Concept:The standard enthalpy of formation (\( \Delta H_f \)) is strictly defined as the heat evolved or absorbed when exactly
one mole of a compound is formed. Extensive properties like enthalpy scale linearly with molar quantities.
Formula:$$ 2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)} \quad \Delta H = -484 \text{ kJ} $$
Solution:- The problem states that producing 2 moles of \( H_2O \) releases \( 484 \text{ kJ} \) of heat (meaning \( \Delta H = -484 \text{ kJ} \)).
- To find the heat of formation for 1 mole, divide the entire reaction equation and its enthalpy by 2.
- $$ \Delta H_f = \frac{-484 \text{ kJ}}{2} = -242 \text{ kJ/mol} $$
- The negative sign must be retained because the word "evolved" indicates an exothermic process.
Why other options are incorrect:Option D is the enthalpy for two moles, not one. Option A has the wrong sign (implies endothermic). Option B results from incorrectly dividing by 4 instead of 2.
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