Review of Equilibrium Concepts
This document reviews basic equilibrium concepts already covered in class.
Consider the reversible reaction between colorless dinitrogen tetraoxide and brown nitrogen
dioxide.
N2O4  2NO2
If you place dinitrogen tetraoxide gas in a closed container, initially, only the forward
reaction is occurring. N2O4 gas decomposes into nitrogen dioxide. The concentration of
dinitrogen tetraoxide is going down and the concentration of nitrogen dioxide is increasing.
The contents of the container change from colorless to brown.
But the reverse reaction can also occur. As nitrogen dioxide molecules form, the reverse
reaction begins to occur as well. NO2 molecules recombine to form N2O4 molecules.
Eventually the intensity of the brown color stops increasing, indicating that the concentration
of nitrogen dioxide has stopped changing. The forward and reverse reactions are occurring
at the same rate; this is referred to as equilibrium.
Remember from the kinetics unit, that for most reactions, the rate of reaction is related to
reactant concentration. We’re seeing that here. As the concentration of N2O4 goes down, the
rate of the forward reaction goes down. As the amount of nitrogen dioxide increases, the rate
of the reverse reaction increases.
There comes a point when the rates of the forward and reverse reactions have become equal.
The amounts of all reactants and products are now constant. This, again, is referred to as
chemical equilibrium.
Equilibrium – a system that proceeds in the forward and reverse
directions at the same rate.
Equilibrium does NOT mean that amounts of reactants and products are equal. In our
previous example, at equilibrium, the concentrations of N2O4 and NO2 are not equal, but the
concentrations are constant.
At equilibrium, the RATES of the forward and reverse reactions are equal. Both reactions are
still occurring, but because they occur at the same rates, the amounts of the reactants and
products are constant.
Some equilibrium systems heavily favor the products. At equilibrium, there’s a lot of the
products in the vessel and not much of the reactants. Other systems heavily favor the
reactants; only a small amount of the products are present at equilibrium. Other equilibrium
systems are closer to the middle. A bit more product than reactant at equilibrium, for
example. We need a method for describing where the equilibrium rests.
For that purpose, we have a numerical value called the equilibrium constant.
Equilibrium Constant – an expression and resulting numerical value
that describes the position of equilibrium for a given reaction.
This is symbolized by Keq. We can also call this Kc or Kp depending on whether the data is
provided as concentrations or partial pressures.
The equilibrium constant expression is always the product of the concentrations of the
products over the product of the concentrations of the reactants, with each concentration
raised to a power equal to its coefficient in the balanced reaction.
aW + bX 
cY + dZ
Kc = [Y]c[Z]d
[W]a[X]b
Note that only solutions and gases appear in the equilibrium constant expressions. Solids
and pure liquids are NOT included.
The numerical value of the equilibrium constant tells us the about the
position of the given equilibrium system.
For example, the equilibrium system between hydrogen gas, bromine gas and hydrogen
bromide gas has a large equilibrium constant, 1.9 x 1019.
What does that mean? Think about how we write the equilibrium constant expression . . . . .
with products on top and reactants on bottom. A large equilibrium constant values means
that the numerator (the amounts of products) is greater than the denominator (the amounts
of reactants) at equilibrium. This is an equilibrium system that favors the products.
What does it mean if an equilibrium constant is small, like 4.1 x 10-31 for the system
containing nitrogen, oxygen and nitrogen monoxide gases?
A small equilibrium constant indicates that the reactants are favored at equilibrium – high
concentrations of reactants and low concentrations of products at equilibrium.
To summarize:
K>1
K<1
K1
equilibrium favors products
equilibrium favors reactants
amounts of reactants and products
nearly equal at equilibrium
The value of the equilibrium constant is NOT an indication of how fast or slowly equilibrium
is established.
Here’s an example problem:
Br2(g) + Cl2(g)  2 BrCl(g)
In an experiment conducted at 100oC, the equilibrium concentrations of the gases were as
follows: [Br2] = 0.023 M, [Cl2] = 0.12 M, [BrCl] = 0.14 M. Write the equilibrium constant
expression and determine the value of the equilibrium constant at this temperature.
Solution:
Kc =
[BrCl]2
[Br2][Cl2]
=
(0.14)2
(0.023)(0.12)
=
7.1
A system at equilibrium has a constant K value at constant temperature. Look at the
equilibrium system between hydrogen, iodine, and hydrogen iodide gases at 445oC.
In scenario (a), we start with only the reactants present initially. The system will reach an
equilibrium with the concentrations of hydrogen and iodine equal to 0.11 M and that of HI
equal to 0.78 M. When these numbers are subbed into the equilibrium constant expression, K
equals 50.
If instead we start with only HI present initially (scenario (b)), when equilibrium is
established, the equilibrium concentrations of the three gasses are all different than in the
previous scenario. But when these concentrations are subbed into the equilibrium constant
expression, the K value is still 50.
In fact, no matter what the initial concentrations of reactants and products, the quotient of the
concentrations will always be the same.
The value of K is constant at a constant temperature.
For a given reaction at a given temperature, the equilibrium concentrations of reactants and
products depend on the initial concentrations. The equilibrium constant is always the same,
though, with ANY set of initial concentrations.
If we use the same expression as K (products over reactants, coefficients as exponents), but
use NON-equilibrium conditions, we call the expression and value the reaction quotient, Q.
The reaction quotient, Q, is the ratio of product to reactant
concentrations AT ANY POINT IN THE REACTION.
This graph illustrates a reversible reaction between reactant A and product B.
For a given reaction at a given temperature, there is only one value for K. Q, on the other
hand, can equal many different values, depending on the concentrations of A and B at that
moment. Q is useful because the value of Q at any moment compared to K reflects how the
current conditions compare to equilibrium conditions. A system is at equilibrium when Q
and K are equal values.
Let’s look at a few example conditions on the graph. K for this system is shown on the graph
to equal 1.45. That is a constant for this reaction at a constant temperature. Consider a vessel
containing 0.75 M A and 0.25 M B (green box and circle). Q (still products over reactants) is
0.25 divided by 0.75, or 0.33. This is a value less than K. To get the Q value up to K, we need
to increase the numerator (the product concentrations) and decrease the denominator (the
reactant concentrations). In other words, the reaction needs to proceed further towards the
products to reach equilibrium.
Look at the system when the concentration of A is 0.25 M and that of B is 0.75 M (purple box
and circle). This time, Q equals 3. This time, the Q value is greater than the K value. The
value of Q needs to be lowered if Q is to equal K. The numerator (the products) needs to
decrease and the denominator (the reactants) needs to increase. The reverse reaction must
occur to reach equilibrium. The system will proceed toward the reactants to reach
equilibrium.
When only A is present, only the forward reaction is a possibility. No need to calculate Q to
see that. It is possible, though, to calculate a Q value and compare it to K as we did before.
Zero molar for products over 1 molar for reactants equals zero. That’s a number less that K,
so the reaction must proceed toward the products to reach equilibrium.
We could calculate lots of Q values for this reaction. Any combination of concentrations that
results in a value of Q that is less than K tells us that the reaction will proceed toward the
products to reach equilibrium. If Q is calculated to be greater than K, the reaction will
proceed toward reactants to reach equilibrium. Only when the calculated value of Q is equal
to the known value of K, is the reaction at equilibrium.
To summarize:
Q=K
Q<K
Q>K
system is at equilibrium
system proceeds towards products
to react equilibrium
system proceeds towards reactants
to react equilibrium
Example:
The equilibrium constant for the formation of nitrosyl chloride from nitric oxide and chlorine,
2NO(g) + Cl2(g)  2NOCl(g)
is 6.5 x 104 at 35oC. In which direction will the reaction proceed to reach equilibrium if the
current concentrations are as follows:
(a) [NO] = 0.011 M
[Cl2] = 0.0085 M
[NOCl] = 0.59 M
(b) [NO] = 0.11 M
[Cl2] = 0.035 M
[NOCl] = 0.019 M
Solution:
For each set of concentrations, calculate Q and compare its value to K.
(a)
Q = [NOCl]2
=
(0.59)2
= 3.4 x 105
2
2
[NO] [Cl2]
(0.011) (0.0085)
Q > K. Numerator (product amount) too large and denominator (reactant
amounts) too small compared to equilibrium conditions. Reaction proceeds
towards reactants to reach equilibrium.
(b)
Q=
[NOCl]2
[NO]2[Cl2]
=
(0.019)2
(0.11)2(0.0035)
= 8.5
Since Q < K, reaction proceeds towards the products to reach equilibrium.
K (and also Q) can be expressed and calculated using either concentrations or partial
pressures. Calculating K with concentrations is more common, but for gaseous reactions, K
can also be calculated using partial pressures. In either case, the process for writing the
expression is the same – products over reactants, coefficients as exponents. The notation
used is slightly different, though. Brackets refer to concentration, so brackets are NOT used
in a Kp expression.
For example, for the reaction:
2NO(g) + O2(g)  2NO2(g)
we can write an expression for Kc and for Kp that are identical except for the notation
referring to concentration vs. partial pressure.
Kc =
[NO2]2
[NO]2[O2]
Kp =
(PNO2)2
(PNO)2(PO2)
While the values of Kc and Kp are not usually equal to each other, their values can be
interpreted identically. Kc for the nitrogen monoxide reaction above is 5.4 x 1013 and Kp is
2.2 x 1012. Both values are large. Both values would indicate that the system favors the
product at equilibrium.