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Loading home…Chemical equilibrium is the state of a reversible reaction in which the forward rate equals the reverse rate.
At equilibrium, macroscopic concentrations of reactants and products remain constant.
The system is dynamic: molecules still react in both directions.
Equilibrium can be reached from either reactant-rich or product-rich starting mixtures.
It is characterised by a constant Kc (or Kp) at a fixed temperature.
Changing conditions can shift the position of equilibrium according to Le Chatelier’s principle.
At the microscopic level, reactant collisions still form products and product collisions reform reactants.
Dynamic means both processes continue at equal rates, so amounts appear unchanged.
Le Chatelier’s principle: if a system at equilibrium is disturbed, it shifts to partially oppose the disturbance.
Homogeneous equilibrium has all species in the same phase (e.g. all gases).
Heterogeneous equilibrium involves more than one phase (e.g. solid CaCO₃ ⇌ CaO(s) + CO₂(g)).
Pure solids and pure liquids usually do not appear in Kc expressions.
The Haber process manufactures ammonia: N₂ + 3H₂ ⇌ 2NH₃ (ΔH negative).
Ammonia is needed for fertilizers and many chemicals.
High pressure favours fewer gas moles and increases NH₃ yield.
Moderate temperature balances a usable rate against equilibrium yield (exothermic).
An iron catalyst speeds attainment of equilibrium without changing K.
Unreacted gases are recycled to improve overall conversion.
Suppose products are removed from A ⇌ B at equilibrium.
Forward collisions temporarily outnumber reverse ones because [B] fell.
Net conversion of A into B occurs until rates become equal again.
If more A is added, forward rate rises until a new balance is reached.
Kc remains the same if temperature is unchanged; only concentrations adjust.
The shift continues until Q equals K again.
Kc = [C]c [D]d[A]a [B]b, using equilibrium concentrations.
Concentrations are usually in mol dm⁻³.
A large Kc means products predominate at equilibrium.
A small Kc means reactants predominate.
Kc depends only on temperature for a given reaction.
Stoichiometric coefficients appear as exponents in the expression.
In homogeneous equilibrium all reactants and products are in one phase.
Example: H₂(g) + I₂(g) ⇌ 2HI(g).
In heterogeneous equilibrium more than one phase is present.
Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g).
For the carbonate case, Kc (or Kp) depends only on CO₂ concentration (or pressure).
Solids have constant activity and are omitted from the K expression.
Chemical equilibrium occurs in reversible reactions under closed conditions.
Forward and reverse rates become equal.
Macroscopic properties such as colour, pressure, and concentration become constant.
The equilibrium is dynamic at the molecular level.
It can be approached from either side of the reaction.
A catalyst does not change the equilibrium position; it only shortens the time to reach it.
The equilibrium constant Kc is the value of the concentration quotient at equilibrium.
For aA + bB ⇌ cC + dD, Kc = [C]c[D]d[A]a[B]b.
Kc is constant at constant temperature.
If Q < Kc, the forward reaction is favoured until equilibrium.
If Q > Kc, the reverse reaction is favoured.
If Q = Kc, the system is at equilibrium.
Kc has units depending on Δn of dissolved/gaseous species, or may be dimensionless in some conventions.
For A ⇌ B with Δn = 0, Kc is often treated as having no units.
Large Kc indicates a product-favoured equilibrium.
Small Kc indicates a reactant-favoured equilibrium.
Kc does not tell the rate of reaching equilibrium.
Changing pressure or concentration changes Q, not Kc (T fixed).
Only temperature changes the value of Kc.
Clear Kc reasoning is central to equilibrium calculations and Le Chatelier predictions.
Le Chatelier’s principle: a system at equilibrium responds to a stress by shifting to reduce that stress.
Concentration: adding a reactant drives the forward reaction; removing a product also drives forward.
Adding a product drives the reverse reaction.
Pressure (gases): increasing pressure favours the side with fewer gas moles.
Decreasing pressure favours more gas moles.
If Δn(g) = 0, pressure changes do not shift the composition significantly.
Temperature: for an exothermic forward reaction, heat is like a product; cooling favours products.
Heating an exothermic equilibrium favours reactants (K decreases).
For an endothermic forward reaction, heating favours products (K increases).
A catalyst speeds both directions equally.
Catalysts do not change Kc or the equilibrium yield.
They are vital industrially because they make equilibrium reachable in practical time.
Inert gas at constant volume does not change partial pressures of reactants, so no shift.
Inert gas at constant pressure dilutes and can shift toward more moles of gas.
Exam answers should specify the reaction and whether ΔH is positive or negative.
Always separate kinetic effects (rate) from equilibrium position (K).
Industrial processes combine pressure, temperature, and catalysts thoughtfully.
Le Chatelier is qualitative; exact yields need K and stoichiometry.
State clearly whether a change alters K or only the position (composition) at fixed K.
Industrial examples (Haber, Contact) reward this precise language.
Clear definitions followed by explained examples meet typical long-question mark schemes.
Clear definitions followed by explained examples meet typical long-question mark schemes.
The Haber process synthesises ammonia: N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
The forward reaction is exothermic.
Nitrogen comes from air; hydrogen from natural gas or other sources.
Stoichiometry shows 4 gas moles → 2 gas moles, so high pressure favours NH₃.
Typical industrial pressures are very high (historically around hundreds of atmospheres).
High pressure improves equilibrium yield of ammonia.
Because the reaction is exothermic, low temperature favours NH₃ in K terms.
Too low a temperature makes the rate impractically slow.
A compromise temperature (moderate, often cited near 400–450 °C) is used.
An iron catalyst (with promoters) increases rate without changing K.
Ammonia is removed by cooling/liquefaction, shifting equilibrium forward.
Unreacted N₂ and H₂ are recycled.
Le Chatelier explains why pressure and continuous product removal help.
Kinetics explains why a catalyst and adequate temperature are essential.
The process is foundational for fertilizers and global food production.
Safety and energy cost constrain how extreme the conditions can be.
Students should link equation, ΔH, Δn, catalyst, and recycling in one answer.
Haber is a classic case study of applied equilibrium chemistry.
Quote the balanced equation, sign of ΔH, and Δn(g) at the start of the answer.
Then apply Le Chatelier to pressure, temperature, and product removal in turn.
Clear definitions followed by explained examples meet typical long-question mark schemes.
Clear definitions followed by explained examples meet typical long-question mark schemes.
The Contact process produces sulphuric acid, a major industrial chemical.
Sulphur is burned in air to form SO₂: S + O₂ → SO₂.
The key equilibrium is 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (exothermic).
Δn(g) is negative, so high pressure favours SO₃, but atmospheric or slight excess pressure often suffices.
Excess air (O₂) and a vanadium(V) oxide catalyst are used.
A moderate temperature (about 450 °C) balances rate and yield.
Too high a temperature lowers K for this exothermic step.
SO₃ is absorbed not directly in water (mist problem) but in concentrated H₂SO₄.
Oleum (H₂S₂O₇) forms and is then diluted carefully to give H₂SO₄.
Le Chatelier supports using excess O₂ and removing SO₃ into absorption.
The catalyst ensures equilibrium is approached quickly.
Multiple catalyst beds with cooling improve conversion.
Sulphuric acid is used in fertilizers, chemicals, and metal processing.
Environmental control of SO₂ emissions is essential.
Contact process answers should name catalyst, temperature compromise, and absorption step.
Linking each condition to K, rate, or safety shows exam-level understanding.
The process remains a standard FBISE long-question topic.
Clear stepwise description plus equilibrium reasoning scores well.
Name V₂O₅, the SO₂/SO₃ equilibrium, and oleum absorption as three must-include points.
Connect each operating condition to either rate or equilibrium yield.
Clear definitions followed by explained examples meet typical long-question mark schemes.
Clear definitions followed by explained examples meet typical long-question mark schemes.
For gas-phase equilibria, concentrations and partial pressures are related by the ideal gas law.
For each gas, p = cRT (with consistent units), so c = pRT.
Substituting into Kc for aA + bB ⇌ cC + dD gives a relation to Kp.
Define Δn = (c + d) − (a + b) for gas moles only.
The standard result is Kp = Kc (RT)Δn.
R must match the pressure and concentration units used.
If Δn = 0, then (RT)Δn = 1, so Kp = Kc.
Example: H₂ + I₂ ⇌ 2HI has Δn = 0, hence Kp = Kc.
If Δn > 0, Kp > Kc (numerically, depending on units of R and T).
If Δn < 0, Kp < Kc in the same unit system.
Kp uses partial pressures; Kc uses concentrations.
Both are constant at constant temperature.
Heterogeneous equilibria may involve only the gaseous species in Kp.
Students should compute Δn carefully from the balanced equation.
A ⇌ B (gases) also has Δn = 0 and no units difference between Kp and Kc forms.
Temperature affects both Kp and Kc; their ratio still follows (RT)^Δn.
This relationship is frequently tested in MCQs and numericals.
State assumptions: ideal gases and equilibrium conditions.
Correct Δn identification is the key step in most problems.
Show Δn from the balanced equation before substituting into Kp = Kc(RT)Δn.
Clear definitions followed by explained examples meet typical long-question mark schemes.
Clear definitions followed by explained examples meet typical long-question mark schemes.