Chapter 7 — Equilibrium

Fig 1 — Educational Diagram: Introduction and basic concepts

What is chemical equilibrium?
Chemical equilibrium occurs in a reversible reaction when the rates of the forward and reverse reactions become equal, so the macroscopic properties (concentrations, pressure, colour) remain constant with time. Equilibrium is dynamic: reactions continue but with no net change.

Reversible reactions and closed system
Only reversible reactions (A ⇌ B) can reach equilibrium. Equilibrium must be established in a closed system (no matter exchange with surroundings); otherwise, concentrations do not remain constant.

Homogeneous and heterogeneous equilibria
A homogeneous equilibrium has all species in the same phase (e.g. gases). A heterogeneous equilibrium has species in different phases (e.g. solid + gas).

Writing the equilibrium constant
For a general reaction: aA + bB ⇌ cC + dD, the equilibrium constant in terms of molar concentrations (Kc) is

Kc = [C]^c [D]^d / ([A]^a [B]^b)

Pure solids and pure liquids are omitted from the expression. For gaseous systems you can write the equilibrium constant in terms of partial pressures (Kp):

Kp = (P_C)^c (P_D)^d / ((P_A)^a (P_B)^b)

Relation between Kp and Kc
Kp and Kc are related by the gas constant and temperature:

Kp = Kc (RT)^{Δn}

where Δn = (moles of gaseous products) − (moles of gaseous reactants), R = 0.08314 L·bar·mol⁻¹·K⁻¹ (or 0.08206 L·atm·mol⁻¹·K⁻¹ depending on pressure units), and T is in kelvin.

Reaction quotient Q
At any moment, we can form the reaction quotient Q (same algebraic form as K but using instantaneous concentrations or partial pressures). Comparison of Q to K predicts direction:

Q < K: reaction proceeds forward (to produce more products).
Q = K: system is at equilibrium.
Q > K: reaction proceeds in reverse (to produce more reactants).

Dependence of equilibrium
Equilibrium position (value of K) depends only on temperature. Concentration, pressure and catalysts change the position of equilibrium (i.e., shift the composition) but do not change K (except temperature). Le Chatelier's principle gives a qualitative prediction of the direction of shift when a system is disturbed: the system shifts to counteract the change.

Example: N2O4 ⇌ 2NO2 (g)
If initial N2O4 is added, forward reaction increases NO2 until rates equal. Increase in pressure (by decreasing volume) shifts equilibrium toward fewer gas moles (N2O4) and vice versa. Increasing temperature for an endothermic forward reaction shifts equilibrium to products.

Degree of dissociation (α)
For dissociation AB ⇌ A + B with initial concentration c of AB and degree of dissociation α, at equilibrium: [AB] = c(1−α), [A] = cα, [B] = cα. Then

Kc = [A][B]/[AB] = (cα)(cα)/(c(1−α)) = c α^2/(1−α)

For a weak monobasic acid HA ⇌ H+ + A− with initial concentration c and small α, Ka ≈ c α^2. Hence α ≈ √(Ka/c) when α ≪ 1.

Key points to remember

• Equilibrium is dynamic and reached only in closed systems for reversible reactions.
• K (Kc or Kp) is constant at a given temperature and tells about the composition at equilibrium: large K ⇒ products favored; small K ⇒ reactants favored.
• Use Q vs K to predict the direction of spontaneous change.
• Le Chatelier's principle predicts the qualitative shift on changing concentration, pressure, or temperature.

Fig 2 — Educational Diagram: Introduction to Equilibrium

What is Chemical Equilibrium?
A chemical equilibrium is reached in a closed system when the rates of the forward and reverse reactions become equal so that the concentrations (or partial pressures) of reactants and products remain constant with time. Equilibrium is dynamic: microscopic processes continue, but there is no net change in composition.

Reversible reactions and example equation
Many reactions are reversible and are written with a double arrow, e.g. N2(g) + 3 H2(g) ⇌ 2 NH3(g). In a closed vessel, after some time the rate of formation of NH3 equals the rate of its decomposition back to N2 and H2, and the system attains equilibrium.

Characteristics of equilibrium

• Dynamic balance: rate_forward = rate_reverse.
• Macroscopic properties (concentration, pressure, colour) remain constant at equilibrium.
• Equilibrium position depends on temperature, concentrations/pressures, but the equilibrium constant (at a given temperature) is fixed for a particular reaction.
• Heterogeneous equilibrium: pure solids and pure liquids do not appear in the equilibrium expression.

Equilibrium expressions
For a general reaction aA + bB ⇌ cC + dD the equilibrium constant in terms of concentration is written as:
Kc = [C]^c [D]^d / ([A]^a [B]^b).
For gases, the partial pressure form is Kp, and the two are related by Kp = Kc (RT)^{Δn}, where Δn = (c + d) − (a + b).

Reaction quotient and direction
The reaction quotient Q has the same form as K but with instantaneous concentrations/pressures. Comparing Q with K tells the direction the reaction will proceed: if Q < K, forward reaction proceeds; if Q > K, reverse reaction proceeds; if Q = K, the system is at equilibrium.

Why only temperature changes K?
K depends on the standard free energy change (ΔG°) and therefore on temperature. Changing concentration or pressure shifts the equilibrium position (Le Chatelier’s principle) but does not change K at a given temperature.

Physical equilibria
Equilibrium concepts also apply to physical processes such as liquid ↔ vapour (evaporation = condensation) and saturated solution (dissolution = crystallization).

Fig 3 — Educational Diagram: Homogeneous and heterogeneous equilibria

Equilibrium — brief recap
A chemical equilibrium is reached in a reversible reaction when the rate of the forward reaction equals the rate of the reverse reaction. At equilibrium the macroscopic properties (concentration, pressure, colour) remain constant though reactions continue microscopically (dynamic equilibrium).

Homogeneous equilibrium
When all reactants and products are present in the same phase (all gases or all liquids/solutions), the equilibrium is called homogeneous. Example reactions: gas-phase Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), or acid dissociation in aqueous phase: CH₃COOH(aq) ⇌ CH₃COO⁻(aq) + H⁺(aq).

Heterogeneous equilibrium
When reactants and products are in different phases (solid + gas, liquid + gas, or solid + liquid + gas), the equilibrium is heterogeneous. Typical examples: decomposition of calcium carbonate: CaCO₃(s) ⇌ CaO(s) + CO₂(g), and the reaction between a metal oxide and gas: Fe₃O₄(s) + 4CO(g) ⇌ 3Fe(s) + 4CO₂(g).

How equilibrium constants are written
For homogeneous equilibria, the equilibrium constant in terms of concentration (K_c) is expressed using molar concentrations of all species (raised to their stoichiometric coefficients):
K_c = [products]^coeff / [reactants]^coeff.

For heterogeneous equilibria, concentrations (or partial pressures) of pure solids and pure liquids are omitted from the equilibrium expression because their activity is essentially constant (taken as 1). Only species whose amounts/pressures/concentrations can change (usually gases and solutes) appear.

Examples of equilibrium expressions
1) Homogeneous gas equilibrium (Haber):
K_p = (P_NH3)² / (P_N2)(P_H2)³ (use partial pressures for gases).
2) Heterogeneous equilibrium (CaCO₃):
CaCO₃(s) ⇌ CaO(s) + CO₂(g) ⇒ K_p = P_CO2 (pure solids not included).

Relation between K_p and K_c
For reactions involving gases, K_p and K_c are related by:
K_p = K_c(RT)^Δn, where Δn = (moles of gaseous products) − (moles of gaseous reactants), R is the gas constant and T is temperature in kelvin.

Important points
- Pure solids and pure liquids are omitted from equilibrium constant expressions (their activity ≈ 1).
- Equilibrium constants are temperature-dependent but independent of initial concentrations or pressure (for a given T).
- The same reaction can be treated as homogeneous or heterogeneous depending on phases present.

Practical implications / real-life contexts
Heterogeneous equilibria are common in industrial processes (e.g., calcination, metallurgy) and natural systems (soil–gas interactions). Homogeneous equilibria are central to solution chemistry (acid–base equilibria) and gas-phase industrial reactions (Haber process).

Fig 4 — Educational Diagram: Types of Equilibrium

What is equilibrium? In chemistry, equilibrium is the state of a reversible process where the macroscopic properties remain constant because the forward and reverse processes occur at equal rates. Most chemical equilibria are dynamic (microscopically active though macroscopically unchanged).

Main types of equilibrium

• Physical (phase) equilibrium — equilibrium between different physical states (phases) of the same substance. Example: liquid ⇌ vapor in a closed container at a fixed temperature (saturated vapor pressure), or solid ⇌ liquid at the melting point. At equilibrium the rate of evaporation equals the rate of condensation.
• Chemical equilibrium — equilibrium in a reversible chemical reaction where concentrations of reactants and products remain constant. Example: N2(g) + 3H2(g) ⇌ 2NH3(g) (Haber process). Chemical equilibrium is dynamic: forward and reverse reaction rates are equal.
• Homogeneous vs heterogeneous equilibrium — homogeneous: all reactants and products are in the same phase (e.g., N2O4(g) ⇌ 2NO2(g)); heterogeneous: reactants/products are in different phases (e.g., CaCO3(s) ⇌ CaO(s) + CO2(g)). In heterogeneous equilibria the activities of pure solids and pure liquids are taken as unity and are omitted when writing equilibrium constants.
• Ionic (solution) equilibria — equilibria involving ions in solution, such as acid–base dissociation (HA ⇌ H+ + A−), salt hydrolysis, solubility equilibria and complexation. These are usually homogeneous equilibria in the aqueous phase.
• Dynamic vs static equilibrium — in chemistry, equilibria are generally dynamic (continuous microscopic activity). Static equilibrium (no microscopic movement) is uncommon for chemical processes; the term is more used in mechanics.

Important practical notes: For gas-phase equilibria you can express equilibrium in terms of concentrations (Kc) or partial pressures (Kp). For heterogeneous equilibria, omit pure solids/liquids from the equilibrium expression. Temperature changes can shift equilibria (Le Chatelier’s principle); pressure and concentration changes also affect equilibria involving gases or species in solution.

Fig 5 — Educational Diagram: Law of chemical equilibrium (Law of mass action)

Introduction
The law of chemical equilibrium, or the law of mass action (formulated by Guldberg and Waage), states that for a reversible chemical reaction in a closed system at constant temperature the rate of the forward reaction equals the rate of the backward reaction at equilibrium. For the general homogeneous reaction

aA + bB ⇌ cC + dD

if the forward and backward reactions are elementary, their rates depend on reactant concentrations raised to the power of their stoichiometric coefficients. At equilibrium:

rate_forward = rate_backward

Let rate_forward = k_f [A]^a [B]^b and rate_backward = k_r [C]^c [D]^d. Equating them gives

k_f [A]^a [B]^b = k_r [C]^c [D]^d

Rearranging gives the equilibrium constant expression (K):

K_c = [C]^c [D]^d / [A]^a [B]^b = k_f / k_r

Key points

• Equilibrium is dynamic: molecules continue to react, but macroscopic concentrations remain constant because forward and reverse rates are equal.
• K (equilibrium constant) is temperature dependent but independent of initial concentrations and pressure (except through temperature and for K_p/K_c relations).
• For heterogeneous equilibria, concentrations of pure solids and pure liquids are taken as unity and are omitted from the K expression.
• K_c uses molar concentrations; K_p uses partial pressures for gases. They are related by K_p = K_c (RT)^{Δn}, where Δn = (moles gas products) − (moles gas reactants).
• Magnitude of K: if K ≫ 1, products are favored at equilibrium; if K ≪ 1, reactants are favored; if K ≈ 1, appreciable amounts of both are present.

Thermodynamic link
Standard Gibbs free energy change is related to K by: ΔG° = −RT ln K. Thus K reflects the thermodynamic driving force at a given temperature.

Limitations & activity
The law in the simple concentration form strictly applies using activities rather than concentrations. For dilute solutions and ideal gases activities ≈ concentrations/standard concentration or partial pressure/standard pressure, so concentration expressions are used in practice.

Summary
The law of mass action gives a quantitative relationship between concentrations (or pressures) of species at equilibrium and connects kinetics (k_f, k_r) with thermodynamics (K, ΔG°).

Fig 6 — Educational Diagram: Law of Mass Action

Definition: The Law of Mass Action (Guldberg and Waage) states that the rate of a chemical reaction at any instant is directly proportional to the product of the active masses (concentrations or partial pressures) of the reactants, each raised to a power equal to its stoichiometric coefficient (for an elementary reaction). At equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, giving an equilibrium constant expression.

Mathematical statement (for an elementary reaction): For aA + bB ⇌ cC + dD,

rate_forward = k_f [A]^a [B]^b
rate_reverse = k_r [C]^c [D]^d

At equilibrium, rate_forward = rate_reverse, therefore

k_f [A]^a[B]^b = k_r [C]^c[D]^d ⟹ K_c = [C]^c[D]^d / [A]^a[B]^b = k_f / k_r

Activities vs concentrations: The rigorous form uses activities (a_i) rather than concentrations: K = Π a_products^(coeff) / Π a_reactants^(coeff). For ideal dilute solutions, activities ≈ concentrations; for gases, activities ≈ partial pressures (P_i/P°).

Relation between Kp and Kc: For gaseous reactions, K_p = K_c (R T)^{Δn}, where Δn = (c + d) − (a + b) and R is the gas constant, T temperature in K.

Significance: The law links kinetics and equilibrium (K = k_f/k_r) and provides the standard equilibrium-constant expressions used to predict composition at equilibrium. It also underpins derived concepts such as reaction quotient Q and Le Chatelier's principle (how equilibria respond to changes).

Limitations & important notes:

• Exponents in the rate law equal stoichiometric coefficients only for elementary reactions (single-step). For complex mechanisms, rate laws must be determined experimentally.
• Use activities (not raw concentrations) for accurate thermodynamic K in non-ideal systems.
• Equilibrium constant depends only on temperature (not on pressure or concentration); K = k_f/k_r and k_f, k_r are temperature dependent.

Fig 7 — Educational Diagram: Equilibrium Constant Expressions

Definition
At chemical equilibrium, for a general reversible reaction aA + bB ⇌ cC + dD the equilibrium constant (law of mass action) is defined as a ratio of products to reactants each raised to the power of their stoichiometric coefficients:

Kc = [C]^c[D]^d / ([A]^a[B]^b)

Here [X] denotes the molar concentration of species X at equilibrium. This expression applies to homogeneous equilibria (all species in the same phase).

Kp (equilibrium constant in terms of partial pressures)
For gaseous reactions the equilibrium constant can be written using partial pressures:

Kp = (P_C^c P_D^d)/(P_A^a P_B^b)

Relation between Kp and Kc
Using ideal‑gas behaviour (P_i=n_iRT/V or P_i=[i]RT) you obtain:

Kp = Kc (RT)^Δn, where Δn = (c + d) − (a + b)

Heterogeneous equilibria
Pure solids and pure liquids have essentially constant concentrations and are omitted from the equilibrium expression. Example: for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp (or Kc) involves only P_CO2 (or [CO₂]).

Reaction quotient Q
Q has the same mathematical form as K but uses instantaneous (not necessarily equilibrium) concentrations or pressures. Comparison of Q with K predicts the direction of net reaction:

If Q < K → forward reaction favored (proceed to products)
If Q > K → reverse reaction favored (proceed to reactants)
If Q = K → system at equilibrium

Temperature dependence
K depends only on temperature. The van't Hoff relation links K at two temperatures:

ln(K₂/K₁) = −ΔH°/R (1/T₂ − 1/T₁)

Units and dimensionless form
Formally equilibrium constants should be written using activities (dimensionless). In school problems concentrations or pressures are used directly and K is often reported without units. For rigorous work use activity-based K.

How to write an equilibrium-constant expression (stepwise)

1. Write the balanced chemical equation.
2. Include only species whose concentrations/pressures change (omit pure solids/liquids).
3. For Kc use equilibrium molar concentrations [ ]. For Kp use partial pressures P_i.
4. Raise each term to the power of its stoichiometric coefficient and form products of products over products of reactants.

Notes for problem solving
To compute K from initial and equilibrium data often you set up an ICE table (Initial, Change, Equilibrium) and solve for unknown equilibrium concentrations/pressures, then substitute into the K expression.

Fig 8 — Educational Diagram: Equilibrium constant Kc

Definition: For a chemical reaction at equilibrium, the equilibrium constant Kc is a number that expresses the ratio of the product concentrations to reactant concentrations (each raised to the power of its stoichiometric coefficient) when the system is at equilibrium. It is a measure of the extent to which a reaction proceeds under a given temperature.

General expression: For the reaction aA + bB ⇌ cC + dD (all species in the same phase, usually aqueous),

Kc = [C]^c [D]^d / ([A]^a [B]^b)

Here [X] means the molar concentration of species X at equilibrium. Kc is written from balanced stoichiometry: changing the stoichiometric coefficients changes the powers in Kc. If the reaction is reversed, Kc becomes 1/(original Kc); if the equation is multiplied by n, Kc is raised to the power n.

Reaction quotient (Qc): Qc has the same form as Kc but uses instantaneous concentrations (not necessarily at equilibrium). Comparison of Qc with Kc predicts the direction of spontaneous shift: Qc < Kc (forward reaction proceeds), Qc > Kc (reverse proceeds), Qc = Kc (equilibrium).

Heterogeneous equilibria: Pure solids and pure liquids do not appear in the Kc expression because their concentrations (activity) are effectively constant. Example: CaCO3(s) ⇌ CaO(s) + CO2(g) gives Kc = [CO2].

Dependence and properties:

• Kc depends only on temperature. Changing temperature changes Kc.
• Changing concentrations or pressure shifts the equilibrium position but does not change Kc (except as a result of a temperature change).
• Catalysts do not change Kc; they only help reach equilibrium faster.
• Kc is often treated as dimensionless (using activities); when quoted using concentrations it may appear to have units depending on Δn, but modern convention reports it as dimensionless.

Relation to Kp (gaseous equilibria): For reactions involving gases, Kp (in terms of partial pressures) relates to Kc by Kp = Kc(RT)^(Δn), where Δn = moles of gaseous products − moles of gaseous reactants, R is gas constant and T is temperature in kelvin.

Temperature dependence (Van 't Hoff): The Van 't Hoff equation links K at two temperatures to the standard enthalpy change: ln(K2/K1) = −ΔH°/R (1/T2 − 1/T1). This shows how Kc changes with temperature depending on whether the reaction is endothermic or exothermic.

Interpretation of magnitude: Large Kc (≫1) means products are favored at equilibrium; small Kc (≪1) means reactants are favored; Kc ≈ 1 means significant amounts of both are present.

Fig 9 — Educational Diagram: Equilibrium constant Kp and relation to Kc

Definition (Kp): For a gaseous equilibrium aA(g) + bB(g) ⇌ cC(g) + dD(g), the equilibrium constant in terms of partial pressures (K_p) is

K_p = (P_C)^c (P_D)^d / [(P_A)^a (P_B)^b]

Definition (Kc): The equilibrium constant in terms of molar concentrations is

K_c = [C]^c [D]^d / ([A]^a [B]^b)

Relation between pressures and concentrations: For ideal gases, partial pressure P_i and concentration [i] (mol L⁻¹) are related by the ideal‑gas law

P_i = [i] R T

Derivation of Kp in terms of Kc (substitute P = [ ]RT into K_p):

K_p = ([C]RT)^c ([D]RT)^d / ([A]RT)^a ([B]RT)^b

Collect concentration terms and RT terms:

K_p = ([C]^c [D]^d / [A]^a [B]^b) · (RT)^{(c+d)−(a+b)}

So

K_p = K_c (RT)^Δn

where Δn = (c + d) − (a + b) = change in moles of gas between products and reactants.

Important consequences:

• If Δn = 0, then K_p = K_c (numerically equal).
• K_p and K_c are temperature dependent (they change with temperature) but for a given temperature they are constants independent of total pressure. The equilibrium partial pressures/concentrations do depend on pressure but the constants do not.
• Units: K_p may carry units that depend on Δn. Strict thermodynamic practice uses dimensionless forms (pressures divided by standard pressure or fugacities), but for CBSE problems we often keep units explicitly.
• The reaction quotient Q has the analogous relation Q_p = Q_c (RT)^Δn and comparing Q_p to K_p predicts the direction of shift.

Temperature and Le Châtelier: Because K depends on temperature (van't Hoff relation), increasing temperature favors endothermic direction (K changes accordingly). Pressure changes shift the position of equilibrium for heterogeneous gas reactions according to Le Châtelier, but do not change K_p at constant T.

Practical note: Use R = 0.08206 L·atm·K⁻¹·mol⁻¹ when pressures are in atm and concentrations in mol·L⁻¹. Ensure consistent units when calculating K_p from K_c.

Fig 10 — Educational Diagram: Relation between Kp and Kc

Definitions: For a homogeneous gaseous equilibrium aA + bB ⇌ cC + dD,

Kc = ([C]^c [D]^d)/([A]^a [B]^b) where concentrations [ ] are in mol L^-1.
Kp = (p_C^c p_D^d)/(p_A^a p_B^b) where p_i are partial pressures (usually in atm).

Derivation using the ideal gas law: for ideal gases p_i = [i]RT. Substitute into Kp:

Kp = ( ([C]RT)^c ([D]RT)^d / ([A]RT)^a ([B]RT)^b )

Simplify the powers of RT:

Kp = Kc (RT)^{(c+d)-(a+b)}

Define Δn = (moles of gaseous products) − (moles of gaseous reactants) = (c + d) − (a + b). Thus

Kp = Kc (RT)^{Δn}

Important notes:

• Δn counts only gaseous species. Pure solids and liquids do not appear in Kp or Kc expressions and do not change Δn.
• If Δn = 0, then Kp = Kc (common case), i.e., Kp and Kc have the same numerical value at a given temperature.
• Both Kp and Kc depend on temperature. The relation contains an explicit T term, so converting between Kc and Kp requires the temperature (in K) and the gas constant R in consistent units (commonly R = 0.082057 L·atm·mol^-1·K^-1).
• Thermodynamically, equilibrium constants are dimensionless when referred to standard states; in practice for CBSE problems Kc and Kp are often given with units and used numerically with the above relation.

When to use: Use this relation whenever you need to convert Kc to Kp or vice versa for equilibria involving gases and you know the temperature.

Fig 11 — Educational Diagram: Reaction Quotient (Q) and Direction of Reaction

What is the Reaction Quotient (Q)?
For a general reversible reaction aA + bB ⇌ cC + dD, the reaction quotient Q measures the current ratio of product to reactant activities (or concentrations/partial pressures) in the same algebraic form as the equilibrium constant K:

Q_c = [C]^c[D]^d / ([A]^a[B]^b)   (using concentrations)
Q_p = (P_C^cP_D^d) / (P_A^aP_B^b)   (using partial pressures)

Q is calculated from the instantaneous concentrations or pressures — it tells you where the system is now compared with the equilibrium position.

Comparison of Q and K (direction of spontaneous change)

• If Q < K: the reaction will proceed in the forward direction (toward products) to reach equilibrium. More reactants → products.
• If Q = K: the system is at equilibrium; no net change.
• If Q > K: the reaction will proceed in the reverse direction (toward reactants) to reach equilibrium.

The reason: the Gibbs free energy change for a non-equilibrium composition is ΔG = ΔG° + RT ln Q. At equilibrium ΔG = 0 and Q = K. If Q < K then ln Q < ln K so ΔG < 0 (forward spontaneous). If Q > K then ΔG > 0 (reverse spontaneous).

Relation between Kp and Kc
If gases are involved, K_p and K_c are related by the change in mole number of gas, Δn = (c + d) − (a + b):

K_p = K_c(RT)^Δn

Practical notes for Class 11 level

• Use concentrations for Q_c and partial pressures for Q_p.
• Strictly, K and Q are defined with activities (dimensionless). In most problems you may substitute concentrations (in mol L⁻¹) or pressures (in bar/atm) consistently.
• Q changes with time as the reaction proceeds; K is constant at a given temperature.

Fig 12 — Educational Diagram: Writing equilibrium expressions and conventions

What is an equilibrium expression?
For a chemical reaction at equilibrium, the equilibrium constant (K) is written as the ratio of the activities of products to reactants, each raised to the power of its stoichiometric coefficient:

K = Π (a_products)^{ν} / Π (a_reactants)^{ν}

In common school practice this is approximated using concentrations ([ ]) for solutes (Kc) or partial pressures (P) for gases (Kp), with these important conventions:

• Use stoichiometric exponents: Each concentration or pressure term is raised to the power equal to the species' coefficient in the balanced equation.
• Exclude pure solids and pure liquids: Their activities are effectively constant and are omitted from the K expression (they are incorporated into K).
• State symbols: Always indicate states (g, aq, s, l). Only include (aq) and (g) species explicitly in Kc or Kp formulas.
• Reverse and scaled reactions: For the reverse reaction K_rev = 1/K_fwd. If the balanced equation is multiplied by n, the new K = (K_original)^n. If two reactions are added, the overall K = product of the individual Ks.
• Activities vs concentrations/pressures: The rigorous form uses dimensionless activities (a = [C]/C° or P/P°). In practice Kc is written with molar concentrations and Kp with pressures; remember that strictly K is temperature dependent and dimensionless when using activities referenced to standard states.
• Reaction quotient Q: Q has the same form as K but uses instantaneous concentrations/pressures. Compare Q and K to predict net reaction direction (Q < K forward, Q > K reverse, Q = K at equilibrium).

Relation between Kp and Kc:

Kp = Kc (R T)^{Δn}

where Δn = (sum of stoichiometric coefficients of gaseous products) − (sum of coefficients of gaseous reactants), R is the gas constant (0.08206 L·atm·K−1·mol−1 if P in atm and concentrations in mol·L−1), and T is temperature in K.

Practical notes for students: Always write the balanced chemical equation first, identify phases, decide whether to use Kc or Kp, omit pure solids/liquids, and put stoichiometric coefficients as exponents. Remember K depends on temperature but not on starting concentrations.

Fig 13 — Educational Diagram: Manipulation of Equilibrium Constants

Basic idea: The equilibrium constant (K) for a chemical reaction is a number (at a given temperature) that relates the activities (often approximated by concentrations or partial pressures) of products and reactants at equilibrium. K is characteristic of the reaction and the temperature.

General expression (for a reaction aA + bB ⇌ cC + dD):
K_c = [C]^c [D]^d / ([A]^a [B]^b) (using molar concentrations)
K_p = (P_C)^c (P_D)^d / ((P_A)^a (P_B)^b) (using partial pressures)

Rules for manipulation:

• Reversing a reaction: If the given reaction is reversed, the equilibrium constant becomes the reciprocal. If original K = K, for reversed reaction K' = 1/K.
• Multiplying coefficients (scaling a reaction): If every stoichiometric coefficient is multiplied by a factor n, the new constant is K' = K^n.
• Adding reactions: When two or more reactions are added to give an overall reaction, the overall equilibrium constant is the product of the equilibrium constants of the individual reactions (K_total = K1 × K2 × ...).
• Combining the above: Use reciprocals and powers appropriately when reactions are reversed or scaled before adding.
• K_p and K_c relation: K_p = K_c (R T)^{Δn_gas}, where Δn_gas = (moles gaseous products) - (moles gaseous reactants), R is the gas constant (0.08314 L·bar·K⁻¹·mol⁻¹ or 0.08206 L·atm·K⁻¹·mol⁻¹ depending on units) and T is absolute temperature.
• Heterogeneous equilibria: Pure solids and pure liquids do not appear in the equilibrium-constant expression (their activity ≈ 1). Only gases and solutes in solution are included.

Temperature dependence (Van't Hoff): The value of K varies with temperature. The Van't Hoff equation in differential form is
ln(K2/K1) = -ΔH°/R (1/T2 - 1/T1)
where ΔH° is standard enthalpy change of the reaction. This shows K increases with T for endothermic reactions (ΔH° > 0) and decreases with T for exothermic reactions (ΔH° < 0).

Reaction quotient Q: Q is calculated like K but for non‑equilibrium concentrations/pressures. Comparison of Q and K predicts the direction the system will shift: Q < K → proceeds forward; Q > K → proceeds backward; Q = K → at equilibrium.

Practical tips: When manipulating reactions algebraically, always track whether you reversed or scaled a reaction and apply reciprocal or power rules to its K before multiplying to get the net K.

Fig 14 — Educational Diagram: Reaction quotient (Q)

Definition: The reaction quotient Q is a number that has the same algebraic form as the equilibrium constant K but is calculated with the instantaneous (not necessarily equilibrium) concentrations or partial pressures of reactants and products. Q tells us how far the system is from equilibrium and in which direction the reaction will proceed to reach equilibrium.

General expression (for the reaction aA + bB ⇌ cC + dD):

Qc = [C]^c [D]^d / ([A]^a [B]^b) (using concentrations in mol L⁻¹)

Similarly, for gases using partial pressures:

Qp = (P_C)^c (P_D)^d / ((P_A)^a (P_B)^b)

Key points:

• Q has the same mathematical form as K (Kc or Kp) but is evaluated at any moment, not necessarily at equilibrium.
• Compare Q with the equilibrium constant K to predict the direction of spontaneous shift:
• If Q < K, the forward reaction (reactants → products) is favored; system shifts right.
• If Q = K, the system is at equilibrium; no net shift.
• If Q > K, the reverse reaction (products → reactants) is favored; system shifts left.
• For heterogeneous equilibria (solids or pure liquids present), the activities of pure solids and pure liquids are constant and omitted from the Q expression. Only gaseous and aqueous species appear.
• Strictly speaking Q and K are calculated from activities and are dimensionless; in class problems concentrations or partial pressures are usually used directly.

Thermodynamic relation:

ΔG = ΔG° + RT ln Q. At equilibrium ΔG = 0 and Q = K, so ΔG° = −RT ln K. The sign of ΔG tells the spontaneous direction and is directly related to how Q compares with K.

Practical interpretation: Q is a snapshot. If you know the current concentrations (or pressures), compute Q and compare with the appropriate K to know which way the reaction must move to reach equilibrium. Le Chatelier's principle describes how changes in concentration, pressure or temperature change Q (and K if T changes) and thus the position of equilibrium.

Fig 15 — Educational Diagram: Calculations using equilibrium constants

Overview

Calculations using equilibrium constants involve using the equilibrium expression (K_c, K_p, K_x, K_a, etc.) to relate concentrations or partial pressures of reactants and products at equilibrium. The usual steps are: write the balanced equation, write the equilibrium-constant expression, set up an ICE (Initial–Change–Equilibrium) table, substitute equilibrium expressions into the constant, and solve for the unknown(s).

Key ideas

• For a general reaction aA + bB ⇌ cC + dD,

  K_c = [C]^c[D]^d / ([A]^a[B]^b)

• For gaseous reactions in terms of partial pressures, K_p = (P_C^cP_D^d)/ (P_A^aP_B^b).
• Relation between K_p and K_c:

  K_p = K_c(RT)^Δn, where Δn = (c + d) − (a + b), R = 0.08206 L·atm·K⁻¹·mol⁻¹, and T is in kelvin.

• Heterogeneous equilibria: concentrations of pure solids and pure liquids are omitted from the equilibrium expression.
• Reaction quotient Q has the same form as K. If Q < K, reaction proceeds forward; if Q > K, reaction goes backward; if Q = K, system is at equilibrium.

ICE-table method (step-by-step)

1. Write balanced equation and K expression.
2. List initial concentrations/pressures (I).
3. Express changes using a variable (x) according to stoichiometry (C).
4. Write equilibrium concentrations/pressures in terms of x (E).
5. Substitute E into the K expression and solve the resulting algebraic equation for x.
6. Check approximations (if any) and compute final concentrations/pressures.

Quadratic and approximation

Often substitution leads to a quadratic (or higher) equation. If K is very small or very large such that x is much smaller than initial concentration I, you can use the approximation x << I and neglect x in denominators. A common check: if x/I < 5% the approximation is acceptable.

Worked example 1 — (dimerisation) 2NO₂ ⇌ N₂O₄

Given: initial [NO₂] = 0.100 M, [N₂O₄] = 0, K_c = 0.25 at the temperature.

Let x = [N₂O₄] formed at equilibrium. Then [NO₂] = 0.100 − 2x.

K_c = [N₂O₄] / [NO₂]² = x / (0.100 − 2x)² = 0.25.

Solving: x = 0.25(0.100 − 2x)² → gives quadratic. Solving the quadratic yields x ≈ 0.0025 M.

Equilibrium concentrations: [N₂O₄] ≈ 0.0025 M, [NO₂] ≈ 0.095 M.

Worked example 2 — relation K_p and K_c (Haber reaction)

For N₂ + 3H₂ ⇌ 2NH₃, given K_c = 6.0×10⁻⁵ at T = 500 K. Find K_p.

Here Δn = (2) − (1 + 3) = −2. Use K_p = K_c(RT)^Δn = K_c/(RT)².

R = 0.08206 L·atm·K⁻¹·mol⁻¹, so RT = 0.08206 × 500 = 41.03. Then (RT)² ≈ 1683.5.

K_p = 6.0×10⁻⁵/1683.5 ≈ 3.6×10⁻⁸.

Checks and tips

• For the reverse reaction, K_reverse = 1 / K_forward.
• If you multiply a reaction by n, the new K = (K_old)ⁿ. If you add reactions, multiply their K values.
• When using partial pressures expressed via mole fraction y_i and total pressure P_tot: P_i = y_iP_tot. Then K_p may be written in terms of mole fractions and P_tot as needed.

Real-life relevance

Calculations with equilibrium constants are essential in industrial chemistry (e.g., optimizing conditions for the Haber process for ammonia synthesis), environmental chemistry (CO₂ equilibria, acid–base equilibria in natural waters), and chemical engineering (design of reactors where yields depend on equilibrium position).

Fig 16 — Educational Diagram: Units and Dimensionality of K

What is K? For a general homogeneous reaction aA + bB ⇌ cC + dD the equilibrium constant in concentration form is

Kc = [C]^c[D]^d / ([A]^a[B]^b)

Here square brackets denote molar concentration (mol L⁻¹). Define Δn = (c + d) − (a + b) (the change in number of moles of gas or species whose concentrations are used).

Units of Kc

• If Δn = 0, the powers of concentration cancel and Kc is dimensionless (no net units).
• If Δn ≠ 0, Kc carries units derived from (mol L⁻¹)^Δn. In other words:

[Kc] = (mol L⁻¹)^Δn = mol^Δn L^−Δn.

Units of Kp

• When the equilibrium constant is written in terms of partial pressures (Kp), its units are (pressure unit)^Δn (for example atm^Δn or Pa^Δn).
• The relation between Kp and Kc for ideal gases is

Kp = Kc (RT)^Δn

where R is the gas constant and T the absolute temperature. This relation shows how units shift consistently: (pressure)^Δn = (concentration × RT)^Δn.

Role of activities and dimensionless K

Strict thermodynamics uses activities (dimensionless) instead of concentrations/pressures. Activities are concentrations (or pressures) divided by standard states (c° = 1 mol L⁻¹, p° = 1 bar). Using activities makes the equilibrium constant truly dimensionless:

K = Π (a_products)^ν / Π (a_reactants)^ν

In practice textbooks and problems often report Kc or Kp with apparent units for clarity; remember the rigorous K (from thermodynamics) is dimensionless.

Heterogeneous equilibria

Pure solids and pure liquids are assigned activity = 1 and do not appear explicitly in K expressions. Thus Δn should count only the species whose concentrations or pressures appear (usually gases and solutes).

Summary

• Compute Δn = (sum of stoichiometric coefficients of products appearing in K) − (sum for reactants).
• Units of Kc = (mol L⁻¹)^Δn; units of Kp = (pressure unit)^Δn.
• Prefer reporting K as dimensionless using activities (divide by standard concentration/pressure).

Fig 17 — Educational Diagram: Degree of dissociation (α) and relation with equilibrium constant

Definition: Degree of dissociation (α) is the fraction of the original molecules that dissociate at equilibrium. If n0 is the initial moles of a substance and n dissociate, α = n/n0. It is dimensionless and 0 ≤ α ≤ 1.

Simple molecular dissociation (A B ⇌ A + B): Consider a substance AB of initial molar concentration c that dissociates as AB ⇌ A + B. If fraction α dissociates, at equilibrium:

• [AB] = c(1 − α)
• [A] = [B] = cα

The equilibrium constant Kc is

Kc = [A][B] / [AB] = (cα)(cα) / (c(1 − α)) = c α² / (1 − α).

This gives a quadratic relation for α:

c α² + Kc α − Kc = 0,

so the positive root is

α = [−Kc + sqrt(Kc² + 4 c Kc)] / (2 c) = [sqrt(Kc(Kc + 4c)) − Kc] / (2c).

Useful approximation: If dissociation is small (α ≪ 1) so that (1 − α) ≈ 1, then Kc ≈ c α² and

α ≈ sqrt(Kc / c).

Electrolyte dissociation (Ostwald's dilution law): For a weak acid/base HA ⇌ H+ + A− with initial concentration c and acid dissociation constant Ka, identical algebra gives Ka = c α² / (1 − α). For dilute solutions (α ≪ 1), α ≈ sqrt(Ka / c). This relation is called Ostwald's dilution law in its approximate form.

Gaseous dissociation and Kp: For N2O4(g) ⇌ 2 NO2(g), if initial pressure of N2O4 is P0 and fraction α dissociates, total moles change and partial pressures at total pressure P are:

• p(N2O4) = P (1 − α) / (1 + α)
• p(NO2) = P (2α) / (1 + α)

The equilibrium constant in pressure terms is

Kp = p(NO2)² / p(N2O4) = P · 4 α² / (1 − α²)

From this, α can be solved; for example, increasing total pressure P shifts equilibrium toward N2O4 (lower α) in accordance with Le Chatelier's principle because dissociation increases the number of moles of gas.

Relation between Kp and Kc: Kp = Kc (RT)^{Δn}, where Δn = moles of gaseous products − moles of gaseous reactants. Use this to convert between Kc-based α formulae and Kp-based ones when gases are involved.

Practical notes:

• α increases when equilibrium constant (extent of dissociation) increases (e.g., with temperature if dissociation is endothermic).
• For electrolytes, conductivity measurements give α experimentally via α = Λ / Λ∞, where Λ is molar conductivity at given dilution and Λ∞ is limiting molar conductivity.

Fig 18 — Educational Diagram: Heterogeneous Equilibria: Concentration and Pressure Terms

What is a heterogeneous equilibrium? A heterogeneous equilibrium involves reactants and products in two or more different phases (solid, liquid, gas). Examples: a solid in equilibrium with its vapor, a solid reacting with a gas, or a gas dissolved in a liquid in equilibrium with the gas phase.

Equilibrium expressions — which species appear? In equilibrium constant expressions (Kc or Kp) only species whose concentrations or partial pressures change with the extent of reaction are included. Pure solids and pure liquids have essentially constant concentrations (their densities do not change appreciably), so their activity is taken as unity and they are omitted from the mathematical expression of the equilibrium constant. Gaseous and dissolved species are included.

Kc and Kp for heterogeneous equilibria:

• Kc uses molar concentrations [ ] (mol L-1) for solutes and may be written for gaseous species if expressed in concentration units.
• Kp uses partial pressures (p, usually in bar or atm) for gaseous species.
• For a heterogeneous reaction the equilibrium constant includes only gaseous and aqueous species; pure solids/liquids are omitted because their activity = 1.

General example and rule — for reaction aA(s) + bB(g) ⇌ cC(g) + dD(l):

• Kc = [C]^c / [B]^b (D and A omitted if pure liquid/solid)
• Kp = (p_C)^c / (p_B)^b (only gaseous partial pressures appear)

Relation between Kp and Kc: For reactions involving gases, Kp is related to Kc by the formula Kp = Kc (RT)^{Δn}, where Δn = (sum of moles of gaseous products) - (sum of moles of gaseous reactants), R is the gas constant and T is temperature in kelvin. Δn counts only gaseous species.

Why solids/liquids omitted? The thermodynamic activity of a pure solid or pure liquid in its standard state is defined as 1. Since their 'concentration' does not change during the reaction (as long as some amount of each pure phase is present), including them would multiply the equilibrium expression by 1 and is unnecessary.

Le Chatelier and pressure effects: Changes in total pressure (or volume) affect equilibria only when the number of moles of gas changes (Δn ≠ 0). If Δn = 0, changing pressure does not shift the equilibrium. In heterogeneous systems pressure changes affect only the gaseous side of the equilibrium.

Activity note (brief): More generally, equilibrium constants are expressed in terms of activities. For ideal gases activity ≈ p/p°, and for dilute solutions activity ≈ [C]/C°. For pure solids and liquids activity = 1.

Fig 19 — Educational Diagram: Factors Affecting Equilibrium (Le Chatelier's Principle)

Le Chatelier's Principle (qualitative statement): If a dynamic equilibrium is disturbed by changing concentration, pressure, volume or temperature, the system shifts in the direction that tends to reduce that disturbance.

General reaction and equilibrium constant: For a reaction aA + bB ⇌ cC + dD, the equilibrium constant (concentration form) is

Kc = [C]^c [D]^d / ([A]^a [B]^b)

Reaction quotient Q: At any moment, the reaction quotient has the same algebraic form as Kc:

Q = [C]^c [D]^d / ([A]^a [B]^b)

• If Q < K, reaction proceeds forward (to products).
• If Q > K, reaction proceeds backward (to reactants).
• If Q = K, system is at equilibrium.

How different factors affect equilibrium:

• Change in concentration: Adding a reactant (or removing a product) makes Q < K, so equilibrium shifts to the right (forms more products). Removing a reactant or adding a product shifts left. This re-establishes K (Kc unchanged at constant temperature).
• Change in pressure / volume (gaseous systems): Only changes that alter partial pressures of gaseous species matter. Decreasing volume (increasing total pressure) favors the side with fewer moles of gas; increasing volume favors the side with more moles of gas. If total moles of gas are equal on both sides, pressure/volume changes have no effect on position of equilibrium (K unchanged).
• Adding an inert gas: At constant volume, adding an inert gas does not change partial pressures of reactants/products and has no effect on equilibrium. At constant pressure, adding an inert gas increases volume and can shift equilibrium as a pressure/volume change would.
• Change in temperature: Temperature changes affect the equilibrium constant K. For an exothermic reaction (ΔH < 0), increasing temperature shifts equilibrium to the left (favoring reactants) and decreases K. For an endothermic reaction (ΔH > 0), increasing temperature shifts equilibrium to the right (favoring products) and increases K. Thus temperature is the only factor that changes the numerical value of K.
• Catalyst: A catalyst increases the rates of both forward and reverse reactions equally, so it speeds attainment of equilibrium but does not change the equilibrium position or K.
• Common-ion effect / ionic equilibria: Adding a species that supplies an ion already present in an equilibrium (e.g., adding NaCl to AgCl ⇌ Ag+ + Cl−) shifts the equilibrium to reduce the added ion (here, to the left), lowering solubility. This is an application of Le Chatelier's principle to solubility equilibria.

Thermodynamic connections (quantitative):

• ΔG° = −RT ln K connects standard free energy change and K. If temperature changes, K changes according to the van't Hoff relation.
• van't Hoff (integrated/differential form): ln(K2/K1) = −ΔH°/R (1/T2 − 1/T1). This shows how K varies with temperature for approximately constant ΔH°.
• Kp & Kc relation (gaseous reactions): Kp = Kc (RT)^{Δn}, where Δn = (c + d) − (a + b) is change in moles of gas.

Limitations and remarks:

• Le Chatelier's principle predicts direction of shift qualitatively, not rates or exact new concentrations.
• It applies to equilibria in closed systems; changing temperature changes K itself (not merely shifting position).
• Statistical/thermodynamic relations (ΔG, van't Hoff) give quantitative predictions.

Fig 20 — Educational Diagram: Le Chatelier's principle

Definition and context
Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions (concentration, pressure, volume or temperature), the system shifts in a direction that tends to counteract the disturbance and restore a new equilibrium.

Dynamic equilibrium
For a reversible reaction at equilibrium, e.g. aA + bB ⇌ cC + dD, the forward and reverse rates are equal. Small changes (stresses) move the position of equilibrium but do not stop the reaction from being reversible.

Qualitative explanations of common stresses

• Change in concentration: Adding a reactant drives the equilibrium toward products to consume the added reactant; removing a product drives it toward products, etc. (Q vs K idea: if Q < K the forward reaction proceeds; if Q > K the reverse proceeds.)
• Change in pressure/volume (for gases): Increasing pressure (or decreasing volume) shifts equilibrium toward the side with fewer moles of gas; decreasing pressure shifts toward the side with more moles of gas. (Only applies when total gas moles differ on two sides.)
• Change in temperature: Treated as adding or removing heat. For an exothermic forward reaction (ΔH < 0), increasing temperature shifts equilibrium toward reactants (reverse); for an endothermic forward reaction (ΔH > 0), increasing temperature shifts toward products. Importantly, temperature changes change the equilibrium constant K.
• Catalyst: A catalyst speeds up both forward and reverse reactions equally, so equilibrium is reached faster but the position of equilibrium (value of K and concentrations at equilibrium) is unchanged.

Why K generally does not change
Concentration or pressure changes shift the position of equilibrium but do not change the equilibrium constant K (except when temperature changes). K depends only on temperature (and standard state definitions).

Thermodynamic connections
K is related to Gibbs free energy: ΔG° = −RT ln K. Temperature dependence of K is given by the van't Hoff relation: ln(K2/K1) = −ΔH°/R (1/T2 − 1/T1).

Important practical implications
Le Chatelier's principle is used to optimize industrial processes (e.g., Haber process), control yields in chemical synthesis (removing product to drive reaction), and to understand buffering and biological equilibria (e.g., CO2/HCO3− balance in blood).

Fig 21 — Educational Diagram: Effect of pressure and volume

Summary: For a gaseous equilibrium, changing the pressure or volume (at constant temperature) can shift the position of equilibrium. Le Chatelier's principle predicts the direction: an increase in pressure (or decrease in volume) favours the side with fewer moles of gas; a decrease in pressure (or increase in volume) favours the side with more moles of gas. If the total number of moles of gas is the same on both sides, changing pressure/volume does not affect the equilibrium position.

Why this happens (short explanation): For a general reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), define Δn = (c + d) - (a + b), the change in moles of gas. The equilibrium constant in terms of pressure Kp is related to Kc by:

• Kp = Kc (RT)^{Δn} (so Kp depends on Δn and temperature, but not on pressure).

When volume is decreased (total pressure increases), partial pressures of all gases increase. Because the reaction quotient Qp = (P_C^c P_D^d)/(P_A^a P_B^b) changes, the system shifts to reduce the change — i.e., toward the side with fewer gas molecules if pressure is increased (Δn < 0 is favoured) and toward the side with more gas molecules if pressure is decreased (Δn > 0 is favoured). If Δn = 0, Qp does not change with pressure/volume and the position of equilibrium is unaffected.

Special case — addition of an inert gas:

• At constant volume: adding an inert gas (e.g., N2, Ar) raises total pressure but does not change the partial pressures of reacting species (their mole fractions remain the same). Therefore the equilibrium position is not affected.
• At constant pressure: adding an inert gas forces the system to expand (volume increases) so partial pressures of reactants/products decrease. The equilibrium shifts as if volume were increased: toward the side with more gas moles.

Heterogeneous equilibria: Reactions involving solids or pure liquids (e.g., CaCO3(s) ⇌ CaO(s) + CO2(g)) are not significantly affected by pressure/volume changes except through the partial pressure of the gas component (e.g., CO2). Solids and liquids do not appear in Kp/Kc expressions and their amounts/volumes are effectively constant.

Important conceptual points:

• Changing pressure/volume moves the position of equilibrium only when gaseous mole numbers differ between sides (Δn ≠ 0).
• Kp (at fixed temperature) is constant and does not change when you change pressure or volume — only temperature changes Kp.
• Le Chatelier's principle predicts the direction of shift but does not give the new concentrations quantitatively — use equilibrium expressions to calculate new values.

Fig 22 — Educational Diagram: Temperature Dependence of K

Concept: The equilibrium constant K for a chemical reaction depends on temperature. Temperature changes shift the position of equilibrium and change the numerical value of K according to the reaction enthalpy (ΔH°).

Thermodynamic relations: Standard free energy and K are related by ΔG° = −RT ln K. Combining ΔG° = ΔH° − TΔS° with this gives
ln K = −ΔH°/(RT) + ΔS°/R
where R is the gas constant (8.314 J mol⁻¹ K⁻¹), T is absolute temperature (K), ΔH° and ΔS° are standard enthalpy and entropy changes.

van't Hoff equation (differential form):
d(ln K)/dT = ΔH°/(R T²)
This shows how ln K changes with temperature. If ΔH° is assumed constant over the temperature range, we can integrate:

Integrated van't Hoff equation:
ln(K₂/K₁) = (ΔH°/R) (1/T₁ − 1/T₂)
Equivalent form: ln(K₂/K₁) = −(ΔH°/R) (1/T₂ − 1/T₁).

Sign interpretation (Le Chatelier's principle consistent):

• If reaction is exothermic (ΔH° < 0): increasing T decreases K (equilibrium shifts toward reactants).
• If reaction is endothermic (ΔH° > 0): increasing T increases K (equilibrium shifts toward products).

Graphical form: From ln K = −ΔH°/(RT) + ΔS°/R, a plot of ln K versus 1/T is a straight line with slope = −ΔH°/R and intercept = ΔS°/R. This is commonly used to determine ΔH° and ΔS° experimentally.

Practical notes:

• In industrial processes (e.g., Haber synthesis of NH₃), temperature is chosen as a compromise between favorable K (low T for exothermic reactions) and sufficient reaction rate (higher T for faster kinetics).
• If ΔH° varies with T (nonconstant heat capacity), corrections using ΔC_p (Kirchhoff's law) are needed for accurate predictions.
• For gas-phase equilibria, Kp and Kc are related by Kp = Kc(RT)^Δn; temperature affects both Kc and Kp via Kc's temperature dependence and the explicit T in (RT)^Δn.

Fig 23 — Educational Diagram: Degree/Extent of Dissociation and Equilibrium

Definition: Degree or extent of dissociation (alpha, α) is the fraction of the original molecules that dissociate into ions or fragments at equilibrium. If x out of n0 molecules dissociate, α = x/n0. 0 ≤ α ≤ 1.

Simple dissociation (molecular): Consider a reversible reaction AB ⇌ A + B. If initial concentration of AB is c and fraction dissociated is α, then at equilibrium:

• [AB] = c(1 - α)
• [A] = cα
• [B] = cα

The equilibrium constant Kc is

Kc = [A][B] / [AB] = (cα)(cα) / (c(1 - α)) = c α^2 / (1 - α).

Rearranging gives Ostwald-type relation for a 1:1 dissociation:

α^2 /(1 - α) = Kc / c

Solving for α:

α = sqrt( Kc / (c + Kc) ).

When α << 1 (weak dissociation), α ≈ sqrt( Kc / c ).

Weak acid dissociation (HA ⇌ H+ + A-): If initial concentration = c and degree = α, then

• [HA] = c(1 - α), [H+] = cα, [A-] = cα
• Ka = [H+][A-] / [HA] = c α^2 / (1 - α)

So α = sqrt( Ka / (c + Ka) ) and approximately α ≈ sqrt( Ka/c ) when α << 1 (Ostwald's dilution law).

Electrolyte dissociation and conductivity: For a weak electrolyte, molar conductivity at concentration c is Λ = α Λ0, where Λ0 is molar conductivity at infinite dilution. Combining with Ka expression gives Ostwald form in terms of conductivities:

Ka = c Λ^2 / (Λ0 (Λ0 - Λ)).

Gaseous dissociation (example: A2 ⇌ 2A). Start with pure A2 at initial pressure p. If fraction α dissociates, equilibrium partial pressures are:

• P_A2 = p(1 - α)
• P_A = 2 p α

For Kp = (P_A)^2 / P_A2, we get

Kp = 4 p α^2 / (1 - α) => α^2 /(1 - α) = Kp / (4 p).

Solved approximately for small α: α ≈ sqrt( Kp / (4 p) ).

Factors affecting degree of dissociation:

• Concentration/dilution: For weak electrolytes/weak acids, α increases on dilution (because denominator c decreases, Ostwald law).
• Pressure (gases): For reactions that produce more moles of gas (e.g., A2 ⇌ 2A), increasing pressure shifts equilibrium toward fewer moles (less dissociation); α decreases with increasing external pressure.
• Temperature: If dissociation is endothermic, increasing temperature increases α (Le Chatelier). If exothermic, increasing temperature decreases α.
• Nature of substance: Strong electrolytes/strong acids/bases are almost fully dissociated (α ≈ 1); weak ones have small α.

Use in calculations: Knowing K (Kc, Ka or Kp) and initial concentration (or initial pressure) allows solution for α via the relations above. For small α, approximations (neglecting 1 - α ≈ 1) simplify algebra; always check the validity of approximations (e.g., α < 0.05).

Practical importance: Degree of dissociation determines acidity/basicity (pH), conductivity of electrolyte solutions, colour and composition of equilibria in gases (e.g., NO2/N2O4), industrial equilibrium yields (Haber process considerations), and many temperature-dependent phenomena in chemistry and engineering.

Fig 24 — Educational Diagram: Effect of temperature and thermodynamic connection

Overview
Temperature is a key factor that affects chemical equilibrium. According to Le Chatelier's principle, if temperature is changed, the equilibrium shifts to oppose that change: raising temperature favors the endothermic direction, lowering temperature favors the exothermic direction. The quantitative thermodynamic connection is provided by Gibbs free energy and the Van't Hoff equation, which relate the equilibrium constant (K) to temperature (T) and the reaction enthalpy (ΔH°).

Thermodynamic relations

• Standard Gibbs free energy and equilibrium constant: ΔG° = -RT ln K. This connects the thermodynamic favorability to K.
• Gibbs free energy and entropy/enthalpy: ΔG° = ΔH° - TΔS°. Combining with ΔG° = -RT ln K gives a direct link between K, ΔH° and ΔS°.

Van't Hoff equation (differential and integrated forms)

Differential (Van't Hoff):

(d ln K) / (dT) = ΔH° / (R T^2)

Integrated (assuming ΔH° ≈ constant over the temperature range):

ln(K2 / K1) = -ΔH° / R (1/T2 - 1/T1)

Interpretation: if ΔH° > 0 (endothermic), d(ln K)/dT > 0 so K increases with T. If ΔH° < 0 (exothermic), K decreases as T increases.

Practical implications and interplay of ΔH° and ΔS°

• If ΔH° < 0 (exothermic): increasing temperature reduces K and shifts equilibrium toward reactants; decreasing temperature increases K and favors products.
• If ΔH° > 0 (endothermic): increasing temperature increases K and shifts equilibrium toward products.
• Because ΔG° = ΔH° - TΔS°, when ΔH° and ΔS° have the same sign the temperature can switch the spontaneity (and thereby K). For example, if ΔH° > 0 and ΔS° > 0, reaction may become spontaneous only above a certain T.

Assumptions and limitations

• The integrated Van't Hoff equation assumes ΔH° is approximately constant over the temperature range used. For large ranges, ΔH° may vary and the approximation may fail.
• Temperature also affects kinetics (reaction rates). A higher temperature can increase rate even if it thermodynamically disfavors product formation (common industrial trade-off).

How to use these relations

• Use ΔG° = -RT ln K to compute K from thermodynamic data (or vice versa).
• Use the integrated Van't Hoff form to estimate how much K will change between two temperatures if ΔH° is known.

Short derivation outline of Van't Hoff (for reference): Start from ΔG° = -RT ln K and ΔG° = ΔH° - TΔS°. Differentiate ΔG° = -RT ln K with respect to T and combine with d(ΔG°)/dT = -ΔS° to obtain (d ln K)/(dT) = ΔH°/(RT^2). Integrate to get ln(K2/K1) = -ΔH°/R (1/T2 - 1/T1).

Fig 25 — Educational Diagram: Effect of catalyst

Definition: A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. It provides an alternative reaction pathway with a lower activation energy.

What a catalyst does:

• Speeds up both the forward and reverse reactions by providing an alternative mechanism with lower activation energy (Ea).
• Does not change the thermodynamic properties of the reaction: the standard Gibbs free energy change (ΔG°) and the equilibrium constant (K) remain unchanged.
• Allows the system to reach equilibrium faster but does not change the position of equilibrium (i.e., equilibrium concentrations or partial pressures remain the same).

Why equilibrium position is unchanged: The equilibrium constant for a reaction A ⇌ B is related to the rate constants of the forward (kf) and reverse (kr) elementary steps by K = kf/kr. A catalyst increases both kf and kr by providing new pathways, but it does so in a way that preserves the ratio kf/kr (because ΔG° between reactants and products is unchanged). Therefore K remains constant.

Molecular/energy picture: On a potential-energy vs reaction-coordinate diagram a catalyst lowers the activation energy peaks for both forward and reverse processes. The energy difference between reactants and products (ΔG) is the same with or without catalyst, so the relative heights of reactants and products do not change—only the barrier heights change.

Practical consequences: Catalysts are essential in industry and biology to make reactions occur at useful rates and temperatures (e.g., enzymes in metabolism, heterogeneous catalysts in chemical plants). They reduce energy use and increase throughput, but never change the final equilibrium composition predicted by thermodynamics.

Fig 26 — Educational Diagram: Relations between K and Degree of Dissociation

What is degree of dissociation (α)?
Degree of dissociation, α, is the fraction of the original substance that dissociates at equilibrium (0 ≤ α ≤ 1). If initial concentration (or pressure) of a substance A is c (or P) and α fraction dissociates, equilibrium amounts are expressed in terms of α.

General idea: The equilibrium constant (Kc or Kp) can be written in terms of α and the initial concentration (or pressure). Solving that relation gives α as a function of K and initial amount. Two important cases are common in Class 11:

• Solution (weak electrolyte) case: For HA ⇌ H+ + A− with initial concentration c and degree α, at equilibrium [HA] = c(1 − α), [H+] = cα, [A−] = cα. So Ka = [H+][A−]/[HA] = (cα)(cα)/[c(1 − α)] = cα^2/(1 − α).
• Gaseous dissociation (mole change) cases:
  • For A ⇌ B + C (start with pure A at initial pressure P): at equilibrium pA = P(1 − α)/(1 + α), pB = Pα/(1 + α), pC = Pα/(1 + α). Then Kp = (pB·pC)/pA = P·α^2/(1 − α^2).
  • For A ⇌ 2B (start with pure A at initial pressure P): pA = P(1 − α)/(1 + α), pB = 2Pα/(1 + α). Then Kp = pB^2/pA = 4P·α^2/(1 − α^2).

Useful approximations:
For weak electrolytes (α ≪ 1) we approximate (1 − α) ≈ 1, giving Ka ≈ cα^2 ⇒ α ≈ √(Ka/c). This shows α increases on dilution (smaller c) — a standard experimental observation.

Exact algebraic relation (solution case HA ⇌ H+ + A−):
Ka = cα^2/(1 − α) → rearranged to the quadratic cα^2 + Ka·α − Ka = 0. The physically meaningful root is
α = [−Ka + √(Ka^2 + 4cKa)]/(2c).

Exact relations for the gas cases (starting with pure A at initial pressure P):

• A ⇌ B + C: Kp = P·α^2/(1 − α^2) → α^2 = Kp/(P + Kp) → α = √(Kp/(P + Kp)).
• A ⇌ 2B: Kp = 4P·α^2/(1 − α^2) → α^2 = Kp/(4P + Kp) → α = √(Kp/(4P + Kp)).

Physical implications and trends:

• For a given Ka (intrinsic property), α increases when c decreases (dilution favours dissociation) — explains why weak acids dissociate more on dilution.
• For gaseous dissociation producing more moles (e.g., A ⇌ 2B), increasing pressure shifts equilibrium toward reactant (Le Châtelier): α decreases with rising initial pressure P.
• If K is large (stronger tendency to dissociate), α approaches 1; if K is very small, α ≪ 1.

Measurement link: For electrolytes, molar conductivity at concentration c, Λm, is related to α: Λm = α·Λm° (Λm° = molar conductivity at infinite dilution). Thus α = Λm/Λm° can be used experimentally to get K (via Ka formula above).

Fig 27 — Educational Diagram: Applications and industrial examples

Chemical equilibrium principles (Kc, Kp, Q, Le Chatelier's principle, effect of temperature/pressure/catalyst) are widely used in industry to maximize product yield, choose operating conditions and design reactors. Industrial practice balances thermodynamics (what equilibrium composition favours) and kinetics (how fast equilibrium is reached). Typical strategies to improve yield are: change temperature or pressure, use excess of one reactant, remove product continuously, or use a catalyst to speed up the approach to equilibrium (a catalyst does not change the equilibrium constant).

Key industrial ideas illustrated by equilibrium theory:

• For exothermic reactions, lowering temperature shifts equilibrium toward products; for endothermic, raising temperature favors products (Le Chatelier).
• Increasing pressure favors the side with fewer moles of gas (if Δn ≠ 0), used in high‑pressure processes like the Haber process.
• Removing a product (or adding a reactant) shifts equilibrium to produce more product — used in esterification where water is removed, or in gas-phase syntheses where product is condensed out.
• Catalysts increase reaction rates so processes can be run at less extreme temperatures/pressures while maintaining acceptable throughput; catalysts do not change K.

Common industrial examples demonstrate these principles in practice (Haber, Contact, methanol and ester syntheses). Process design always balances yield (thermodynamics), rate and cost (energy, materials, reactor design).

Fig 28 — Educational Diagram: Calculations and Problem-Solving Techniques

Overview: Calculations in chemical equilibrium involve using the equilibrium constant (Kc or Kp), the reaction quotient (Q), stoichiometry and algebra (often ICE tables) to find equilibrium concentrations/pressures, percent dissociation, or to predict the direction of reaction. Key ideas: write a balanced equation, form the K expression, set up changes using stoichiometry, substitute into K, solve for the unknowns, and check approximations.

Step-by-step procedure:

• Write the balanced chemical equation: aA + bB ⇌ cC + dD.
• Write the equilibrium expression for Kc (concentrations) or Kp (partial pressures): Kc = [C]^c[D]^d/[A]^a[B]^b.
• Set up an ICE table (Initial, Change, Equilibrium). Represent change by a variable (usually x) using stoichiometric ratios.
• Substitute equilibrium expressions into the K equation to get an equation in x. Solve algebraically (linear or quadratic). If the quadratic is messy and x is small relative to initial concentrations, use the approximation x << initial to simplify.
• Convert between Kc and Kp when gases are involved: Kp = Kc (RT)^{Δn}, where Δn = moles of gaseous products − moles of gaseous reactants.
• Check results: concentrations must be non-negative; verify approximation by ensuring x/initial < 5% (or recalculate exactly if not satisfied).
• Use Q to predict direction: if Q < K, reaction proceeds forward; if Q > K, it proceeds backward; if Q = K, it's at equilibrium.

Special points:

• Heterogeneous equilibria: omit pure solids and pure liquids from K expressions (their activities ≈ 1).
• Percent dissociation/ionization: for A ⇌ products, percent dissociation = (amount dissociated / initial amount) × 100% = (x/initial)×100%.
• Common-ion effect: presence of an ion common to an equilibrium shifts it and reduces dissociation; calculations proceed by including the common-ion initial concentration in the ICE table.
• Temperature dependence: K depends on temperature (Le Chatelier tells qualitative shift when T changes; for numerical change, use thermodynamic relations if available).

ICE table template:

Tips for solving:

• Keep units consistent (M for Kc, atm for Kp). Use R = 0.08206 L·atm·K−1·mol−1 when converting Kc ↔ Kp.
• For gas mixtures, partial pressure Pi = yi × Ptotal or Pi = ni RT / V.
• If quadratic arises: ax^2 + bx + c = 0, use x = [-b ± sqrt(b^2 - 4ac)]/(2a). Choose physically meaningful root.
• Always re-evaluate approximations: if x/initial > ~0.05 (5%), solve the full quadratic instead of using the small-x approximation.

Fig 29 — Educational Diagram: Applications and Examples

Overview: Chemical equilibrium is the state in which the rates of the forward and reverse reactions are equal so that macroscopic properties (concentrations, pressure, color) remain constant. This dynamic balance is quantified by equilibrium constants and is governed qualitatively by Le Chatelier's principle, which predicts how an equilibrium responds to external changes (concentration, pressure, temperature, catalyst).

Why it matters: Understanding equilibrium allows prediction and control of yields in industrial chemical processes, explains behavior of gases and solutions, and underpins biological systems (buffers, respiration). Engineers and chemists shift equilibria to maximize product formation or to maintain safe operating conditions.

Key ideas:

• Equilibrium constant (K): For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentrations (Kc) is Kc = [C]^c [D]^d / ([A]^a [B]^b) (only gases/aqueous species included). Kp is defined analogously for partial pressures.
• Reaction quotient (Q): Q has the same form as K but uses instantaneous concentrations/pressures. Comparing Q and K predicts the direction of net reaction: if Q < K the forward reaction proceeds; if Q > K the reverse proceeds.
• Effect of concentration and pressure: By Le Chatelier's principle, adding reactant (or removing product) shifts equilibrium toward products. For gaseous equilibria, increasing pressure (by decreasing volume) shifts equilibrium toward the side with fewer moles of gas.
• Effect of temperature: Temperature changes alter K. For an exothermic reaction (ΔH < 0) raising temperature decreases K (shifts left); for endothermic (ΔH > 0) raising temperature increases K (shifts right). The van't Hoff relation links ln K to 1/T.
• Effect of catalyst: Catalysts speed up attainment of equilibrium (increase both forward and reverse rates) but do not change K or equilibrium concentrations.
• Common-ion and ionic equilibria: Adding an ion common to a sparingly soluble salt decreases its solubility (common-ion effect). Similar principles control weak acid/base dissociation and buffer action.

Typical applications:

• Industrial synthesis optimization (Haber process for NH3, Contact process for SO3) — conditions (pressure, temperature, catalysts) are chosen to maximize yield while considering kinetics and economics.
• Control of solubility and precipitation in analytical chemistry and water treatment (use of common-ion effect and selective precipitation).
• Biological buffers (blood bicarbonate system CO2 + H2O ⇌ H2CO3 ⇌ H+ + HCO3−) maintain pH by shifting equilibria when acid/base is added.
• Gas solubility in liquids (Henry's law): solubility changes with partial pressure and temperature — relevant for carbonation of drinks, respiration, and decompression sickness.

How to approach equilibrium problems:

1. Write a balanced chemical equation and the expression for K (Kc or Kp).
2. Use initial concentrations/pressures and an unknown change (x) to write equilibrium values (ICE table).
3. Substitute into the K expression and solve for x (approximate quadratic methods may be used when appropriate).
4. Check units and whether approximations are valid; interpret the result in chemical terms (shift direction, percent dissociation, yield).

Practical tip: Always check whether the species are gases or in solution to use Kp or Kc. For gas-phase equilibria that involve change in moles of gas, use Kp = Kc (RT)^(Δn), where Δn = moles of gaseous products − moles of gaseous reactants.

Fig 30 — Educational Diagram: Limitations of law of mass action and role of activities

Law of mass action (brief): For a general reaction aA + bB ⇌ cC + dD the law of mass action (as usually stated in school texts) gives an equilibrium expression in terms of concentrations: Kc = [C]^c [D]^d / ([A]^a [B]^b). For gases it is often written in terms of partial pressures as Kp.

Limitations of the simple concentration/pressure form:

• The simple Kc or Kp expressions are strictly valid only for ideal dilute solutions or ideal gases. Real systems show non‑ideal behaviour at higher concentrations or pressures.
• In ionic solutions (electrolytes), inter‑ionic forces cause the effective reactive 'concentration' to differ from the analytical concentration. Thus use of analytical concentrations can give wrong equilibrium predictions.
• For heterogeneous equilibria involving pure solids or pure liquids the concentrations of those pure phases do not change with extent of reaction; treating them like concentrations is misleading (thermodynamically they are taken as activity = 1).
• The Kc/Kp expressions ignore effects of medium (solvent), ionic strength, and specific interactions (complexation, association) that change the effective reactivity of species.
• The common classroom relation Kp = Kc(RT)^{Δn} assumes ideal gas behaviour (activity of gas = P/po). For real gases this relation fails unless fugacity (not pressure) is used.

Role of activities (correct thermodynamic formulation):

Thermodynamically the equilibrium constant is defined in terms of activities a_i (dimensionless effective concentrations):

K = Π a_i^{ν_i}

where ν_i are stoichiometric coefficients (positive for products, negative for reactants). Activities account for non‑ideality and are defined according to the phase:

• For solutes (standard state = 1 mol L⁻¹): a_i = γ_i · (c_i / c°), where γ_i is the activity coefficient and c° = 1 mol L⁻¹.
• For solvents (standard state often pure solvent): a_solvent ≈ x_solvent · γ_solvent (often ≈ 1 for a pure solvent).
• For gases (standard state = 1 bar or 1 atm depending on convention): a_gas = f_gas / p° = (φ P)/p°, where f is fugacity and φ is the fugacity coefficient. For an ideal gas φ = 1 so a_gas = P/p°.
• For pure solids and pure liquids in heterogeneous equilibria, a = 1 (so they do not appear explicitly in the equilibrium expression).

Practical consequences:

• Measured equilibrium concentrations must often be corrected by activity coefficients to obtain the true thermodynamic constant K (independent of ionic strength and medium).
• In concentrated electrolyte solutions or high pressure gas reactions the predictions made using Kc/Kp without activities can be significantly in error.
• Acid–base equilibria: pKa determined from concentrations differs from thermodynamic pKa unless activities are used. In practice pH measurements are often reported as "operational" pH, and activity corrections are needed for precise thermodynamic analysis.

How activity coefficients are estimated: For dilute ionic solutions the Debye–Hückel limiting law gives an approximate relation for the mean ionic activity coefficient γ±:

log₁₀ γ± = −A |z+ z−| √I

where I is the ionic strength, z+ and z− are ionic charges and A ≈ 0.51 at 25 °C (for water). For higher ionic strengths extended Debye–Hückel or Pitzer models are used.

Summary: The law of mass action must be applied using activities (not raw concentrations or pressures) to obtain thermodynamically correct equilibrium constants. For ideal/very dilute systems activities ≈ concentrations (or pressures) and the simple Kc/Kp forms are good approximations; for non‑ideal systems activity corrections are essential.