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Chemical Thermodynamics

Gibbs Free Energy

J. Willard Gibbs was the first person to be awarded a Ph.D. in science from an American University (Yale, 1863)

Spontaneous reactions often have:

A negative enthalpy (release of heat energy, DH < 0).
An increase in entropy (increase in disorder, DS > 0)

The spontaneity of a reaction appears to involve two thermodynamic properties: enthalpy and entropy

Furthermore, spontaneous reactions are those that go downhill in energetic terms. In other words, the final state has a lower energy content than the initial state

Gibbs came up with an equation, combining both enthalpy and entropy contributions, that provided a means to describe energy content and therefore a means to evaluate the spontaneity of a reaction when that energy content changes. The energy contents of a substance was termed the Gibbs Free Energy and it was defined by the Gibbs Free Energy equation:

G = H - T*S

The free energy of a substance = stored heat energy - inherent disorder at a reference temperature

H is enthalpy, S is entropy and T is the temperature in Kelvin

- If there is a lot of stored heat energy, then the substance has a lot of free energy
- The more disorder a substance has, the less free energy it has

Changes in a substance (as in a chemical reaction or physical phase change):

DG = DH - T*DS

- If the substance releases heat energy, then the product has a lower value of stored heat energy and DH < 0. Such a change is downhill energetically (i.e. spontaneous)
- If the disorder (entropy) increases, this is also a spontaneous process and DS > 0. Since DS > 0 this means that (-T * DS) < 0.
- Therefore, both the enthalpic (DH) and entropic (-T*DS) terms are negative for processes that are spontaneous

How to quantitate the contribution of the entropic term to the free energy?

In "instant ice packs" a reaction occurs that is spontaneous, and yet is endothermic (i.e. it is "cold" due to absorption of heat energy). The absorption of heat energy is unfavorable and therefore must be "driven" by an increase in entropy (i.e. a large -T*DS term). Therefore, the degree to which heat can be spontaneously absorbed is actually providing us with information regarding the magnitude of the entropic increase (i.e. the entropic increase is what is "driving" the unfavorable heat absorption).

Recall that entropy was defined previously as q/T (i.e. non-mechanical energy transferred at a constant temperature). This definition is just a way of saying that if heat is absorbed but the temperature does not change, then the heat energy is being sucked in due to an entropic increase - and that entropic increases is energetically equal to the heat energy absorbed
In such a case DS is positive, entropy increases, and (-T*DS) is negative (favoring spontaneous reaction)
Heat absorbed at a fixed temperature is essentially an infinite heat capacity, and this is just to point out that entropy has units of J/molK, like heat capacity values. A material with a high heat capacity has the ability to absorb a lot of heat energy with a small temperature change. It can do this because it is able to increase its internal disorder and "soak" up the heat energy and thus the temperature won't rise much. If a material cannot increase its internal disorder, then it will increase temperature in response to even a small amount of added heat energy.

For a process occuring at constant temperature, T:

If DG = 0, it means that the enthalpy change associated with a reaction is equal in magnitude (and opposite in sign) to the entropy change. For example, the process being considered may result in a reduction in entropy (i.e. more order) but it releases an amount of heat energy that exactly counteracts the effects of the TDS term. Such a reaction is in equilibrium (i.e. no net reaction)
If DG < 0. This would occur for not only an exothermic reaction (DH = negative) that overwhelms any unfavorable entropic effect. But potentially also for an endothermic process that has a significant increase in disorder (i.e. DS is large, and thus, TDS is large and negative). In either case, the process being considered has a net driving force (release of energy or increase in entropy) that indicates the reaction is spontaneous
If DG > 0. This will happen if the reaction is highly endothermic (DH positive) and the entropic term is not so great. Or, if the reaction is exothermic (i.e. DH is negative), but the process results in significant increase in order (i.e. the T*DS term ends up being a negative number, so that -T*DS is positive in magnitude). In any case, this is energetically an unfavorable process being considered. To move in the forward direction, energy must be supplied. If energy is not supplied, then the reaction will proceed spontaneously in the reverse direction.

Free energy and chemical reactions at equilibrium.

It would seem there are 4 possible types of reactions or processes with regard to the enthalpic and entropic contribution to the free energy change:

DH = (-), -TDS = (-). Favorable enthalpic change (exothermic) and favorable entropic change (disorder increases)
DH = (+), -TDS = (+). Unfavorable enthalpic change (endothermic) and unfavorable entropic change (disorder decreases)
DH = (-), -TDS = (+). Favorable enthalpic change (exothermic) and unfavorable entropic change (disorder decreases)
DH = (+), -TDS = (-). Unfavorable enthalpic change (endothermic) and favorable entropic change (disorder increases)

If you look at these four types of energy changes, you will notice that 1) and 2) are considering the same process, just from different directions. Likewise, with 3) and 4) (i.e. an exothermic process in one direction is endothermic in the opposite direction). So, in principle, we just have to understand two types of processes.

For type 1) above, both enthalpy and entropy favor the forward direction. There would appear to be no energetic term favoring the reverse direction. Thus, this type of process would be expected to go to completion (i.e. no equilibrium condition because no reverse reaction occurs). Type 2) is the same situation, but viewed from the opposite direction.
For type 3) above, enthalpy favors the forward direction but the energy associated with the entropic change favors the reverse direction. There are, therefore, energetic forces driving the process in opposite directions, and it would seem likely that an equilibrium condition might exist. Type 4) is the same situation, but viewed from the opposite direction.

Consider our old friend, the Haber reaction:

If we start with nothing but NH₃(g) in the sample, the reaction will proceed in the reverse direction to produce H₂(g) and N₂(g) (i.e. DG for the reaction will be a positive value)

N₂(g) + 3H₂(g) ¬ 2NH₃(g)

Or, DG is negative for the following reaction (i.e. the following reaction is spontaneous):

2NH₃(g) ® N₂(g) + 3H₂(g)

If we start with nothing but H₂(g) and N₂(g) in the sample, the reaction will proceed in the forward direction to produce NH₃(g) (i.e. DG for the reaction will be a negative value)

N₂(g) + 3H₂(g) ® 2NH₃(g)

If we start with concentrations of all components such that Q = K_c, then the reaction is at equilbrium and DG = 0.

The reaction wants to be driven in a direction such that DG goes to 0 (i.e. equilibrium)

At equilibrium the free energy of the system is at a minimum. To produce either more product, or more reactants, requires an increase in free energy (i.e. some modification of heat or entropy properties)

Standard Free-Energy Changes

The free-energy term, G, is a state function, thus values can be defined for substances at specific conditions of temperature and pressure known as the standard state. In this case, we will have DG values associated with the formation of compounds from their elemental constituents, known as the standard free energy of formation, DG_f⁰.

Standard conditions include:

1 atm pressure
1M concentration (solutions)
Pure solid (if a solid) or pure liquid (if a liquid)
For elements, the standard free energy of formation, DG_f⁰ of an element in its normal state is 0
There is no standard state for temperature. G will vary with temperature. 298K (i.e. 25°C) is a common temperature chosen for standard reference values of G

Standard free-energy values can be used to calculate the standard free energy change associated with a reaction:

DG⁰ = S n DG_f⁰(products) - S m DG_f⁰(reactants)

What information will the calculation of DG⁰ provide?

DG⁰ < 0, the reaction is spontaneous as written (i.e. goes to the right)
DG⁰ > 0, the reaction will proceed to the left as written
DG⁰ = 0, the reaction is at equilibrium

Calculate DG⁰ for the Haber reaction will all components at standard conditions at 298K

H₂(g) : DG_f⁰(298K) = 0 (note: normal state of element)
N₂(g): DG_f⁰(298K) = 0 (same note)
NH₃(g): -16.66 kJ/mol

N₂(g) + 3H₂(g) ó 2NH₃(g)

DG⁰ = S n DG_f⁰(products) - S m DG_f⁰(reactants)

DG⁰ = 2*(-16.66 kJ/mol) - (0 + (3*0))

DG⁰ = -33.3 kJ/mol

DG⁰ is a negative value, which says that if each component were present at 1 atm (standard conditions for gas) at 298K, the reaction would proceed spontaneously to the right to produce more product (but we know nothing of the rate)