If 5.90 kJ of heat evolved from a neutralization of 156 mmols of a monoprotic acid, what is the molar heat of neutralization (in kJ/mol) of that acid

37.82kJ/mol is the molar heat of neutralization

Explanation:

The neutralization of a monoprotic acid, HX, with NaOH as example is:

HX + NaOH → H₂O + NaX + ΔH

Where ΔH is molar heat of neutralization (The heat that is released when 1 mole of HX is neutralized).

As neutralization of 156mmol = 0.156mol HX release 5.90kJ. When 1 mole is neutralized the released heat is:

ΔH = 5.90kJ / 0.156moles

ΔH = 37.82kJ/mol is the molar heat of neutralization