverse rates, although each rate may remain nonzero.
A water-transfer model makes the balance visible
Use two measuring cylinders and two narrow tubes. Begin with coloured water in the first cylinder and none in the second. Transfer a tube-full in each direction during each cycle. The volume carried by a tube depends on how deeply it is immersed, so the higher water level initially produces a larger transfer. Repeating the cycle gradually establishes levels at which the two opposite transfers are equal.
At that stage, continue transferring water. The levels remain steady even though water is moving both ways. If the tubes have different diameters, the final levels can differ because different column heights can give equal transferred volumes. This is an analogy for equal reaction rates with unequal concentrations. It is a model, so it does not claim that actual molecular reaction rates obey the same tube geometry.
Problem
In the transfer model one tube has twice the cross-sectional area of the other. What condition is needed for equal transferred volumes?
- 1.A water column carried by a tube has volume equal to cross-sectional area multiplied by immersed height.
- 2.If the first area is and the second is , equal carried volumes require .
- 3.Thus . The steady levels need not be equal, just as equilibrium concentrations need not be equal.
Isotopes reveal reactions that concentration measurements hide
Ammonia synthesis provides direct chemical evidence. One mixture reaches equilibrium with ordinary hydrogen, H2, nitrogen and ammonia. Another reaches equilibrium using deuterium, D2, producing deuterated ammonia, ND3. Deuterium is an isotope of hydrogen: it has the same element identity but a different nuclear mass. The mixtures can be followed by mass spectrometry because molecules containing different numbers of D atoms have different masses.
After the equilibrated mixtures are combined under suitable conditions, mixed species such as NH2D, NHD2 and HD appear. The observed composition can remain approximately steady while atoms exchange among molecules. If every reaction had stopped at equilibrium, these mixed molecules could not have formed. The isotope experiment supports continuing forward and reverse reactions. Isotope substitution can slightly change quantitative equilibrium values; the evidence here is the exchange itself.
Problem
After H2/NH3 and D2/ND3 equilibrium mixtures are combined, NH2D is detected. What does this establish?
- 1.NH2D contains both ordinary hydrogen and deuterium in the same ammonia molecule.
- 2.Its formation requires atoms from the original molecules to be redistributed by chemical reactions.
- 3.The result rules out the claim that every reaction stops when the macroscopic equilibrium composition is reached.
Equilibrium can be approached from either direction
For hydrogen reacting with iodine, a mixture initially containing H2 and I2 forms HI until the concentrations level off. Starting instead with HI allows decomposition to form H2 and I2. At the same temperature and volume, with the same total numbers of hydrogen and iodine atoms and the same accessible phases, both preparations approach the same equilibrium composition. Merely saying “same temperature” is insufficient if the atom inventories are different.
On a concentration–time graph, equilibrium is shown by horizontal concentration curves. Their heights may differ. The curve for each species must reflect the balanced equation: forming two moles of HI consumes one mole each of H2 and I2. A concentration graph describes amounts per volume; a rate graph describes how fast they change. Confusing their vertical axes causes many errors.
Reversible binding is also important in living systems. Oxygen binds to haemoglobin in the lungs and can be released where oxygen is needed. Carbon monoxide competes strongly for haemoglobin binding, disrupting oxygen transport. These applications show why changing concentrations can have important consequences in an equilibrium system; they do not mean every biological process is itself at equilibrium.
Equal forward and reverse rates do not imply that each reaction is complete, that all concentrations are equal, or that all molecules have equal energies. A catalyst may help a system reach equilibrium faster without determining its final composition.
Quiz
Which observation alone is consistent with chemical equilibrium in a closed system at fixed conditions?
What does the formation of NH2D after isotope mixtures are combined demonstrate?
On a concentration–time graph, equilibrium is indicated by
In the cylinder analogy, unequal steady water levels show that
Which condition must accompany fixed temperature and volume when comparing equilibria reached from opposite directions?
If forward and reverse rates are both 0.020 mol per minute, the net rate is
Practice Problems
- Draw and label separate qualitative rate–time and concentration–time graphs for a reversible reaction initially containing only reactants.
- Explain why ammonia and its reactants can remain together at equilibrium.
- Describe the two-cylinder activity and the observation that makes it a model of dynamic equilibrium.
- Explain the significance of NH2D and HD in the isotope experiment.
- For H2 + I2 forming 2HI, 0.030 mol of iodine is consumed. How much HI is formed if no other process occurs?
- State the conditions under which reactant-only and product-only preparations can approach the same equilibrium composition.
Key Takeaways
At chemical equilibrium both directions continue at equal rates. Concentrations become constant but generally remain unequal. Transfer models, isotope exchange and approach from opposite directions provide complementary evidence.