Exams › JEE Advanced › Chemistry
The heat of formation of a substance is defined as the enthalpy change when 1 mole of that substance is formed from its constituent elements in their gaseous standard states. Which of the following reactions does NOT represent the heat of formation?
- 1/2 N2(g) + 3/2 H2(g) -> NH3(g)
- C(s) + O2(g) -> CO2(g)
- H2(g) + O2(g) -> H2O(l)
- Xe(g) + 2F2(g) -> XeF4(s)
Correct answer: H2(g) + O2(g) -> H2O(l)
Solution
Option A: 1/2 N2 + 3/2 H2 -> NH3 is balanced and forms 1 mol NH3. Option B: C(s) + O2(g) -> CO2(g) is balanced and forms 1 mol CO2 from elements in standard states. Option C: H2(g) + O2(g) -> H2O(l) is NOT balanced (2 O atoms in, 1 O atom out) and does not represent a valid formation reaction. Option D: Xe(g) + 2F2(g) -> XeF4(s) is balanced and forms 1 mol XeF4. Hence option C does not represent heat of formation.
Related JEE Advanced Chemistry questions
- If the specific heat at constant pressure (Cp) is 10 cal at 1000 K, the initial temperature (T1) is 1000 K, the final temperature (T2) is 100 K, and the mass (m) is 32 g, what is the entropy change (ΔS) under constant pressure?
- For the reaction H2(g) + 1/2 O2(g) → H2O(l), where the enthalpy of formation (ΔHf) is given as −285.9 kJ mol⁻¹, determine the value of ΔHf.
- For the reaction C3H8(g) → 3C(g) + 8H(g), determine the enthalpy change (ΔHf) given that ΔHf for C2H2(g) → 2C(g) + 6H(g) is −620 kJ mol⁻¹.
- The standard enthalpies of formation of CO₂(g), H₂O(l) and glucose(s) at 25°C are -400 kJ/mol, -300 kJ/mol and -1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25°C is -
- Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is(are) -
- During the transformation of liquid water to water vapor at 100°C and standard atmospheric pressure, which of the following is accurate?
⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →