Chemical Equilibrium

 

 

Introduction

                We have all had some exposure to chemical reactions and equilibrium. The 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, which is a ratio of the final and initial concentrations to the power of the number of moles, through the following formula:

 

If the right side of the equation is equal to the equilibrium constant, then the system is at equilibrium. If it is not equal to the equilibrium constant, then the value is called Q, which is the reaction quotient. By analyzing the equation, we can see if Q>K, the system will favor starting material. If Q<K the system will favor products.

 

 We have also just learned about entropy and Gibbs energy.  See the previous cycle's exploration of entropy for a deeper discussion of this topic.   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 in Joules per mole (J/mol), H is enthalpy in Joules per mole (J/mole), T is temperature in Kelvin (K), and S is entropy in Joule per Kelvin per mole (J/K mol), and these are found under constant temperature)

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 in Joules (J), P is pressure Pascals (Pa), and V is volume in liters (L) and this is found under constant pressure)

dU=dw+dq

(where w is work in Joules (J) and q is heat in Joules (J))

dw=-PdV+dw’

(where w’ is non-expansion work, the free energy also in Joules (J))

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 or less than  the free energy of the system, which is non-PV work. This is because no system is 100% efficient, some free energy will be inevitably lost.  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 the molar Gibbs energy 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.

 

Head Ticklers

 

The Clausius-Clapeyron Equation

Systems seek to achieve minimum Gibbs energy; similarly, substances exist in the phase where their Gibbs energy is lowest. This condition is dependent on pressure and temperature.

 

Phase boundaries exist where a substance is in equilibrium between two states, and this boundary can be mathematically explained with the Clausius-Clapeyron equation. 

 

The following section derives the Clausius-Clapeyron equation from the statement of Gibbs energy.  

 

The Gibbs energy of a substance in two different states can be expressed as such: 

 

Since the phase boundaries represent the state of equilibrium between two phases, dG₁ = dG₂, and the two equations can be set equal to each other. 

By rearranging the terms, we get the expression: 

We can further substitute the changes of entropy and volume with 'Δ,'  and rearrange the terms to get the Clausius equation:

where the phase boundary of vaporization is considered. 

 

To the right hand of the equation, we can rewrite entropy in terms of enthalpy:

We approximate that the molar volume of the total is approximately equal to that of the gas only, and that volume contribution from the liquid is negligible. We further take ideal gas approximations, and simplify. 

We can further rearrange the terms of the equation to get the following expression

Take the integral on both sides:, 

and the resulting integrand gives the Clausius-Clapeyron equation for vaporization. This is expressed by the blue line in the graph below.

 

This equation gives the relation between the different pressures and temperatures for a given enthalpy of vaporization. 

 

 

Entropy of Mixing, Chemical Equilibrium, and delG

 

Mixing is a spontaneous process which always results in an increase of entropy. Since Gibbs energy and entropy are related by the thermodynamic potential, G = H - TS, and the enthalpy of mixing is 0 by definition for ideal gases, the entropy of mixing corresponds to a decrease in Gibbs energy at constant temperature. 

 

If reactants and products were completely pure, then there would be no reversibility or equilibrium. 

 

Gibbs energy is the driving force of a reaction, and a system seeks to equilibrate itself at its lowest Gibbs energy. In an "extent of reaction" graph, where Gibbs energy is plotted against the concentration of reactants and products, the direction of reaction, indicated by the red arrows, can be determined by the location of the minimum Gibbs energy. In the case of (a) of the above picture, the minimum is located by pure reactants, and the reverse reaction is favored. The opposite is true for (c), and the forward reaction is favored. In the case of (b), equilibrium exists at about a half/half mixture. 

 

A reaction will never really reach 100 percent completion, since equilibrium is a dynamic process, oscillating between a state of reactants and products.

 

For more information, you may consult the other explorations here.

 

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. Moreover, in the above example, A, B, C, and D do not necessarily have to be reactants. They can be anything from heat to pressure, 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.

 

Pressure + 2SO_2(g) + O_2(g) ⇋ 2SO_3(g)

 

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 (this would also be the case for injecting an inert gas into the system), 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.

 

Note that solids and liquids are not considered in equilibrium reactions as their concentrations remain, approximately the same regardless of the alterations subjected to the system. Another way of thinking about this is that the mass per volume ratio of solids and liquids do not appreciably change compared to the change in concentration of dissolved ions or the partial pressures of gases.

 

Applications of Equilibrium Constants: Acid Dissociation Constants

 

Waiting in line is an inevitable part of everyday life—and the capacity to do so varies between different types of people. How can we find a way to measure someone’s ability to wait in line, and what does it have to do with acid base equilibrium? Let’s take a look at an example.

 

There are eight people waiting in line at the bank.

 

Clearly, people are very different and live very different lives. As a consequence, they probably can’t dedicate the same amount of time waiting in line to do their daily banking: some people might be willing to wait all day, while other may not be able to wait at all. Let’s identify a few individuals in a line to make some observations (see below):

 

 

From above, we have four different individuals—all with a different tolerance for standing in line. While on his lunch hour, the businessman has to stop by the bank to deposit a check in his daughter’s saving account she received for her birthday. Also standing in line is the homeless guy from the street corner who is depositing some money into his checking account. Behind him is The Grandmother, who is going to make a withdrawal from her account to enjoy a night at the casino. Lastly, The Stay-at-home mom is waiting in line with her daughter to make their monthly mortgage payment.

 

Now that we know a little more about our crowd, let’s see how well they do in line. As previously mentioned, the business man is at the bank on his lunch hour, and will still need time to eat before he gets back to the office. He will only stand in line about 10 minutes before he’ll have to leave the bank without depositing his daughter’s check. Because it is lunch hour, The Homeless Guy realizes that he is missing a lot of pan-handling opportunity because the city streets are busier as people go out to eat. He’ll waste only about 18 minutes in line before he returns to the street. The Grandmother knows she will have to wait in line no matter what (the casino is open until 2:00 a.m. anyway)—she’ll wait in line for an hour before she goes off to find another bank. With no job and plenty of snacks for her daughter, The Stay-at-home mom is willing to stake out for quite some time. She will last about 40 minutes before her daughter starts crying—prompting her to leave for the day. We can sum up these observations as follows:

 

The Businessman	10 minutes
The "Dispossessed" Man	18 minutes
The Grandmother	60 minutes
The Stay-at-home mom	40 minutes

 

So let’s say the average wait time at the bank is 20 minutes. Who gets to do their banking? We can see that The Businessman and The Homeless Guy will only last 10 and 18 minutes respectively—so they’ll will be gone before the can be helped. The Grandmother and The Stay-at-home mom will have lasted, and have the chance to take care of their banking.

 

Now let’s say we take our four people and transplant them at the nearest McDonalds, where the average wait time is about 5 minutes. Looking at the table above, we can see that everyone will be helped at McDonalds.

 

If we move everyone to the Social Security office, where the average wait time is 50 minutes (on a good day), we can see that only the Grandmother will be satisfied—all others will have left while waiting in line. I think we can all agree that no one will be helped at the DMC.

 

We can make some general observations from these different scenarios. Whether or not our friends are helped depends entirely on what environment they are in. Their internal ‘wait time’ is set and does not change based on where they are.

 

Acids and bases behave in a similar manner. We measure acids and bases based on how willing they are to donate a proton. This property of acids and bases are much like the internal “wait times” of our friends above. Acids and bases will only give up their proton(s) in certain environments in the same way our friends will only be helped in certain environments based on how long the wait time is.

 

The tendency for acids to give up their proton is measured by en equilibrium constant: the acid dissociation constant k_a. The acid dissociation constant is calculated the same way any other equilibrium constant is calculated:

 

 

Where HA is a general acid and A^- is a general base. The general acid dissociates in solution to produce the general base and a proton. The relative concentrations of each of these determine the k_a of the acid. It is often more convenient to use the base-ten logarithm to compare acids because the k_a values are often very large or very small. When we run the k_a through the base-ten logarithm, we call the result the pk_a:

 

 

Unlike pH (the negative base-ten log of proton concentration in aqueous solution), pk_a can be negative (in which case HA almost never has its proton in aqueous solution) to above 50 (in which HA always has its proton). But what about acids with pk_a values in the range of 1 to 14 (the allowed pH values of water)? How do we know when they have their proton and when they do not?

 

To answer this question, we can return to our example. We know that for people waiting in line there is a range of wait times that they will be helped at the bank, as well as a range of wait times at which they won’t be helped (i.e. they leave before they reach the front of the line). The same concept is valid for acids and bases. For any given acid, there are a range of pH values that the acid will have its proton and a range that they will not have their proton. This pH value above which the acid does not have its proton and below which it has its proton happens to be its k_a (for acids with k_a values between 1 and 14). For instance, for The Stay-at-home mom who’s “pk_a” is 40 minutes will not be helped at wait times over 40 minutes (e.g. she does not have her proton), and will be helped at times below 40 minutes (e.g. she will have her proton).

Another way to look at this is to consider the pk_a of an acid the half-on / half-off point of the acid. When pH = pk_a we can say that half the acid molecules have their proton and half do not. If we increase the pH above the half way point, the solution becomes more basic and more of the acid molecules become bases (they lose their proton). Inversely, a decrease in pH at this point means that more acid molecules exist as the acid (they have their proton).

 

Acids and bases with pka values around 7 (neutral) are especially useful in biological processes. These so-called 'weak acids (or bases)' have both their acid and base form present at pH 7--thus they are resistant to changes in pH. If an acid is added to the solution, the protons are quickly absorbed by the base form of the buffer and sequestered as a weak base (virtually no change in pH). Likewise, if a strong base is added to the solution, it quickly takes a proton from the weak acid, thus resulting in little pH change. This is an important process in biological processes considering the function of many enzymes and the process of many vital reactions are very sensitive to pH.

 

 

 

 

Questions:

 

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

 

5A_(s)+6B  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, their concentrations do not change. Thus, they are neglected when we consider the equilibrium of the system.
2. (A) There will be a net reaction towards the products since the reaction quotient is larger than the equilibrium constant.
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.

 

References:

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

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