Hess's Law — Enthalpy calculations
Thermodynamics & Thermochemistry
9
JEE Qs
8%
Hard
75
min
Systematically manipulate and sum thermochemical equations by focusing on the target reaction's reactants and products to correctly determine the overall enthalpy change, paying close attention to signs and stoichiometric factors.
🧮 Key Formulas
✅ Key Points for JEE
- 1Hess's Law states that the overall enthalpy change for a reaction is independent of the pathway taken, as enthalpy (ΔH) is a state function.
- 2Thermochemical equations can be treated algebraically: they can be reversed (changing the sign of ΔH) or multiplied/divided by a factor (multiplying/dividing ΔH by the same factor).
- 3To apply Hess's Law, manipulate a series of known thermochemical equations such that their algebraic sum yields the target reaction, ensuring compounds not in the target reaction cancel out.
- 4Carefully identify the target reactants and products, and their stoichiometric coefficients and physical states in the final equation to guide the manipulation of intermediate equations.
- 5The physical states (s, l, g) of reactants and products are crucial and must match between the given equations and the target equation for accurate ΔH calculations.
⚠️ Common Mistakes
- ✕Incorrectly changing the sign of ΔH when reversing a thermochemical equation.
- ✕Failing to multiply or divide the ΔH value by the stoichiometric factor when scaling an equation.
- ✕Not ensuring all intermediate species (those not in the target reaction) cancel out to arrive precisely at the target reaction.
- ✕Ignoring or misinterpreting the physical states of substances, which can lead to errors in cases involving phase transitions.
📝 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