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 

one can conceive of the reverse process 

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 

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. 
H_{2}(g) + I_{2}(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 onestep bimolecular mechanism, then 
Rate Forward Reaction = R_{f }= k_{f }[AB][CD] 
Similarly, assuming that the reverse reaction also occurs by a onestep mechanism 
Rate Reverse Reaction = R_{r }= k_{r }[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, E_{f }represents the activation energy for the forward reaction and E_{r }that for the reverse; since the activated complex is identical for both directions it follows that 
H = E_{r } E_{r} 
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 E_{f }= E_{r}. 
Returning to the rate law constant expression, we know that at equilibrium 
R_{f }= R_{r} 
therefore k_{f}[AB]_{e}[CD]_{e} = k_{r}[AD]_{e}[CB]_{e} 
where the subscript e denotes equilibrium concentrations. By rearrangement we can obtain the expression 
[AD]_{e}[CB]_{e}_{ }= K_{f}
= a constant Keq, the Equilbrium Constant [AB]_{e}[CD]_{e} K_{r} 
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 singlestep 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 R_{f }= 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 R_{1 }= k_{1}[AB] 
step 2 A + B > AB R_{2 }= k_{2}[A][B] 
step 3 A + C > AC R_{3 }= k_{3}[A][C] 
step 4 AC > A + C R_{4 }= k_{4}[AC] 
Now since R_{1 }= R_{2} at equilibrium, 
k_{1}[AB] = k_{2}[A][B] 
and since R_{4 }= R_{3 }at equilibrium, 
k_{4}[AC] = k_{3}[A][C] 
By equating the ratio of the lefthand sides of the two equations to the ratio of the righthand sides, we obtain 
or upon rearrangement 
(Note that the concentrations of the intermediate species cancel out.) 
Therefore, even though the reaction proceeds by a multistep 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 K_{eq} 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. 
Calculation of K_{eq }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 K_{eq }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 PCl_{5 }is heated in a 10.0 L container, an equilibrium is established is which 0.25 moles of Cl_{2 }is present. 

a) What is the number of moles of PCl_{5 }and PCl_{3}
at equilibrium? b) What are the equilibrium concentration of all three components?
PCl_{5 } <=====> _{
}PCl_{3} +
Cl_{2} 
From the balanced equation and the coefficients it should be obvious that everything works on a onetoone basis. So there should be 0.25 moles of PCl_{3} and 0.25 moles of PCl_{5} should have been consumed, so there are 0.400.25 moles = 0.15_{ }moles of PCl_{5 }left. 
Therefore [PCl_{5}] = 0.15 moles = 0.015 moles/L
10.0 L 
[PCl_{3}] and [Cl_{2}] both are 0.25 moles each. Therefore their concentrations are identical as well. 
[PCl_{3}] = [Cl_{2}] = 0.25 moles = 0.025
moles/L 10.0 L 
Problem #2 
When 1.00 mole of NH_{3 }gas and 0.40 moles of N_{2
}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 NH_{3}(g) <======>
3 H_{2}(g) + N_{2}(g) 
a) What are the number of moles of H_{2} and N_{2}
at equilibrium?
b) What is the concentration in moles/L of each species?

Problem #3 
A mixture of H_{2} and I_{2} is allowed to react st 448^{o}C. When equilibrium is established, the concentrations of the participants are found to be [H_{2}] = 0.46 mol/L, [I_{2}] = 0.39 mol/L, and [HI] = 3.0 mol/L. Calculate the value of the K_{eq }at 448^{o}C. H_{2}(g) + I_{2}(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 448^{o}C,
the equilibrium concentrations of I_{2} and H_{2} 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 448^{o}C H_{2}(g) + I_{2}(g) <=====> 2 HI(g) a) How many moles of HI are present at equilibrium when 1.0 moles of H_{2} is mixed with 1.0 moles of I_{2} in a 0.50 L container and allowed to react at 448^{o}C? b) How many moles of H_{2} and I_{2} are left unreacted? c) If the conversion of H_{2} and I_{2} 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 K_{eq} equation based on the balanced reaction. You should always make sure that the reaction is balanced first.
The concentrations at the start are: [H_{2}] = 1.0 mol/0.50 L = 2.0 mol/L [I_{2}] = 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.
H_{2 }+ I_{2}
<=====> 2 HI Let 'x' be the number of moles of H_{2 }per litre consumed. finish 2.0  x 2.0
 x
2x 
Stop and confirm the above. If 'x' amount of H_{2} is consumed then 'x' amount of I_{2} will also be consumed because the H_{2 }and I_{2} react on a 1:1 basis. Also if 'x' amount of H_{2} (or for that matter I_{2}) 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 K_{eq }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 
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? [H_{2}] = 2.0 mols/ 1  x = 2.0 mols/L  1.56 mols/L = 0.44 mols/L [I_{2}] = 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 H_{2}
and I_{2} 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
[H_{2}] = [I_{2}] = 0.5 L * 0.44 mols/L = 0.22 moles Therefore 0.22 moles of H_{2} and 0.22 moles of I_{2}
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 H_{2} to react with I_{2}. Theoretically if both the H_{2} and I_{2} 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 = 1.56 moles/L * 100 = 78% yield (note that the moles/L units cancel) OR Percentage Yield = 3.12 moles * 100 = 78% yield (note that the units still cancel) 
Problem #6 
How many moles of HI are present at equilibrium when 2.0 moles of H_{2} is mixed with 1.0 moles of I_{2} in a 0.50 L container and allowed to react at 448^{o}C. At this temperature K_{eq} = 50. 
Solution First the equation: H_{2} + I_{2} <=====> 2 HI Then the concentrations: [H_{2}] = 2.0 moles/0.50 L = 4.0 moles/L [I_{2}] = 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.
H_{2} + I_{2 }
<====> 2 HI 
We will subtract 'x' from both H_{2} and I_{2} because this is the amount of reactant that will convert into product '2x'. 
Now substitute these 'x' factors into the K_{eq }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 = 4x^{2}
8.0  6.0x + x^{2} 400  300x + 50x^{2 }= 4x^{2} 46x^{2 } 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 H_{2
}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.
H_{2} + I_{2 }
<======> 2 HI 4.01.9
2.01.9
2(1.9) To finally answer the question: moles = concentration * volume 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 H_{2}, and 1.5 moles of I_{2} are placed in a 1.0 L container at 448^{o}C, 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 K_{eq} (50 in this case) then the system
is already at equilibrium and stable.
If Q > K_{eq} then the [product] is too high and must decrease. Therefore the reverse reaction is favoured. If Q < K_{eq} then the [product] is still to small and must increase, therefore the forward reaction is favoured. H_{2} + I_{2 }<=======> 2 HI
therefore Q =
(3.0 mol/L)^{2}
= 9.0 = 3 Therefore Q < K_{eq} since 3 < 50. The forward
reaction is favoured to occur until the ratio of product:reactants = 50.

InClass Followup Problems 
1. When 0.040 moles of PCl_{5 }is heated
to 250^{o}C in a 1.0 L vessel, an equilibrium is established in
which the equilibrium concentrations of Cl_{2 }is 0.025 mol/L. Find
the equilibrium constant, K_{eq }, at 250^{o}C for the reaction:
PCl_{5}(g) <======> PCl_{3}(g) + Cl_{2}(g) Note that the initial, not the equilibrium concentration for PCl_{3} is given. The amount of PCl_{5 }reacted is the same as the amount of Cl_{2} formed. 2. Assume that the analysis of another equilibrium mixture of the system from above shows that the equilibrium concentration of PCl_{5} is 0.012 mol/L and that of Cl_{2 }is 0.049 mol/L. What is the equilibrium concentration of PCl_{3} at 250^{o}C? 3. How many moles of PCl_{5} must be heated in a 1.0 L flask at 250^{o}C 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 PCl_{5}, 0.60 mole of Cl_{2}, and 0.60 mole of PCl_{3} are placed in a 1.0 L flask and heated to 250^{o}C. If so, which reaction takes place? 
Go to the Equilibrium Calculations Worksheet 
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

Predict the probability of H_{2}O(l) > H_{2}
(g) + ½O_{2}(g) at 25^{o}C and 1500^{o}C.

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 > CO_{2}
+ H_{2}O (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 K_{eq} equation.

3. Formation of a slightly ionized product.
NaCl(s) + KBr(s) H_{2}O> 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 K_{eq} Values Using the Nernst Equation 
The E^{o} 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 E^{o} 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 K_{eq}. ie. 0 = E^{o}  0.059/n log [C]^{c}[D]^{d} / [A]^{a}[B]^{b } but ^{ }K_{eq} = [C]^{c}[D]^{d} / [A]^{a}[B]^{b} Therefore 0 = E^{o}  0.059/n log^{ }K_{eq} log K_{eq} = n E^{o} / 0.059 K_{eq} = 10 ^{(nEo
/ 0.059)} 
So the equilibrium constant for any redox reaction can be calculated using it's standard halfcell reduction potential. If the E^{o} value is +ve then Keq > 1 and a reaction with an E^{o} 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 AgNO_{3} + Zn^{o} > Zn(NO_{3})_{2} + 2 Ag^{o} 2. Write the half cell reactions:
2 Ag^{+1} + 2e^{1} > 2 Ag^{o}
E^{o} = +0.80 V 3. Solve for K_{eq}
K_{eq} = 10 ^{(nEo / 0.059)} 
The extremely large value of K_{eq} indicates a reaction that is essentially complete. 
Ex #2: Calculate the K_{eq} value for the reaction
between Zn and Copper(II) sulphate. Is the reaction complete?
Zn^{o} + CuSO_{4} > ZnSO_{4} + Cu^{o} Zn^{o}
> Zn^{+2} + 2e^{1}
E^{o} = +0.76 V 
K_{eq} = 10 ^{(nEo / 0.059)} = 10 ^{(2)(1.10)/0.059} = 10^{37.29} 
Therefore it is essentially complete. 