Concept:The standard enthalpy of atomization (\( \Delta H_{at}^\circ \)) is specifically defined as the enthalpy change when
1 mole of gaseous atoms is formed from the element in its standard state.
Formula:$$ \frac{1}{2} H_{2(g)} \rightarrow H_{(g)} \quad \Delta H = +\text{ve} $$
Solution:- Hydrogen's standard state is the diatomic gas, \( H_{2(g)} \).
- To form exactly 1 mole of \( H_{(g)} \) atoms, we must break half a mole of \( H-H \) bonds. The balanced equation is: \( \frac{1}{2} H_{2(g)} \rightarrow H_{(g)} \).
- Because bond breaking always requires energy, the process is endothermic and the sign must be positive (\( +218 \text{ kJmol}^{-1} \)).
Why other options are incorrect:Option D has the right equation but a negative sign (implies atomization releases energy, which is false). Options B and C describe the decomposition of water, not the atomization of elemental hydrogen.
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