AP Chemistry Notes: Chapter 15 Chemical Equilibrium
Many reactions do not run to completion. But, reaction concentrations may cease to change. The system
contains a mixture of reactants and products.
A chemical equilibrium has been reached.
There are physical equilibria as well: ex. constant vapor pressure in a closed container with a volatile
liquid, or solution saturation.
15.1 The Concept of Equilibrium
A chemical equilibrium is a dynamic state. The forward reaction has not stopped. Rather, the reverse
reaction rate has caught up. Both rates are now equal, so overall concentrations cease to change.
Forward reaction: A  B r = kf[A]
Reverse reaction: B  A
r = kr[B]
Beginning with all A and no B, [A] decreases as does the forward rate. [B] increases as does the
reverse rate. Eventually, the rates become equal and kf[A] = kr[B].
If we rearrange the equation, we get kf/kr = [B]/[A] = a constant.
The Equilibrium Constant is a ratio of product to reactant concentrations.
Reversibility is indicated by a set of double arrows: ⇆ (The ones in your text look nicer than this,
but most word processors do not have the truly correct symbol.)
A⇆ B
Ex. N2(g) + 3H2(g) ⇆ 2NH3(g)
15.2 The Equilibrium Constant
The equation above is for the Haber process for synthesizing ammonia from its elements. It is a
tremendously important industrial reaction and is used to produce the majority of the world’s fixed
nitrogen. The principal use is for agricultural fertilizers, although there are lots of others as well.
The interesting thing about equilibria and reversible reaction, is that the equilibrium state can be reached
from either direction. We can start with all reactants, all products, or a mixture of the two and the system
will still come to equilibrium with a fixed value for the constant.
In the Haber process, N2(g) and H2(g) are mixed at about 200 atmospheres and 500ºC. The forward
reaction does not go to completion. Again, equilibrium can be reached from either direction. See the
diagrams below:
Equilibrium
achieved
H2
P
NH3
N2
time
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By changing the starting concentrations and examining the concentrations at equilibrium, we can learn
more about the equilibrium system.
In 1864, Guldberg and Waage postulated their Law of Mass Action. Rather than a long statement, a
simple example is best: for aA + bB ⇆ pP + qQ
Kc =
Here, [ ] = molar concentration, however, equil. constant expressions may be written with
pressures for gases as well.
Special Notes:
Do not confuse equilibrium constant ÒKÓ with the rate constant "k."
Products are always on top, reactants are always on the bottom.
The value of K at a given temperature does not depend on initial concentrations.
The Law of Mass Action applies only to systems at equilibrium.
There may also be other substances present, but as long as they do not consume the
materials in question, they do not affect K (ex. catalysts or inert gases.)
K does vary with temperature.
K is typically expressed without units. We are usually more interested in the value of
the number than in the units.
In the space below, write the equilibrium constant expressions for:
the Haber process equation and
the following equilibrium: N2O4(g) ⇆ 2NO2(g)
Use concentrations rather than pressures for now.
Table 5.1 on P. 539 in your text shows the results of several experiments with the system in (2) above.
Note the values of K in each experiment. What do you conclude from this information?
When reactant and product materials are expressed using molar concentrations ([ ]), Kc is used.
In the space below, write the equilibrium expressions for the above equations in terms of pressure, and
using Kp instead of Kc.
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Usually, equilibrium equations are written with the products of interest on the right. However, the
equations are reversible. Reversing an equilibrium equation results in the reciprocal of K (1/K).
If the numerical value of K is small (<<1) the reaction lies far to the left, that is, there are more
reactants in the system than products, or we say, "The reactants are favored."
Conversely, if K is large (>>1), the products are favored.
Ex. production of phosgene: CO(g) + Cl2(g) ⇆ COCl2(g) Kc = 4.57x 109
Ex. N2(g) + O2(g) ⇆ 2NO(g) Kc = 1 x 10-30
K must be calculated using experimental values for concentrations.
Also, if K is known, equilibrium [ ]’s can be calculated if just one value is known. This is a major
type of problem for this unit and they get really fun!!
15.3 Heterogeneous Equilibria
In homogeneous equilibria, like the examples we have seen so far, all reactants and products are in the
same phase.
In heterogeneous equilibria, the materials are not all in the same phase. As it turns out, only gases and
dissolved materials are really subject to any appreciable concentration change as reactions occur.
Solids and liquids by comparison have concentrations that are so high, that they are
essentially constant. Therefore, solids and liquids are left out of equilibrium constant
expressions. Either they are present, in which case equilibrium can be reached, or they are not.
Ex. CaCO3(s) ⇆ CaO(s) + CO2(g)
Kc = [CO2] or Kp = PCO2
15.4 Calculating Equilibrium Constants
Sometimes we do not know all of the concentrations of products and reactants at equilibrium, so if we
know initial concentrations and K, we can calculate the others. To do this we use an ICE Table.
ICE stands for Initial, Change, and Equilibrium. These tables are very commonly used in this unit
and for all types of equilibrium problems. You should become very familiar with them.
ICE tables look like this:
Reactants
⇆
Products
Initial
Change
Equilibrium
See example problem 15.7 on p. 546 regarding use of an ICE table.
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It is possible to relate Kc to Kp through a simple equation: Kp = Kc(RT)n
where n is the change in the number of moles of gas:
total moles product gases - total moles reactant gases, T is Kelvin temp. and R is the universal gas
constant in the appropriate units.
(See P 548, Example problem 15.5.)
15.5 Applications of Equilibrium Constants
Predicting reaction direction: If a system is not at equilibrium, it is possible to project which way the
reaction will go to reach equilibrium (toward the product side or toward the reactant side.)
This is a simple process: Just write the K expression and plug in the current concentration or pressure
values. Calculate the resulting value, which is called the Reaction Quotient (Q).
Q is calculated exactly the same way as K.
Then compare Q with the actual K value at that temperature. If
Q > K, there is too much product and the equilibrium will shift to the left <--Q = K, the system is at equilibrium and will not shift.
Q < K, there is too much reactant and the equilibrium will shift to the right --->
Determining Equilibrium Concentrations (or "And You Thought There Wasn’t a Use for the
Quadratic Equation!!")
It is possible using the equilibrium constant expression to calculate the concentrations of reactants and/or
products at equilibrium.
Here is an example problem from the book:
H2(g)
Initial
Change
Equilibrium
1.000 M
+
Kc = [HI]2 / [H2][I2]
I2(g)
⇆ 2HI(g)
2.000 M
0M
-x
-x
+2x
1.000 - x M
2.000 - x M
2x M
At 448ºC, Kc has a value of 50.5.
Substituting the Equil. concentration values into Kc: (2x)2 / (1.000 - x)(2.000 - x) = 50.5
If this is beginning to look suspiciously like something from a recent math class, it’s because IT
IS!!
4x2 = 50.5(x2 - 3.000x + 2.000) ---> 46.5x2 - 151.5x + 101.0 = 0
x = -(-151.5)±√ (-151.5)2 - 4(46.5)(101.0)
2(46.5)
= 2.323 or 0.935
Typically one of the options from the quadratic equation will be automatically eliminated. In this
case it is the 2.323 value, since 1.000 M -2.323 gives a negative concentration value, which is
impossible. So we choose x = 0.935.
[H2] = 1.000 - 0.935 = 0.065 M
[I2] = 2.000 - 0.935 = 1.065 M
[HI] = 2x = 1.870 M
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The results can be double checked by plugging the values above into the equilibrium expressions.
If the resulting answer is 50.5, the problem has been done correctly.
On the bright side, it will be only a rare case when you will actually need to use the quadratic
equation to solve problems of this type. We will talk about why in the near future.
15.6 Factors Affecting Equilibrium : Le Chatelier’s Principle
Le Chat’s Principle simply stated is: When subjected to a stress, an equilibrium system will shift in order
to relieve that stress.
Put more simply still, an equilibrium will shift toward a decrease and away from an increase.
What constitutes a stress?
A change in a reactant or product concentration. An equilibrium will shift to use up something
that is being added, or to replace something that is being lost.
A pressure change in systems containing gases. If pressure is increased by reducing volume, the
equilibrium will shift toward the side with the fewest moles of gas to relieve the pressure. If P is
decreased, a shift toward the side with more gas moles will occur.
**Note: the addition of an inert gas does not alter the partial pressures of any of the reacting gases
(even though the total pressure increases) so the equilibrium is not affected.
A temperature change. All reactions are exothermic in one direction and endothermic in the
other. Raising the temperature favors the endothermic reaction and hinders the exothermic one.
Remember, temperature is the only factor that affects the value of K.
Energy can be added to any equation and treated just like any other component with respect to Le
Chat’s Principle when predicting shifts in response to stress.
N2(g) + 3H2(g) ⇆ 2NH3(g) + Energy
Here, raising the temp. causes a shift away from the
energy, or to the left. [NH3] decreases and [N2] and [H2] both increase.
Be sure you can predict not only which way an equilibrium will shift in response to a stress,
but what will happen to the [ ]’s of all materials as a result of the shift.
Effect of a Catalyst
Catalysts affect forward and reverse reactions equally and so adding a catalyst allows the reaction system
to come to equilibrium more quickly, but does not change the composition of the equilibrium mixture.
K remains the same.
The important effect of a catalyst is in that it may allow the reaction mixture to reach equilibrium at a
lower temperature.
Back to our friend Fritz Haber. Haber faced two major problems:
The reaction pressure had to be quite high, around 200 atm. Not an easy thing to do in 1912.
The reaction temperature had to be very high to force effective collisions between nitrogen and
hydrogen molecules. However, ammonia decomposes rapidly at high temps. In equilibrium
terms, K becomes rapidly smaller at temp. rises.
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Haber’s solution lay in finding a suitable catalyst which would allow the mixture to come to
equilibrium at a lower temperature where K and the amount of ammonia in the mixture was
greater. The catalyst Haber eventually used was a mixture of iron with metal oxides.
Equilibrium with a suitable amount of ammonia can be reached at 400-500ºC and pressures of
200-600 atmospheres. The equilibrium mixture is piped into a chamber where rapid cooling
condenses the ammonia to a liquid while the nitrogen and hydrogen remain gaseous and are
pumped back into the hot reaction chamber.
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