The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to H+ ions; that is, =ΔHofH+aq0.HClg H2O +H+aq Cl−aq =ΔHorxn−74.7kJmolNote: Reference the Thermodynamic properties of pure substances table for additional information.Part 1 of 2For the following reaction, calculate ΔHof for the Cl− ions. Be sure your answer has the correct number of significant digits.−74.7Part 2 of 2Calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong base (such as KOH) at 25°C. Be sure your answer has the correct number of significant digits.−57.1CheckSave For LaterSubmit Assignment
Question
The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to H+ ions; that is, =ΔHofH+aq0.HClg H2O +H+aq Cl−aq =ΔHorxn−74.7kJmolNote: Reference the Thermodynamic properties of pure substances table for additional information.Part 1 of 2For the following reaction, calculate ΔHof for the Cl− ions. Be sure your answer has the correct number of significant digits.−74.7Part 2 of 2Calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong base (such as KOH) at 25°C. Be sure your answer has the correct number of significant digits.−57.1CheckSave For LaterSubmit Assignment
Solution
Part 1 of 2
The enthalpy of formation (ΔHof) for Cl- ions can be calculated from the given reaction:
HCl(g) -> H2O + H+(aq) + Cl-(aq)
The enthalpy change for this reaction (ΔHorxn) is given as -74.7 kJ/mol. Since the enthalpy of formation of H+ ions is defined as zero, the enthalpy of formation of Cl- ions is equal to the enthalpy change of the reaction. Therefore, ΔHof for Cl- ions is -74.7 kJ/mol.
Part 2 of 2
The enthalpy of neutralization is the heat change when one mole of an acid reacts with one mole of a base to form one mole of water. For a strong monoprotic acid like HCl and a strong base like KOH, the reaction is:
HCl(aq) + KOH(aq) -> KCl(aq) + H2O(l)
The enthalpy change for this reaction is typically around -57.1 kJ/mol at 25°C. This is the enthalpy of neutralization.
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