| 
  • If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old.

  • Finally, you can manage your Google Docs, uploads, and email attachments (plus Dropbox and Slack files) in one convenient place. Claim a free account, and in less than 2 minutes, Dokkio (from the makers of PBworks) can automatically organize your content for you.

View
 

Chemical Equilibrium

Page history last edited by storeyr 7 years, 4 months ago

 

Introduction

 

     Chemical equilibrium is the state of a reaction at which the concentrations of products and reactants of a reaction no longer vary with time. This does not mean that the reaction has stopped or that the concentrations of products and reactants are equivalent, merely that both the forwards and reverse reactions have achieved a constant rate. We can call that dynamic equilibrium. Equilibrium is important for study simply because it occurs everywhere, from the metabolic pathways of the body to the conversion of carbon monoxide into carbon dioxide gas and solid carbon.

 

Chemical Equilibrium and Gibbs Free Energy

 

     Equilibrium is defined as the point where the forward and reverse reaction rates are equal. Typically, entropy is used to help measure equilibrium or how likely a reaction may proceed. However, there is a drawback to using entropy, because it is a state function of the universe. This means it is a property that only relates to the current states of the universe and not the path to achieve those states. This gives rise to the concept of Gibbs Free Energy (G). Note that entropy is used to describe the system as well as the surroundings of that system (i.e. the universe), while Gibbs free energy is used to describe just the system itself.

     This free energy is considered a driving force for a reaction to obtain equilibrium. Also being a state function of the system means that the components that are needed to solve for this quantity are all state functions. The equation is as follows:

ΔG = ΔH - TΔS

where H is the enthalpy of a reaction, T is the temperature, and S is the entropy change of a reaction.

 

This equations gives results of ΔG if a system is kept at a constant temperature and pressure.  Looking at this equation, one may conclude what drives a reaction: either enthalpy or entropy. With this, temperature is the major factor in determining this because the T factor is in front of the change of entropy value. At low temperatures, enthalpy drives the reaction. At high temperature, entropy drives the reaction.

 

Figure 2: This table shows the different possible ΔG values based off of the components, ΔH and ΔS

v

ΔG can also be interpreted as the transfer of energy from one system to another system. Lower energy systems are more favorable because the energy of the system tends to decrease.  Likewise, ΔG is also able to predict if a reaction will be spontaneous. The reaction can be summarized with these following results of ΔG:

  • ΔG < 0 means the reaction will decrease in energy - which is favorable and will drive the reaction towards the products.

  • ΔG = 0 means the reaction will have no net change in energy - which means that the reaction is at equilibrium and both sides of the reaction are favored.

  • ΔG > 0 means the reaction will increase in energy - which is unfavorable and will drive the reaction towards the reactants.

 

Equilibrium Constant, Reaction Quotient, and ΔG

 

In order to explain the equilibrium constant, start with a reaction in equilibrium (here we will use a generic reaction):  aA + bB ⇌ cC + dD.

So then the equilibrium constant becomes K0 = ({C}c{D}d)/({A}a{B}b)  where the {N} signifies the thermodynamic activities (where thermodynamic activities can roughly be thought of as the "effective concentrations") of the species.  It can also be written as a function of the concentrations of each species in equilibrium, Kc=([C]c[D]d)/([A]a[B]b) such that Kc=K0.

 

Similar to the equilibrium constant, the reaction quotient (denoted Q) is calculated in the same way, but can be used to describe reactions that are not in equilibrium.  At equilibrium, the reaction quotient and the equilibrium constant are equivalent.  The reaction quotient can also be written as a function of the chemical activities of the species or the concentrations of species in the reaction (not necessarily at equilibrium).

Qr=([C]c[D]d)/([A]a[B]b) or  ({C}c{D}d)/({A}a{B}b)

 

The equilibrium constant becomes very helpful when calculating the ΔG, or change in Gibbs Free Energy, in the reaction in equilibrium.  In general, when the value of K is known, the value of ΔG can be easily calculated. Given the equation for ΔG= ΔG°+RTln(Q), as the reaction approaches equilibrium ΔG will approach 0 and the reaction quotient will approach the value of the equilibrium constant.  Therefore we have 0= ΔG°+RTln(Keq).  Solving for K in this expression will give us the equation relating the two values: K=exp[-ΔG/(RT)].

 

Chemical Potentials and Equilibrium Condition

 

Chemical potential (µ)  is related to Gibbs Free Energy. In fact, at constant pressure, the chemical potential is equal to the molar Gibbs Free Energy and is the partial change in energy per change in particle number. Similar to Gibbs Free Energy, G, the sum of all chemical potentials in a reaction multiplied by the change in amount is equal to zero when the system is in equilibrium and temperature and pressure are constant. Therefore the chemical potential of the reactants must balance the chemical potential of the products in equilibrium. Gibbs Free Energy and chemical potential are related as follows:

 

dG=-SdT+VdP+udN → u=(dG/dN)T,P

 

The first equation can be simplified to the second equation because we know that temperature and pressure are constant, therefore dT=0 and dP=0. 

 

The equilibrium can also be determined by the product of the gas constant (R) and temperature (T) and the natural log of the reactant quotient (Q) divided by the equilibrium constant (K), given as  

 

ΔG=RTln(Q/K)

 

Using this formula for the equilibrium the equilibrium constant can be obtained.

 

Deriving Clausius-Clapeyron

 

The Clausius-Clapeyron equation is a very useful equation where it relates vapor pressure of a solution or substance to the phase transition temperature. This equation is:

ln(Pf/Pi) = ΔHvap/R (1/Ti - 1/Tf)

It is important to know how to derive this. The following is the proof of how the equation is derived:

Walking through the steps, we start with the Clapeyron equation that shows how transition temperatures change with pressure. At phase changes, ΔG is zero, which means that ΔH = TΔS. The reason it is zero is because phase transitions are similar to normal reactions and they are at equilibrium because it is equally changing into different phases, therefore ΔG is zero. Also, we assume the volume of gases to be much larger than volumes of liquids. Next, that volume is substituted with the ideal gas law equation and gives us a relationship of pressure and temperature with the molar heat of vaporization. Taking the integral of both sides, we can get to the final equation by assuming that the molar heat of vaporization is independent of T at relatively small temperature differences.

 

Entropy of Mixing  

 

The entropy of mixing is defined as the increase in total entropy when several separate systems (each in thermodynamic equilibrium) are mixed with no chemical reaction taking place and allowed to come to a new state of equilibrium.  As the substances are mixed and entropy increases, the Gibbs Free Energy decreases. To calculate the change in entropy of this process, we must first keep in mind the general equation for entropy:

ΔS=nRln(V2/V1).

Adapting that equation for the change in entropy of each gas as its allowed to come to equilibrium we get two separate ΔS equations, for components A and B in our gas mixture:

 

For component A:  ΔS=nARln[(VA+VB)/VA]

For component B: ΔS=nBRln[(VA+VB)/VB]

 

Adding the two components together will give us the total change in entropy for the mixing of the two gases:  ΔSTOT= ΔSA+ ΔSB= nARln[(VA+VB)/VA]  +   nBRln[(VA+VB)/VB]  which can be further simplified by substituting the moles of A or B for the volume of A or B, since they’re proportional.  

This gives us:  nARln[(nA+nB)/nA] +  nBRln[(nA+nB)/nB]   where (nA+nB)/nA or (nA+nB)/nB are the inverses of the mole fractions of A and B respectively.  Taking the inverses and factoring out the common constants from each of the two terms gives us a new equation for the entropy of mixing, which is a bit simpler than just the addition of the entropies for the separate components:

ΔmixS= -nR(xAln(xA)+xBln(xB))

 

Figure 2: This system is an example of chemical equilibrium.

 

Common Pitfalls (Le Chatelier’s Principle)

 

Equilibrium is impacted by physical changes.The reactions in equilibrium can shift to accommodate these changes. For example, if you have a reaction:

aA(s) + bB(g) ⇌ cC(g) + dD(g)

and the pressure of the reaction vessel is decreased, the reaction will go towards, or favor, the products. This principle is called Le Chatelier’s Principle and it applies to more than the pressure of a reaction. It also applies to temperature and concentration.

 

In terms of temperature, depending on if the reaction is exothermic or endothermic, then there is a heat component on the products side and reactants side. Since Le Chatelier's principle is basically the rebalancing of certain reactions. Heat can be treated as another "reactant" or "product".

 

The impact of concentration on equilibrium is fairly obvious. If you add more of the product species to the reaction vessel, the reaction will shift to favor the reactants. Alternatively, if more reactant species are added, the reaction will shift towards the products.

 

The impact of temperature on equilibrium depends on the reaction itself; if the reaction is exothermic and temperature is raised, there will be a shift towards reactants, but if the temperature is lowered the products will be favored. If the reaction is endothermic and the temperature is raised, there will be a shift towards the products, but if the same reaction is exposed to a lower temperature, there will be a shift towards the products.

     

Decreasing the pressures of the reactants and products would cause a shift towards the side with the highest number of moles of gas, while increasing the pressure of the system would cause a shift towards the side with the lowest number of moles in the gas phase. 

 

In biochemistry in particular, enzymes or catalysts may be used to speed up a reaction, and they can also shift equilibrium depending on how the forward and backwards reactions are impacted. This has to do with their activation energies. 

 

 

Concept Questions

1. Which definition best describes equilibrium?

a. The state at which the concentrations of products = the concentration of reactants

b. The state at which the rates of both the forwards and backwards reactions are zero

c. The state at which there are no more reactants left in the reaction container

d. The state at which the rates of the forwards and backwards reactions no longer vary with time.

 

2. What is chemical potential?

a. A constant used for equilibrium in ideal gases

b. The partial change in energy per change in particle number

c. How many different reactions a given reagent can undergo

d. An expression for the number of microstates in a system

 

3. Changing which of the following will not cause a shift in equilibrium?

a. Temperature

b. Pressure

c. Inserting a catalyst

d. None of the above 

 

4. A reaction that is exothermic and has a positive change in entropy will be:

a. spontaneous

b. non-spontaneous

c. spontaneous and then non-spontaneous

d. non-spontaneous and then spontaneous

 

References

 

http://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_Equilibrium_Constants.htm

 

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Entropy/Entropy_of_Mixing

 

http://www.science.uwaterloo.ca/~cchieh/cact/applychem/gibbsenergy.html

 

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Free_Energy/Gibb's_Free_Energy

 

Answers to Concept Questions  

 

1.d  2.b  3.d   4.a

 

 

Comments (0)

You don't have permission to comment on this page.