Chemical Equilibrium

Chapter Objectives

  1. Describe the condition of equilibrium in a reversible reaction.
  2. Describe how equilibrium concentrations are established experimentally.
  3. Write the equilibrium constant expression in terms of concentrations, Kc for a reaction and use the value of Kc and the concentrations of all species but one to determine the equilibrium concentration of that species.
  4. Derive K values for situations where chemical equations are reversed, multiplied through by constant coefficients, or added together.
  5. Assess the relative importance of the forward and reverse reactions from the magnitude of an equilibrium constant.
  6. Write an equilibrium constant expression in terms of partial pressures of gases, Kp; and relate a value of Kp to the corresponding value of Kc.
  7. Know that the concentrations of pure liquids and solids are omitted from equilibrium constant expressions.
  8. Calculate a numerical value of an equilibrium constant if equilibrium conditions are given.
  9. Predict the direction in which a reaction proceeds toward equilibrium by comparing the reaction quotient, Qc to Kc.
  10. Use the ideal gas law and Dalton's law of partial pressures in working with Kp expressions.
  11. Make qualitative predictions of how equilibrium conditions change when an equilibrium mixture is disturbed. That is, apply Le Châtelier's principle.
  12. Calculate the final equilibrium condition in a reversible reaction from a given set of initial conditions.
  13. Calculate the new equilibrium concentrations or partial pressures after an equilibrium system has adapted to changed conditions.
Chemical Equilibrium is a Dynamic State

The concept of a dynamic equilibrium is central to quantitative discussion of most chemical phenomena.  A dynamic equilibrium
is a state in which there appears to be nothing happening, a state in which there is no net change, but is also a state in which
chemical reactions are taking place, often at rapid rates.  These reactions, however, are opposites of each other so that the net
change is no change at all.  Any chemical system at equilibrium is a dynamic system in which real reactions are occurring with
real speed. And, should that system be altered in some way be external means, the reactions will operate so as to bring it back
to equilibrium again.

A dynamic equilibrium, like a balanced teeter-totter in a school playground, is a condition which can be expressed in either of
two ways.  One of these ways is to say the the condition of equilibrium is one in which the driving forces of the system are equal
in both of the opposite directions. If two children of equal mass sit on a teeter-totter equally distant from its center, each applies
a driving force to rotate the teeter-totter but the forces are equal but opposed and the teeter-totter does not move.  The
alternative is to say that the condition of equilibrium is one in which the rates of movement of the system are equal in both of the
opposite directions.  Again, if the two ends of the teeter-totter are moving equally in opposed directions they are not having any
net movement at all.  On the teeter-totter, of course, up and up are the opposed directions rather than one side up and one side
down.

These two expressions of equilibrium appear also as two alternative statements of the chemical principle called the law of mass
action. In its thermodynamic form, the law of mass action is:

The driving force of a chemical reaction is directly proportional to the active masses of the reactants.

The alternative kinetic form is:

The rate of a chemical reaction is directly proportional to the active masses of the reactants.

Since dynamic equilibrium is the state in which both the driving forces, and the rates, of opposed reactions are equal, the
concept has been approached in both ways. In either case, however, the concept of active mass appears.

The Law of Mass Action

Active mass, or reactive mass, or chemical activity is the amount of substance which is reactive within the volume in which the
reaction is occurring. A dynamic equilibrium requires that both the forward and reverse processes be considered, and so it is usually indicated by a double arrow or double-headed arrow () rather than the single arrow of a chemical reaction. Even so, the equilibrium is written with the reactants on the left and products on the right in the same manner as an ordinary chemical reaction. In the equilibrium CaCO3(s)  CaO(s) + CO2(g), calcium carbonate is considered to be the reactant while calcium oxide and carbon dioxide are considered to be the products.

The following example gives an experimental rational for the mathematical description of the law of mass action in terms of molar concentration of reactants and products.  For the reaction:

The following data from 3 experiments were collected:
 
Exp# [CO]i [H2]i [CH3OH]i [CO]eq [H2]eq [CH3OH]eq
1 1.000 1.000 0.000 0.911 0.822 0.089
2 0.000 0.000 1.000 0.753 1.506 0.247
3 1.000 1.000 1.000 1.380 1.760 0.620

Fitting the data to various expressions of the ratio of products to reactants leads to an observation.
 
[CH3OH]
____________________ 
[CO]+[H2]
[CH3OH]
____________________
[CO]+(2 x [H2])
[CH3OH]
____________________
[CO]+[H2]2
[CH3OH]
____________________
[CO][H2]
[CH3OH]
____________________
[CO](2 x [H2])
[CH3OH]
____________________
[CO][H2]2
0.051 0.035 0.056 0.119 0.060 0.145
0.109 0.066 0.082 0.218 0.109 0.145
0.197 0.127 0.138 0.255 0.128 0.145

While the equilibrium concentrations of the individual components of a reaction depend on the initial conditions, the ration of the sixth column gives a constant value.

This resulting constant, Kc, is called the equilibrium constant for the equilibrium, and is always written with the concentration of the products (right side) in the numerator and the concentrations of the reactants (left side) in the denominator. If the equilibrium reaction were written in the reverse direction, as it could be, the value of the equilibrium constant would be the reciprocal of its value for the equilibrium written for its original direction.

The form of any equilibrium constant is completely specified by the stoichiometry of the equilibrium to which it refers. If the
equilibrium is that of hydrogen and iodine with hydrogen iodide,

H2(g) + I2(g)  2 HI(g)

then

Kc = [HI]2 / [H2][I2], while if the form

1/2 H2(g) + 1/2 I2(g)  HI(g)

is preferred then

Kc = [HI] / [H2]1/2[I2]1/2

The numeric value of the constant always depends on the form of the equilibrium to which it refers; in the example above, the
first constant is the square of the second.

Any equilibrium however complex can be described by an equilibrium constant if its stoichiometry is known; for

6 A + 7 B + 8 C  2 D + 3 E + 4 F

the equilibrium constant expression would be

Kc = [D]2[E]3[F]4 / [A]6[B]7[C]8

Qualitatively, the magnitude of K gives you some idea of the extent of the reaction.  If K>>1, the reaction goes in the direction of products and if K<<1, reactants are favored.  A quantitative description of the direction a reaction mixture not at equilibrium can be achieved by considering the reaction quotient, Q.  Q has the same form as the equilibrium constant expression.  However, the concentrations (or partial pressures) in the expression are not the equilibrium value, but the initial values.  There are several possibilities:

Q > K  The reaction will go toward reactants (to the left) to increase reactants and decrease products (Q goes to K).

Q < K  The reaction will go toward products (to the right) to increase products and decrease reactants (Q goes to K).

Q = K  The reaction is at equilibrium.

Gas Phase Reactions

When a reaction occurs entirely in the gas phase, the concentrations of the gaseous reactants and products can be replaced by their partial pressures. These partial pressures are the partial pressures which actually prevail at equilibrium, not the partial pressures of the gases which may have been mixed initially.

Example. The equilibrium constant for the reaction H2(g) + Cl2(g)  2 HCl(g) is written as:

Kc = [HCl]2 / [H2][Cl2]
or

Kp = P(HCl)2 / P(H2) x P(Cl2)

The relationship between Kc and Kp is derived from the ideal gas law

Kp = Kc(RT)n , where n = total moles of gaseous product - total moles of gaseous reactant

Equilbrium Calculations

Calculation of K

a.  It is trivial to calculate K when all the equilibrium concentrations (or partial pressures) of reactants and products are known (or measured).  The values are plugged into the equilibrium constant expression, and K is calculated.

Example:  Given

2 CO(g)  +  2 H2(g)  CH4(g)  +  CO2(g)

At 298 K, the following equilibrium concentrations of the four compounds are determined experimentally:

[CO] = 4.3 x 10-6 M; [H2] = 1.15 x 10-5 M; [CH4] = 5.14 x 104 M; and [CO2] = 4.12 x 104 M.

Kc = [CH4][CO2] / [CO]2[H2]2 = (5.14 x 104)(4.12 x 104 M) / (4.3 x 10-6)2(1.15 x 10-5)2 = 8.66 x 1029

b.  It is usually easier to calculate K from a knowlege of initial conditions, stoichiometery of the reaction and one equilibrium concentration (or pressure).  It is very useful to organize your data in a table that describes the initial situation, the change and equilibrium position.

Example:  Given

2 HI(g)   H2(g) + I2(g)

A 5.00 L flask was filled with 4.00 moles of HI.  At 458 oC, the equilibrium mixture was found to contain 0.442 moles of I2.

[HI]i = 4.00 mole/5.00 L = 0.800 M
[I2]eq = 0.442 mole/5.00 L = 0.0884 M
 
[HI] [H2] [I2]
Initial 0.800 0 0
Change -2x +x +x
Equilibrium 0.800 - 2x = 0.623 x = 0.0844 x = 0.0844

Kc = [H2][I2] / [HI]2 = (0.0844)(0.0844) / (0.623)2 = 0.0201

Calculating Equilibrium Concentrations (or pressures)

Example: Given

CO(g)  +  H2O(g)  H2(g)  +  CO2(g)

A 50.0 L flask is filled with 1.00 mole of CO and 1.00 mole H2O and heated to 1000 oC.  The value of Kc at this temperature is 0.58.

[CO]i = 1.00 mole/50.0 L = 0.0200 M
[H2O]i = 1.00 mole/50.0 L = 0.0200 M
 
[CO] [H2O] [H2] [CO2]
Initial 0.0200 0.0200 0 0
Change -x -x +x +x
Equilibrium 0.0200 - x 0.0200 - x x x

Kc = 0.58 = [H2][CO2] / [CO][H2O] = (x)(x) / (0.0200-x)(0.0200-x)

x = 0.0086 M and - 0.063 M(physically impossible)

[H2] = [CO2] = 0.00865 M
[CO] = [H2O] = 0.0200 - 0.00865 = 0.0114 M

Henri Le Chatelier (1850-1936)

Le Chatelier's Principle

The state of dynamic equilibrium, in which a chemical system is reacting to form and to destroy products, is a stable state. If a
system is not in equilibrium, the driving force toward equilibrium is greater than the driving force away from equilibrium and the
system has a net driving force toward equilibrium. Alternatively, if a system is not in equilibrium the rate of the reaction toward
equilibrium is greater than the rate of the opposite reaction so there is a net movement always toward the equilibrium state. In
other words, a system at equilibrium will tend to stay at equilibrium and a system not at equilibrium will tend to more toward
equilibrium.

This principle was first clearly enunciated by Henri Le Chatelier (1850-1936) and bears his name. Its elegant formulation is:

If, to a system at equilibrium, a stress be applied, the system will react so as to relieve the stress.

Examples of Le Chatelier's Principle:

Added Reactants:

The effect of adding reactants to a system at chemical equilibrium is to increase the concentration or partial pressures of the
products; the equilibrium concentrations of the reactants will be less than the sum of the original concentrations and those added
due to the equilibrium reaction.

Example. The equilibrium for the formation of hydrogen iodide, H2(g) + I2(g)  2 HI(g), is governed by the equilibrium
constant Kp = P(HI)2 / P(H2) x P(I2). If the equilibrium partial pressures of hydrogen and iodine are taken to be, or made to be, exactly 1.0 on some pressure scale (atmospheres, bars, torr, pascals, or kilopascals) then the equilibrium partial pressure m of hydrogen iodide is given by K = m2/(1.0)2

If an additional partial pressure z of a reactant, hydrogen, is added, then the partial pressure of hydrogen would become 1.0 +
z. The partial pressure of hydrogen having increased, there is a stress on the equilibrium which is relieved by removal of some
partial pressure of hydrogen, y, through the chemical reaction. The equilibrium constant would then give

K = (m + 2y)2/(1.0 + z - y)(1.0 - y)

because m + 2y is the new equilibrium partial pressure of hydrogen iodide, 1.0 + z - y is the new equilibrium partial pressure of
hydrogen, and 1.0 - y is the new equilibrium partial pressure of iodine. The new equilibrium partial pressure of hydrogen is m +
2y because two moles of HI are produced for every mole of H2 which reacts.

The effect of adding a reactant is to decrease the concentrations of the reactants (as well as the concentration of the reactant
added, though only to the extent of part of the original increase) and to increase the concentrations of all products. The
equilibrium shifts to the right when a component is added on the left, or reactant, side.

Added Products

The effect of adding products to a system at equilibrium is the reverse of adding reactants. The equilibrium will shift to the left,
increasing the reactants. For the example used above, let z now represent the added partial pressure of HI. Then Kp = P(HI)2 / P(H2) x P(I2) with the equilibrium state before the addition being as before K = m2/(1.0)2.The partial pressure of hydrogen iodide after addition and re-equilibration is then m + z - 2y, where y is again the change in partial pressure of hydrogen or iodine due to reaction. These partial pressures are both originally taken as 1.0 (atmospheres, bars, torr, pascals, or kilopascals), so

K = (m + z - 2y)2/(1.0 + y)2

The effect of addition of product is to shift the equilibrium to the left, back toward the reactants. The final concentrations of HI,
I2, and H2 will all be higher after addition and re-equilibration than before the addition of HI.

These equations have been written in terms of partial pressures but would be exactly the same if the addition of products, and
reactants, were measured in moles per cubic meter.

Total Pressure

All chemical equilibria are affected to some degree by pressure, but in most cases the equilibrium constant varies only slightly
with pressure and this is the only effect of pressure on the equilibrium. When gases are involved in the equilibrium, however, the
effect of pressure can be much more significant. This greater effect can be understood in terms of the principle of Le Chatelier. Several examples are given below.

The Haber process for ammonia synthesis is based on the equilibrium N2(g) + 3 H2(g)  2 NH3(g). In this
equilibrium, there are four moles of reactants on the left yielding two moles of product on the right. Since by Avogadro's
hypothesis equal volumes of gases contain equal numbers of molecules, four volumes yield two volumes. An increase in total
pressure will be a force tending to decrease the volume of the system described by this equilibrium, and the equilibrium will
respond by reacting so as to reduce the total volume - thereby increasing the proportion of ammonia. Conversely, a lower total
pressure will tend to increase the proportion of reactants. For this reason, the synthesis of ammonia is normally carried out at
high pressure. The same equilibrium, in reverse, is used to prepare reducing atmospheres for heat-treating furnaces. When this
is done,the ammonia is dissociated at atmospheric pressure in order to increase the extent of dissociation to hydrogen and
nitrogen.

The carbonated beverages are obtained by dissolving carbon dioxide in water. The equilibrium can be written as
CO2(g)  CO2(aq). Since carbon dioxide in gaseous form is significantly affected by pressure, increasing the CO2 pressure in the bottling plant causes dissolution; when the pressure cap is removed, effervescence occurs as the equilibrium shifts to produce carbon dioxide gas from solution.

The heating of limestone to produce lime involves the equilibrium CaCO3(s)  CaO(s) + CO2(g). Production will
be adversely affected unless the carbon dioxide gas is removed as formed, so provision for ventilation must be made.

Gas phase reactions are not significantly affected by pressure when equal volumes, or moles, of gas are found on the reactant
and product sides. Thus there will be no significant shift in the equilibrium H2(g) + Cl2(g)  2 HCl(g) when the pressure is
changed while the equilibrium 2 H2(g) + O2(g)  2 H2O(g) will be affected by changes in pressure.

Temperature

All chemical equilibria are affected by temperature and for most equilibria the effect of temperature is significant. Equilibrium
constant values are therefore always given at a particular temperature, often 250 oC.

Le Chatelier's principle permits a simple qualitative explanation of the effect of temperature on chemical equilibria. An exothermic reaction is one in which heat is produced, as a product, while an endothermic reaction is one in which heat is
absorbed, or consumed, as a reactant. Heat, considered as a product or a reactant, acts as a stress just as would the addition
of moles of a substance which is a reactant or a product. When heat is a product, the addition of heat which must occur as the
temperature is raised drives the equilibrium backward, while if heat is a reactant then the addition of heat drives the equilibrium
forward.

Example. The heat of formation of HI(g) is +135.1 kJ/mole. The heat of the formation reaction H2(g) + I2(g)  2 HI(g) is
therefore +270.2 kJ and the reaction is endothermic. The effect of adding heat to the equilibrium:

heat + H2(g) + I2(g)  2 HI(g)

is therefore to drive the equilibrium to the right. Increasing the temperature will therefore form additional hydrogen iodide.

Example. The heat of formation of ammonia is -46.11 kJ/mole. The heat of the formation reaction N2(g) + 3 H2(g) 
2 NH3(g) is -92.22 kJ/mole and the reaction is exothermic. The effect of adding heat to the equilibrium:

N2(g) + 3 H2(g)  2 NH3(g) + heat

is then to drive the equilibrium to the left. Increasing the temperature will therefore increase the extent of dissociation of
ammonia to nitrogen and hydrogen.