Step 1: Understand the definition of standard state and standard enthalpy of formation.
Standard State: The standard state of a substance is a specific set of conditions chosen as a reference point for thermodynamic properties. By international convention, the standard pressure (\( P^\ominus \)) is 1 bar (\( 10^5 \) Pa). The temperature is not specified as part of the definition of standard state, although thermodynamic data are often tabulated at a standard temperature of 298.15 K (\( 25^\circ C \)). For a pure gas, the standard state is the pure gas at a pressure of 1 bar behaving ideally. For a pure liquid or solid, it is the pure substance at a pressure of 1 bar. For a solute in solution, it is a solution with a molality of 1 mol/kg behaving ideally.
Standard Enthalpy of Formation (\( \Delta_f H^\ominus \)): The standard enthalpy of formation of a compound is the change of enthalpy that accompanies the formation of 1 mole of the substance in its standard state from its constituent elements in their standard states. The standard enthalpy of formation of an element in its most stable allotropic form at the specified temperature (and 1 bar pressure) is zero.
Step 2: Evaluate each statement.
(1) The term 'standard state' implies that the temperature is \( 0^\circ C \).
This statement is incorrect. The standard state specifies a pressure of 1 bar, but the temperature is not fixed at \( 0^\circ C \) (273.15 K). While 273.15 K is a common reference temperature (especially in gas laws), it is not inherently part of the definition of standard state in thermodynamics.
(2) The standard state of pure gas is the pure gas at a pressure of 1 bar and temperature 273 K.
This statement is incorrect. The standard state of a pure gas is defined at a pressure of 1 bar, but the temperature is not fixed at 273 K. The temperature can be any specified value, although 298.15 K is most commonly used for tabulated thermodynamic data.
(3) \( \Delta_f H_{298}^\ominus \) is zero for O(g).
This statement is incorrect. \( O(g) \) is atomic oxygen, which is not the most stable allotropic form of the element oxygen at 298 K and 1 bar. The most stable allotropic form of oxygen under these conditions is diatomic oxygen, \( O_2(g) \). Therefore, \( \Delta_f H_{298}^\ominus \) for \( O(g) \) is not zero; it is the enthalpy change for the formation of \( O(g) \) from \( O_2(g) \), which requires energy to break the bond in \( O_2 \).
(4) \( \Delta_f H_{500}^\ominus \) is zero for \( O_2(g) \).
This statement is correct. \( O_2(g) \) is the most stable allotropic form of the element oxygen at 500 K and 1 bar (and at any temperature under standard pressure). By definition, the standard enthalpy of formation of an element in its most stable allotropic form at the specified temperature and standard pressure is zero.
Step 3: Identify the correct statement.
Based on the analysis, the correct statement is (4).