Concept:The standard enthalpy of neutralization (\( \Delta H_n \)) is the energy released (approx. \( -57.4 \text{ kJ} \)) for the formation of
exactly 1 mole of water from \( H^+ \) and \( OH^- \). To double this energy, a reaction must produce exactly
2 moles of water from strong acid-base components.
Formula:$$ H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)} \quad \Delta H \approx -57.4 \text{ kJ} $$
Solution:- The reference equation \( HCl + NaOH \rightarrow NaCl + H_2O \) generates 1 mole of water, releasing \( \sim 57.4 \text{ kJ} \).
- To yield "twice" the enthalpy change (\( 2 \times 57.4 = 114.8 \text{ kJ} \)), the reaction must yield 2 moles of water.
- Look at Option C: \( H_2SO_4 \) provides 2 moles of \( H^+ \), and \( Mg(OH)_2 \) provides 2 moles of \( OH^- \), generating exactly \( 2H_2O \). Therefore, the enthalpy change will be double.
Why other options are incorrect:Options A and D only produce 1 mole of water. Option B is a reaction with a carbonate, which involves the formation of \( CO_2 \) and not a simple strong acid-base double neutralization, so the thermochemistry differs.
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