I was studying Thermodynamics from my Chemistry textbook (page 181-182, section 6.6 (a)).

I have a question regarding the following text from my book.

Now let us examine the following reactions: $$\begin{align} \ce{1/2 N2(g) + O2(g) &-> NO2(g)} & \Delta_\mathrm{r}H^⦵ &= \pu{+33.2 kJ mol-1} \\ \ce{C (graphite, s) + 2 S(l) &-> CS2(l)} & \Delta_\mathrm{r}H^⦵ &= \pu{+128.5 kJ mol-1} \end{align}$$These reactions though endothermic, are spontaneous.

I vistied the Wikipedia page on Standard Gibbs free energy of formation to check the values of $\Delta_\mathrm{f} G^⦵$ for $\ce{NO2}$ and $\ce{CS2}$ and verify if the above stated reactions were truly spontaneous. It turns out that $\Delta_\mathrm{f} G^⦵$ for $\ce{NO2}$ is $\pu{51.3 kJ mol-1}$; and $\Delta_\mathrm{f} G^⦵$ for $\ce{CS2}$ is $\pu{67.1 kJ mol-1}$.

I understand that as the change in entropy involved in the formation of $\ce{CS2}$ is positive, this reaction can be spontaneous at high temperatures even though it is not at $298 \pu K$.

This is not the case with the formation of $\ce{NO2}$ though. The change in entropy associated with its formation is negative and so $\Delta_\mathrm{f} G^⦵$ will always be positive. This means that this reaction will never be spontaneous.

Then why does my textbook state so? Is that statement (bold text) wrongly stated or am I understanding it wrong?

  • 1
    $\begingroup$ If $\Delta G°$ is positive, the reaction is not spontaneous. If $\Delta G°$ is negative, the reaction is spontaneous, but it can be rather slow, so slow that it does not appear to happen even for years. Your inverse reactions, namely the decompositions of $\ce{NO2}$ or of $\ce{CS2}$ are good examples of this extremely slow behavior. $\endgroup$
    – Maurice
    Mar 14 at 12:33
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    $\begingroup$ Hi, please note my edits, particularly the MathJax usage. \ce{...} is supposed to be used only for chemical formulae and equations. $\Delta G$ isn't a chemical formula or equation, you shouldn't use it there, as it leads to the $G$ being incorrectly written upright. Then, your physical quantities with units (like x kJ/mol) are much better typeset using $\pu{x kJ mol-1}$ rather than a mixture of \text and \ce. You may find chemistry.meta.stackexchange.com/q/86/16683 useful as a guide. $\endgroup$ Mar 14 at 12:50
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    $\begingroup$ Note that $\Delta G_\mathrm{f}(T)=\Delta H_\mathrm{f}(T)-T\Delta S_\mathrm{f}(T)$ and that for $\Delta S_\mathrm{f}(T) \gt 0$ and high enough T, those reactions may become spontaneous in spite of being endothermic. $\endgroup$
    – Poutnik
    Mar 14 at 13:12
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    $\begingroup$ @Silica19 Compare the respective $\Delta G_\mathrm{f}^°$ and $\Delta H_\mathrm{f}^°$ values to get the sign of $\Delta S_\mathrm{f}^°$. Additionally, for high temperatures, the standard values are not that important, but $\Delta H_\mathrm{f}(T)$ and $\Delta S_\mathrm{f}(T)$, using integration of $\int{\Delta C \mathrm{d}T}$ resp. $\int{\frac{\Delta C}{T} \mathrm{d}T}$ $\endgroup$
    – Poutnik
    Mar 14 at 14:42
  • $\begingroup$ @Poutnik I've added details about sign of entropy in my question accordingly. $\endgroup$
    – Silica19
    Mar 14 at 19:30


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