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CHAPTER 15
Chemical Equilibrium
Equilibrium
Consider a system consisting of a mixture of gases at some
temperature T where the following general and reversible chemical
reaction is taking place.
a A(g) + b B(g)  c C(g) + d D(g)
where A, B, C, and D are chemical substances (reactants or products), a,
b, c, and d are stoichiometric coefficients (numbers that balance the
reaction) and “  “ indicates that the reaction can proceed in both the
forward and reverse direction.
We define chemical equilibrium for this system as the state where
the concentrations of reactants and products no longer change with time.
Example
Consider the following reaction taking place in the gas phase
N2O4(g)  2 NO2(g)
In both of the above systems the concentrations of NO2 and N2O4
eventually reach a constant value. At this point the system is at equilibrium. Note that this is a dynamic equilibrium. Reactants continue to be
converted into products, and products into reactants, but at equilibrium
the two processes are in balance.
Equilibrium Constant
Let us return to our general gas phase reaction
a A(g) + b B(g)  c C(g) + d D(g)
The equilibrium constant for this reaction is defined as the ratio
of the concentrations (partial pressures) of the products divided by the
concentration (partial pressures) of the reactants, at equilibrium, raised to
their stoichiometric coefficients.
For reactions of gases we can define two equilibrium constants,
KC (in terms of concentration; molarity) and Kp (in terms of partial
pressure; atmospheres).
KC = [C]c [D]d
[A]a [B]b
Kp = (pC)c (pD)d
(pA)a (pB)b
Examples:
N2O4(g)  2 NO2(g)
2 H2(g) + O2(g)  2 H2O(g)
Significance of the Equilibrium Constant
The significance of the equilibrium constant lies in the fact that
for a chemical reaction taking place at a particular temperature T, the
equilibrium constant (KC or Kp) has a particular numerical value. This
means that no matter what the starting concentrations or partial pressures
for the system, at equilibrium the ratio of the concentrations (partial
pressures) of the products divided by the concentrations (partial
pressures) of the reactants, raised to their appropriate powers, will be
equal to KC (or Kp). This gives us a method for determining the
concentrations (or partial pressures) of reactants and products that will be
present in the system when it has achieved equilibrium.
Comments
1) Equilibrium constants have no units. That is because what
appears in the expressions for the equilibrium constant is actually the
concentration or partial pressure divided by our standard concentration or
partial pressure unit.
N2O4(g)  2 NO2(g)
KC = {[NO2]/(1. M)}2 =
{[N2O4]/(1.M)}
[NO2]2
[N2O4]
Kp = {(pNO2)/(1. atm)}2 =
{(pN2O4)/(1. atm)}
(pNO2)2
(pN2O4)
2) For gas phase reactions the value for KC and Kp are related.
From the Dalton’s law of partial pressures
pAV = nART
nA = [A] = pA
V
RT
Divide both sides by VRT
and so pA = (RT) [A]
For our generic reaction
a A(g) + b B(g)  c C(g) + d D(g)
Kp = (pC)c (pD)d = {RT [C] }c {RT [D] }d = [C]c [D]d (RT)[(c+d)-(a+b)]
(pA)a (pB)b {RT [A] }a {RT [B] }b
[A]a [B]b
or
Kp = KC (RT)n where n = (c + d) - (a + b) = change in the number
of moles of gas per mole of reaction.
3) The way in which the equilibrium constant is written depends
on how the balanced reaction is written.
Example:
N2O4(g)  2 NO2(g)
KC = [NO2]2
[N2O4]
2 NO2(g)  N2O4(g)
KC = [N2O4] = 1
[NO2]2
KC
2 N2O4(g)  4 NO2(g)
KC = [NO2]4 = (KC)2
[N2O4]2
One needs to specify the reaction before writing KC or Kp.
4) For heterogeneous equilibrium, the following do not appear in
the expression for the equilibrium constant (though they must be present
for equilibrium to be achieved).
solids
liquids
solvents
What will appear in the equilibrium constant is
gases
solutes
Examples: Write KC and Kp for the following reactions
a) CaCO3(s)  CaO(s) + CO2(g)
b) NH3(aq) + H2O()  NH4+(aq) + OH-(aq)
5) We can get equilibrium constants for new reactions by finding
the appropriate combinations of equilibrium constants for old reactions.
Example: Given the equilibrium constants for the reactions
2 NO(g)  N2(g) + O2(g)
KC1 = 2.3 x 1024
2 NO(g) + O2(g)  2 NO2(g)
KC2 = 6.4 x 109
find the numerical value for KC3, the equilibrium constant for the
reaction
N2(g) + 2 O2(g)  2 NO2(g)
KC3 = ?
Finding the Numerical Value for KC and Kp
If we have a chemical reaction and know the concentrations or
partial pressures of all the reactants and products we can substitute into
the definition of KC or Kp to find their value.
Example: Consider the reaction
PCl5(g)  PCl3(g) + Cl2(g)
The equilibrium concentrations of reactants and products,
measured at T = 450. K, are as follows:
[PCl3] = 0.0150 M
[PCl5] = 0.0083 M
What are the numerical values for KC and Kp?
[Cl2] = 0.0320
Using the Equilibrium Constant
There are several ways in which the equilibrium constant for a
reaction can be used. The most common example is using the initial
conditions for a system, the balanced chemical reaction, and the
equilibrium constant to find concentrations of reactants and products at
equilibrium.
Example: The equilibrium constant for the reaction
PCl5(g)  PCl3(g) + Cl2(g)
is KC = 0.0578 at T = 450. K.
In a particular system at T = 450. K we initially have 0.100 M
PCl5. No PCl3 or Cl2 are initially present. What are the concentrations
of PCl5, PCl3, and Cl2 when the system reaches equilibrium?
We will use the “ICE” method - write down initial, change, and
equilibrium concentrations in terms of the smallest possible number of
unknowns.
General Trends in KC (Kp)
Consider our general gas phase reaction
a A(g) + b B(g)  c C(g) + d D(g)
KC = [C]c [D]d
[A]a[B]b
We make the following general statements
If KC >> 1 then products will be greatly favored over reactants at
equilibrium.
If KC << 1 then reactants will be greatly favored over products at
equilibrium
If KC ~ 1 then both reactants and products will be expected to be
present at significant concentrations at equilibrium.
The same general trends apply if we use Kp instead of KC.
Example: The equilibrium constant for the reaction
2 H2(g) + O2(g)  2 H2O(g)
is KC = 2.4 x 1047 at T = 500. K.
A system initially has 0.100 M H2O. No H2 or O2 are initially
present. What are the concentrations of H2, O2, and H2O when the
system reaches equilibrium?
Reaction Quotient
Consider our general gas phase reaction
a A(g) + b B(g)  c C(g) + d D(g)
The reaction quotient for this reaction (Q) is defined as the ratio
of the concentrations (partial pressures) of the products divided by the
concentration (partial pressures) of the reactants raised to their
stoichiometric coefficients.
For reactions of gases we can define two reaction quotients, QC
(in terms of concentration) and Qp (in terms of partial pressure).
QC = [C]c [D]d
[A]a [B]b
Qp = (pC)c (pD)d
(pA)a (pB)b
QC and KC
The definitions of QC and KC are similar. The key difference is
that QC is defined for whatever conditions happen to be present in the
system, while KC is defined only for a system at equilibrium (and so at a
particular temperature has only one specific value).
We use QC as follows:
a A(g) + b B(g)  c C(g) + d D(g)
QC = [C]c [D]d
KC = [C]c [D]d
[A]a [B]b
[A]a [B]b
If QC = KC, the system is at equilibrium, and so will not change.
If QC > KC, reaction must shift from right to left (form more
reactants) to reach equilibrium.
If QC < KC, reaction must shift from left to right (form more
products) to reach equilibrium
Le Chatlier’s Principle
Consider a system initially at equilibrium. We apply a change to
the system. What will happen? How will the system respond to the
change?
The general way in which systems respond to changes is
governed by Le Chatlier’s principle - “A system initially at equilibrium
will respond to a change in a way that minimizes or opposes the effects
of the change.”
Note that in general a prediction we make using Le Chatlier’s
principle can also be made using the reaction quotient. (The one
common exception is the effect of changes in temperature on a system.)
Example: Consider the following reaction
N2(g) + 3 H2(g)  2 NH3(g)
Hrxn = - 92.2 kJ/mol
QC =
[NH3]2
[N2] [H2]3
We initially have a system at equilibrium. Predict the effect of each of
the following changes to the system:
1) Addition of H2 to the system.
2) Addition of NH3 to the system.
3) Double the volume at constant temperature.
4) Increase in temperature by 20. K
5) Addition of a catalyst.
Equilibrium Calculation – A Second Example
Consider the reaction
N2O4(g)  2 NO2(g)
KC = [NO2]2 = 0.0046 at
[N2O4]
T = 25 C
A system at T = 25 C initially has 0.0100 M NO2 and 0.0100 M N2O4.
What will the concentrations of these substances be when the system
achieves equilibrium?
Thermodynamic Equilibrium Constant and
Reaction Quotient
We have previously discussed KC (equilibrium constant in terms
of concentration) and Kp (equilibrium constant in terms of pressure), and
the corresponding terms QC and Qp.
The thermodynamic equilibrium constant (K) and thermodynamic reaction quotient (Q) are defined in terms of the following
standard states:
gases - in terms of a standard pressure of 1.00 atm
solids, liquids, solvents - do not appear
solutes - in terms of a standard concentration of 1.00 mol/L
For reactions involving only gases K = Kp, while for reactions
involving only solutes K = KC.
Example: Give the expression for K (thermodynamic equilibrium
constant) for the following reactions
2 H2(g) + O2(g)  2 H2O(g)
2 H3O+(aq) + CO32-(aq)  CO2(g) + 3 H2O(l )
Free Energy Change for non-Standard Conditions
For a chemical reaction where all the reactants and products are
present at standard concentration (1.00 atm pressure for gases, 1.000 M
for solutes, and ideal behavior) the free energy change is Grxn. What
about the case where standard concentrations are not present?
For a chemical reaction taking place at constant temperature and
pressure, but any conditions of concentration, the following relationship
applies
Grxn = Grxn + RT ln Q
Grxn - free energy change for the actual concentrations present
Grxn - free energy change for standard concentrations
Q - thermodynamic reaction quotient
Example: Consider the reaction
2 H2(g) + O2(g)  2 H2O(l)
What is Grxn for a) standard conditions (pH2 = pO2 = 1.00 atm),
and b) for pH2 = 0.0100 atm, pO2 = 0.0050 atm.
Free Energy and Equilibrium
Our general expression for the change in free energy for a
chemical reaction takes on a particularly interesting form when we are at
equilibrium.
In general
G = Grxn + RT ln Q
At equilibrium, G = 0. , and Q = K. Substituting, we get
0 = Grxn + RT ln K
Grxn = - RT ln K
ln K = - Grxn/RT
Notice what this says. Based solely on thermodynamic data we
can find the numerical value for the equilibrium constant for a chemical
reaction.
Example: Based on thermodynamic data find K for the process
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
K = [NH4+] [OH-]
[NH3]
End of Chapter 15
“Science is built up of facts, as a house is built of stones; but an
accumulation of facts is no more a science than a heap of stones is a
house.” - Henri Poincaré, Science and Hypothesis .
“Quite ordinary people can be good at science. To say this is not
to deprecate science but to appreciate ordinary people.” - Peter Medawar