Equilibrium Class 11 Notes: Kc, Kp, pH and Le Chatelier

These equilibrium class 11 notes give you the whole Chapter 6 in one revision pass: dynamic equilibrium, the equilibrium law (Kc), Kp and the reaction quotient Q, Le Chatelier’s principle, and ionic equilibrium — acids and bases, Kw, the pH scale, Ka and Kb.

Key definitions, every formula, four solved numericals with fresh numbers and the mistakes that cost marks are all here, each linked to its NCERT page.

Start with the revision map to see how the chapter splits into its two halves (chemical equilibrium and ionic equilibrium). Then work through Kc, Q, Kp and Le Chatelier’s principle before moving to pH and Ka/Kb. Finish with the common-mistakes table and exam notes — those two target exactly where Chapter 6 answers usually lose marks.

What Equilibrium Class 11 Notes Cover: A Revision Map

The chapter builds in two clear halves. The first half (pp 169-189) is chemical equilibrium — rates, the equilibrium law, Kc and Kp, then Le Chatelier’s principle. The second half (pp 189-199) is ionic equilibrium — electrolytes, acid-base theories, Kw, pH and the ionization constants Ka and Kb.

The unit objectives (p 169) also list buffer solutions and solubility product, but this revision page covers the grounded content through weak-base ionization. Every page reference below comes from the rationalised NCERT text; you can open the official NCERT Chapter 6 Equilibrium PDF to verify any detail directly.

Why Equilibrium Is Dynamic: Physical and Chemical Equilibrium

Equilibrium does not mean the reaction has stopped. It means the forward and reverse rates are equal, so concentrations stay constant while both reactions keep running (NCERT, p. 169). This is why it is called a dynamic equilibrium, and why the mixture of reactants and products is called an equilibrium mixture.

A good image: a tank with an inlet pipe and an outlet pipe. Water pours in and drains out at the same rate, so the water level never changes — yet exchange is continuous. The water level is the concentration; the constant inflow and outflow are the two reactions. That is dynamic equilibrium.

The chapter first shows equilibrium in physical processes (phase changes), where one measurable property holds steady (NCERT, p. 170). The four cases, summarised from the book’s Table 6.1:

Three key examples to quote in answers:

• Solid-liquid: ice and water at 273 K and 1 atm are in equilibrium; the temperature at which the two coexist at 1 atm is the normal melting point (or freezing point) (NCERT, p. 170).
• Liquid-vapour: the constant pressure of vapour above a liquid is its equilibrium vapour pressure; the temperature at which liquid and vapour balance at 1.013 bar is the normal boiling point. More volatile liquids have higher vapour pressure and lower boiling point (NCERT, p. 171).
• Solid-vapour: iodine, camphor and NH4Cl sublimate to give an equilibrium vapour (NCERT, p. 171).

Two more physical cases matter: a saturated solution is confirmed dynamic by the radioactive-sugar experiment — dropping radioactive sugar into a saturated non-radioactive solution spreads radioactivity into both phases (NCERT, p. 172).

And Henry’s law governs a gas dissolving in a liquid: the mass of gas dissolved is proportional to the pressure of the gas above the solvent — which is why an opened soda water bottle goes flat (NCERT, p. 172).

General characteristics of physical equilibria (NCERT, p. 173):

• Possible only in a closed system at a given temperature.
• Opposing processes run at the same rate — dynamic but stable.
• All measurable properties of the system stay constant.

For chemical equilibrium, the same idea holds: when the forward and reverse reaction rates match, concentrations become constant (NCERT, p. 173). Equilibrium can be reached from either direction — starting from pure H2 + I2 or from pure HI gives the same equilibrium mixture for H2(g) + I2(g) ⇌ 2HI(g) (NCERT, p. 175).

Haber’s deuterium experiment proves reactions keep running: mixing equilibrium mixtures made with H2 and with D2 scrambles the isotopes into NH3, NH2D, NHD2 and ND3 — impossible if the reaction had truly stopped (NCERT, p. 173-174).

Writing Kc: The Equilibrium Law and Its Rules

The equilibrium law (also the law of mass action, since old chemists called concentration “active mass”) states: at a given temperature, the product of product concentrations raised to their coefficients, divided by product of reactant concentrations raised to their coefficients, is a constant (NCERT, p. 176).

For a general reaction \( aA + bB \rightleftharpoons cC + dD \):

\[ K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b} \]

Why coefficients become exponents: in six experiments on H2(g) + I2(g) ⇌ 2HI(g) at 731 K, the unsquared ratio \( [HI]/([H_2][I_2]) \) wandered from 790 to 1970, while the squared ratio \( [HI]^2/([H_2][I_2]) \) stayed flat at about 46.4-47.6 in every case (NCERT, p. 176). The data itself forces the exponents to match the coefficients.

Rules that govern writing Kc (NCERT, p. 177):

• Concentrations in the expression are equilibrium values (the “eq” subscript is dropped by convention).
• Stoichiometric coefficients become the exponents — writing \( [NH_3]/([N_2][H_2]) \) instead of \( [NH_3]^2/([N_2][H_2]^3) \) is wrong.
• Phase symbols (s, l, g) are ignored when writing the expression.
• The reverse reaction has \( K’ = 1/K_c \).
• Multiplying the balanced equation by \( n \) gives \( K_c^n \).

Because rearranging the equation changes the number, you must always state which balanced equation a quoted K belongs to (NCERT, p. 177). For units: Kc and Kp are treated as dimensionless by using standard states (1 M for solutes, 1 bar for gases); otherwise Kc carries units such as (mol/L)^Δn (NCERT, p. 181).

Kp and Kc: Gaseous Equilibria and the RT Link

For gases it is often more convenient to write the equilibrium constant in partial pressures, called Kp. Start from the ideal gas law \( pV = nRT \), so \( p = (n/V)RT = cRT \): at fixed temperature, pressure of a gas is proportional to its concentration (NCERT, p. 178). With \( R = 0.0831 \) bar L mol⁻¹ K⁻¹.

For \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \), substituting \( p = [gas]RT \) into Kp cancels every RT factor, giving Kp = Kc. That happens only when \( \Delta n = 0 \).

The general relation (NCERT, p. 179):

\[ K_p = K_c (RT)^{\Delta n}, \quad \Delta n = (\text{moles of gaseous products}) – (\text{moles of gaseous reactants}) \]

Count gases only — solids and liquids never enter \( \Delta n \). Book’s Kp values show strong temperature dependence: for \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \), Kp drops from \( 6.8 \times 10^5 \) at 298 K to \( 3.6 \times 10^{-2} \) at 500 K; for \( N_2O_4 \rightleftharpoons 2NO_2 \), Kp rises from 0.98 to 1700 over the same span (NCERT, p. 179).

Express pressure in bar because the standard state for pressure is 1 bar (1 bar = 10^5 Pa). The ideal gas link and its units come from the Some Basic Concepts of Chemistry notes if you need a refresh.

Heterogeneous Equilibria: Why Pure Solids and Liquids Stay Out of K

A heterogeneous equilibrium involves more than one phase — for example H2O(l) ⇌ H2O(g), or a solid in contact with its saturated solution (NCERT, p. 180).

The rule that removes solids and pure liquids from K: the molar concentration of a pure solid or pure liquid is constant, independent of how much is present, so it is absorbed into the equilibrium constant (NCERT, p. 180).

The classic example is thermal dissociation of calcium carbonate:

\[ CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g) \]

Since both [CaCO3(s)] and [CaO(s)] are constant, they fold into K, leaving (NCERT, p. 180-181):

\[ K_p = p_{CO_2} \]

Experimentally, at 1100 K the pressure of CO2 in equilibrium with CaCO3(s) and CaO(s) is \( 2.0 \times 10^5 \) Pa, so \( K_p = (2.0 \times 10^5\ \text{Pa})/(10^5\ \text{Pa}) = 2.00 \) — dimensionless after dividing by the 1 bar standard state (NCERT, p. 181). Two more grounded examples:

• Nickel purification: \( Ni(s) + 4CO(g) \rightleftharpoons Ni(CO)_4(g) \), give \( K_c = [Ni(CO)_4]/[CO]^4 \).
• \( Ag_2O(s) + 2HNO_3(aq) \rightleftharpoons 2AgNO_3(aq) + H_2O(l) \), give \( K_c = [AgNO_3]^2/[HNO_3]^2 \).

The reminder that loses marks: for the heterogeneous equilibrium to exist, the pure solids or liquids must actually be present at equilibrium, however small the amount — but their concentrations or partial pressures never appear in the expression (NCERT, p. 181).

Using K: Extent, Direction (Q) and Equilibrium Concentrations

The equilibrium constant does three practical jobs (NCERT, p. 182). First, its magnitude predicts extent of reaction:

• \( K_c \gt 10^3 \): products predominate; the reaction runs nearly to completion. Example: \( H_2 + Cl_2 \rightleftharpoons 2HCl \) at 300 K has \( K_c = 4.0 \times 10^{31} \) (NCERT, p. 182).
• \( K_c \lt 10^{-3} \): reactants predominate; products barely form. Example: \( N_2 + O_2 \rightleftharpoons 2NO \) at 298 K has \( K_c = 4.8 \times 10^{-31} \) (NCERT, p. 182).
• \( 10^{-3} \lt K_c \lt 10^3 \): both reactants and products are present in comparable amounts. Example: \( H_2 + I_2 \rightleftharpoons 2HI \) at 700 K has \( K_c = 57.0 \) (NCERT, p. 182).

Second, K predicts direction through the reaction quotient Q. Q has the identical form as K but uses concentrations at any instant, not necessarily equilibrium (NCERT, p. 183):

• \( Q \lt K \): net reaction runs forward (left to right).
• \( Q \gt K \): net reaction runs reverse (right to left).
• \( Q = K \): the mixture is at equilibrium, no net change.

A finish-line image: K is the finish line; Q is where the reaction stands at this moment. If Q is before the line (Q < K), the reaction keeps running forward to reach it; if Q has overshot (Q > K), it must run backward; once Q lands exactly on K, the race is over — no net reaction.

Third, K finds equilibrium concentrations by the ICE method (NCERT, p. 184):

1. Write the balanced equation.
2. Build a table: Initial concentrations, Change to reach equilibrium in terms of \( x \), Equilibrium concentrations.
3. Substitute equilibrium concentrations into the K expression and solve for \( x \).
4. If a quadratic arises, reject the root that gives a physically impossible (negative, or larger-than-initial) concentration.
5. Check the answer by substituting back into the equilibrium equation.

K, Q and Gibbs Energy: The Thermodynamic Connection

As you saw in Unit 5 thermodynamics, equilibrium is linked to free energy change (NCERT, p. 185):

\[ \Delta G = \Delta G^\ominus + RT \ln Q \]

At equilibrium \( \Delta G = 0 \) and \( Q = K \), which gives:

\[ \Delta G^\ominus = -RT \ln K, \qquad K = e^{-\Delta G^\ominus / RT} \]

The interpretation (NCERT, p. 185):

• \( \Delta G^\ominus \lt 0 \Rightarrow K \gt 1 \): forward reaction is spontaneous, products dominate.
• \( \Delta G^\ominus \gt 0 \Rightarrow K \lt 1 \): reactants dominate, only a minute amount of product forms.

If the free-energy reasoning feels shaky, the Class 11 Thermodynamics notes on this site re-teach the \( \Delta G \) sign conventions from Unit 5.

Le Chatelier’s Principle: How Equilibrium Responds to Stress

Le Chatelier’s principle: a change in any factor determining the equilibrium conditions of a system will cause the system to shift so as to counteract or reduce that change (NCERT, p. 186). It applies to all physical and chemical equilibria. The consolidated factor table — everything examiners ask about at once:

Why each rule holds:

• Concentration (NCERT, p. 186): adding H2 to H2 + I2 ⇌ 2HI makes \( Q_c = [HI]^2/([H_2][I_2]) \) drop below Kc, so the reaction runs forward until Q returns to K. Removing CO2 from the CaCO3 kiln keeps Q below K and drives lime production to completion.
• Pressure (NCERT, p. 187): compression raises all partial pressures; the system relieves the stress by moving toward the side with fewer gas moles. Solids and liquids are ignored because their volume is nearly pressure-independent.
• Inert gas (NCERT, p. 188): at constant volume, adding argon changes no partial pressures, so equilibrium is undisturbed.
• Temperature (NCERT, p. 188): unlike other stresses, temperature changes the value of Kc itself. For the exothermic \( N_2 + 3H_2 \rightleftharpoons 2NH_3 \) (\( \Delta H = -92.38 \) kJ mol⁻¹), raising temperature shifts the equilibrium left and lowers ammonia yield.
• Catalyst (NCERT, p. 189): it lowers the activation energy for forward and reverse by exactly the same amount, speeds both rates equally, and never appears in the balanced equation or in K — so composition is unaffected.

Two lab demonstrations to quote. The blood-red \( Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons [Fe(SCN)]^{2+}(aq) \) equilibrium: adding thiocyanate deepens the red (shift right); adding oxalic acid (binds Fe³⁺ as the stable \( [Fe(C_2O_4)_3]^{3-} \)) or HgCl2 (binds SCN⁻ as \( [Hg(SCN)_4]^{2-} \)) removes ions and shifts the equilibrium left, fading the colour (NCERT, p. 187).

And \( 2NO_2(g) \rightleftharpoons N_2O_4(g) \), \( \Delta H = -57.2 \) kJ mol⁻¹: a freezing mixture makes the brown NO2 fade (forward, colourless N2O4), hot water intensifies the brown (reverse) (NCERT, p. 188).

Industrial payoffs: Haber’s ammonia synthesis runs at about 500 °C, 200 atm with an iron catalyst — low temperature favours yield but slows the rate, so the catalyst and high pressure buy both speed and yield (NCERT, p. 189).

The contact process for sulphuric acid uses V2O5 (or platinum) because even with a huge K, the SO2 → SO3 oxidation is otherwise too slow; but a catalyst is useless when K itself is tiny (NCERT, p. 189).

A real-life application the chapter opens with: equilibria involving O2 and haemoglobin control oxygen transport from lungs to muscles. Where partial pressure of O2 is high (lungs), O2 binds; in tissues where pO2 is low, O2 is released.

Carbon monoxide binds haemoglobin far more strongly than O2, effectively removing the oxygen carrier from the equilibrium and blocking oxygen delivery — which is why CO is toxic (NCERT, p. 169). Le Chatelier’s principle explains the whole chain: raising pO2 in the lungs pushes O2 onto the carrier; CO hijacks the same site.

Ionic Equilibrium: Electrolytes and Three Concepts of Acids and Bases

An electrolyte is a substance whose aqueous solution conducts electricity; a non-electrolyte (like sugar solution) does not. Faraday split electrolytes into strong (almost completely ionised, e.g. NaCl ~100%) and weak (partially dissociated, e.g. acetic acid under 5%) (NCERT, p. 189). Ionic equilibrium is the equilibrium established between ions and unionised molecules in such a solution.

Why NaCl dissolves so completely: water’s dielectric constant is high (~80), so the electrostatic pull between Na⁺ and Cl⁻ is reduced by a factor of about 80, and the ions are further stabilised by hydration (NCERT, p. 190).

The three acid-base concepts — compare them side by side, including each one’s limitation:

Conjugate pairs differ by exactly one proton (NCERT, p. 191). The conjugate base of an acid is the acid minus one H⁺; the conjugate acid of a base is the base plus one H⁺. Examples from the book’s problems:

• Conjugate bases: HF → F⁻; H2SO4 → HSO4⁻; HCO3⁻ → CO3²⁻ (NCERT, p. 192).
• Conjugate acids: NH2⁻ → NH3; NH3 → NH4⁺; HCOO⁻ → HCOOH (NCERT, p. 192).
• Water is amphiprotic: with HCl it is a base (accepts proton to form H3O⁺); with NH3 it is an acid (donates a proton) (NCERT, p. 191).

A strong acid has a very weak conjugate base, and a strong base has a very weak conjugate acid, because the stronger acid donates its proton to the stronger base (NCERT, p. 193).

Memory device for the three concepts — A-B-L: Aqueous for Arrhenius (works only in water), B for proton (Brønsted-Lowry is all about the proton), L for Lone pair (Lewis is all about the electron lone pair).

Water, Kw and the pH Scale

Water self-ionises because one water molecule donates a proton and another accepts it (NCERT, p. 194):

\[ H_2O(l) + H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq) \]

The ionic product of water is (NCERT, p. 194):

\[ K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}\ \text{M}^2 \text{ at 298 K} \]

Water is a pure liquid, so its concentration is constant and already folded into Kw. In pure water \( [H^+] = [OH^-] = 1.0 \times 10^{-7} \) M. Kw is temperature dependent, being an equilibrium constant. The molarity of pure water is 55.55 M, so only about 2 molecules in every 10^9 are dissociated — equilibrium lies overwhelmingly toward undissociated water (NCERT, p. 194).

The pH scale is a log scale for hydronium-ion concentration (NCERT, p. 194):

\[ pH = -\log[H^+] \]

• Acidic: \( [H_3O^+] \gt [OH^-] \), so pH < 7.
• Neutral: \( [H_3O^+] = [OH^-] \), pH = 7.
• Basic: \( [H_3O^+] \lt [OH^-] \), pH > 7.

Taking the negative log of the Kw expression gives (NCERT, p. 195):

\[ pK_w = pH + pOH = 14 \]

Because the scale is logarithmic, one pH unit means a 10× change in \( [H^+] \), and a 100× change in \( [H^+] \) shifts pH by 2 units. Familiar values from the book’s table (NCERT, p. 195): gastric juice ~1.2, lemon juice ~2.2, milk 6.8, human blood 7.4, sea water (and egg white) 7.8. Measure pH with pH paper (accuracy ~0.5) or a pH meter (precision ~0.001).

The dilution trap: a \( 1 \times 10^{-8} \) M HCl solution has pH ≈ 6.98, not 8, because at such extreme dilution the H⁺ contributed by water itself (\( 9.5 \times 10^{-8} \) M of OH⁻, giving pH 6.98) can no longer be ignored (NCERT, p. 195-196).

Weak Acids and Weak Bases: Ka, Kb and Degree of Ionisation

For a weak acid HA dissociating as \( HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq) \), the acid ionization constant is (NCERT, p. 196):

\[ K_a = \frac{[H^+][A^-]}{[HA]} = \frac{c\alpha^2}{1-\alpha} \]

where \( c \) is the initial acid concentration and \( \alpha \) is the degree of ionization (the fraction that ionises). Larger Ka means a stronger acid; \( pK_a = -\log K_a \) (NCERT, p. 196). Strong acids (HCl, HBr, HI, HNO3, H2SO4, HClO4) ionise almost completely; typical weak acids are HF (\( 3.5 \times 10^{-4} \)), acetic acid (\( 1.74 \times 10^{-5} \)) and HCN (\( 4.9 \times 10^{-10} \)).

For a weak base MOH, the base ionization constant mirrors it exactly (NCERT, p. 198):

\[ K_b = \frac{[M^+][OH^-]}{[MOH]} = \frac{c\alpha^2}{1-\alpha}, \qquad pK_b = -\log K_b \]

Ammonia has \( K_b = 1.77 \times 10^{-5} \); amines (methylamine, quinine, nicotine) are very weak bases with tiny Kb (NCERT, p. 198-199). The measure of strength is (NCERT, p. 197):

\[ \text{Percent dissociation} = \frac{\text{concentration dissociated}}{\text{initial concentration}} \times 100 \]

When to simplify: if K is small, \( x \) is negligible next to \( c \), so \( 1-\alpha \) (or \( c-x \)) ≈ 1 (or \( c \)) and \( [H^+] = \sqrt{K_a \cdot c} \) — the HOCl-style problem works this way.

When the acid is strong enough that the approximation fails, you must solve the full quadratic — the HF problem (\( K_a = 3.2 \times 10^{-4} \), 0.02 M) keeps the \( 1-\alpha \) term and gets \( \alpha = 0.12 \) (NCERT, p. 197).

Common-ion effect: in a solution that is 0.1 M NH3 and 0.2 M NH4Cl, the added NH4⁺ (from the salt) suppresses NH3 ionisation, giving \( [OH^-] \approx 0.88 \times 10^{-5} \) M and pH = 8.95 (NCERT, p. 199). The same ICE structure applies — the salt simply adds an initial NH4⁺ concentration that is no longer zero.

(Buffer solutions and solubility product are listed in the unit objectives, p 169, but lie beyond this grounded extract.)

Key Terms at a Glance: Equilibrium Definitions Table

One-stop glossary, written for quick scanning. The textbook page after each term is where the book defines it.

Equilibrium Formula Sheet: Symbols, Units and When to Use Each

Every formula in one table. The warning that applies to all: Kc and Kp are taken as dimensionless using standard states (1 M, 1 bar).

Worked Examples: Four Equilibrium Numericals Solved Step by Step

Name the method first, then follow the steps — units at every substitution. All numbers here are fresh; the patterns mirror the book’s problems.

Example 1: Calculating Kc from equilibrium concentrations

Example 2: Kp from Kc using Δn

Example 3: Predicting direction with Q

Example 4: Weak-acid pH and percent dissociation

Common Mistakes in Equilibrium (and the Correct Version)

Exam Notes: What Earns the Mark in Equilibrium

Observed patterns from how the textbook structures its problems — the steps that conventionally earn credit:

• Writing the Kc expression exactly from the balanced equation is the first earned mark — coefficients as exponents, solids and pure liquids omitted.
• For Kp conversions, compute Δn from gaseous species only, then apply \( K_p = K_c(RT)^{\Delta n} \). Stating Δn explicitly earns the step.
• In quadratic problems, the mark-earning step is rejecting the impossible root (negative, or larger than the initial concentration) with a one-line justification, exactly as the book’s CO/H2O and N2O4 problems do (NCERT, p. 184).
• For pH, show the log-splitting step: \( -\log(3.8 \times 10^{-3}) = -(\log 3.8 + \log 10^{-3}) = -\{0.58 + (-3.0)\} = 2.42 \) (NCERT, p. 195). That intermediate line is where credit is given.
• Quote the \( 10^3 \) and \( 10^{-3} \) thresholds when judging extent of reaction (NCERT, p. 182).
• Form conjugate pairs by removing or adding one proton — HF → F⁻; NH3 → NH4⁺ (NCERT, p. 192).
• State that temperature is the only factor that changes K; concentration, pressure and catalyst do not.
• The Haber compromise reasoning: low temperature favours yield but slows the rate, so an iron catalyst plus about 200 atm pressure is used (NCERT, p. 189).
• Descriptive answers often reward figure-based reasoning — equilibrium reached from either direction, or the NO2/N2O4 colour switch with temperature.

Reading the Graphs: What Each Equilibrium Figure Shows

No image assets are available from the grounding, so here is what each named figure demonstrates, in words, based only on the book’s captions.

• Fig 6.1 (p 170): a manometer U-tube with mercury records the pressure inside a closed box as water evaporates. The mercury level climbs, then holds constant — the steady reading is the equilibrium vapour pressure of water at that temperature.
• Fig 6.2 (p 173): as a reaction A + B ⇌ C + D proceeds, reactant concentration falls and product concentration rises until both level off — the point where forward and reverse rates become equal.
• Fig 6.3 (p 174): two measuring cylinders joined by glass tubes transfer coloured water; after levels steady, transfer continues — visual proof that equilibrium is dynamic.
• Fig 6.4 (p 174): Haber’s N2 + 3H2 ⇌ 2NH3 — the mixture’s composition becomes constant at equilibrium even though unreacted H2 and N2 are still present.
• Fig 6.5 (p 175): H2 + I2 ⇌ 2HI reached starting from pure reactants or pure HI — the same equilibrium mixture results from either direction.
• Fig 6.6 (p 183): extent of reaction vs magnitude of Kc — large Kc means products dominate, small Kc means reactants dominate.
• Fig 6.7 (p 183): predicting direction — comparing Q with K tells whether the net reaction moves left, right, or not at all.
• Fig 6.8 (p 186): adding H2 to the H2 + I2 ⇌ 2HI mixture jumps the H2 level, then all three concentrations re-settle as equilibrium re-establishes.
• Fig 6.9 (p 188): NO2/N2O4 in three baths — brown NO2 fades in the freezing mixture (forward, colourless N2O4 favoured) and intensifies in hot water (reverse).
• Fig 6.10 (p 190): NaCl dissolving — Na⁺ and Cl⁻ ions leave the crystal and are stabilised by hydration with polar water molecules.
• Fig 6.11 (p 195): pH paper with four strips that show different colours at the same pH, giving ~0.5 accuracy.

NCERT Activities That Prove Equilibrium Is Dynamic

These experiments back the words “dynamic” and “shift” — quick to revise and quotable in answers.

• Two-cylinder transfer (p 174, Fig 6.3): coloured water is transferred between two measuring cylinders; the levels become constant while the transfer continues. The ‘level’ stands for concentration, so the process is dynamic.
• Haber’s deuterium experiment (p 173-174): mixing equilibrium mixtures made with H2 and with D2 scrambles isotopes into NH3, NH2D, NHD2 and ND3 — forward and reverse reactions clearly keep running at equilibrium.
• Fe³⁺ + SCN⁻ blood-red equilibrium (p 187): added thiocyanate deepens the red; oxalic acid (binds Fe³⁺) or HgCl2 (binds SCN⁻) removes ions and shifts the equilibrium left, fading the colour.
• 2NO2(g) ⇌ N2O4(g) (p 188, Fig 6.9): the brown colour fades in a freezing mixture and intensifies when heated — a direct visual of temperature shifting an exothermic equilibrium.
• Cobalt complex (p 188-189): \( [Co(H_2O)_6]^{3+} \rightleftharpoons [CoCl_4]^{2-} + 6H_2O \) — blue at room temperature, turning pink when cooled in a freezing mixture.

Equilibrium in One Page: Quick Revision Recap

The whole chapter as a flow:

• Equilibrium in a closed system is dynamic — forward and reverse rates equal, concentrations constant.
• The equilibrium law gives Kc: coefficients become exponents over equilibrium concentrations.
• For gases, Kp = Kc(RT)^Δn; Δn counts gaseous species only.
• Compare Q with K to predict direction; the ICE method finds equilibrium concentrations.
• Le Chatelier’s principle predicts shifts for concentration, pressure and temperature; a catalyst changes only speed, never K.
• Ionic equilibrium: acids and bases by Arrhenius, Brønsted-Lowry, Lewis; conjugate pairs differ by one proton.
• Kw = 1.0 × 10⁻¹⁴, pH + pOH = 14; pH < 7 acid, > 7 base at 25 °C.
• Weak acids/bases: Ka = cα²/(1−α), pKa, percent dissociation; strong acid → weak conjugate base.

Must-memorise numbers:

For full chapter context and other units, browse the Class 11 Chemistry notes hub on this site.

Frequently Asked Questions About Equilibrium

Why is chemical equilibrium called dynamic even though the concentrations stay constant?

Because the forward and reverse reactions never stop — they run at equal rates, so the concentrations happen to stay constant (NCERT, p. 169). The proof is Haber’s deuterium experiment: isotope scrambling in NH3 is impossible unless reactions continue at equilibrium (NCERT, p. 173-174).

When is Kp equal to Kc?

Only when \( \Delta n = 0 \), that is, when the moles of gaseous products equal the moles of gaseous reactants, as in \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \). The RT factors in Kp cancel one-for-one, leaving Kp = Kc (NCERT, p. 178-179).

What is the difference between the reaction quotient Q and the equilibrium constant K?

They have the same algebraic form, but Q uses concentrations at any arbitrary instant while K uses only equilibrium concentrations. Comparing them — Q < K runs forward, Q > K runs reverse, Q = K at equilibrium — predicts the reaction’s direction (NCERT, p. 183).

Why do pure solids and pure liquids not appear in the equilibrium constant expression?

Because the molar concentration of a pure solid or pure liquid is constant, independent of the amount present, so it is absorbed into K. For CaCO3(s) ⇌ CaO(s) + CO2(g), that leaves Kp = pCO2 (NCERT, p. 180-181).

Does a catalyst change the equilibrium constant?

No. A catalyst lowers the activation energy for forward and reverse by the same amount, so it speeds both equally, shifts nothing, and never appears in the balanced equation or in K (NCERT, p. 189).

What is the pH of a 1 × 10⁻⁸ M HCl solution, and why is it not 8?

About 6.98. At that extreme dilution the H⁺ from water’s own ionisation (\( 9.5 \times 10^{-8} \) M) is comparable to the acid’s \( 10^{-8} \) M, so both sources must be added: solving \( (10^{-8} + x)(x) = 10^{-14} \) gives pOH 7.02 and pH 6.98 (NCERT, p. 195-196).

Reference: NCERT Class 11 Chemistry textbook, chapter Equilibrium.