# Explain why spontaneity depends on temperature?

Mar 19, 2016

Spontaneity is most easily defined by a thermodynamic equation you've probably seen before:

$\setminus m a t h b f \left(\Delta G = \Delta H - T \Delta S\right)$

where:

• $\Delta G$ is the Gibbs' free energy, which tells you when a reaction is spontaneous, non-spontaneous, or at equilibrium.
• $\Delta H$ is the enthalpy, which is the heat flow $q$ at a constant pressure.
• $\Delta S$ is the entropy, which is directly proportional to the number of microstates that can pertain to a system, i.e. loosely speaking, a measure of "disorder".

Since $T$ is a variable in the function for $\Delta G$, and since a reaction is spontaneous when $\Delta G < 0$, at equilibrium when $\Delta G = 0$, and nonspontaneous when $\Delta G > 0$, spontaneity depends on temperature.

EXAMPLE: ENTROPY

For example, let's consider this reaction:

$2 {\text{Al"(s) + 6"HCl"(aq) -> 2"AlCl"_3(aq) + 3"H}}_{2} \left(g\right)$

You can see that there are more $\text{mol}$s of gas on the products side, so $\textcolor{g r e e n}{\Delta S > 0}$; the entropy of a gas is significantly higher than that of a solid or liquid.

So significant that in many General Chemistry classes, a general rule of thumb is used where the sign of the entropy can often be found from the $\text{mol}$s of gas on each side.

EXAMPLE: ENTHALPY

The following bonds were broken:

Therefore, the total enthalpy of the reaction is approximately:

$\textcolor{g r e e n}{\Delta {H}_{\text{rxn") = sum_i DeltaH_("broken",i) - sum_j DeltaH_("made}} , j}$

$= \left(\text{431 kJ/mol") - ("436 kJ/mol" + "502 kJ/mol}\right)$

$= \textcolor{g r e e n}{- \text{507 kJ/mol}}$

EXAMPLE: GIBBS' FREE ENERGY

Interestingly enough, since $\Delta H < 0$ and $\Delta S > 0$, what we have is a situation where the reaction is spontaneous all the time.

Remember that $T > 0$ if in $\text{K}$, and we know right now that $\Delta S > 0$. Therefore, we have:

$\textcolor{b l u e}{\Delta G = - | \Delta H | - | T | \cdot | \Delta S | < 0}$

Therefore, this reaction is spontaneous for all temperatures.

However, had $\Delta H$ been positive, then the spontaneity would depend on the temperature; if so, then:

• If $T$ is high enough, the reaction is spontaneous because $| \Delta H | < | T | \cdot | \Delta S |$ and thus $\Delta G < 0$.
• If $T$ is low enough, then the reaction is nonspontaneous because $| \Delta H | > | T | \cdot | \Delta S |$ and thus $\Delta G > 0$.
• If $\Delta H = T \Delta S$, the reaction is at equilibrium because $\Delta G = 0$.