Question 9.5.5: Standard Heat of a Neutralization Reaction 1. Calculate ΔH°r......
Standard Heat of a Neutralization Reaction
1. Calculate ΔH^{°}_{r} for the reaction
H_{3}PO_{4}(aq, r = ∞) + 3NaOH(aq, r = 50) → Na_{3}PO_{4}(aq, r = ∞) + 3H_{2}O(l)
2. If 5.00 mol of NaOH dissolved in 250 mol of water is neutralized completely at 25°C with dilute phosphoric acid, what is the attendant enthalpy change?
1. H_{3}PO_{4}(aq): Δ\hat{H}^{°}_{f} = – 309.3 kcal/mol = – 1294 kJ/mol [from p. 2-188 of Perry’s Chemical Engineers’ Handbook (see Footnote 2)].
NaOH (aq, r = 50): (\Delta \hat{H}^{°}_{f})_{NaOH(aq)} = (\Delta \hat{H}^{°}_{f})_{NaOH(s)} + (\Delta \hat{H}^{°}_{s}) (r=50) \\ \left. \Large{\Downarrow} \right. \begin{matrix} Table B.1 (\Delta \hat{H}^{°}_{f}) \\Table B.11 (\Delta \hat{H}^{°}_{s})\end{matrix} \\= (-426.6 – 42.51) kJ/mol =- 469.1 kJ/mol
Na_{3}PO_{4}(aq): Δ\hat{H}^{°}_{f} = – 471.9 kcal/mol = – 1974 kJ/mol (from p. 2-192 of Perry’s Chemical Engineers’ Handbook).
2. If 5 mol of dissolved NaOH is neutralized, then
\Delta H(25°C) = \begin{array}{c|c} – 130.1 kJ& 5.00 mol NaOH \\ \hline 3.00 mol NaOH&\end{array} = \boxed{- 217 kJ}