ThermochemistryHard

Question

Enthalpy of neutralization of oxalic acid is −25.4 kcal/mol using strong base, NaOH. Enthalpy change for the process H2C2O4(aq) → 2H+ (aq) + C2O42–(aq) is about

Options

A.2.0 kcal
B.−11.7 kcal
C.1.0 kcal
D.4.0 kcal

Solution

$\Delta H = ( - 25.4) - 2 \times ( - 13.7) = 2.0\text{ kcal/mol}$

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