Chemical Equilibrium And Ionic Equilibrium: Important Notes and Questions for NEET by PW

Chemical Equilibrium and Ionic Equilibrium are important Chemistry topics for NEET preparation, covering both conceptual understanding and numerical problem-solving. From equilibrium constants and reaction shifts to pH calculations, buffer solutions and solubility product, you need to understand the underlying principles and practise applying formulas to different situations.

To revise these concepts, PW offers Chemical Equilibrium and Ionic Equilibrium Full Chapter notes for NEET. These notes cover the important definitions, relationships, memory tips and worked-out questions to help you strengthen your understanding and practise numerical problems before the examination.

Chemical Reactions And Equilibrium

A chemical reaction converts one chemical species into another. Particles collide continuously, exchange energy, and may break existing bonds. New bonds form after old bonds are broken.

(*Memory Tip: New bonds form only after existing bonds are broken.)

Irreversible And Reversible Reactions

A reversible reaction is written as:

Reactants ⇌ Products

Initially, the forward reaction is faster because reactants are present. As products form, the backward reaction begins. Equilibrium is reached when:

Rate of forward reaction = Rate of backward reaction

Equilibrium is dynamic. Both reactions continue, but equal rates produce no net change.

At equilibrium:

• Reaction rates are equal.

• Concentrations become constant.

• Reactant and product concentrations are not necessarily equal.

• The reaction does not stop.

(*Memory Tip: At equilibrium, rates are equal, concentrations become constant, and concentrations are not necessarily equal.)

Equilibrium Constant And Active Mass

For the balanced reaction:

aA + bB ⇌ cC + dD

The equilibrium constant is:

K_eq = (active mass of C)^c × (active mass of D)^d / [(active mass of A)^a × (active mass of B)^b]

The powers are the stoichiometric coefficients in the balanced chemical equation.

Active mass may be represented by:

Only equilibrium concentrations or pressures are used. Pure solids and pure liquids are not included as variable terms.

Kc And Kp

• Kc uses equilibrium molar concentrations.

• Kp uses equilibrium partial pressures of gases.

Their relationship is:

Kp = Kc(RT)^Δn

Here:

Δn = gaseous product moles − gaseous reactant moles

Only gaseous species are counted in Δn.

Changes In Equilibrium Constants

For a reaction with equilibrium constant K:

• Reversing the reaction gives 1/K.

• Multiplying the reaction by n gives K^n.

• Adding reactions gives the product of their constants.

• Subtracting a reaction places its constant in the denominator.

For example, if the required reaction is formed by adding reaction 2 and three times reaction 3, while subtracting reaction 1:

K_required = K2 × K3^3 / K1

The equilibrium constant changes only with temperature. Changes in pressure, volume, or concentration may shift equilibrium but do not change K at a fixed temperature.

Le Chatelier’s Principle

When a system at equilibrium is disturbed, it shifts in the direction that opposes the imposed change.

Concentration

For:

A + B ⇌ C + D

• Increasing a reactant concentration shifts equilibrium forward.

• Increasing a product concentration shifts equilibrium backwards.

Pressure And Volume

For gaseous reactions:

• Increasing pressure favours the side with fewer gaseous moles.

• Decreasing pressure favours the side with more gaseous moles.

• Increasing volume favours the side with more gaseous moles.

• Decreasing volume favours the side with fewer gaseous moles.

For:

PCl3(g) + Cl2(g) ⇌ PCl5(g)

Increasing pressure favours PCl5 because the product side has fewer gaseous moles.

When only solids and liquids are involved, pressure favours the phase with greater density.

Temperature

Increasing temperature favors the endothermic direction.

• Endothermic reaction: K increases when temperature increases.

• Exothermic reaction: K decreases when temperature increases.

(*Memory Tip: T-increase, endo-increase; T-increase, exo-decrease.)

For the exothermic formation of ammonia:

N2 + 3H2 ⇌ 2NH3

Maximum product formation is favoured by high pressure and low temperature.

Catalyst

A catalyst:

• Increases the rate of forward and backward reactions.

• Helps equilibrium form faster.

• Does not shift equilibrium.

• Does not change the equilibrium constant.

Gibbs Energy And Equilibrium

The important relationship is:

ΔG° = −2.303RT log Kc

For natural logarithms:

ΔG° = −RT ln Kc

The general relation is:

ΔG = ΔG° + RT ln Q

Use 2.303 only with logarithm to base 10. A positive ΔG° generally corresponds to a small K, while a negative ΔG° indicates that products are more favorable.

Acid–Base Theories

Arrhenius Theory

• Acid produces H+ in water.

• Base produces OH− in water.

Examples include HCl as an acid and NaOH as a base.

Brønsted–Lowry Theory

• Brønsted acid: Proton donor.

• Brønsted base: Proton acceptor.

For example:

CO3^2− + H+ → HCO3−

Carbonate is a Brønsted base because it accepts a proton.

Lewis Theory

• Lewis acid: Electron-pair acceptor.

• Lewis base: Electron-pair donor.

BF3, BCl3, and AlCl3 act as Lewis acids because they are electron-deficient. A species with an available lone pair can act as a Lewis base.

Conjugate Acid–Base Pairs

When an acid loses H+, it forms its conjugate base. When a base gains H+, it forms its conjugate acid.

Examples:

• HCl and Cl−

• H2O and OH−

• HF and F−

• HSO4− and H2SO4

(*Memory Tip: Acid loses H+ to form a conjugate base; base gains H+ to form a conjugate acid.)

Ostwald’s Dilution Law

For a weak acid:

HA ⇌ H+ + A−

If its initial concentration is C and degree of dissociation is α:

• [HA] = C(1 − α)

• [H+] = Cα

• [A−] = Cα

For a weak electrolyte:

Ka = Cα²

Therefore:

α = square root of (Ka/C)

Dilution lowers C and increases the degree of dissociation.

pH, pOH, And Kw

pH = −log[H+]

pOH = −log[OH−]

The ionic product of water is:

Kw = [H+][OH−]

At 25°C:

• Kw = 10^−14

• [H+] = [OH−] = 10^−7 M in pure water

• pH = 7

At 25°C:

pH + pOH = 14

Neutrality depends on equal H+ and OH− concentrations. Therefore, pH 7 represents neutrality specifically at 25°C.

Neutralization And Salt Hydrolysis

For strong acid–strong base mixtures, calculate moles or millimoles first:

Millimoles = molarity × volume in mL

Then:

1. Apply the reaction ratio.

2. Identify the excess reactant.

3. Divide excess amount by total volume.

4. Calculate pH or pOH.

Salt character depends on the parent acid and base:

Buffer Solutions

A buffer resists a significant pH change when a small amount of acid or base is added.

Acidic Buffer

Contains a weak acid and its salt with a strong base.

pH = pKa + log(salt concentration / acid concentration)

Basic Buffer

Contains a weak base and its salt with a strong acid.

pOH = pKb + log(salt concentration / base concentration)

At 25°C:

pH = 14 − pOH

Solubility Product

For a sparingly soluble salt:

Ni(OH)2(s) ⇌ Ni^2+ + 2OH−

Ksp = [Ni^2+][OH−]^2

The solid is excluded from the expression. The powers come from the balanced dissociation equation.

If a common ion is supplied by a strong electrolyte, its concentration may dominate the equilibrium concentration. This lowers the solubility of the sparingly soluble salt.

(*Memory Tip: If one mole of a salt produces two moles of an ion, its concentration is 2s, not s.)*

Important Worked-Out Questions on Chemical and Ionic Equilibrium

Practise these solved examples to apply the formulas and concepts from the chapter. Try each question yourself before reading its solution.

Question: In a reversible reaction, the forward reaction rate is 4 mol L⁻¹ s⁻¹ and the backward reaction rate is also 4 mol L⁻¹ s⁻¹. Has equilibrium been established?

Solution:

Condition for equilibrium:

Forward rate = Backward rate

Here:

4 = 4 mol L⁻¹ s⁻¹

Therefore, the rates are equal, and the system is at equilibrium if these rates persist under the given conditions.

Answer: Dynamic equilibrium has been established.

Question: Write the equilibrium constant expression for:

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Solution:

The equilibrium constant expression is:

Kc = [NH₃]² / ([N₂][H₂]³)

The concentration of each species is raised to the power of its stoichiometric coefficient.

Answer: Kc = [NH₃]² / ([N₂][H₂]³)

Question: For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), determine Δn and write the relationship between Kp and Kc.

Solution:

Gaseous product moles = 2

Gaseous reactant moles = 1 + 3 = 4

Therefore:

Δn = 2 − 4 = −2

Using the Kp–Kc relationship:

Kp = Kc(RT)⁻²

Answer: Δn = −2 and Kp = Kc/(RT)².

Question: A reaction has an equilibrium constant K₁ = 2. If the reaction is reversed, determine the new equilibrium constant.

Solution:

For a reversed reaction:

K₂ = 1/K₁

Substituting:

K₂ = 1/2

Answer: K₂ = 0.5.

Question: Consider the reaction:

PCl₃(g) + Cl₂(g) ⇌ PCl₅(g)

What happens when pressure is increased at constant temperature?

Solution:

Gaseous moles on the reactant side = 1 + 1 = 2

Gaseous moles on the product side = 1

Increasing pressure favours the side with fewer gaseous moles.

Therefore, equilibrium shifts towards PCl₅.

Answer: The forward reaction is favoured, increasing the tendency towards PCl₅ formation.

Question: A catalyst is added to a reversible reaction at constant temperature. What happens to the equilibrium constant?

Solution:

A catalyst increases the rates of both the forward and backward reactions. It helps the system reach equilibrium faster but does not change the equilibrium position or the equilibrium constant.

Answer: The equilibrium constant remains unchanged.

Question: For a reaction, K = 100. Express ΔG° in terms of RT.

Solution:

Using:

ΔG° = −2.303RT log K

Substituting K = 100:

ΔG° = −2.303RT log 100

Since log₁₀ 100 = 2:

ΔG° = −2.303RT × 2

ΔG° = −4.606RT

Answer: ΔG° = −4.606RT.

The negative value indicates that products are thermodynamically favoured under standard conditions.

Question: In the reaction CO₃²⁻ + H⁺ → HCO₃⁻, identify the Brønsted–Lowry base.

Solution:

A Brønsted–Lowry base accepts a proton.

Here, CO₃²⁻ accepts H⁺ to form HCO₃⁻.

Answer: CO₃²⁻ is the Brønsted–Lowry base.

Question: A weak acid has Ka = 10⁻⁵ and initial concentration C = 0.1 M. Calculate its approximate degree of dissociation using Ostwald’s dilution law.

Solution:

For a weak acid:

α = √(Ka/C)

Substituting:

α = √(10⁻⁵/0.1)

α = √(10⁻⁴)

α = 10⁻² = 0.01

Converting into a percentage:

Percentage dissociation = 0.01 × 100 = 1%

Answer: The approximate degree of dissociation is 0.01, or 1%.

Question: Calculate the pH of a strong acid solution with [H⁺] = 10⁻³ M.

Solution:

Using:

pH = −log[H⁺]

Substituting:

pH = −log(10⁻³)

pH = 3

Answer: pH = 3.

To practise more questions and sharpen your understanding of Chemical Equilibrium and Ionic Equilibrium, watch the Chemical Equilibrium & Ionic Equilibrium Full Chapter NEET revision video by PW. Use it to revise important concepts, revisit formulas and strengthen your approach to equilibrium-based numerical problems.

Chemical Equilibrium and Ionic Equilibrium require you to understand equilibrium expressions, reaction shifts, acid–base behaviour and numerical calculations. Revise the formulas, practise the 10 worked-out questions and revisit concepts where you make mistakes. 

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