Thermodynamically Favored

Thermodynamically Favorable Meaning

Before exploring what drives reactions to be thermodynamically favorable, let's review the laws of thermodynamics, entropy (S), and enthalpy (H).

There are three laws of thermodynamics. The first law states that the universe has a constant energy. In other words, energy cannot be created nor destroyed, only transferred between the system and the surroundings.

$$\Delta E_{universe}\cup =\Delta E_{system}+\Delta E_{surroundings}=0$$

The second law of thermodynamics tells us that, over time, the entropy of the universe will increase. Now, the third law of thermodynamics states that the entropy of a perfect crystal at zero K is 0.

Entropy vs. Enthalpy

Let's look at the definitions of entropy and enthalpy. Entropy (S) is a measurement of the disorder (randomness) of a system. The larger this value, the greater the randomness a system has. The change in entropy is given by the following equation.

$$\Delta S^{o}=S^{o}_{products}-S^{o}_{reactants}$$

Enthalpy (H), is referred to as the heat energy that a reaction has. Basically, if the change in enthalpy (ΔH) is less than zero, then the reaction will be exothermic. If ΔH surpasses zero, the reaction is considered endothermic.

The formula for change in enthalpy is shown below.

$$\Delta H^{o}=\Sigma n\Delta H^{o}_{products}-\Sigma n\Delta H^{o}_{reactants}$$

Now, let's define thermodynamically favorability.

Being thermodynamically favorable means that the reaction can proceed without any assistance from outside the system. For example, NaCl dissolving in water is a thermodynamically favored process

A thermodynamically favored process is thermodynamically favored in one direction, and non-thermodynamically favored in the other direction. But, what does this mean? Let's use water as an example. Water can go from liquid to solid at -10 °C, but it cannot go from solid to liquid at this temperature. So,

• ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)$is thermodynamically favored at -10 °C.
• ${\mathrm{H}}_2\mathrm{O}\left(\mathrm{s}\right)\to {\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$is non-thermodynamically favored at -10 °C.

A non-thermodynamically favored process is considered a process that needs assistance from outside the system. These reactions can be forced to proceed if we apply an external source of energy to it.

Gibbs Free Energy

The standard Gibbs Free Energy change is used by chemists as a measure of thermodynamic favorability. The equations below can be used to calculate ΔG°.

$$\Delta G^{o}_{reaction}=\Sigma [G^{o}_{f}(products)]-\Sigma [G^{o}_{f}(reactants)]$$

$$\Delta G^{o}_{reaction}=\Delta H^{o}-T\Delta S^{o}$$

Where:

• ΔG° is the standard Gibbs Free Energy change.
• ΔG°_f is the standard free energy of formation.
• ΔH° is the change in enthalpy.
• T is the temperature.
• ΔS° is the change in entropy.

When dealing with Gibbs Free Energy, we can say that a reaction will be thermodynamically favored if ΔG° < 0 and there is no need for an external source of energy.

Thermodynamically Favored Reactions

Think about exothermic and endothermic reactions, and choose which one of these you think is most often thermodynamic favored. If you guessed exothermic reactions, then you are right!

Exothermic reactions are usually thermodynamically favored because they release energy to the surroundings. However, some endothermic reactions can also be thermodynamically favored. For example, evaporation is an endothermic process that is thermodynamically favored.

• The enthalpy change (ΔH) for an endothermic reaction is always positive.

$$H_{2}O_{(l)}\rightarrow H_{2}O_{(g)}\ \ \ \ \Delta H^{o}=+40.7kJ$$

$$C_{3}H_{8\ (l)}\rightarrow C_{3}H_{8\ (g)}\ \ \ \ \ \Delta H^{o}=+16.7kJ$$

Increases in entropy (S) are also thermodynamically favorable. The general rule is that if the entropy of the universe (thermodynamic system plus its surroundings) is positive, then the reaction is thermodynamically favored. If ΔS_universe is negative, then the reaction is not thermodynamically favored.

$$\Delta S_{universe}=\Delta S_{system}+\Delta S_{surroundings}$$

When does entropy increase? Entropy increases when:

• During a reaction, the number of molecules increases.
• Temperature increases.
• A gas is formed from a liquid or solid.
• There is an increase in volume.
• The number of moles of gas increases because of a chemical reaction.

Thermodynamic Favorability Chart

To make things simpler, let's make a chart showing thermodynamic favorability, considering changes in entropy and changes in enthalpy.

Thermodynamically Favorable vs Kinetically Favorable

Let's look at the difference between reactions that are thermodynamically favorable and those that are kinetically favorable.

A reaction that is thermodynamically favorable usually occurs at slow rates and higher temperatures in a reversible reaction, and form stable products known as thermodynamic products.

Kinetically favorable reactions are the opposite; they are fast, can be carried out at low temperatures, and lead to the formation of unstable products. The products of kinetically favored reactions are called kinetic products.

As an example, let's use the addition of hydrogen bromide (HBr) to 1,3 butadiene. This reaction is commonly seen in organic chemistry courses, and it is an example of conjugated hydrohalogenation, or hydrohalogenation of dienes.

• At low temperature, the major product formed is the 1,2-product. This is the kinetic product.
• At high temperature, the major product formed is the 1,4-product. This is the thermodynamic product, and it is more stable than the kinetic product.

Thermodynamically Favored Reaction Examples

To top it off, let's look at an example involving a thermodynamically favored reaction. This first example asks us to determine if the following reaction is thermodynamically favored.

Now, I hope that you feel more familiar with the concept of thermodynamic favorability!