Equilibria
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Readings for this section

Petrucci: Chapter 15

Chemical Equilibrium 2008-03-23

General Discussion

• Often (in our Stoichiometric calculations), we consider that reactions go to completion, i.e., that one or more of the reactants is completely used up.
• This assumption is not always valid and in reality, is never reality.
• Nothing is 100% in this world. There is no 'pure' substance, no 'complete' reaction, in a closed container
• Any chemical system in a closed container will always reach a state of equilibrium. Sometimes that equilibrium state may be such that the container has almost no products or almost no reactants but it is still an equilibrium.

Consider the system:        2 NO₂(g) N₂O₄(g)

Reactants are reddish brown, products are colourless. Both temperature and pressure affect this equilibrium. Higher pressures (by compressing the container containing the reaction mixture) will favour the production of product (colourless) at the expense of the amount of reactant (coloured) hence, the equilibrium can be observed to change as the reaction vessel is compressed or expanded (as in a piston moving up and down in a glass cylinder.)

Definition: Equilibrium:

State of a reaction mixture at which the forward reaction rate is equal to the reverse rate.

Note that there is no reference to the amounts of reactants necessary to achieve this state.

Let's look at a simple example:
The equilibrium between reactant A and product B:       A B

If there is initially, no B then there can be no reverse reaction Kinetic Molecular theory tells us that the rate of reaction depends on the collision frequency and that the collision frequency of molecules A depends on how many molecules there are in the container. That seems to make sense. Therefore, as the amount of A is diminished from the initial situation (used up in the forward reaction) the forward rate will drop. Similarly, as the amount of B is increased due to the forward reaction, then the backward rate will increase from zero. eventually, a point will be reached where the forward and reverse rates are equal. A is being produced and used up at equal rates and so is B. Thus, although the reaction continues unabated, we see no overall changes to the amount of either A or B.

Fig. 1a:

Here we see that as time progresses from zero. the forward rate (f) diminishes and the reverse rate (r) increases. Eventually (only after infinite time) the two are equal and the chemical system has reached equilibrium.

Fig. 1b:

Here we see the effect of the changing rates on the amount of reactant (A) and product (B). Initially there is no B and some amount of A. These amounts change with time until equilibrium is reached at which time the amounts of A and B is constant in time.

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Equilibrium Constant

Consider the reaction        jA + kB lC + mD

The law of mass action (equil. const) gives us the following:

Note:  the square brackets [] indicate concentration in mol/L (M)  the subscript c on the K indicates this value derived using concentration units.

More generally, one can replace [A] by C_A where the capital C means Concentration in whatever units.

If the reaction of interest is a gas phase one then we could replace the concentration C with pressure P. thus we would now have defined a new type of equilibrium constant K_p as follows:

Sometimes mixed units are used, for example combining such things as the vapour pressure of a chemical and concentrations in the reaction mixture. These kinds of equilibrium constants would only occur in rare kinds of dual phase reactions.

Activities

In reality, the equilibrium constants do not have units even though those discussed above obviously do. Hence the necessity to label them with subscripts c or p. True Thermodynamic equilibrium constants do not use actual measures of amounts of a substance but rather they use the activity of the substance. The activity of a substance is related to some standard measure (concentration, pressure, etc) of that substance via

Relative activity  (a)        =         measure/std measure.

For example, if concentrations are used, the std concentration in M of an aqueous solution is 1. Thus a substance A with concentration [A] (in M) would have relative activity a_A = [A] / 1 M (note in this case, the units were merely cancelled out by this process). Similarly, if gaseous substance A has partial pressure of P_A then the relative activity is a_A = P_A/P_Aº where P_Aº = 100 kPa.  Thus, the thermodynamic equilibrium constant is written using activities:

Since the individual activities are uniless, so too is the K constant.  This is the kind of K constant we need in thermodynamic equations such as we will see in the thermodynamics section later.  It is generally a good habit to start using this kind of K constant exclusively.

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Haber Process

Consider the process of converting hydrogen and nitrogen into ammonia. The simple looking gas phase reaction is useful to illustrate several points.

3 H₂ + N₂  2 NH₃

At room temperature the rate of this reaction is quite slow. It will seem to the casual observer that no reaction is occurring. At more elevated temperatures, the reactions speeds up and equilibrium is reached more quickly.

Example: At 127 ºC the3 following set of equilibrium concentrations were measured for a particular reaction mixture.
 

What is the equilibrium constant K?

What if we had written the balanced chemical equation in terms of one mole of product? (often done).
We would have:

3/2 H₂ + 1/2 N₂  NH₃

What is the equilibrium constant K' now?

We could have also written the equation as:

NH₃  3/2 H₂ + 1/2 N₂

giving us

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Heterogeneous Equilibria

I mentioned in the last lecture that some equilibria can involve materials in different phases. We will discuss herein, how to handle some of these cases.

Consider the case of calcium carbonate decomposing to lime and carbon dioxide:

CaCO₃(s)  CaO(s) + CO₂(g)

The law of mass action would give us:

     a(CO2)a(CaO)
K  = ------------
       a(CaCO3)

However, the concentrations of the solids calcium carbonate and lime do not have an effect on the equilibrium.  Their activities are simply one so they do not contribute.

Your own experience will be able to tell you this as well. Have you ever added sugar to a drink (say coffee) to sweeten it up. If you are like most people, there was a time when sweet was the only taste you liked (childhood). In this case, you added as much sugar as you could before your mother or father took back the sugar dispenser and invariably, there was always some sugar left undissolved in the bottom of your glass (bowl). No matter how much sugar you added, only a certain amount of it actually dissolved. Although this is more of a physical process than a chemical process, the idea is the same. The amount of sugar in solution (concentration) was not affected by adding more, once there was already some solid in the bottom of the glass.

Since the relative activities of a pure liquid or solid are 1, it follows that we can make the approximation

K = a(CO₂).

Another example, this time involving solids and gases follows:

NH₄HS (s)  NH₃ (g) + H₂S (g)

gives us

     a(NH3) a(H2S)
K = ------------
       a(NH4HS)

K = a(NH₃) a(H₂S),  since the relative activity of pure NH₄HS solid is 1.

For dilute aqueous solutions we also make the approximation that the relative activity of the water is 1 but note that this is only a good approximation for dilute solutions.

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Applications of the equilibrium Constant

The equilibrium constants can be tabulated and used as necessary.

We can, for example, decide whether a particular reaction system will react nearly to completion (very large K) or almost not at all (very small K) or somewhere in between. The larger the value of K, the more the reaction will proceed to the right (more products). The smaller the value of K, the more the reaction will find equilibrium to the left (more reactants). We will look into this concept further in later lectures.

Reaction Quotient

One use is to determine in advance whether a particular chemical system will be at equilibrium or not. To do this, we must introduce another concept, closely related to the equilibrium constant.

For a chemical system whether in equilibrium or not, we can calculate the value given by the law of mass action. Only if the system is at equilibrium will this value be the equilibrium constant. In general, we call this value the reaction quotient Q. Taking again the Haber equilibrium, we can write the reaction quotient for the system at any time.

3 H₂ + N₂  2 NH₃

       a(NH3)2
Q = --------------
    a(H2)3 * a(N2)

In this case, we are not necessarily using equilibrium concentrations and hence we calculated not the equilibrium constant K but the reaction quotient Q. We can now compare the two values as follows:

• Q < K: This implies that the reaction is not at equilibrium and that the reaction must proceed to the right (make more products) to get to equilibrium.
• Q = K: at this point the amount of products and reactants is just right to meet equilibrium conditions.
• Q > K: implies that there are too many products. The reaction will proceed to the left to produce more reactant (and lower the among of product) until Q = K.

NOTE: if you are not using the relative activities to calculate Q and K, you must compare a Q_c with a K_c, a Q_p with a K_p etc.

Example

Consider the Haber equilibrium at 500ºC. K = 3.5 * 10^-7. Predict which way the equilibrium will shift for each of the following cases.

	p(NH3)	p(H2)	p(N2)
a.	4.16 kPa	8.3 kPa	403.5 kPa
b.	0.83 kPa	1470 kPa	6.20 kPa
c.	0.416 kPa	416 kPa	2.08 × 104 kPa

 

Solution

a.

       a(NH3)2
Q = --------------
    a(H2)3 × a(N2)

       (4.16/100)2
  = ---------------------
    (8.3/100)3 (403.5/100)

  = 0.75

Since Q > K the reaction will proceed to the left 
until equilibrium.

b.

           (0.83/100)2
Q  = ------------------------
      (1470/100)3 (6.20/100)

    = 3.50 × 10-7.

Since Q = K the reaction is at equilibrium.

c.

           (.416/100)2
Q  = ------------------------
     (416/100)3 (208×104/100)

    = 1.1557 × 10-13

Since Q < K the reaction will proceed to the 
right until equilibrium.

Calculating Equilibrium Pressures and Concentrations:

Typically, we will be asked to find equilibrium concentrations (or pressures) given initial values. The basic technique for solving these problems is always the same. I will attempt to introduce you to the steps involved. Note, however that the exact execution of the steps may vary from problem to problem.

1. Determine the balanced chemical equation for the equilibrium in question. Be sure you used the correct formula (for example, does the K constant represent an equation where there is one mole of products etc...)
2. Set up the conditions (amounts of reactants and products present) not taking into account the equilibrium process itself, i.e., pretend you can artificially turn the equilibrium reaction off until you are ready.
3. Now turn on the equilibrium. amount x will react using up reactants and producing products. In some problems x is given, in others it is to be determined.
4. Do the calculations to find the unknown by substituting into the equilibrium constant expression for the balanced chemical reaction your wrote in step 1.

I will illustrate these steps by solving the following problem.

The reaction for the formation of gaseous hydrogen fluoride from hydrogen and fluorine has an equilibrium constant K of 1.15 * 10². 200.0 kPa  of each cpd are added to a 1.5 L flask. Calculate the equilibrium partial pressures of all the species.

                                               Balanced chemical equation

                        H2(2)   +   F2(g)     2 HF (g)  

initial Rel. act.       2.000       2.000         2.000  

No other preparations so now we can let the reaction go to equilibrium.
Let x be the amount of reaction (in rel act).

change (x)               -x          -x            +2x

equil. conc. (M)        2.000-x     2.000-x       2.000+2x  

Substitute these values into the equilibrium equation

                         a(HF)2           (2.000+2x)2
    K = 1.15 * 102 =  -----------   =  ------------------
                       a(H2) a(F2)      (2.000-x)(2.000-x)   

In this case, we can take the square root of both sides of the equation since this is easy to do. In other cases, you may need to do some other mathematical trick to make it work.

                          2.00 + 2x
        (1.15 * 102)1/2 = ----------
                          2.00 - x  

After some algebra, we get x = 1.528. Thus, the equilibrium rel. activities can now be calculated:

   a(H2) = a(F2) = 2.000 - 1.528   = 0.472
   a(HF)        = 2.000 + 2*1.528 = 5.056.  

Now, convert back to pressures from relative activities

   p(H2) = p(F2) = 0.472 * 100 = 47.2 kPa
   p(HF)        = 5.056 * 100 = 505.6 kPa  

This case was actually unusual in that K_c = K_p = K. and thus, we could equally well have used partial pressures or concentrations directly in the equilibrium expression using the same numerical equilibrium constant.

We can do a check by recalculating K from these calculated pressures to see if the correct value is determined (it is).

This example was simple to solve only because we fortuitously chose initial amounts of reactants which made it possible to simplify the expression using a square root of the entire equation. Many times, these calculations require the use of the quadratic formula to obtain a solution.

Systems with small K constants

In some cases, we can take advantage of certain conditions to make some assumptions about the answer which will in turn make it easier to calculate the answer. We should always check any assumptions we make by calculation. I will illustrate the point with the following example

Gaseous NOCl decomposes incompletely to form NO and Cl₂ gases.

Given initially that 1.0 mol of NOCl is introduced into a 2.0 L container at 300 K with no NO or Cl₂ present, what are the equilibrium pressures?

PV = nRT

a(NOCl) = 1.2×10³/100 = 12

The balanced chemical equation is:

                     2 NOCl (g)  2 NO (g)  +  Cl2 (g)  

Initial activity        12               0           0  

Change (x)             -2x              +2x          +x

Equilibrium activ.    12-2x             2x           x  

and so we can substitute these equilibrium activities into the equilibrium expression:

                   a(NO)2 a(Cl2)      (2x)2 * x
K = 4.0 * 10-11  = -------------  =  -----------
                     a(NOCl)2          (12-2x)2

Because the value K is at least two orders of magnitude smaller than the activity 12, we can try the following assumption.

We assume that the extent of reaction is small (since K is small) and hence x is small. We assume, in fact that x is so small that the denominator (12-2x) can be approximated by (12). (or another way of saying it is: assume that 2x « 0.12). Thus, we can now write

                      (2x)2 * x          4x3
Kc = 4.0*10-11     ~  -----------   = --------
                         (12)2          (12)2

x3 =  4.0*10-11  * (12)2 / 4  =  1.44 * 10-9 .

x = 1.13×10-3 .

Now check the validity of our assumption.

our assumption was   12 - 2*x ~ 0.012
results show         12 - 2×.00113 = 11.98 = 12 (subscript means not significant figure)

0.01198 is .17% different from 0.012. The arbitrary cut-off for these types of calculations is about 5% so this is a good assumption.

We can now calculate the equilibrium activities.

a(NOCl) = 12 - 2x ~ 12 (rounded to two sig. figs)
a(NO)   =  2x  =  0.0023
a(Cl2)  =   x  =  0.0011.

p(NOCl) = 12 * 100  = 1200 kPa
p(NO)   = 0.0023 * 100 = 0.23 kPa 
p(Cl2)  = 0.0011 * 100 = 0.11 kPa 

Again, we can check to see that our work is correct by substituting these into the equilibrium expression to see if we get the correct value for K (we do, within about 5%).

This technique of assuming that there was a amount of reaction (small x) because of a small K value made the problem quite a bit easier to solve. This is valid most of the time if the K constant is at least two order of magnitudes smaller than the activities. You can see that in this case, we were borderline. Our answers were accurate to about 2 sig. figs. If we had need more precision on our numerical answers, we could not have made the assumption.

Systems with large  K constants

When the K constant is large, we can simplify the calculations by adding one extra step before using the Ice table.  Since K is large, almost 100% of the reaction proceeds before reaching equilibrium.  In this case, the amount of reaction (done the normal way) would be almost exactly the same as the amount of (the limiting) reagent and our calculations would be very difficult.  If we first, pretend the reaction goes to completion and then allow it to reverse towards equilibrium, we get results that are easier to work with.

Consider the combustion reaction of 100 kPa of methane in excess oxygen (80kPa) to produce carbon monoxide and hydrogen gas.

CH₄(g) + 1/2 O₂(g) ---> CO (g) + 2 H₂(g)          K = 1.4x10¹⁵

Lets do this first the normal way.

                CH4(g)   +   1/2 O2   --->  CO(s)   +   2 H2(g)

I  (pressures)  100kPa        80kPa          0            0
   (activities)   1.00         0.80          0            0

C                 -x           -1/2x        +x           +2x

E                 1.00-x       .8-1/2x       x            2x

Thus, we get the following K expression after substituting these activities

Thus, x = 0.9999999999999984. 

Since only two sig. figs are valid, we would have to round up to 1.0 and from which we would have to assume that all CH₄ was used up (1 - x = 0).  But we know that there is a measurable K constant so there must be a measurable amount of CH₄ at equilibrium.

Let's try this again, first allowing all the limiting reagent (CH₄ in this case) to be used up.

                CH4(g)   +   1/2 O2   --->  CO(s)   +   2 H2(g)

I  (pressures)  100kPa        80kPa          0            0
      react 100%)   0         30kPa        100kPa       200kPa
                             (excess)
   (activities)     0         0.30         1.00          2.00

C  (now let it     +x         +1/2x        -x           -2x
    come back 
   to equilibrium)

E                 1.00-x       .8-1/2x       x            2x

Here, we assume that X  is small.  In other words, we assume that 1.00 – X  = 1.00 and 2.00 – 2X  = 2.00 and .3 + 1/2X  = .3. Thus we were able to simplify the expression.

We get X = 1.6x10^-14.  Obviously, our assumption about X  being small is valid.

Now, we can easily calculate the amounts of all the chemicals.

CH₄ is 100 kPa * 1.6x10^-14 = 1.6x10^-12 kPa.
O₂ is 30 kPa, CO₂ is 100 kPa and H₂ is 200 kPa.

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Le Châtelier's Principle

A chemical system at equilibrium has many factors affecting the position of that equilibrium. The amount of each of the components, the temperature, the pressure (in gas phase reactions especially), presence of radiation, etc. We must be able to analyze these conditions and determine if one of them is changed what affect that will have on the position of the equilibrium.

The guiding light through all this is Le Châtelier's Principle (LP). It states:

If a change (stress) is applied to a system at equilibrium the system will adjust to try to reduce the change.

Consider again the Haber process for the production of NH₃. Experience shows that the pressure and temperature both affect the equilibrium. Higher pressures give higher amounts of NH₃ and higher temperatures give lower amounts of NH₃ at equilibrium. looking at this data alone, one might think that high pressure, low temperature is a good set of conditions to produce high yields of ammonia. Unfortunately, at low temperatures, the rate of reaction is slower. A balance must be reached between the thermodynamic requirements for high yield and the kinetic requirements for achieving the results rapidly.

We will look at kinetics later. We will focus on the thermodynamic properties of the mixture as an example.

The Effect of a Change in Concentration

Consider the Haber process (3 H₂ + N₂  2 NH₃ ) at equilibrium with the following concentrations:

[N₂] = 0.399 M,        [H₂] = 1.197 M,       [NH₃] = 0.202 M.

What effect will the addition of N₂ to the system such that its concentration raises (instantaneously) by 1.000 M.

We can guess from LP that if addition of N₂ is the change then the equilibrium will shift to try to use up some of the extra N₂. The equilibrium will shift to the right. Calculation of Q will confirm this quantitatively.

         [NH3]2            (0.202 M)2
Qc = ------------  =  --------------------  =  1.70 * 10-2 M-2
     [H2]3 * [N2]     (1.197 M)3 * 1.399 M  

Calculate K_c from the equilibrium concentrations.

         [NH3]2            (0.202 M)2
Kc = ------------  =  --------------------  =  5.96 * 10-2 M-2
     [H2]3 * [N2]     (1.197 M)3 * 0.399 M  

It's obvious that Q_c < K_c and hence the equilibrium will shift to the right to produce more products (increasing Q until it is equal to K again). This is just what LP predicted qualitatively.

NOTE: In the previous example, I used K_c and Q_c rather than the thermodynamic values K and Q.  I illustrate several points with this: a) you can use concentration units in the gas phase (with caution), b) you can compare the subscripted K and Q values just as you can the thermodynamic ones as long as you are consistent. 

The caveat in item a) is that you cannot easily convert a concentration of a gas into its activity since the standard concentration of 1M applies only to solute activities in aqueous solutions.  You would need to convert the concentration to pressures and then use the gas standard pressure.

Effect of a Change in Pressure

There are three ways to affect a change in the pressure of a gaseous reaction mixture.

1. Add or remove a gaseous reactant or product.
2. Add an inert gas (not involved in the reaction).
3. Change the volume of the container.

We already considered the first option above.

The second will have no effect on a gas phase reaction (according to the assumptions made in deriving the ideal gas law). Since the partial pressures (or concentrations) of the individual gas components of the reaction do not change, an inert gas will not change the equilibrium.

If the volume of the container is changed, the individual pressures will change (and so will their concentrations). recall n_i/V = P_i/RT. If n_i is constant then when the volume V changes, the concentration (n_i/V) will change and so will P_i.

Qualitatively, we can use LP to predict the effect on an equilibrium if we re word the original definition slightly.

If a change in pressure is applied to a gaseous system at equilibrium the system will adjust to try to reduce the change in pressure.

Consider again the Haber process (3 H₂ + N₂  2 NH₃ ). We see that there are 4 moles of gas molecules in the reactant side and two moles of gas molecules in the product side (they're all gases). Thus, if the equilibrium shifts to the right the pressure will drop and if it shifts to the right the pressure will rise.

Now consider a system at equilibrium (at some equilibrium pressure = sum of all partial pressures). If we squeeze the container so as to increase the pressure (of each partial pressure) we will change the equilibrium. LP predicts that the equilibrium will shift to the right to reduce the extra pressure. The new equilibrium will be reached where more products are present than before.

Exercise:

What will happen to the following reactions if the pressure is increased (by reducing the volume).

1. P₄(s) + 6 Cl₂ (g)  4 PCl₃ (l) [answer1]
2. PCl₃(g) + Cl₂(g)  PCl₅(g) [answer2]
3. PCl₃(g) + 3 NH₃(g)  P(NH₂)₃(g) + 3 HCl(g) [answer3]

Effect of a change in temperature

The changes discussed so far have changed the equilibrium position but have not had effect on the equilibrium constant itself. Temperature changes will almost always cause a change in the equilibrium constant. To think about this using LP, we need to remember that all reactions involve the breaking of bonds (absorption of energy) and/or the formation of new ones (release of energy). It is exceedingly rare to find a chemical reaction where the energy absorbed is exactly equal to the energy released. In general we can assume that there is always some net change in energy involved in a chemical reaction. We need to consider this net change as a reactant or product to use LP.

Consider again the Haber process but this time note that energy is released during the process (exothermic).

3 H₂(g) + N₂(g) 2 NH₃(g) + energy.

Now it's easy to use LP to predict that if the temperature (kinetic energy) of the system is increased, the equilibrium will shift to the left to try to use up the excess energy. Since nothing else was changed, the new equilibrium position will have a different equilibrium constant than the old one. Since the new equilibrium has less products than the old one, the new K constant will be smaller than the old one (as T rises, K drops).

 A reaction which consumes energy (endothermic) will have the opposite effect.

energy + CaCO₃(s)  CaO(s) + CO₂(g)

If the Temperature of this system initially at equilibrium (Kº) is increased the reaction will shift to the right (to use up some of the extra energy), establishing a new equilibrium with a new K>Kº.

We can represent this temperature dependence using the van't Hoff equation. A rigorous derivation will be given for this equation later in this course.

Example:

The equilibrium constant for the Haber process

3/2 H₂ + 1/2 N₂  NH₃

is 668 at 300 K and 6.04 at 400 K. What is the average enthalpy of reaction for the process in that temperature range?

DH º = -47 kJ/mol.

Looking up the value for DH º at 298.15 K, we find -46.11 kJ/mol, acceptably close to our estimated average value.

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Real Gases

Up to this point, we've always been using the ideal gas approximations when discussing equilibrium constants, etc, where gases were involved. In reality, there are few gases which behave in an ideal way except at pressures which are relatively low. Higher pressures result in properties of gases which deviate from ideal behaviour quite appreciably.

Consider again the Haber process (at 723 K).

3 H₂(g) + N₂(g) 2 NH₃(g)

The K_p value we calculate using observed partial pressures would be

         (PNH3obs)2
Kpobs = ------------------
     (PH2obs)3 * (PN2obs)

Since the observed pressure deviates from ideal pressure noticeably at high pressure, this calculated value will be in some error and the error will increase as pressure increases. See the table below for a set of pressures and observed equilibrium constants to illustrate this point.

 We can correct for this error using a more complete definition for activities. This will not be discussed further here in first-year Chemistry. You will see discussions on this topic if you take upper year Physical Chemistry courses later in your career.

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Michael J. Mombourquette.
Copyright © 1998
Revised: March 23, 2008.