The correct statement amongst the following is :

<p>Option A: </p> <p><p>It claims that the standard state for a pure gas is defined at 1 bar and 273 K. </p></p> <p><p>In modern thermochemistry, the standard state generally means the pure substance at 1 bar, but the standard temperature most commonly used is 298 K (25°C), not 273 K (0°C). </p></p> <p><p>Thus, Option A is incorrect.</p></p> <p>Option B: </p> <p><p>It states that the standard enthalpy of formation $\Delta_f H^\circ_{500}$ is zero for $O_2(g)$. </p></p> <p><p>By convention, the standard enthalpy of formation for an element in its most stable form is defined as zero. Even though typical tables list $\Delta_f H^\circ_{298}$ values, if one were to evaluate the formation enthalpy at any temperature where the element remains in its standard state, the reaction </p> <p>$O_2(g) \rightarrow O_2(g)$ </p> <p>has no change in enthalpy. </p></p> <p><p>Therefore, by definition, $\Delta_f H^\circ_{500}(O_2(g)) = 0$. </p></p> <p><p>This makes Option B correct.</p></p> <p>Option C: </p> <p><p>It claims that $\Delta_f H^\circ_{298}$ is zero for $O(g)$ (atomic oxygen). </p></p> <p><p>However, the standard state of oxygen is diatomic $O_2(g)$, not atomic oxygen. Since $O(g)$ is not the most stable form of oxygen under standard conditions, its formation enthalpy is not zero. </p></p> <p><p>Hence, Option C is incorrect.</p></p> <p>Option D: </p> <p><p>It asserts that the term "standard state" implies that the temperature is 0°C (273 K). </p></p> <p><p>In fact, "standard state" typically specifies a standard pressure (1 bar) and a chosen temperature—commonly 298 K for tabulated thermodynamic data—not necessarily 0°C. </p></p> <p><p>Thus, Option D is also incorrect.</p></p> <p>To summarize:</p> <p>The only correct statement is Option B.</p> <p>Therefore, the correct answer is: Option B.</p>