# Enthalpy of neutralization of strong acid and strong base differing from enthalpy of formation of water from ions

If the neutralization between strong acid and bases has enthalpy of neutralization of around $$\pu{-57.1 kJ mol^-1}$$, why is it that when I try to calculate the enthalpy of formation of water from $$\ce{H+}$$ and $$\ce{OH-}$$ ions I get approximately $$\pu{-55.8 kJ mol^-1}$$ using the reference values?

Why is there this difference in enthalpy if the reaction between say $$\ce{NaOH}$$ and $$\ce{HCl}$$ is only between $$\ce{H+}$$ and $$\ce{OH-}$$ ions, while $$\ce{Na}$$ and $$\ce{Cl}$$ remain dissociated?

\begin{align} \ce{H2(g) + 1/2 O2(g) &-> H2O(l)} &\quad \Delta_\mathrm{f,1}H &= \pu{-285.8 kJ mol-1}\tag{1}\\ \ce{1/2 H2(g) &-> H+(aq) + e-} &\quad \Delta_\mathrm{f,2}H &= 0 \tag{2}\\ \ce{1/2 H2(g) + 1/2 O2(g) + e- &-> OH-(aq)}&\quad \Delta_\mathrm{f,3}H &= \pu{-230 kJ mol-1} \tag{3} \end{align}

Computing $$\Delta_\mathrm{f,1}H - \Delta_\mathrm{f,2}H - \Delta_\mathrm{f,3}H$$ gives us the reaction $$\ce{H+ (aq) + OH- (aq) -> H2O (l)}$$ with the associated enthalpy change $$\Delta_\mathrm{f}H(\ce{H2O(l)}) = \pu{-55.8 kJ mol-1}.$$

• Good first question, there are some things to consider. Firstly there might be the possibility that the different answers are due to differences in the values measured for a given thing by different authors. Secondly it looks like you are using a Hess's law method, the thing is that errors might build up there. Lastly what do you mean by "strong acids and strong bases", do you mean an infinitely dilute solution of both or do you mean something like 4 M nitric acid and 5 M sodium hydroxide ? Jul 6 '18 at 4:22
• @NuclearChemist, regarding the Hess's law method, what are the possible errors that can build up in there? The strong acids and base I am referring to are dilute solution of strong HCl and NaOH, e.g. 2M HCl, but not infinitely dilute. (I am not sure what infinitely dilute means) Jul 6 '18 at 4:28
• Well if you have A forming B, and the measured delta H is 100 +/- 1 kJ mol-1. If you measure A to C, C to D, D to E and E to B all with an error of +/- 1 kJ mol-1. Under some conditions the effect of the errors will build up so this Hess's law based method will give a worse answer than a more direct measurement of A to B. Jul 6 '18 at 4:48 