Start learning →

The standard enthalpy of formation $\Delta_f H^\circ$ of any element in its most stable form at the standard state is:

A$0$
B$-57.1\,\text{kJ mol}^{-1}$
C$+285.8\,\text{kJ mol}^{-1}$
DEqual to its bond enthalpy
Answer & Solution
Correct answer: A. $0$
By convention, the standard enthalpy of formation of an element in its most stable allotropic form at the standard state (typically 1 bar, 298 K) is taken to be zero. So $\Delta_f H^\circ(\mathrm{O_2, g}) = 0$, $\Delta_f H^\circ(\mathrm{C, graphite}) = 0$, etc. This convention provides a reference relative to which the formation enthalpies of compounds are measured.
Practise this question with Quill — JEE Advanced →

Keep practising Unit 6A1: Heat Transfer and Calorimetry

Browse all Unit 6A1: Heat Transfer and Calorimetry questions for JEE Advanced, or explore more Chemistry topics.

Start a Unit 6A1: Heat Transfer and Calorimetry lesson →