pling a) Calculate the standard heat of the neutralization reaction between dilu
ID: 859003 • Letter: P
Question
pling a) Calculate the standard heat of the neutralization reaction between dilute hydrochloric acid and dilute potassium hydroxide, given that the heat of solution of potassium hydroxide is -54.01 kJ/mol, the heat of formation of potassium hydroxide is -426.84 kJ/mol, the heat of solution of potassium chloride is 18.426 kJ/mol, and the heat of formation of potassium chloride is -436.584 kJ/mol. Number kJ mol b) Calculate the heat of reaction for the reaction of gaseous HCl and solid KOH to form liquid water and solid Number kJ mol AHExplanation / Answer
a) HCl(aq) + KOH(aq) ----> KCl(aq) + H20(L)
Heat of formation of KOH(s) = -426.84 Kj/mol
Heat of solution of KOH = -54.01 kj/mol
Heat of formation of HCl(g) = -92.31 Kj/mol
Heat of solution of HCl = ?74.89 kJ/mol
Heat of formation of KCl(s) = -436.584 Kj/mol
Heat of solution KCl = 18.426 kj/mol
Heat of formation of H2O(l) = -285.83 Kj/mol
DHrxn = (-436.584+18.426)+(-285.83) - (-426.84-54.01)+(-92.31?74.89)
= -55.938 kj/mol.
(or)
net ionic equation is
H+(aq) + OH-(aq) -----> H2O(l)
The reaction is the same in each case of a strong acid and(HCl) a strong alkali(KOH), the enthalpy change is similar.
so that
DHrxn = -57.9 kj/mol
b) HCl(aq) + KOH(s) ----> KCl(s) + H20(L)
Heat of formation of KOH(s) = -426.84 Kj/mol
Heat of formation of HCl(g) = -92.31 Kj/mol
Heat of formation of KCl(s) = -436.584 Kj/mol
Heat of formation of H2O(l) = -285.83 Kj/mol
DHrxn = (-436.584)+(-285.83)-(-92.31)+(-426.84)
= -722.414 +519.15
= -203.264 kj/mol
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