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20.6: Temperature and Free Energy

Difficulty Level: At Grade Created by: CK-12
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Practice Temperature and Free Energy
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The production of steel is nonspontaneous at low temperatures, but spontaneous at higher temperatures

How is steel produced?

Iron ore (Fe2O3) and coke (an impure form of carbon) are heated together to make iron and carbon dioxide. The reaction is non-spontaneous at room temperature, but becomes spontaneous at temperature above 842 K. The iron can then be treated with small amounts of other materials to make a variety of steel products.

Temperature and Free Energy

Consider the reversible reaction in which calcium carbonate decomposes into calcium oxide and carbon dioxide gas. The production of CaO (called quicklime) has been an important reaction for centuries.

\begin{align*}\text{CaCO}_3(s) \rightleftarrows \text{CaO}(s) +\text{CO}_2(g)\end{align*}

The \begin{align*}\Delta H^\circ\end{align*} for the reaction is 177.8 kJ/mol, while the \begin{align*}\Delta S^\circ\end{align*} is 160.5 J/K • mol. The reaction is endothermic with an increase in entropy due to the production of a gas. We can first calculate the \begin{align*}\Delta G^\circ\end{align*} at 25°C in order to determine if the reaction is spontaneous at room temperature.

\begin{align*}\Delta G^\circ=\Delta H^\circ -T \Delta S^\circ=177.8 \ \text{kJ}/ \text{mol} - 298 \ K (0.1605 \ \text{kJ} / \text{K} \cdot \text{mol})=130.0 \ \text{kJ} / \text{mol}\end{align*}

Since the \begin{align*}\Delta G^\circ\end{align*} is a large positive quantity, the reaction strongly favors the reactants and very little products would be formed. In order to determine a temperature at which \begin{align*}\Delta G^\circ\end{align*} will become negative, we can first solve the equation for the temperature when \begin{align*}\Delta G^\circ\end{align*} is equal to zero.

\begin{align*}0 &=\Delta H^\circ - T \Delta S^\circ \\ T &=\frac{\Delta H^\circ}{\Delta S^\circ}= \frac{177.8 \ \text{kJ} / \text{mol}}{0.1605 \ \text{kJ} / \text{K} \cdot \text{mol}}=1108 \text{ K}=835^\circ \text{ C}\end{align*}

So at any temperature higher than 835°C, the value of \begin{align*}\Delta G^\circ\end{align*} will be negative and the decomposition reaction will be spontaneous.

At high temperatures, reactions with positive Delta S can become thermodynamically favorable

This lime kiln in Cornwall was used to produce quicklime (calcium oxide), an important ingredient in mortar and cement.

Recall that the assumption that \begin{align*}\Delta H^\circ\end{align*} and \begin{align*}\Delta S^\circ\end{align*} are independent of temperature means that the temperature at which the sign of \begin{align*}\Delta G^\circ\end{align*} switches from being positive to negative (835°C) is an approximation. It is also important to point out that one should not assume that absolutely no products are formed below 835°C and that at that temperature decomposition suddenly begins. Rather, at lower temperatures, the amount of products formed is simply not great enough to say that the products are favored. When this reaction is performed, the amount of products can be detected by monitoring the pressure of the CO2 gas that is produced. Above about 700°C, measurable amounts of CO2 are produced. The pressure of CO2 at equilibrium gradually increases with increasing temperature. Above 835°C, the pressure of CO2 at equilibrium begins to exceed 1 atm, the standard-state pressure. This is an indication that the products of the reaction are now favored above that temperature. When quicklime is manufactured, the CO2 is constantly removed from the reaction mixture as it is produced. This causes the reaction to be driven towards the products according to LeChâtelier’s principle.


  • The influence of temperature on free energy is described.



Read the material at the link below and answer the following questions:


  1. Is an exothermic reaction where entropy increases spontaneous or nonspontaneous?
  2. Give an example of a process where a process proceeds spontaneously only below a certain temperature.
  3. If \begin{align*}\Delta H > 0\end{align*} and \begin{align*}\Delta S < 0\end{align*}, can the reaction ever be spontaneous?



  1. If you increased the pressure of CO2 in the quicklime reaction, what would happen to the equilibrium?
  2. Why do we calculate the situation where \begin{align*}\Delta G\end{align*} is zero?
  3. At temperatures below 835°C, is any product formed?

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Difficulty Level:

At Grade



Date Created:

Jun 25, 2013

Last Modified:

Sep 09, 2015
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