Standard Enthalpy — Formation, combustion, neutralization
Thermodynamics & Thermochemistry
9
JEE Qs
8%
Hard
75
min
Master writing and manipulating thermochemical equations, meticulously applying Hess's Law with correct stoichiometric coefficients and signs for formation and combustion enthalpies.
🧮 Key Formulas
✅ Key Points for JEE
- 1Standard state refers to a specific set of conditions: 1 bar pressure, specified temperature (usually 298 K), and 1 M concentration for solutions.
- 2The standard enthalpy of formation (ΔfH°) for an element in its most stable reference state is defined as zero (e.g., O2(g), C(graphite), Br2(l), H2(g)).
- 3Hess's Law enables calculation of standard enthalpy of reaction (ΔrH°) using either standard enthalpies of formation (products - reactants) or standard enthalpies of combustion (reactants - products).
- 4Standard enthalpy of combustion (ΔcH°) is always exothermic (negative) as it involves the complete burning of one mole of a substance in excess oxygen.
- 5The enthalpy of neutralization (ΔneutH°) for a strong acid-strong base reaction is approximately -57.3 kJ/mol, as it primarily involves the formation of one mole of water from H+ and OH- ions.
⚠️ Common Mistakes
- ✕Failing to balance chemical equations correctly or using incorrect stoichiometric coefficients in Hess's Law calculations.
- ✕Incorrectly assigning ΔfH° = 0 to elements not in their most stable standard state (e.g., O3(g), C(diamond)) or to compounds.
- ✕Making sign errors while applying Hess's Law, especially when subtracting reactant values or dealing with reversed reactions.
- ✕Confusing the application of Hess's Law for formation enthalpies (products - reactants) with combustion enthalpies (reactants - products).
- ✕Ignoring the specified phase of a substance (e.g., H2O(g) vs H2O(l)) when using tabulated enthalpy values.
📝 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