Thermodynamics

The Hydrophobic Effect and Entropy vs. Enthalpy

By Curtis Bustos

Required equation & Key terms:
∆G = ∆H – T∆S used to determine whether a chemical-reaction process is spontaneous or not.
 

∆G = change in free energy
∆H = change in enthalpy (heat)
T = temperature in Kelvin (˚C + 273 = Kelvin (K)). Temperature affects ∆S.
∆S = change in entropy (disorder)
Spontaneous = favorable (-∆G) reaction.
Non-spontaneous = non-favorable (+∆G) reaction.
Exergonic = -∆G
Endergonic = +∆G
Favorable = ê∆H and é∆S. If ∆H is low and ∆S is high, then ∆G will be negative.
Non-favorable = é∆H and ê∆S. If ∆H is high and ∆S is low, then ∆G will be positive.

Order = more water molecules are required to “cage” or block non-polar molecules (see forward reaction).

Disorder = less water molecules are required to cage or block non-polar molecules and result in more freedom to move throughout a solution (see reverse reaction).

Making bonds = favorable ∆H.

Breaking bonds = non-favorable ∆H.

Understanding the hydrophobic effect & entropy vs. enthalpy consists of simply being familiar with terms and being able to algebraically manipulate ∆G = ∆H – T∆S. The reasoning behind +/- ∆G will then become apparent. 


Overall reaction:

Oil + Water ⥃ Solution

1) Forward reaction - mixing oil and water (initial) results in a non-spontaneous reaction:

Oil + Water à non-spontaneous reaction (+∆G)

Experiment shows a ê∆S or an increase-in-order (forward reaction). MORE water molecules are required to orderly “cage” the non-polar molecules. Remember, an increase in order = ê∆S. ê∆S contributes toward a non-favorable reaction.  

The forward reaction represents +∆G. The reverse reaction represents -∆G.

*NOTE – these figures assume that oil and water have equal densities.

The forward reaction is also the result of a decrease in ∆H. This makes sense as more water-water bonds are being formed to “cage” the non-polar molecules. SO, IF ∆H IS DECREASING (FAVORABLE), THEN WHY ISN’T THE FORWARD REACTION EXERGONIC? THERE IS ONLY ONE MORE CONTRIBUTING FACTOR, AND IT ISN’T HEAT. THE CONTRIBUTING FACTOR MUST HAVE A RELATIVELY LARGER ê∆S (UNFAVORABLE) THAN ê∆H (FAVORABLE)!


For example:
If,
there is a decrease in both enthalpy and entropy, we will need to consider their relative magnitudes to determine ∆G.  


And, 
∆G = ∆H (-1000 kJ/mol) - T∆S (-1200 kJ/mol • K) à  ∆G = ê∆H (-1200 kJ/mol) - Tê∆S (-1400 kJ/mol • K) à ∆G = +200 kJ/mol (non-spontaneous) with temperature being negligible when T > ∆H/∆S. Note - when T < ∆H/∆S, the reaction is spontaneous due to a major temperature influence.

Once again - forming water-water bonds is enthalpically favorable (-∆H). Nevertheless, while the formation of water-water bonds is enthalpically favorable, a relatively large ê∆S (order) promotes an endergonic reaction and prevents an exergonic reaction (in the forward direction) from occurring.

2) Reverse reaction - mixing oil and water (final) results in a spontaneous reaction:

Oil + Water ß spontaneous reaction (-∆G)

This suggests a é∆S or a decrease-in-order (reverse reaction). LESS water molecules are required to orderly “cage” the non-polar molecules. It takes less water molecules to cage one relatively-larger ball of oil than it does to cage five relatively-small balls of oil. In other words, after the five oil-balls aggregate into one, the remaining water molecules are allowed to be disordered and freely interact with other random water molecules as opposed to being obligated to interact with five orderly-cages. Remember, a decrease in order aka disorder = é∆S. é∆S contributes toward a favorable reaction. 

The forward reaction represents +∆G. The reverse reaction represents -∆G.

*NOTE – these figures assume that oil and water have equal densities.

Additionally, in the reverse reaction, there is an increase in ∆H. Bonds are being broken as fewer H-bonds are required after water reduces and minimizes the surface-area of the “caged” non-polar molecules. SO, IF ∆H IS INCREASING (UNFAVORABLE), THEN WHY ISN’T THE REVERSE REACTION ENDERGONIC? THERE CAN ONLY ONE ADDITIONAL CONTRIBUTING FACTOR, AND IT ISN’T HEAT. THE CONTRIBUTING FACTOR MUST HAVE A RELATIVELY LARGER é∆S (FAVORABLE) THAN é∆H (UNFAVORABLE)!



For example:
If
there is an increase in both enthalpy and entropy, we will need to consider their relative magnitudes to determine ∆G.  


And 
∆G = ∆H (1000 kJ/mol) - T∆S (1200 kJ/mol • K) à  ∆G = é∆H (1200 kJ/mol) - Té∆S (1400 kJ/mol • K) à ∆G = -200 kJ/mol (spontaneous) with temperature being negligible when T > ∆H/∆S. Note - when T < ∆H/∆S, the reaction is non-spontaneous due to a major temperature influence.

Once again - breaking water-water bonds is enthalpically unfavorable (+∆H). Nevertheless, while breaking water-water bonds is enthalpically unfavorable, a relatively large é∆S (disorder) causes an exergonic reaction (in the reverse direction) to occur.

Entropy ∆S is driving BOTH the forward (unfavorable) and reverse (favorable) reaction!

Think about our previous examples:

∆G = ∆H – T∆S

-          Forward reaction (endergonic or +∆G) consisted of ê∆H (bond forming FAVORABLE) and ê∆S (increased order - UNFAVORABLE).

-          Reverse reaction (exergonic or -∆G) consisted of é∆H (bond breaking UNFAVORABLE) and é∆S (decreased order - FAVORABLE).

The main contributor to the endergonic reaction is ∆S and the main contributor to the exergonic reaction is ∆S. Therefore, ∆S controls the forward and reverse direction of the chemical reaction! If ∆S is unfavorable, the reaction will be endergonic and if ∆S is favorable, the reaction will be exergonic.

Please take a moment and also remember that entropy (∆S) applies to the solvent (water) and not the solute (oil). The entropy of the solute is relatively minor for this reaction and the hydrophobic effect. Therefore, the hydrophobic effect is the result of the entropy of water!