Bond Enthalpy — Calculating ΔH from bond energies
Thermodynamics & Thermochemistry
9
JEE Qs
8%
Hard
60
min
Always start by drawing accurate Lewis structures for all species in the balanced equation to correctly count and identify all bonds broken and formed.
🧮 Key Formulas
✅ Key Points for JEE
- 1Bond enthalpy (or average bond enthalpy) is the energy required to break one mole of a particular type of bond in the gaseous state, hence its value is always positive.
- 2The formula implicitly handles bond breaking as endothermic (+ve) and bond formation as exothermic (-ve) by summing energies of broken bonds and subtracting energies of formed bonds.
- 3Accurately draw Lewis structures for all reactants and products to correctly identify and count the number and types of bonds broken and formed.
- 4Bond enthalpy calculations provide an estimated ΔH_reaction because average bond enthalpies (derived from many different molecules) are used, not specific bond dissociation energies for the exact molecule.
- 5Ensure the chemical equation is balanced before counting bonds; stoichiometric coefficients apply to the number of bonds broken/formed.
⚠️ Common Mistakes
- ✕Reversing the formula, i.e., subtracting reactant bond enthalpies from product bond enthalpies (products - reactants instead of reactants - products).
- ✕Incorrectly drawing Lewis structures, leading to miscounting the number of bonds or mistaking single bonds for double/triple bonds (e.g., C=O vs C-O).
- ✕Not balancing the chemical equation, which leads to incorrect multiplication of bond enthalpies by stoichiometric coefficients.
📝 Practice Questions
See allQ32.Given are statements for certain thermodynamic variables, (A) Internal energy, volume (V) and mass (M) are extensive variables. (B) Pressure (P), temperature (T) and density ( ρ ) are intensive variables. (C) Volume (V), temperature (T) and density ( ρ ) are intensive variables. (D) Mass (M), temperature (T) and internal energy are extensive variables. Choose the correct answer from the options given below : (1) (B) and (C) Only (2) (C) and (D) Only (3) (D) and (A) Only (4) (A) and (B) Only
Q64.Ice at −5∘C is heated to become vapor with temperature of 110∘C at atmospheric pressure. The entropy change associated with this process can be obtained from 2025 (23 Jan Shift 1) JEE Main Previous Year Paper (1) (2) (3) (4)
Q53.A liquid when kept inside a thermally insulated closed vessel at 25∘C was mechanically stirred from outside. What will be the correct option for the following thermodynamic parameters ? (1) ΔU < 0, q = 0, w > 0 (2) ΔU = 0, q = 0, w = 0 (3) ΔU > 0, q = 0, w > 0 (4) ΔU = 0, q < 0, w > 0
Q31.Water of mass m gram is slowly heated to increase the temperature from T1 to Tz The change in entropy of the water, given specific heat of water is 1Jkg−1 K−1 , is : (1) m ln ( T2T1 ) (2) zero (3) m ln ( T1T2 ) (4) m (T2 −T1)
Q74.Niobium ( Nb ) and ruthenium (Ru) have " x " and " y " number of electrons in their respective 4 d orbitals. The value of x + y is ______ -.
Q70.The correct stability order of the following species/molecules is: (1) q > r > p (2) r > q > p (3) q > p > r (4) p > q > r Q71. 1 The standard enthalpy and standard entropy of decomposition of N2O4 to NO2 are 55.0 kJ mol−1 and 175.0 J/K/mol respectively. The standard free energy change for this reaction at 25∘C in J mol−1 is ______ (Nearest integer)
NCERT Chapters
- Class 11 Chemistry Ch 6: Thermodynamics