Question

During a neutralization reaction, a student combined 100 ml of 0.5 M HCl with 100 ml of 0.5 M of NaOH and recorded a change of temperature from 24.3 to 26.7. Determine ∆H for the given reaction (knowing that Ccal=335 J/°C):
NaOH(aq) + HCl(aq) →NaCl(aq) + H2O(L)

Answer 1

∆H = -335 × (26.7 - 24.3) / (100 × 0.5 / 1000) = -16080 J / mol = -16.08 kJ / mol

Answer: -16.08 kJ / mol.