# What is an example of a free energy practice problem?

Dec 31, 2014

Most Gibbs free energy problems revolve around determining a reaction's spontaneity or the temperature at which a reaction either is or isn't spontaneous.

For example, determine if this reaction is spontaneous under standard conditions; knowing that the reaction's change is enthalpy is $\Delta {H}^{\circ} = - 144$ $\text{kJ}$, and its change in entropy is $\Delta {S}^{\circ} = - 36.8$ $\text{J/K}$.

$4 K C l {O}_{3 \left(s\right)} \to 3 K C l {O}_{4 \left(s\right)} + K C {l}_{\left(s\right)}$

We know that $\Delta {G}^{\circ} = \Delta {H}^{\circ} - T \cdot \Delta {S}^{\circ}$ for standard state conditions, which imply a pressure of 1 atm and a temperature of 298 K, so

$\Delta {G}^{\circ} = - 144 \cdot {10}^{3} \text{J" - 298 "K} \cdot \left(- 38.6 \frac{J}{K}\right) = - 133$ kJ"

If $\Delta {G}^{\circ} < 0$, the reaction is said to be spontaneous, so this reaction is spontaneous under standard conditions.

Let's determine for what temperatures this reaction will be spontaneous. In other words, we need to find a temperature at which the reaction stops being spontaneous.

A reaction is no longer spontaneous when $\Delta G > 0$, i.e. when $\Delta H - T \cdot \Delta S > 0$. So,

$\Delta H - T \cdot \Delta S > 0 \to \Delta H > T \cdot \Delta S \to \frac{\Delta H}{\Delta S} > T$

We get $\frac{- 144 \cdot {10}^{3} \text{J}}{- 38.6 \frac{J}{K}} = 3732 > T$, which means that this reaction is spontaneous for any $T < \text{3731 K}$.

Here's a link to some more Gibbs free energy examples;

http://www.chem.fsu.edu/chemlab/chm1046course/gibbs.html