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In this chemistry problem, $\Delta H$ and $\Delta S$ are given at $298\text{ K}$, and it asks for the temperature at which the reaction becomes spontaneous:

For the reaction $2\text{SO}_3 \rightarrow 2\text{SO}_2 + \text{O}_2$, $\Delta H_r^\circ = +198 \text{ kJ/mol}$ and $\Delta S_r^\circ = 190 \text{ J / mol}\cdot\text{K}$. At what temperature will the forward reaction will become spontaneous?

I understand to solve this problem; the Gibbs free energy equation, $$\Delta G = \Delta H - T \Delta S,$$ can be used to determine the temperature by assuming $\Delta H$ and $\Delta S$ are constant and solving for $T$ when $\Delta G = 0$. This gives $T = 1042\text{ K}$.

However, I don't understand why is this assumption valid. Isn't it true that $\Delta S_r$ would change at different temperatures? Why is it that the value for $\Delta S$ for this reaction under standard conditions can be used to calculate $\Delta G$ at $1042 \text{ K}$?

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  • $\begingroup$ yes, $\Delta$G will change with temperature, the values given are likely for 298K but I'm not checking. there is also the matter that the reaction is assumed to occur at 1 atm and no consideration is given to an equilibrium mixture of the two gases. $\endgroup$
    – A.K.
    Oct 17, 2017 at 3:00
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    $\begingroup$ as an aside, spontaneous is a poor word for thermodynamic reactions given that some "spontaneous" reactions are kinetically hindered from reacting and require a catalyst or concentrated heat for activation. From this some people get the impression that the reaction will proceed the instant that the system reaches the temperature calculated from $\Delta$G, which is not true (think flash point vs. autoignition temperatures). a more proper term is "thermodynamically favored". $\endgroup$
    – A.K.
    Oct 17, 2017 at 3:06
  • $\begingroup$ You are correct that entropy and enthalpy should change with temperature. Are you sure enthalpy and entropy are given at 298K? Contrary to what many students (and even teachers!) seem to assume, the standard state for standard molar enthalpy and entropy takes no account of temperature. Related: en.wikipedia.org/wiki/Standard_enthalpy_of_formation $\endgroup$
    – Tyberius ♦
    Oct 17, 2017 at 3:29
  • $\begingroup$ @Tyberius Yes, the values are given at 298K. Could you elaborate on how standard molar enthalpy and entropy take no account of temperature? I thought that entropy would change greatly at different temperatures. $\endgroup$
    – Eric Zhang
    Oct 17, 2017 at 5:39
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    $\begingroup$ @EricZhang I don't mean to suggest that enthalpy and entropy don't depend on temperature. I was trying to say that standard enthalpy/entropy doesn't specify a temperature. You can have standard enthalpies/entropies at different temperatures. $\endgroup$
    – Tyberius ♦
    Oct 19, 2017 at 16:05

1 Answer 1

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The question implicitly assumes that $\Delta H^\circ$ and $\Delta S^\circ$ are independent of temperature.

The temperature dependences depends on the change in heat capcity ($\Delta C_p$) are given by $\Delta H^\circ(T_2) = \Delta H^\circ(T_1) + \Delta C_p(T_2-T_1)$ and $\Delta S^\circ(T_2) = \Delta S^\circ(T_1) + \Delta C_p \ln(T_2/T_1)$. (These equations assume $\Delta C$ is constant.)

So if $T_2-T_1$ or $\Delta C_p$ is small then $\Delta H^\circ$ and $\Delta S^\circ$ will be fairly constant.

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