📖 Chapter CN06 · NDA Class 11–12 Level🎯 NDA Level : High Priority
This chapter spans three connected topics that appear regularly across NDA papers. Thermodynamics gives the energy picture of reactions; equilibrium explains why reactions don't always go to completion; redox explains electron flow between species. Average students who can classify exo/endothermic reactions, apply Le Chatelier's principle to three stresses, and assign oxidation states correctly can score consistently from all three areas.
📌 What to expect in NDA (based on 2022–2025 pattern): (1) System vs surroundings; open/closed/isolated systems; sign of ΔH for exo/endothermic; (2) First law of thermodynamics: ΔU = q + w; enthalpy H = U + PV; (3) Le Chatelier's principle — effect of concentration, pressure, and temperature on equilibrium position; (4) Equilibrium constant Kc expression; relationship between Kc and reaction quotient Q; (5) Oxidation = loss of electrons (OIL); reduction = gain of electrons (RIG); oxidation states; (6) Balancing redox reactions by half-reaction method; identifying oxidising and reducing agents.
Reversible reactions, Kc, Le Chatelier's principle
③ Redox Reactions
Oxidation states, OIL RIG, balancing, agents
1. Thermodynamics
1.1
System, Surroundings & Basic Concepts
The universe = system + surroundings — every thermodynamic problem starts here
Thermodynamics studies energy changes accompanying chemical and physical processes. The system is the part of the universe under study; everything else is the surroundings. The boundary separates them.
OPEN SYSTEM
Energy & Matter Exchange
Exchanges both energy AND matter with surroundings
Example: Open beaker of hot water
Steam (matter) and heat (energy) both escape
Most biological systems are open
CLOSED SYSTEM
Energy Only Exchange
Exchanges energy but NOT matter
Example: Sealed cylinder with a piston
Heat crosses boundary; gas cannot escape
Most lab reactions in sealed vessels
ISOLATED SYSTEM
No Exchange
Neither energy nor matter exchanged
Example: Thermos flask (ideal)
The universe itself is an isolated system
ΔU_total of universe = 0 (1st law)
STATE FUNCTIONS
Path-Independent
Value depends only on current state, not path taken
U (internal energy), H (enthalpy), S (entropy)
q (heat) and w (work) are NOT state functions
ΔH = H_products − H_reactants
⚛ First Law of Thermodynamics & Enthalpy
First Law: ΔU = q + w
ΔU = change in internal energy
q = heat absorbed by system (+q = heat in; −q = heat out)
w = work done ON system (+w = compression; −w = expansion)
Work: w = −PΔV (for expansion against constant external pressure)
Enthalpy: H = U + PV
ΔH = ΔU + PΔV (at constant pressure)
ΔH = q_p (heat at constant pressure = enthalpy change)
Sign convention (crucial for NDA):
Exothermic: ΔH < 0 (heat released; products at lower energy)
Endothermic: ΔH > 0 (heat absorbed; products at higher energy)
Hess's Law: ΔH for a reaction = sum of ΔH for each step
(enthalpy is a state function — path doesn't matter)
ΔH_overall = ΔH₁ + ΔH₂ + ΔH₃ + …
Enthalpy of formation (ΔHf°): enthalpy change when 1 mole of compound forms from elements in standard state. Elements in standard state have ΔHf° = 0 by definition.
🔥 Exothermic Reactions (ΔH < 0)
Heat released to surroundings; temperature of surroundings rises
Dissolution of NH₄NO₃ in water (instant cold packs)
Evaporation, melting of ice are also endothermic
Fig. 1 — Enthalpy diagrams: Exothermic (left, ΔH < 0 — products lower than reactants, heat released) and Endothermic (right, ΔH > 0 — products higher than reactants, heat absorbed). Dashed curve = activation energy barrier.
📝 TOPIC-WISE PYQ
Thermodynamics — NDA Pattern Questions
Q1. For an exothermic reaction, the enthalpy change ΔH is:
(a) Positive (b) Negative (c) Zero (d) Infinite
Answer: (b) Negative
In an exothermic reaction, heat is released to the surroundings. Products have less energy than reactants, so ΔH = H_products − H_reactants < 0. Examples: combustion (ΔH = −890 kJ/mol for CH₄), neutralisation (ΔH = −57 kJ/mol). Remember: Exothermic = exit heat = negative ΔH.
Q2. A thermos flask is an example of which type of system?
(a) Open system (b) Closed system (c) Isolated system (d) Adiabatic open system
Answer: (c) Isolated system
An ideal thermos flask exchanges neither matter (it's sealed) nor energy (vacuum insulation prevents heat transfer) with the surroundings. This is the definition of an isolated system. In practice, a thermos only approximates this — some heat exchange occurs slowly. The universe itself is considered the ultimate isolated system.
Q3. Which of the following processes is endothermic?
(a) Combustion of coal (b) Neutralisation (c) Melting of ice (d) Rusting of iron
Answer: (c) Melting of ice
Melting requires energy input (ΔH = +6.01 kJ/mol for ice → water). Heat is absorbed from the surroundings, which is why a glass of ice water feels cold to touch. All the other options are exothermic: combustion (heat released), neutralisation (ΔH = −57 kJ), rusting (slow oxidation, heat released). Photosynthesis and dissolution of NH₄NO₃ are other common endothermic examples.
🧠 TRICKY QUESTIONS
Thermodynamics — Conceptual Traps
Q. A student says: "Since combustion is exothermic, it does not need activation energy." Is this correct?
Answer: Incorrect — all reactions need activation energy, including exothermic ones.
Activation energy (Ea) is the minimum energy required to start a reaction — to break bonds in reactants so new bonds can form. It is independent of whether the overall reaction is exo or endothermic. Even though combustion ultimately releases large amounts of energy (ΔH ≪ 0), you still need to provide a spark or ignition to get it started. The enthalpy diagram for exothermic reactions still shows an "energy hill" (Ea) that must be overcome before the reaction descends to the lower-energy products. Without Ea, petrol would spontaneously combust in air — it doesn't, because Ea acts as a barrier.
Q. ΔH and ΔU are both energy changes — when are they equal?
Answer: ΔH = ΔU when Δn_gas = 0 (no change in moles of gas).
ΔH = ΔU + PΔV. For ideal gases, PΔV = Δn_gas × RT.
If Δn_gas = 0 (same moles of gas on both sides): ΔH = ΔU + 0 = ΔU.
Example: H₂(g) + Cl₂(g) → 2HCl(g) — both sides have 2 moles of gas → ΔH = ΔU.
Example where they differ: N₂(g) + 3H₂(g) → 2NH₃(g) — Δn_gas = 2−4 = −2 → ΔH ≠ ΔU.
For reactions involving only solids/liquids: ΔV ≈ 0 → ΔH ≈ ΔU.
2. Chemical Equilibrium
2.1
Reversible Reactions & Equilibrium Constant Kc
Dynamic equilibrium — forward and reverse rates become equal
Many chemical reactions do not go to completion but reach a state of dynamic equilibrium, where the forward and reverse reactions occur at equal rates. The concentrations of reactants and products remain constant at equilibrium (though not necessarily equal).
⚛ Equilibrium Constant & Reaction Quotient
For: aA + bB ⇌ cC + dD
Equilibrium constant: Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ
(square brackets = molar concentration at equilibrium)
Kp = (pC)ᶜ(pD)ᵈ / (pA)ᵃ(pB)ᵇ (in terms of partial pressures)
Relationship: Kp = Kc(RT)^Δn where Δn = (c+d) − (a+b)
If Δn = 0: Kp = Kc
Reaction Quotient Q:
Same expression as Kc but using NON-equilibrium concentrations
Q < Kc → reaction proceeds forward (→) to make more products
Q > Kc → reaction proceeds backward (←) to make more reactants
Q = Kc → system is at equilibrium
Large Kc (>> 1): products favoured; reaction nearly complete
Small Kc (<< 1): reactants favoured; very little product formed
Pure solids and pure liquids are NOT included in Kc expressions — their concentrations are taken as constant (activity = 1). For heterogeneous equilibria: CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂] only.
Fig. 2 — Dynamic equilibrium: Forward rate starts high and decreases; reverse rate starts at zero and increases; they meet and become equal at t_eq. Inset shows concentrations reaching constant values (not zero) at equilibrium.
2.2
Le Chatelier's Principle
"If a system at equilibrium is disturbed, it shifts to oppose the disturbance" — the single most tested concept in equilibrium
Le Chatelier's principle lets us predict the direction of equilibrium shift when concentration, pressure, or temperature is changed — without calculating Kc.
① CONCENTRATION CHANGE
Add more reactant: equilibrium shifts → forward (makes more product)
Remove a reactant: equilibrium shifts ← backward
Add more product: equilibrium shifts ← backward
Remove a product: equilibrium shifts → forward
Kc does NOT change with concentration
Application: removing NH₃ in Haber process keeps reaction moving forward
② PRESSURE CHANGE (gases only)
Increase pressure: shifts toward side with fewer moles of gas
Decrease pressure: shifts toward side with more moles of gas
If Δn_gas = 0: pressure change has NO effect
Haber process: N₂ + 3H₂ ⇌ 2NH₃ (Δn = 2−4 = −2)
High pressure → shifts right (fewer moles) → more NH₃ ✓
Adding inert gas at constant volume: no effect on equilibrium
③ TEMPERATURE CHANGE
Increase T: shifts toward endothermic direction (absorbs extra heat)
Decrease T: shifts toward exothermic direction (releases heat to compensate)
Temperature is the ONLY factor that changes Kc
Haber (exothermic): lower T → more NH₃ but slow rate → compromise ~450°C
Exothermic: Kc decreases as T increases
Endothermic: Kc increases as T increases
📝 NDA Worked Application — Haber Process: N₂ + 3H₂ ⇌ 2NH₃ ΔH = −92 kJ/mol
Concentration: Remove NH₃ as it forms → Q < Kc → shift right → more NH₃ produced ✓ (used industrially)
Temperature: Exothermic (ΔH < 0). Lower T → shift right → more NH₃. But low T = slow rate. Compromise: ~450°C + iron catalyst. Catalyst: speeds equilibrium but does NOT shift it (Kc unchanged).
📌 NDA Critical Point — Catalyst and Equilibrium: A catalyst increases the rate of both forward AND reverse reactions equally. It helps the system reach equilibrium faster but does NOT change the equilibrium position or the value of Kc. A catalyst does NOT change the yield of products — it only saves time. This is one of the most tested misconceptions in NDA.
📝 TOPIC-WISE PYQ
Chemical Equilibrium — NDA Pattern Questions
Q1. For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), increasing the pressure will:
(a) Shift equilibrium left (b) Shift equilibrium right (c) Have no effect (d) Increase Kc
Answer: (b) Shift equilibrium right
Δn_gas = 2 − (1+3) = −2. Right side has fewer moles of gas. Increasing pressure favours the side with fewer gas moles → shifts right → more NH₃. This is why the Haber process operates at ~200 atm. Note: Kc is unchanged (pressure only changes position, not Kc — only temperature changes Kc).
Q2. For the reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), which change will increase the yield of PCl₃?
(a) Increase pressure (b) Add more PCl₅ (c) Decrease pressure (d) Add a catalyst
Answer: (c) Decrease pressure
Δn_gas = (1+1) − 1 = +1. Right side has more moles of gas. Decreasing pressure favours side with more gas moles → shifts right → more PCl₃ and Cl₂. Adding PCl₅ also shifts right but the question asks for the option that works by physical change (not adding material). A catalyst speeds equilibrium but doesn't increase yield.
Q3. For an exothermic equilibrium reaction, increasing temperature will:
(a) Shift equilibrium to the right and increase Kc (b) Shift equilibrium to the left and decrease Kc (c) Have no effect on equilibrium (d) Increase the rate only
Answer: (b) Shift left and decrease Kc
For an exothermic reaction: Reactants ⇌ Products + Heat. Increasing temperature adds heat — system opposes by absorbing it (Le Chatelier) → shifts left (endothermic direction) → less product. Temperature is the only factor that changes Kc. For exothermic reactions: Kc decreases as T increases. For endothermic reactions: Kc increases as T increases.
🧠 TRICKY QUESTIONS
Equilibrium — The Catalyst & Pressure Traps
Q. Adding an inert gas (e.g. He) to a gas-phase equilibrium at constant volume shifts the equilibrium. True or False?
Answer: False — adding inert gas at constant volume does NOT shift equilibrium.
At constant volume, adding He increases total pressure but does NOT change the partial pressures of the reactants or products (they still occupy the same volume). Since Kp depends on partial pressures, and partial pressures are unchanged, Q = Kp and there is no shift.
However: if inert gas is added at constant total pressure, the system must expand → partial pressures of all gases decrease → effectively like decreasing pressure → shifts toward more moles of gas side. This distinction is a classic NDA and competitive exam trap.
Q. For H₂(g) + I₂(g) ⇌ 2HI(g), increasing pressure has no effect. Why?
Answer: Because Δn_gas = 0 for this reaction.
Δn_gas = 2 (HI) − (1 H₂ + 1 I₂) = 2 − 2 = 0. When there is no change in moles of gas, pressure change shifts neither side. This reaction is often used in NDA as the "no pressure effect" example. Other similar cases: N₂ + O₂ ⇌ 2NO (Δn=0), CO + H₂O ⇌ CO₂ + H₂ (Δn=0) — all immune to pressure changes.
3. Redox Reactions
3.1
Oxidation, Reduction & Oxidation States
OIL RIG — the most important mnemonic in all of chemistry
OXIDATION — OIL
Loss of Electrons
Oxidation Is Loss (of electrons)
Oxidation state increases
The species oxidised is the Reducing Agent (gives electrons away)
Example: Fe²⁺ → Fe³⁺ + e⁻ (Fe is oxidised; Fe is reducing agent)
Mg → Mg²⁺ + 2e⁻ (Mg is oxidised)
Zn → Zn²⁺ + 2e⁻ (in voltaic cells)
REDUCTION — RIG
Gain of Electrons
Reduction Is Gain (of electrons)
Oxidation state decreases
The species reduced is the Oxidising Agent (takes electrons)
Example: Fe³⁺ + e⁻ → Fe²⁺ (Fe³⁺ is reduced; Fe³⁺ is oxidising agent)
Cu²⁺ + 2e⁻ → Cu (copper deposits in electroplating)
MnO₄⁻ + 5e⁻ → Mn²⁺ (KMnO₄ acts as oxidising agent)
💡 Memory: OIL RIG — Oxidation Is Loss; Reduction Is Gain (of electrons). And: the Reducing Agent is itself Oxidised (it gives away electrons and gets oxidised). The Oxidising Agent is itself Reduced (it takes electrons and gets reduced). These two pairs always go together in a redox reaction.
⚛ Rules for Assigning Oxidation States
Rule 1: Free element (uncombined) has oxidation state = 0
Examples: Na(s), O₂(g), Cl₂(g), Fe(s), S₈(s) → all = 0
Rule 2: Monatomic ion = its charge
Na⁺ = +1; Mg²⁺ = +2; Cl⁻ = −1; Fe³⁺ = +3
Rule 3: In compounds — fixed values:
H = +1 always (except in metal hydrides: NaH, H = −1)
O = −2 always (except in peroxides H₂O₂: O = −1; OF₂: O = +2)
F = −1 always (most electronegative; never positive)
Alkali metals (Group 1) = +1 always
Alkaline earth (Group 2) = +2 always
Rule 4: Sum of oxidation states = overall charge of species
Neutral molecule: sum = 0
Ion: sum = charge of ion
Example — find Cr in K₂Cr₂O₇:
2(+1) + 2(Cr) + 7(−2) = 0 → 2 + 2Cr − 14 = 0 → Cr = +6
Common oxidation states to memorise: Mn (+2,+4,+7); Fe (+2,+3); Cu (+1,+2); Cr (+3,+6); S (−2,+4,+6); N (−3, 0, +1 to +5); Cl (−1, +1, +3, +5, +7).
H₂SO₄: 2(+1) + S + 4(−2) = 0 → S = +6 ✓
HNO₃: +1 + N + 3(−2) = 0 → N = +5 ✓
KMnO₄: +1 + Mn + 4(−2) = 0 → Mn = +7 ✓
K₂Cr₂O₇: 2(+1) + 2Cr + 7(−2) = 0 → Cr = +6 ✓
Na₂O₂: 2(+1) + 2O = 0 → O = −1 (peroxide!) ✓
NH₄⁺: N + 4(+1) = +1 → N = −3 ✓
3.2
Balancing Redox Reactions — Half-Reaction Method
Split into oxidation half and reduction half, balance electrons, then add
🔧 Half-Reaction Method — Step by Step: Step 1: Assign oxidation states to all atoms. Identify what is oxidised and reduced. Step 2: Write separate half-reactions for oxidation and reduction. Step 3: Balance atoms other than H and O first. Step 4: Balance O by adding H₂O; balance H by adding H⁺ (in acidic) or OH⁻ (in basic). Step 5: Balance charge by adding electrons (e⁻). Step 6: Multiply half-reactions so electrons cancel. Add the two half-reactions. Step 7: Simplify and verify: atoms and charge must balance on both sides.
📝 Worked Example — Fe²⁺ oxidised by KMnO₄ in acidic solution
Q1. In the reaction: 2Mg + O₂ → 2MgO, which species is the reducing agent?
(a) MgO (b) O₂ (c) Mg (d) Both Mg and O₂
Answer: (c) Mg
Mg: 0 → +2 (loses 2 electrons) = oxidised = reducing agent.
O₂: 0 → −2 (gains 2 electrons) = reduced = oxidising agent.
Mg burns to give MgO (bright white flash — used in flares). Reducing agent = species that loses electrons and causes reduction of the other reactant.
Q2. What is the oxidation state of nitrogen in NH₄NO₃?
(a) +5 only (b) −3 only (c) −3 in NH₄⁺ and +5 in NO₃⁻ (d) 0 throughout
Answer: (c) −3 in NH₄⁺ and +5 in NO₃⁻
NH₄NO₃ contains two different N atoms:
NH₄⁺: N + 4(+1) = +1 → N = −3.
NO₃⁻: N + 3(−2) = −1 → N = +5.
This is a classic NDA question — the same compound can contain the same element in two different oxidation states. NH₄NO₃ is also an important industrial explosive (ammonium nitrate).
Q3. In the reaction: Fe + CuSO₄ → FeSO₄ + Cu, which of the following is correct?
(a) Fe is oxidised; CuSO₄ is the reducing agent (b) Fe is reduced; Cu²⁺ is the oxidising agent (c) Fe is oxidised; Cu²⁺ is the oxidising agent (d) Fe is reduced; Fe is the reducing agent
Answer: (c) Fe is oxidised; Cu²⁺ is the oxidising agent
Fe: 0 → +2 (loses 2e⁻) → oxidised → Fe is the reducing agent.
Cu²⁺: +2 → 0 (gains 2e⁻) → reduced → Cu²⁺ is the oxidising agent.
This is a displacement reaction: Fe is more reactive than Cu (higher in reactivity series), so Fe displaces Cu from its salt solution. Blue colour of CuSO₄ fades; reddish copper deposits.
Q4. In the reaction: Cl₂ + 2KBr → 2KCl + Br₂, the species that is oxidised is:
(a) Cl₂ (b) KBr (c) Br⁻ (d) K⁺
Answer: (c) Br⁻
Br⁻: −1 → 0 (loses 1e⁻) = oxidised; Br⁻ is the reducing agent.
Cl₂: 0 → −1 (gains 1e⁻) = reduced; Cl₂ is the oxidising agent.
Cl₂ is a stronger oxidising agent than Br₂, so it can displace Br⁻ from solution. This is the halogen displacement series: F₂ > Cl₂ > Br₂ > I₂ (decreasing oxidising power), meaning each halogen displaces the halide ions below it.
🧠 TRICKY QUESTIONS
Redox — Oxidation State & Agent Traps
Q. In the reaction: 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O, what type of redox reaction is this, and what is the oxidation state of Cl in each product?
Answer: Disproportionation (comproportionation) reaction.
Cl₂ starts at 0 oxidation state. In the products:
NaCl: Cl = −1 (reduced from 0)
NaClO₃: Cl = +5 (oxidised from 0)
The same element (Cl) is simultaneously oxidised (0 → +5) and reduced (0 → −1) in the same reaction. This is called disproportionation (self-oxidation-reduction). It is also why bleaching powder (Ca(OCl)Cl) has both Cl in +1 and −1 states. NDA tests oxidation state calculations and identification of this special redox type.
Q. H₂O₂ acts as an oxidising agent with KI but as a reducing agent with KMnO₄. How is this possible?
Answer: H₂O₂ is amphoteric in redox — it can both gain and lose electrons.
In H₂O₂: O is in −1 oxidation state (intermediate between −2 and 0).
With KI (weaker reducing agent): H₂O₂ + 2KI + H₂SO₄ → K₂SO₄ + I₂ + 2H₂O. O: −1 → −2 (gains electrons = reduced → H₂O₂ acts as oxidising agent).
With KMnO₄ (strong oxidising agent): 2KMnO₄ + 5H₂O₂ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 5O₂ + 8H₂O. O: −1 → 0 (loses electrons = oxidised → H₂O₂ acts as reducing agent).
Intermediate oxidation state = can go either way depending on the partner.
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