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Thermochemistry Lecture 5 |Bond Energy| Hess’s Law| Born–Haber Cycle JEE ADVANCED+MAINS | Harshit..|

Автор: Dr. IIT – Harshit Mishra

Загружено: 2026-01-19

Просмотров: 250

Описание: 🔥 Thermochemistry Lecture 5 | JEE Main & Advanced | NEET
Enthalpy of Atomisation • Bond Energy • Hess’s Law • Born–Haber Cycle (Advanced Level)

In this Thermochemistry Lecture–5, we dive deep into high-weightage numerical and conceptual problem solving from Enthalpy of Atomisation and Bond Dissociation Energy, followed by a complete advanced-level treatment of Hess’s Law and the Born–Haber Cycle.
This lecture is specially designed for JEE Advanced aspirants, while remaining fully relevant for JEE Main and NEET students.

📌 What you will learn in this lecture:
🔹 Part 1: Mixed Advanced Questions

We start with mixed illustrations and numericals involving:

Enthalpy of atomisation of elements & molecules

Average bond energy vs bond dissociation energy

Relation between bond energy and bond length

Stability comparison using bond enthalpy

Multi-bond systems (O₂, N₂, CO₂, etc.)

All questions are solved using proper thermochemical sign conventions, energy cycles, and exam-oriented shortcuts, exactly the way they are required in JEE Advanced.

🔹 Part 2: Hess’s Law (Concept + Applications)

Hess’s Law is explained from fundamental logic to advanced application level, including:

Why enthalpy is a state function

Construction of Hess cycles

Calculation of reaction enthalpy using indirect paths

Linking Hess’s Law with bond energy and lattice energy

PYQ-type multi-step reactions

Special focus is given to how JEE twists Hess’s Law questions.

🔹 Part 3: Born–Haber Cycle (Advanced & Numerical Heavy)

The Born–Haber Cycle is taught in full depth, including:

Step-by-step construction of the cycle

Role of atomisation, ionisation enthalpy, electron affinity

Lattice enthalpy (formation vs dissociation)

Numerical problem solving for ionic solids (NaCl, MgO, CaF₂ etc.)

JEE Advanced conceptual traps & sign mistakes

This part builds a strong base for Thermodynamics, Electrochemistry, and Solid State Chemistry.

🎯 Who should watch this lecture?

✔ JEE Advanced 2026/2027 aspirants
✔ JEE Main students targeting 90+ percentile
✔ NEET students wanting conceptual clarity
✔ Students struggling with thermochemistry numericals

📘 DPP: JEE PYQs (Thermochemistry)

Topic: Enthalpy of Atomisation, Bond Energy, Hess’s Law & Born–Haber Cycle

Q1. (JEE Advanced 2012)

Which of the following has the highest enthalpy of atomisation?
(A) O₂
(B) N₂
(C) F₂
(D) Cl₂

Q2. (JEE Main 2015)

Average bond enthalpy of O–H bond is maximum in:
(A) H₂O
(B) H₂O₂
(C) H₃O⁺
(D) OH⁻

Q3. (JEE Advanced 2016)

Bond dissociation energy of O₂ is less than that of N₂ because:
(A) O₂ has larger atomic size
(B) O₂ has lower bond order
(C) O₂ has higher electronegativity
(D) O₂ is paramagnetic

Q4. (JEE Main 2018)

Which statement is correct regarding Hess’s Law?
(A) Enthalpy depends on reaction path
(B) Enthalpy depends only on initial and final states
(C) It is applicable only to exothermic reactions
(D) It is valid only for constant pressure

Q5. (JEE Advanced 2013)

Using Hess’s Law, the enthalpy change for the reaction
C(graphite) → C(diamond) is:
(A) Positive
(B) Negative
(C) Zero
(D) Cannot be determined

Q6. (JEE Main 2019)

Which step is always endothermic in a Born–Haber cycle?
(A) Lattice formation
(B) Electron gain enthalpy
(C) Ionisation enthalpy
(D) Hydration enthalpy

Q7. (JEE Advanced 2020)

Which factor increases lattice enthalpy the most?
(A) Increase in ionic size
(B) Decrease in ionic charge
(C) Increase in ionic charge
(D) Decrease in electronegativity

Q8. (JEE Main 2021)

Enthalpy of atomisation of a diatomic molecule represents:
(A) Energy to break all bonds
(B) Energy to form ions
(C) Energy to convert one mole of molecule into gaseous atoms
(D) Energy of sublimation

Q9. (JEE Advanced 2014)

Which cycle is used to calculate lattice enthalpy indirectly?
(A) Carnot cycle
(B) Born–Haber cycle
(C) Otto cycle
(D) Rankine cycle

Q10. (JEE Main 2023)

If bond energy of A–B is high, the molecule is:
(A) Less stable
(B) More reactive
(C) More stable
(D) Paramagnetic

🚀 SEO Keywords

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🔖 Hashtags

#Thermochemistry
#JEEAdvanced
#JEEMains
#NEET
#EnthalpyOfAtomisation
#BondEnergy
#HessLaw
#BornHaberCycle
#ChemistryNumericals
#IITJEE

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