Introduction:

                We have all had some exposure to chemical reactions and equilibrium. An equilibrium is the point within a system in which the movement of products to reactants, and reactants back to products are equal. In other words, the net movement of the reaction is zero. If we are given the following general reaction:

We can calculate an equilibrium constant through the following formula:

 We have also just learned about entropy and Gibbs energy.  For a reminder about these, see the previous cycles exploration of entropy.  Gibbs energy can also be defined as the free energy, or the energy available to do non pressure-volume work. A derivation of this will follow.

Basics:

Gibbs Energy and Free Energy

As discussed above, Gibbs energy is actually the free energy available to do work (other than pressure volume work.) We will derive this.

We know that the equation for Gibbs energy is:

dG=dH-TdS

(where G is Gibbs energy, H is enthalpy, T is temperature, and S is entropy)

We will make a couple substitutions from other formulas that have been discussed in previous explorations

dH=dU+PdV

(where U is total internal energy, P is pressure, and V is volume)

dU=dw+dq

(where w is work and q is heat)

dw=-PdV+dw’

(where w’ is non-expansion work, the free energy)

All together this leads to

dG=-PdV+dw’+dq+PdV-TdS

We quickly see that a couple terms cancel and we are left with

dG=dw’+dq-TdS

but we know that dq=TdS and therefore we are left with

dG=dw’

In conclusion, we see that the change in Gibbs free energy is equivalent to the free energy of the system.  Furthermore, in previous explorations it was discussed that in order for a system to spontaneously react, the Gibbs free energy must be negative. At equilibrium, there is no net reaction occurring, therefore we want the Gibbs free energy to be zero. This means that the free energy available at equilibrium is zero.  No net work is being done within the system at equilibrium.

The difference between the Gibbs energy and the Standard Gibbs Energy:

Reactions may be conducted at various temperatures and pressures, but there needs to be a standard set of conditions in which measurements are conducted for data reporting. The standard conditions are 1 bar pressure and 25®C.  Gibbs energies found in these conditions are called standard Gibbs energy. Once these standard conditions are varied the Gibbs energy will also vary. The following formula demonstrates the interaction between Gibbs energy at varying pressure and temperature and Standard Gibbs Energy (denoted G®).

G=G®+kTln(P/P®)

Where P® is the reference pressure, in this case 1 bar.

Standard Gibbs Energy and Equilibrium Constant:

Both Standard Gibbs Energy and Equilibrium Constants are constant for a reaction. They are related to each other by the following formula:

Relationship between free energy and chemical potential:

Within a reaction, different reactants may have different Gibbs energy. This is so important, that it has been given its own name and symbol, the chemical potential, µ.  This chemical potential can be used to calculate the Gibbs energy of a reaction through the following formula:

dG= µ_adn_a+ µ_bdn_b+ µ_cdn_c+…

where dn_a is the change in the number of moles of n. Since the reaction always tends towards the smallest Gibbs energy, this equation will always be less than zero. 

Le Chatelier's Principle

Le Chatelier's principle states that when a chemical reaction is at equilibrium and is perturbed from equilibrium, then the reaction will counteract the perturbation to reestablish equilibrium.

For example, if the above reaction is at equilibrium, and more A is added to the reaction, then the concentration of C and D will increase and the concentration of A and/or B will decrease. Le Chatelier's principle is so powerful, that it can be used for any kind of equilibrium reaction. Thus, it can be used to anything from acid-base equilibriums to gas equilibriums. For acid-base problems, the concept of equilibrium can only be applied to weak bases or weak acids as their strong counterparts dissociate completely and exhibit little to no equilibrium-like character. For gases, an extra complication arises, one must account for the partial pressures of the gases, not their molarity, and the total volume of the system. For the following reaction:

2SO_2(g) + O_2(g) ⇋ 2SO_3(g)

There are three moles of gases on the left hand side of the reaction and two moles of gas on the right. If we assume ideal gas behavior, then moles of gas is directly proportional volume and pressure. If we think of pressure as a reactant, then we can add pressure to the left side of the above reaction, as if it were a reactant, since that side of the reaction produces more pressure. Thus, if we increase the volume of the system, then the pressure of the system will decrease. Since this perturbs the equilibrium by lowering pressure, the reaction will shift towards the side of the reaction with the greatest moles of gas to increase the pressure of the system so that it reaches equilibrium once again. For example, since we said pressure was a reactant on the left side of the above equation, decreasing pressure will make the equilibrium shift to the left. The same can be said for changing the pressure of the entire system, if pressure is increased, then the reaction will shift towards the side with less moles of gas. For the above reaction, it would be to the right. Note that if a gas equilibrium reaction had equal amounts of moles of gas on the left and the right side of the equation, changing the volume or pressure of the system will not affect the equilibrium as the partial pressures of the gases involved will be altered equally by the volume/pressure changes.

Questions:

For the following reaction, what will happen if the pressure of the entire system was increased?

5A_(s)+6B_(g) ⇋ 6C_(g) + 17D_(l)

A. There will be a net reaction to the right (towards products)
B. There will be a net reaction to the left (towards reactants)
C. There will be no net reaction in either direction

For a certain reaction, the equilibrium constant is 10^-3. If the reactants were mixed together so that the reaction quotient is equal to 10^-1 initially, in what direction will the reaction occur?

A. There will be a net reaction to the right (towards products)
B. There will be a net reaction to the left (towards reactants)
C. There will be no net reaction

According to the Clausius-Clapeyron equation, does a liquid boil at a higher temperature, relative to its boiling point at sea level, at high or low elevation?

A. The liquid will boil at a higher temperature at high elevation
B. The liquid will boil at a higher temperature at low elevation

Answers:

1. (C) There is no net reaction in either direction because the number of moles of gas is equal on both sides of the reaction. Since solids and liquids do not appreciably change in volume due to pressure change, they are neglected when we consider the equilibrium of the system
2. (A) Since the reaction quotient is larger than the equilibrium constant, then there will be a net reaction towards the products.
3. (B) According to the Clausius-Clapeyron equation, increasing pressure increases the boiling point of a solution. Thus, at lower elevations, atmospheric pressure is greater and a liquid boils at a higher temperature, relative to its boiling point at sea level, at a lower elevation.

References:

http://www.files.chem.vt.edu/RVGS/ACT/notes/chem-eqm.html

http://www.chemteam.info/Equilibrium/Dynamic-Equilibrium.html