JEE Chemistry 2025 Question with Solution

Given: Four statements about standard state and standard enthalpy of formation are given.

Find: The correct statement.

The definition of standard state fixes the pressure at $1 \, \text{bar}$, but it does not fix the temperature. Thermodynamic data are often tabulated at $298.15 \, \text{K}$, but that is not the definition of standard state itself.

Now evaluate each option:

1. Option A is incorrect because standard state does not imply temperature $0^\circ C$.
2. Option B is incorrect because for a pure gas the standard pressure is $1 \, \text{bar}$, but the temperature is not fixed at $273 \, \text{K}$.
3. Option C is incorrect because $O(g)$ is atomic oxygen, not the most stable standard form of elemental oxygen. Therefore, $\Delta_f H_{298}^\ominus$ for $O(g)$ is not zero.
4. Option D is correct because $O_2(g)$ is the standard reference state of oxygen, so its standard enthalpy of formation is zero at the specified temperature.

By definition, the standard enthalpy of formation of an element in its most stable form at the specified temperature and standard pressure is

Hence for $O_2(g)$ at $500 \, \text{K}$,

Therefore, the correct option is D.

Given: Statements about standard state and enthalpy of formation.

Find: Which statement is correct.

Key principle: The standard enthalpy of formation of an element in its most stable allotropic form under standard pressure at the stated temperature is zero.

• A: Incorrect. Standard state does not mean temperature $0^\circ C$.
• B: Incorrect. Standard state for a gas requires pressure $1 \, \text{bar}$, but not a fixed temperature of $273 \, \text{K}$.
• C: Incorrect. $O(g)$ is monatomic oxygen, not the stable reference form at $298 \, \text{K}$.
• D: Correct. $O_2(g)$ is the stable elemental form of oxygen, so its formation enthalpy is zero even when written at $500 \, \text{K}$.

Thus, the correct statement is $\Delta_f H_{500}^\ominus$ is zero for $O_2(g)$, so the correct option is D.