Concept:In thermodynamics, the stability of a compound is directly tied to its standard enthalpy of formation (\( \Delta H_f^\circ \)). The more negative the enthalpy of formation, the more energy was lost, and the deeper the compound sits in a potential energy well.
Formula:$$ \Delta H = -692 \text{ kJmol}^{-1} $$
Solution:- The formation of \( MgO \) releases a massive amount of energy (\( 692 \text{ kJ} \) for every mole).
- Because so much internal energy has been lost to the surroundings, the resulting solid \( MgO \) lattice is in a much lower energy state than the starting elements.
- Lower energy directly correlates with higher thermodynamic stability. Hence, the product is very stable.
Why other options are incorrect:The reaction is clearly exothermic (negative sign), ruling out endothermic. Because the reactants spontaneously lost so much energy to become \( MgO \), they were relatively less stable compared to the product.
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