At a particular temperature, H+(aq) + OH− (aq) → H2O(l); ∆H = −57.1 kJ The approximate heat evolved when 400 ml of 0.2 M H2SO4 is mixed with 600 ml of 0.1 M KOH solution will be:
Correct answer: A. 3.426 kJ
- A. 3.426 kJ
- B. 13.7 kJ
- C. 5.2 kJ
- D. 55 kJ
Explanation
The balanced chemical equation for the neutralization is: H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l). The enthalpy change (∆H) for the neutralization of 1 mole of H+ with OH− is -57.1 kJ. Calculate the moles of H+ and OH−: 0.4 L × 0.2 M = 0.08 mol of H+ and 0.6 L × 0.1 M = 0.06 mol of OH−. The limiting reactant is OH−, with 0.06 mol available to react. Multiply the moles of the limiting reactant by the enthalpy change per mole: 0.06 mol × -57.1 kJ/mol = -3.426 kJ. Therefore, the heat evolved is 3.426 kJ. Other options either miscalculate the stoichiometry or misapply the enthalpy value.
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About Enthalpy
Enthalpy is the heat content of a system measured at constant pressure, and its change indicates whether a reaction or physical process is endothermic or exothermic. The topic covers state functions, enthalpy diagrams, standard enthalpy of formation, combustion and neutralisation, and the relationship between ΔH and internal energy change.
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