I am hoping someone can help me with a clarifying a calculation on a fundamental thermodynamic understanding of reactions. Below is Table 2.5 from Stumm and Morgan's Aquatic Chemistry.

Table 2.5. Influence of Pressure and Temperature on the Energies of Water Ionization a $$ \begin{array}{rrr} \hline p (\pu{bars}) &\log K_\mathrm{w} &\Delta G_\mathrm{w}^\circ (\pu{kJ mol-1})\\ \hline &T = \pu{298.17 K} &\\ 1 &-14.00 & 79.94\\ 200 & -13.92 & 79.48\\ 400 & -13.84 & 79.02\\ 600 & -13.77 & 78.63\\ 800 & -13.70 & 78.22\\ 1000 & -13.63 & 77.83\\ \hline T (\pu{^\circ C}) &\log K_\mathrm{w} &\Delta G_\mathrm{w}^\circ (\pu{kJ mol-1})\\ \hline &p = \pu{1 bar} &\\ 0 & -14.93 & 85.25\\ 10 & -14.53 & 82.97\\ 20 & -14.17 & 80.91\\ 30 & -13.83 & 78.97\\ 50 & -13.26 & 75.71\\ \hline \end{array} $$ aData from Harned and Owen (1958).

I'd like to focus on the second portion of the table, where temperature is varied. $K_\mathrm{w}$ and $\Delta G_\mathrm{w}^\circ$ values are listed, but using the given $K_\mathrm{w}$ values, I calculate different $\Delta G_\mathrm{w}^\circ$ values than what the table offers. Here is a sample calculation for $\pu{50^\circ C}$:

$$K_\mathrm{w} = 10^{-13.26} = 5.49\cdot10^{-14}$$ $$\Delta G_\mathrm{w}^\circ = -RT\ln K$$ $$\Delta G_\mathrm{w}^\circ = -\pu{8.314e-3 kJ mol-1 K-1} \cdot \pu{323.15 K} \ln (\pu{5.49e-14}) = \pu{82.03 kJ mol-1}$$

The table lists $\Delta G_\mathrm{w}^\circ = \pu{75.71 kJ mol-1}$

Not only is this answer substantially different, but the value is heading in the opposite direction of where my calculations would head (more positive vs more negative). Unfortunately, I don't have access to the original source (Harned and Owen).

Can someone please help me to understand my error?


1 Answer 1


TL;DR: It appears that OP is right and has estimated standard Gibbs free energy correctly. The latter book accurately fetched data from the original source, but the calculations were carried out based on a false premise that $\Delta G^\circ$ refers to $\pu{25^\circ C}$ exclusively.

First, a general disclaimer. The uppercase circle ($^\circ$) generally denotes standard state. According to IUPAC, temperature is not declared within, but many authors attribute (negligently) $\pu{298.15 K}$ to Gibbs standard free energy $\Delta G^\circ$. Also check out What is the difference between ∆G and ∆G°?

For the given equilibrium one can use

$$\Delta G_T^\circ = -2.303RT\log K_T, \tag{1}$$

and at $\pu{298.15 K}$ this equation can be rewritten as follows, assuming $[\Delta G_{298.15}^\circ] = \pu{kJ mol-1}$:

$$\Delta G_{298.15}^\circ = -5.71\log K_{298.15}. \tag{2}$$

Now to the references. I didn't find the 3rd edition of The Physical Chemistry of Electrolyte Solutions by Harned and Owen (1958), but I do have an earlier version from 1943 [1]. In chapter 15 "The Ionization and Thermodynamic Properties of Weak Electrolytes" [1, p. 485], table 15-1-1 accumulates values for $K_\mathrm{w}$ determined electrometrically:

In subsequent investigations, the cells without liquid junction, were employed in certain combinations for the determination of the ioniza­tion constant itself, as well as $m_\ce{H}m_\ce{OH}$ and $\gamma_\ce{H}\gamma_\ce{OH}/a_\ce{H2O}$ salt solutions. \begin{align} &\ce{H2 | HX($m$) | AgX-Ag} \tag{I}\\ &\ce{H2 | MOH($m_1$), MX($m_2$) | AgX-Ag} \tag{II}\\ &\ce{H2 | HX($m_1$), MX($m_2$) | AgX-Ag} \tag{III} \end{align}

$$-\log K'_\mathrm{w} \equiv \frac{(\ce{E_{II} - \ce{E_{III}}})F}{2.303RT} - \log\frac{(m_\ce{H}m_\ce{X})_\mathrm{III}}{(m_\ce{X}/m_\ce{OH})_\mathrm{II}} + \\ + \frac{3.629 \times 10^6}{(DT)^{3/2}}\sqrt{\mu} = -\log K_\mathrm{w} + f(\mu) \tag{15-1-10}$$

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So, Stumm and Morgan [2, p. 57] correctly adapted the data from table 15-1-1:

\begin{array}{rr|rr} T (\pu{^\circ C}) &K_\mathrm{w} \times 10^{-14} {^{[1]}} &\log K_\mathrm{w}{^\text{calcd. [1]}} &\log K_\mathrm{w} {^{[2]}}\\ \hline 0 & 0.1139 &-14.9435 &-14.93\\ 10 & 0.292 &-14.5346 &-14.53\\ 20 & 0.6809 &-14.1669 &-14.17\\ 30 & 1.469 &-13.8330 &-13.83\\ 50 & 5.474 &-13.2617 &-13.26 \end{array}

But, what Stumm and Morgan did is they calculated the standard free energy exclusively at $\pu{25^\circ C}$ using the $K_\mathrm{w}$ values for other temperatures; disregarding the fact that equation (2) is only valid for $K_{298.15}$, but it was applied to all values as follows:

$$\Delta G_{298.15}^\circ {^{[2]}} = -5.71\log K_T, \tag{2a}$$

which is $\color{red}{\text{incorrect}}$. The latter column is populated with appropriate values calculated with (1), as it was done by OP:

\begin{array}{rr|rr} T (\pu{^\circ C}) &\log K_\mathrm{w}{^\text{calcd. [1]}} &\color{red}{\Delta G_{298.15}^\circ {^{[2], (2a)}} (\pu{kJ mol-1})} &\Delta G_T^\circ {^{(1)}} (\pu{kJ mol-1})\\ \hline 0 & 0.1139 &\color{red}{85.29} &78.14\\ 10 & 0.292 &\color{red}{82.96} &78.79\\ 20 & 0.6809 &\color{red}{80.86} &79.50\\ 30 & 1.469 &\color{red}{78.95} &80.28\\ 50 & 5.474 &\color{red}{75.69} &82.04 \end{array}


  1. Harned, H. S.; Owen, B. B. The Physical Chemistry of Electrolyte Solutions; American Chemical Society Monograph Series; Reinhold Publishing Corporation: New York, 1943; Vol. 95.
  2. Stumm, W.; Morgan, J. J. Aquatic Chemistry: Chemical Equilibria and Rates in Natural Waters, 3rd ed.; Environmental science and technology; Wiley: New York, 1996.
  • $\begingroup$ Thank you for both your post edits (still learning mathjax) and your very detailed answer. I can sleep better knowing this information! $\endgroup$ Sep 1, 2017 at 18:41
  • $\begingroup$ @prof.kvothe No prob at all, no one should ever stop learning:) $\endgroup$
    – andselisk
    Sep 1, 2017 at 18:44
  • $\begingroup$ Indeed, when I plot the new data as a function of temperature I get a slope of 0.0808 and an intercept of 55.81, which both match nicely to the S and H values calculated from std (298 K) tables (-0.08066 and 55.81, respectively). $\endgroup$ Sep 6, 2017 at 17:12

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