Kirchhoff's Law — ΔH vs temperature
Thermodynamics & Thermochemistry
9
JEE Qs
8%
Hard
50
min
Master the calculation of ΔCp/ΔCv for a reaction and correctly apply the integrated Kirchhoff's equation based on whether heat capacities are constant or temperature-dependent.
🧮 Key Formulas
✅ Key Points for JEE
- 1Kirchhoff's Law describes how the enthalpy (ΔH) or internal energy (ΔU) of a reaction changes with temperature.
- 2It relates the rate of change of ΔH with temperature to the change in heat capacity at constant pressure (ΔCp) and similarly for ΔU with ΔCv.
- 3The integrated form involving ∫ΔCp dT or ∫ΔCv dT must be used if the heat capacities are temperature-dependent; otherwise, a simpler linear relationship applies if heat capacities are constant.
- 4ΔCp and ΔCv for a reaction are calculated by summing the molar heat capacities of products minus the sum for reactants, considering their stoichiometric coefficients.
- 5This law is essential for calculating thermodynamic quantities at temperatures other than standard conditions, which is common in advanced problems.
⚠️ Common Mistakes
- ✕Confusing ΔCp with ΔCv and using the wrong one for ΔH or ΔU calculations.
- ✕Incorrectly calculating ΔCp or ΔCv, often by neglecting stoichiometry or sign conventions (products - reactants).
- ✕Assuming ΔCp or ΔCv is constant over a large temperature range or when explicit temperature dependence is provided.
- ✕Errors in integrating the temperature-dependent heat capacity functions or in unit conversions (e.g., R value, Cp/Cv units).
- ✕Not using absolute temperature (Kelvin) for all calculations.
📝 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