AP Chemistry - Chemical Equilibrium
Introduction to Equilibrium - Reverse Reactions and Chemical Equilibrium

In all the preceding work on chemical kinetics, reaction rates and mechanisms, attention has been focused upon chemical reactions which are proceeding in one direction only (forward). You should realize, however that many reactions are reversible, i.e. they can occur in either direction. In particular, for a given forward reaction
 aA + bB ---> lL + mM

 
one can conceive of the reverse process
lL + mM ---> aA + bB

 
Now if no product molecules L & M are present at the beginning of a reaction, the reverse reaction may well be forgotten about until an appreciable concentration of products builds up, since the rate of the reverse reaction will be proportional to some product of the concentrations of L and M. In general, however, the reverse reaction becomes important eventually, since the forward reaction slows down with time (reactant concentrations being reduced by conversion to products) whereas the reverse reaction rate increases with time (as product concentrations build up). Eventually a point in time is reached when the
Rate Forward Reaction = Rate Reverse Reaction
At this point CHEMICAL EQUILIBRIUM is achieved; unless the system is disturbed by a temperature change or by adding excess reactant or product molecules, the equality of rates is maintained as are the "equilibrium" concentrations of all chemical species. The variation of reaction rates with time is illustrated graphically as follows:
In any chemical system, which a state of equilibrium is reached the reaction seems to be stopped. The macroscopic visible properties that we can see are constant. In reality the reactions are still taking place.
H2(g) + I2(g)   <====>   2 HI(g)
Both the forward and reverse reactions are proceeding at the same rate when a reaction is in a state of equilibrium.
The observable properties and concentrations of all participants (species) become constant when a chemical system reaches a states of equilibrium.
A chemical system is said to be in a state of equilibrium if it meets the following criteria:
1. The system is closed.
2. The observable macroscopic properties are constant.
3. The reaction is sufficiently reversible so that observable properties change and then return to the original rate when a factor that affects the rate of the reaction is varied and then restored to it's original value.
Let's consider in detail the forward and reverse processes for some general "atom exchange" reaction
AB + CD ---> AC + BD
Assuming that the forward reaction occurs by a one-step bimolecular mechanism, then
Rate Forward Reaction = Rf = kf [AB][CD]
Similarly, assuming that the reverse reaction also occurs by a one-step mechanism
Rate Reverse Reaction = Rr = kr [AC][BD]
Now according to the "Principle of Microscopic reversibility", the activated complex which must be achieved in the reverse reaction is identical with activated complex for the forward reaction; thus the energy profile for the forward and revers reactions can be illustrated by the same plot:
In the graph above, Erepresents the activation energy for the forward reaction and Ethat for the reverse; since the activated complex is identical for both directions it follows that
H = Er - Er
Thus, if the reaction is exothermic (as shown), the activation energy for the reverse reaction must be greater that for the forward reaction. If the reaction is endothermic, the converse is true. If H = 0 then Ef = Er.
Returning to the rate law constant expression, we know that at equilibrium
Rf = Rr
therefore kf[AB]e[CD]e = kr[AD]e[CB]e
where the subscript e denotes equilibrium concentrations. By rearrangement we can obtain the expression
[AD]e[CB]e  = Kf = a constant Keq,  the Equilbrium Constant
[AB]e[CD]e    Kr
The relationships derived above for a particular reaction type are, in fact, completely general for any reaction regardless of its complexity or mechanism. That is, for any reaction in which the forward and reverse reaction rates have achieved equality (equilibrium as denoted by a double arrow),
aA + bB <------->  lL + mM
then the product of the equilibrium concentration of the products, each raised to its coefficients in the balanced equation, divided by the product of the equilibrium concentrations of the reactants, each raised to its coefficients in the balanced equation, is a characteristic constant for the system.
Before leaving this topic, one paradox" should be cleared up - that is, the appearance of coefficients in the balanced equation as the concentration exponents in K, whereas as I have emphasized earlier in the rate unit, the exponents in the rate laws for the forward and reverse reactions need not equal these coefficients unless the reaction proceeds by a single-step mechanism. The paradox is resolved by considering the particular reaction
AB + C -----> AC + B
Where the reaction mechanism is deduced from the initial rate data is
AB ---> A + B (slow)
A + C ---> AC (fast)
to yield a rate law for the forward direction Rf = k[AB]
Now if true chemical equilibrium has been achieved in the system, it must follow that EQUILIBRIUM HAS ALSO BEEN ESTABLISHED IN EVERY STEP OF THE REACTION. That is, at equilibrium the reverse of each step is also important, so the mechanism is
step 1        AB ---> A + B R1 = k1[AB]
step 2    A + B ---> AB R2 = k2[A][B]
step 3    A + C ---> AC R3 = k3[A][C]
step 4        AC ---> A + C R4 = k4[AC]
Now since R1 = R2 at equilibrium,
                 k1[AB] = k2[A][B]
and since R4 = R3 at equilibrium,
                k4[AC] = k3[A][C]
By equating the ratio of the left-hand sides of the two equations to the ratio of the right-hand sides, we obtain
or upon rearrangement
(Note that the concentrations of the intermediate species cancel out.)
Therefore, even though the reaction proceeds by a multi-step mechanism, we have proved in this case (and can prove it for any other particular case) that the ratio of product to reactant concentrations, each raised to the corresponding coefficient in the balanced overall equation, is still a constant Keq characteristic of the reaction if complete equilibrium has been established. Thus there is no "conflict" between the use of reaction orders which are not equal to balanced equation coefficients and the rule that in the equilibrium constant one always obtains exponents which equal the coefficients.
The properties and concentrations of an equilibrium system are constant because the rates of the forward and reverse reactions are equal so that it appears that no reactions are occurring.
Reversibility of Reactions at Equilibrium
Once a system has reached equilibrium, any factor which causes a change in the rate of either the forward or reverse reaction will disturb or shift the equilibrium. The system will readjust itself so that a new equilibrium is reached and the rates will again become equal.
Le Châtelier's Principle
When a stress is applied to a system at Equilibrium, the system readjusts so as to relieve or offset the stress. Stress is any imposed factor which upsets the balance in rates between the forward and reverse reactions. Stress factors may be changes in concentrations [], total pressure with gases only, volume changes (which cause the pressure changes), and temperature.
Stress and Changes in Concentration [ ]
Increasing the concentration of the reactants will cause a stress which is relieved by an increase in the concentration of product, [product]. If the [product] is increased the [reactants] will increase to offset the stress.
Stress and Changes in total Volume (or related pressure)
eg.     N2(g) + 3 H2(g) <===>  2 NH3(g) + 92.00 kJ
A stress imposed by a decrease in volume is actually a stress caused by the increase in the concentration. The stress is relieved when the system reduces the number of molecules.
ie.    4 molecules ------> 2 molecules
A decrease in total pressure (which is caused by an increase in volume) will shift the reaction to the reverse. A decrease in to total volume of a gaseous system (or the accompanying increase in pressure) shifts an equilibrium in the direction of the fewer molecules as shown by the equation for the reaction.
Stress and Changes in Temperature
In any reaction a temperature increase favours the reaction that absorbs heat. i.e. the endothermic reaction.
eg. N2(g) + 3 H2(g) <=====>  2 NH3(g) + 92.00 kJ
This reaction is exothermic as written. That means the reverse reaction or backward reaction is endothermic. Adding heat (demonstrated as an increase in temperature) will shift the reaction left and more reactants will be formed as the products get used up. A decease in temperature will favour the reaction that produces heat. i.e. the exothermic reaction. The reaction about would switch to the forward direction to produce more heat and product.
Effect of a Catalyst
A catalyst equally favours both the forward and reverse reactions. Therefore a catalyst does not shift the Equilibrium [ ]'s. It simply causes the reaction system to reach Equilibrium in a shorter period of time.
Example Problem
For the reaction below:

     2 SO2(g) + O2(g) <=====>  2 SO3(g) + heat
 

Indicate the favoured reaction either forward or reverse under the stress applied below.
a) increase the [SO2]
The reaction will shift to the forward direction and more SO3will be produced.
b) Partial pressure of SO3 is decreased (Some of the SO3 is removed fro the system).
The system will shift to the right. As the SO3 is removed its concentration drops. The reactants will react to produce more product.
c) Decrease in temperature.
The reaction will shift right. As written the reaction is exothermic. As heat is removed the reaction proceeds to produce more heat.
d) He gas is added at constant volume so that the total pressure is increased.
No change takes place. Dalton Law of Partial Pressures comes into play here. Each gas is treated separately from it's neighbours. Because the volume stays constant then the concentrations don't change. Therefore no changes take place.
e) He gas is added, and the container is allowed to expand so that the total pressure is kept constant.
To the left or the reactants will be favoured. The volume was allowed to increase to maintain the total pressure. Since the volume increased the concentrations must have dropped for all reactants and products. The reaction will react do produce the maximum number of moles of gas to fill the available space. Hence the reactants are favoured.
f) A catalyst is added.
No change. The catalyst favours both reactions, forward and reverse equally.
Follow-up Problem:
For the reaction below:

        heat + 2 H2O(g) <=====>  2 H2(g) + O2(g)

Indicate the direction of the reaction most favoured by each of the following conditions.
a) increase the [H2]
b) the partial pressure of [H2O] is increased
c) the [O2] is increased
d) increase the temperature
e) decrease the volume of the container
f) He gas is added 

The Equilibrium Law
For any general reaction system the value of Keq is called the equilibrium constant for the reaction. for example:
                             3 H2(g) + N2(g) <=====>  2 NH3(g) + heat
                         Keq     [NH3]2
                                      [H2]3[N2]
The above equation is called the Equilibrium Law Expression and it is a general form of the Law of Mass Action.
Law of Mass Action
In a system at equilibrium, at a fixed temperature, the product of the equilibrium concentration of the products divided by the product of the concentrations of the reactants, each being raised to the coefficient of the substance in the equation, must be equal to a constant.
This law can only be applied to ideal gases and solution. (There are some reactions that do go all the way).
Here is another example of an equilibrium law expression.
H2(g) + I2(g) <=====>   2 HI(g) + heat
Write equilibrium law expressions for each of the following reactions:
a) 2 NO(g) + O2(g) <=====>   2 NO2(g)

b) 2 SO2(g) + O2(g) <======>   2 SO3(g)

c) SO2(g) + ½O2(g)  <======>  SO3(g)

d) SO3(g) <======>   SO2(g) + ½O2(g)

e) 4 HCl(g) + O2(g)  <======>  2 H2O(g) + 2 Cl2(g)
 

Some of the equations above are almost carbon copies of each other except for the values of the coefficients. A couple of general rules should help.
Keq(reverse) =          1 
                        Keq(forward)
Any reaction written normally can have a Keq value. But does this Keq value stay the same if the reaction is flipped over and written in reverse. No, if the equation flips the Keq value becomes the inverse.

At any given temperature there will be numerous [  ] values that will satisfy a constant Keq value.

Effect of Temperature on Keq
Changing the temperature favours one reaction over the other. Therefore a shift in the Keq value is caused, as well as a change in the [participants]. A new equilibrium will be reached with it's own Keq value for the reaction at that temperature. At constant temperature, changing the equilibrium concentration does not affect Keq because the rate constants are not affected by the concentration changes.
When the concentration of one of the participants is changed, the concentration of the others vary in such a way as to maintain a constant value for the Keq.
Relationship of Keq to the Extent of Reaction
The magnitude of Keq is a measure of the extent to which a given reaction has or will take place.
A large value of Keq indicates that, at equilibrium there is a high concentration of products and a relatively low concentration of reactants.
A small value of Keq indicates that at equilibrium, the concentration of reactants is still high compared with that of the products.
A Keq value of 1 means that half of the reactants have turned into products. Therefore if the Keq is 1 then there should be a 50.0% yield.
If the Keq is about 1 x 1010 then the reaction has gone almost to completion.
If the Keq is about 1 x 10-10 it is essentially an incomplete reaction.
1.0 x 10-10                                                                 1                                                       1.0 x 1010
       |                                                           |                                                              |
0.0001%    1%                                                                   50%                                                                  99.9%    99.999%
Calculation of Keq and Concentrations
After you have finished this you should be able to:
1) Calculate the equilibrium constant.
2) Calculate the equilibrium concentration of a participant when the value of Keq is known as well as the concentrations of the other participants.
3) Determination of the net direction of a reaction prior to establishing an equilibrium.
4) Calculation of equilibrium []'s when initial []'s and the equilibrium constant are known.
5) Calculation of the % dissociation and the % yield of a reaction.
Example Problems
Problem #1
When 0.40 moles of PCl5 is heated in a 10.0 L container, an equilibrium is established is which 0.25 moles of Cl2 is present.
PCl5(g) <====>  PCl3(g) + Cl2(g)
a) What is the number of moles of PCl5 and PCl3 at equilibrium?
b) What are the equilibrium concentration of all three components?

                PCl  <=====>       PCl3     +      Cl2
start   0.40 moles                    ------          -------
after   0.15 moles               0.25 moles    0.25 moles
 

From the balanced equation and the coefficients it should be obvious that everything works on a one-to-one basis. So there should be 0.25 moles of PCl3 and 0.25 moles of PCl5 should have been consumed, so there are 0.40-0.25 moles = 0.15 moles of PCl5 left.
Therefore [PCl5] = 0.15 moles = 0.015 moles/L
                                 10.0 L
[PCl3] and [Cl2] both are 0.25 moles each. Therefore their concentrations are identical as well.
[PCl3] = [Cl2] = 0.25 moles = 0.025 moles/L
                             10.0 L
Problem #2
When 1.00 mole of NH3 gas and 0.40 moles of N2 gas are placed in a 5.0 L container and allowed to reach an equilibrium at a certain temperature, it is found that 0.78 moles of NH3 is present. The reaction is:

     2 NH3(g)  <======>   3 H2(g) + N2(g)
 

a) What are the number of moles of H2 and N2 at equilibrium?

b) What is the concentration in moles/L of each species?
 

Problem #3
A mixture of H2 and I2 is allowed to react st 448oC. When equilibrium is established, the concentrations of the participants are found to be [H2] = 0.46 mol/L, [I2] = 0.39 mol/L, and [HI] = 3.0 mol/L. Calculate the value of the Keq at 448oC.             H2(g) + I2(g) <=====>  2 HI(g)
Problem #4
Assume that in the analysis of another equilibrium mixture of the same reaction as above, at the same temperature of 448oC, the equilibrium concentrations of I2 and H2 are both 0.50 mol/L. What is the equilibrium concentrations of HI?
 
 
 

 

Problem #5
The equilibrium constant for the reaction represented below is 50 at 448oC

       H2(g) + I2(g) <=====>  2 HI(g)

a) How many moles of HI are present at equilibrium when 1.0 moles of H2 is mixed with 1.0 moles of I2 in a 0.50 L container and allowed to react at 448oC?

b) How many moles of H2 and I2 are left unreacted?

c) If the conversion of H2 and I2 to HI is essentially complete, how many moles of HI would be present?

d) What is the percent yield of the equilibrium mixture?
 

Solution
a) First write the Keq equation based on the balanced reaction. You should always make sure that the reaction is balanced first.

The concentrations at the start are:

[H2] = 1.0 mol/0.50 L = 2.0 mol/L

[I2] = 1.0 mol/0.50 L = 2.0 mol/L

[HI] = 0 moles therefore 0 mol/L

Fill in the numbers under the reaction for the starting concentrations.

                 H2     +     I2      <=====> 2 HI
  start    2.0 M      2.0 M                    0 M

Let 'x' be the number of moles of H2 per litre consumed.

finish    2.0 - x     2.0 - x                      2x
 

Stop and confirm the above. If 'x' amount of H2 is consumed then 'x' amount of I2 will also be consumed because the H2 and I2 react on a 1:1 basis. Also if 'x' amount of H2 (or for that matter I2) reacts then '2x' of HI will be produced because the reactants and product react on a 1:2 basis.
Fill in the finish concentrations into the Keq equation. and solve for 'x'.
50 =            (2x)2
           (2.0 - x)(2.0 - x)
50 =       (2x)2 ___
            (2.0 - x)2
This is going to work nicely because both the top and bottom on the right are squares. There square root the entire right hand side, along with the left.
7.1 =     2x
          2.0 - x

14.2 - 7.1x = 2x

x = 14.2 = 1.56 mol/L
       9.1
 

You are solving for 'x' which is in moles/L therefore you can automatically fill in the units for any value of 'x'.

Now to answer the question. What are the final reactant and product concentrations?

[H2] = 2.0 mols/ 1 - x = 2.0 mols/L - 1.56 mols/L = 0.44 mols/L

[I2] = 2.0 mols/ l - x = 2.0 mols/L - 1.56 mols/L = 0.44 mols/L

[HI] = 2 x = 2 * 1.56 mols/L = 3.12 moles/L
 

b)  From a) you get the number of moles/L of H2 and I2 left unreacted. This is the concentration not the answer to the question. You are asked for the number of moles left unreacted therefore moles = concentration * volume

[H2] = [I2] = 0.5 L * 0.44 mols/L = 0.22 moles

Therefore 0.22 moles of H2 and 0.22 moles of I2 are left unreacted. Note that the L's unit cancels out.
 

c) If you look at the equation again you can see that there is exactly the right amount of H2 to react with I2. Theoretically if both the H2 and I2 where all used up then we should make 2 moles of HI. This is the theoretical yield because it is what could be made in theory if everything went to completion.
d) The percentage yield is calculated by using the actual yield from part a) and the theoretical yield from part b). You may use either moles or moles/L but make sure that they are both the same.

Percentage Yield =    Actual Yield       *  100
                               Theoretical Yield

1.56 moles/L * 100
    2.00 moles/L

= 78% yield (note that the moles/L units cancel)

OR

Percentage Yield = 3.12 moles * 100
                              4.00 moles

= 78% yield (note that the units still cancel)
 

Problem #6
How many moles of HI are present at equilibrium when 2.0 moles of H2 is mixed with 1.0 moles of I2 in a 0.50 L container and allowed to react at 448oC. At this temperature Keq = 50.
Solution
First the equation: H2 + I2 <=====>  2 HI

Then the concentrations:

[H2] = 2.0 moles/0.50 L = 4.0 moles/L

[I2] = 1.0 moles/0.50 L = 2.0 moles/L

[HI] = 0 moles therefore 0 moles/L

Then the Keq equation is:


 

Write down the equation and the start concentrations. Let 'x' be the finish concentrations.

               H2    +      I2   <====> 2 HI
start    4.0 M        2.0 M             0 M
finish   4.0-x          2.0-x               2x
 

We will subtract 'x' from both H2 and I2 because this is the amount of reactant that will convert into product '2x'.
Now substitute these 'x' factors into the Keq equations and solve for 'x'.
50 =         (2x)2
         (4.0 - x)(2.0 - x)

We're not so lucky this time. Multiply this out and get the equation into standard form.
 

50 =        4x2
         8.0 - 6.0x + x2

400 - 300x + 50x2 = 4x2

46x2 - 300x + 400 = 0 (standard form)

Use the quadratic equation to solve for 'x'.

      In this case a = +46, b = -300 and c = +400.
 

The quadratic equation is a mathematical relationship for solving a line to find it's roots. In chemistry the 'roots' represent real life values. One will be realistic and the other will not be.


 

4.7 is to large since we only started with 4 mols/L of H2 to begin with. It is therefore the unreal root. So let x = 1.9 mol/L.

Substitute 'x =1.9 mol/L' back up into the original concentration calculations to see how much is left after reaction.

                H2    +      I  <======>  2  HI
start     4.0 M      2.0 M                      0 M
finish    4.0-x       2.0-x                         2x

          4.0-1.9    2.0-1.9                    2(1.9)
            2.1 M     0.1 M                     3.8 M

To finally answer the question:
The moles of HI present at equilibrium would be:

moles = concentration * volume
         = 3.8 moles/L * 0.5 L
         = 1.9 moles

There would be 1.9 moles of HI in the reaction vessel at equilibrium.
 

Problem #7
When 3.0 moles of HI, 20 moles of H2, and 1.5 moles of I2 are placed in a 1.0 L container at 448oC, will a reaction occur?   If so, which reaction will take place?
Solution
Calculate the concentrations and substitute them into an expression and compute the "experimental concentration quotient" or "Q".
If Q is equal to Keq (50 in this case) then the system is already at equilibrium and stable.

If Q > Keq then the [product] is too high and must decrease. Therefore the reverse reaction is favoured.

If Q < Keq then the [product] is still to small and must increase, therefore the forward reaction is favoured.

            H2 + I2 <=======>   2 HI

therefore Q =          (3.0 mol/L)2
                      (2.0 mol/L)(1.5 mol/L)

                  = 9.0 = 3
                     3.0

Therefore Q < Keq since 3 < 50. The forward reaction is favoured to occur until the ratio of product:reactants = 50.
 

In-Class Follow-up Problems
1.   When 0.040 moles of PCl5 is heated to 250oC in a 1.0 L vessel, an equilibrium is established in which the equilibrium concentrations of Cl2 is 0.025 mol/L. Find the equilibrium constant, Keq , at 250oC for the reaction:

PCl5(g) <======> PCl3(g) + Cl2(g)

Note that the initial, not the equilibrium concentration for PCl3 is given. The amount of PCl5 reacted is the same as the amount of Cl2 formed.

2. Assume that the analysis of another equilibrium mixture of the system from above shows that the equilibrium concentration of PCl5 is 0.012 mol/L and that of Cl2 is 0.049 mol/L. What is the equilibrium concentration of PCl3 at 250oC?

3. How many moles of PCl5 must be heated in a 1.0 L flask at 250oC in order to produce enough chlorine to give an equilibrium concentration of 0.10 mol/L?

4. Will there be a net reaction when 2.5 moles of PCl5, 0.60 mole of Cl2, and 0.60 mole of PCl3 are placed in a 1.0 L flask and heated to 250oC. If so, which reaction takes place?

Go to the Equilibrium Calculations Worksheet
Quantitative Aspects of Le Châtelier's Principle
Consider the effect of changing the concentration of a participant in a system that is already at equilibrium.
SO2(g) + NO2(g) <=====>  SO3(g) + NO(g)
 
Suppose we add SO2 to the system. This produces a non-equilibrium that is a stress. It must be relieved by a reaction that relieves the stress. The increase in the denominator (of the Keq equation for this reaction) means that the experimental reaction quotient, Q , is less that Keq. In order to re-establish equilibrium the numerator must increase. ie. The reaction must shift to the right and increase to yield of products.
At 200oC analysis of the equilibrium mixture shows:

[SO2] = 4.0 M,   [NO2] = 0.50 M

[SO3] = 3.0 M,   [NO] = 2.0 M
 

What is the new equilibrium concentration of NO when 1.5 moles of NO2 is added to the equilibrium mixture above?
Solution
This is a new equation. Find the value of Keq first. Since its a constant at this temperature then it will be valid for the equilibrium mixture and the stressed mixture.
So the value of Keq for this reaction, at this temperature is 3.0
Fill in the start and finish values including an 'x' term.
              SO2(g)    +      NO2(g) <------>  SO3(g)     +     NO(g)
start       4.0 M              0.5 M                  3.0 M               2.0 M
finish      (4.0-x)         (0.5+1.5-x)            (3.0+x)             (2.0+x)
Let's take a second to explain what is happening. When we add NO2 we stress the reaction so that more product is formed. This means we will lose 'x' amount of SO2 more. The NO2 term is 0.5 +1.5-x because we started with 0.5 M, added 1.5 M more to stress it and this amount will then react 'x' amount of itself to produce product. We are lucky here because everything works on a one-to-one basis. This is not always the case. SO3 and NO will each increase by 'x' as more product forms.
Now we will fill in these factors in the Keq equation and solve for 'x'.
3.0 = (3.0 + x)(2.0 + x)
          (4.0 - x)(2.0 - x)
After a few multiplications and additions we get:
2.0x2 -23x +18 = 0
After using the quadratic equation on this formula we get a value for x = 0.75 mol/L
Therefore the [NO] = 2.0 + x = 2.75 mol/L
Follow-Up Problem
How many moles of NO2 would have to be added to the original equilibrium mixture to increase the equilibrium concentration of SO3 from 3.0 to 4.0 moles at the same temperature?
Heterogeneous Reaction Systems
A system that has two for more phases is heterogenous.
eg. CaCO3(s)  + dil. HCl   <===>  CO2(g) + CaO(s)
Increasing the amount of solid present in a heterogenous equilibrium increases the total, surface area at which both the forward and reverse reactions are occurring. The total number of moles of solid reactant being converted into product per unit time increases. However, the total number of moles of product being converted back into reactant increases proportionally, This results in no net change in the total amount of reactant or product.
Therefore Keq for the above reaction is dependant upon the [CO2] only.
ie. the Keq equation for the above reaction is Keq = [CO2] only.
The equilibrium constant expression for a heterogeneous reaction involving gases does not include the concentrations of pure solids.
eg. NH4Cl(s)  <====>  NH3(g) + HCl(g)
Keq = [NH3][HCl]
Factors Related to the Magnitude of Keq
The Tendency or Drive Toward Minimum Energy (Enthalpy) 
and Maximum Disorder (Entropy)
Why are some reactions "complete" and others not???
You should already know this:
In a closed system all reactions, unless prevented by exceedingly high activation energy barriers (slow rates), spontaneously approach an equilibrium state.
Now take a look at:
N2(g) + 3 H2(g) <=====>  2 NH3(g) + heat
There is a drive towards minimum energy, (to the right), since this is the exothermic reaction. There is an opposing drive to the left since the system at the same time tries to establish maximum randomness (entropy). Entropy is favoured because 4 moles are made out of two. The Keq for this reaction is actually a compromise between these two opposing drives.
The relative importance of the drives depends on the temperature. At low temperatures (Low Kinetic Energy), the enthalpy reaction will be favoured. At high temperatures (lots of Kinetic Energy) the drive toward maximum randomness will be favoured.
ie.   Equilibrium favours the exothermic reaction at low temperature and the drive towards maximum randomness at high temperatures.
The likelihood of a reaction occurring is related to the temperature effect and the relative enthalpy and entropy changes associated with the reaction.
Reactions that increase in entropy (products are more random than the reactants) and decrease in enthalpy (endothermic) are spontaneous.
Reactions that decrease in entropy (products are more structured than the reactants) and increase in enthalpy (endothermic) are nonspontaneous.
Here is a list of conditions in which there is an increase in the entropy of a system.
1. When a gas is formed from a solid. The gas is allows to escape from the system in this case so a single arrow is used.
CaCO3(s) + heat -----> CaO(S) + CO2(g)
2. When a gas is evolved from a solution and allowed to escape.
Zn(s) + 2 H+(aq) -----> H2(g) + Zn2+(aq)
3. When the number of moles of gaseous product exceeds the number of moles of gaseous reactant.
2 C2H6(g) + 7 O2 -------> 4 CO2(g) + 6 H2O(g)
4. When any crystal dissolves in water.
NaCl(s)  <=====>  Na+(aq) + Cl-(aq)
For each of these processes, predict whether the entropy increases or decreases.
a) 2 H2(g) + O2(g) <=====> 2H2O(g)

b) 2 SO3(g) <====> 2 SO2(g) + O2(g)

c) MgCO3(s) + 2 H3O+(aq) <====> Mg2+(aq) + 3 H2O + CO2(g)

d) Ag+(aq) + Cl-(aq)  <====> AgCl(s)

e) Cl2(g) <====> 2 Clo(g)

f) NH4NO3(s)  <=====> NH4+(aq) + NO3-(aq)

g) H2O(l)  <====> H2O(g)

h) Mg(s) + 2 H3O+(aq)  <====>  Mg2+(aq) + H2(g) + 2 H2O

i) 2 C2H2(g) + 5 O2(g)   <====>  4 CO2(g) + 2 H2O(g)

j) NH3(g) + HCl(g)  <====>  NH4Cl(s)

Change in Free Energy or Spontaneity of a Reaction
The Second Law of Thermodynamics: The entropy of the universe increases for any spontaneous reaction.
Applied to a chemical system this means that the entropy of a system can increase or decrease but if it does decrease, then the entropy of the surroundings must increase to a greater extent so that the overall entropy change in the universe is positive.
This criteria can be calculated out and is referred to as the G or Gibb's Free Energy. G is related to the enthalpy change, the entropy change and the temperature of the system.
ΔG = ΔH - T ΔS
G can be used to predict the spontaneity of a reaction at constant temperature and pressure. the criteria are listed below:
1. If G is negative, then the reaction is spontaneous (probable) as written.
2. If G is positive, the reaction is improbable as written, but the reverse reaction is probable. Reversing the equation would reverse the sign of G.
3. If G is 0, the system is at equilibrium and there is no net reaction. When G = 0 then H = tS. thus at equilibrium, the entropy factor is balanced by the enthalpy factor.
-Exothermic reactions which are accompanied by an increase in entropy of the system are probable.

-Endothermic reactions accompanied by a decrease in entropy are improbable.

-At very high temperatures, the sign and magnitude of G and the spontaneity of the reaction are determined primarily by the change in entropy.

-At very low temperatures, the sign and magnitude of G and the spontaneity of the reaction are determined primarily by the enthalpy change, H.
 

Thermal Spontaneity of

Effect               H    S    Reaction as Written Comment
Exothermic       -     +       probable no exceptions
Exothermic       -     -       probable at low temps
Endothermic     +    +       probable at high temps
Endothermic     +    -        improbable no exceptions
 

Predict the probability of H2O(l) ------> H2 (g) + ½O2(g) at  25oC and 1500oC.
 
Go to the Equilibrium Factors Worksheet
Reactions That Run To Completion
There are some reactions that are considered to run to completion. They do not set up an equilibrium but run until one of the reactants runs out. They basically fall into 3 basic categories.
1. Formation of a gas that is allowed to escape.

      Sugar -------> CO2 + H2O (not reversible)
 

2. Formation of a precipitate

      2 Tl+ + 2 Cl- -----> 2 TlCl (ppt)

The ppt is a solid that makes the equation heterogenous and essentially eliminates the TlCl from the Keq equation.
 

3. Formation of a slightly ionized product.

     NaCl(s) + KBr(s) --H2O--> Na+(aq) + Cl-(aq) + K+(aq) + Br-(aq)

Remove water and you would get a mixture of four products not just two. You would get NaCl, NaBr, KBr,& KCl
 

Go to the Introduction to Equilibrium Review Worksheet
Determination of Keq Values Using the Nernst Equation
The Eo values and the equilibrium constant are both measures of the tendency for a reaction to take place.  It would be reasonable for us to suspect that they are related.
At equilibrium the value of Eo is 0, since no net flow of elecrtons will be occurring.
Therefore the existing [ ]'s at equilibrium and the logrithm term from the Nernst equation become Keq.
ie.  0 = Eo - 0.059/n log [C]c[D]d / [A]a[B]b          but     Keq = [C]c[D]d / [A]a[B]b
     Therefore   0 = Eo - 0.059/n log Keq

      log Keq = n Eo / 0.059

      Keq = 10 (nEo / 0.059)
 

So the equilibrium constant for any redox reaction can be calculated using it's standard half-cell reduction potential.  If the Eo value is +ve then Keq > 1 and a reaction with an Eo value that is -ve will have a Keq < 1.
ex.   Calculate the Keq value for the reaction between silver nitrate and metallic zinc:  Is the reaction essentially complete or incomplete.

1.  Write the overall reaction:    2 AgNO3  +  Zno  ---->   Zn(NO3)2  +  2 Ago

2.    Write the half cell reactions:     2 Ag+1  +  2e-1  ---> 2 Ago   Eo = +0.80 V
                                                          Zno  ----->  Zn+2   + 2e-1       Eo = +0.76 V
                                                                                                         Eo = +1.56 V

3.   Solve for Keq           Keq = 10 (nEo / 0.059)
                                              =  10 (2)(1.56)/0.059
                                              =  1052.88
 

The extremely large value of Keq indicates a reaction that is essentially complete.
Ex #2:  Calculate the Keq value for the reaction between Zn and Copper(II) sulphate.  Is the reaction complete?

        Zno   +  CuSO4 --->  ZnSO4  + Cuo

        Zno  ---->  Zn+2   + 2e-1       Eo  = +0.76 V
        Cu+2  +  2e-1  ---->  Cuo      Eo = +0.34 V
                                                     Eo = 1.10 V
 

            Keq = 10 (nEo / 0.059)
                   =  10 (2)(1.10)/0.059
                   =  1037.29
Therefore it is essentially complete.