Question:

Which of the following substance has the standard molar enthalpy of formation zero?

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Stable States to Remember: \(O_{2(g)}, H_{2(g)}, N_{2(g)}, Cl_{2(g)}, Br_{2(l)}, I_{2(s)}, S_{8(rhombic)}, P_{4(white)}, C(graphite)\). All these have \(\Delta_f H^\circ = 0\).
Updated On: Jun 24, 2026
  • HCl\(_{(g)}\)
  • H$_2$O\(_{(l)}\)
  • Br$_2$$_{(g)}$
  • CH$_4$$_{(g)}$
  • C(graphite) \textbf{Correct Answer:} (E) C(graphite)
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The Correct Option is

Solution and Explanation

Step 1: Understanding the Concept:
By convention, the standard molar enthalpy of formation (\(\Delta_f H^\circ\)) of an element in its most stable state of aggregation at 298 K and 1 bar is taken as zero.

Step 2: Detailed Explanation:

1. Compounds (HCl, H$_2$O, CH$_4$): These are not elements, so their enthalpies of formation are non-zero.
2. Bromine (Br$_2$): The standard state of bromine at 298 K is liquid. Therefore, \(\Delta_f H^\circ [Br_{2(l)}] = 0\), but \(\Delta_f H^\circ [Br_{2(g)}]\) is positive (heat of vaporization).
3. Carbon (C): Carbon has several allotropes (graphite, diamond, fullerenes). Graphite is defined as the most stable standard state. Hence, \(\Delta_f H^\circ [C(graphite)] = 0\).

Step 3: Final Answer:

C(graphite) has a standard molar enthalpy of formation of zero.
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