A.P. Chemistry
Chapter 13
Equilibrium
13.1 Equilibrium is not static, but is a highly
dynamic state. At the macro level everything
appears to have stopped but at the molecular
level, there is frantic activity. The concentration
of products builds as the reaction proceeds.
There will come a time, in some reactions, where
the products will collide and thus react and
reform the reactants. When both the forward and
reverse reaction occur at the same rate, there is
no change in concentration of reactants and
products, and the reaction is said to be at
equilibrium
A double arrow () is used to show that a
reaction can occur in either direction.
13.2 For the generic equation
jA + kB  mC + nD
(p. 615)
A, B, C, and D represent chemical species, and j, k, m, and n are
their coefficients in the balanced equation. Since the
concentration of the products and reactants remains constant
at equilibrium, we can set up a mass action expression: which
is the concentration of the products (raised to the power of
their coefficients) divided by the concentration of the
reactants (raised to the power of their coefficients),
Kc = [C]m[D]n
[A]j[B]k
(p. 615)
The mass action expression can then be used to find a constant
for the reaction at a given temperature. This constant is
known as the equilibrium expression (Kc or Keq). (p. 617)
The equilibrium expression for a reaction is the reciprocal of that
for the reaction written in reverse.
When the balanced equation for a reaction is multiplied by a
factor n, the equilibrium expression for the new reaction is
the original expression raised to the nth power. (Knew =
(Koriginal)n)
The units for K are determined by the powers of the various
concentration terms, and depend on the reaction being
considered.
What does it mean when the value of Keq is very large?
(products are favored heavily)
What does it mean when the value of Keq is very small?
(reactants are favored heavily)
Equilibrium position- set of equilibrium concentrations. There is
only one equilibrium constant for a particular system at a
particular temperature, but there are an infinite number of
equilibrium positions. This can tell us in what direction the
reaction will proceed. (p. 618)
Example: For the Haber process
3H2(g) + N2(g)  2NH3(g),
Kc = 8.7 at a given temperature, T.
Write the equilibrium expression for the
equation. K = [NH3]2/([H2]3[N2])
Calculate the equilibrium constant if the
reaction is written as:
3/2 H2 + ½ N2  NH3 K’c = (8.7) ½
2NH3  3H2 + N2
K’’c = 1/K = (8.7)-1
13.3 Equilibrium Positions Involving Pressures
Rearrange the Ideal Gas Law to include molarity (use C for concentration)
PV = nRT or P = (n/V)RT = CRT
(p. 619)
Where C = n/V, or the number of moles n of gas per unit volume V. Thus C
represents the molar concentration of the gas. For the Haber process:
The equilibrium expression would look like:
K = [NH3]2
= CNH32
[H2]3[N2]
CH23CN2
Or, in terms of the equilibrium partial pressures of the gases:
(p. 620)
Kp = PNH32
PH23PN2
Since we learned there is a relationship which exists between concentration
and pressure, so there must be a relationship between Kc and Kp:
(p. 621)
Kp = K(RT)/\n
Where /\n is the sum of the coefficients of the gaseous products minus the
sum of the coefficients of the gaseous reactants (see a derivation of this
equation on p. 621)
13.4 Heterogeneous Equilibria
Homogeneous equilibria- where all reactants and
products are all in the same phase (p. 622)
(Haber process is an example of a homogeneous
equilibrium; all substances are in gas phase)
Heterogeneous equilibria- where all reactants and
products are not in the same phase. These
equilibria do not depend on the amounts of pure
solids or liquids (Therefore, pure solids and pure
liquids are not included in the equilibrium
expression)(p. 622)
Example: CaCO3(s)  CaO(s) + CO2(g)
K = [CO2]
13.5 Applications of the Equilibrium Constant (p. 627)
Reaction Quotient (Q) – obtained by applying the law of mass action using
initial concentrations instead of equilibrium concentrations. This will help
us to determine which direction the reaction will proceed in to reach
equilibrium. For example, for the Haber process again, the expression for
the reaction quotient would be
Q = [NH3]2/([H2]3[N2]
From this equation, 3 conditions may occur:
• Q = K. The system is at equilibrium. No shift will occur.
• Q>K In this case, the ratio of initial concentrations of products to initial
concentrations of reactants is too large. To reach equilibrium, a net change
of products to reactants must occur. The system shifts to the left,
consuming products and forming reactants, until equilibrium is reached.
• Q<K In this case, the ratio of initial concentrations of products to initial
concentrations of reactants is too small. The system must shift to the right,
consuming reactants and forming products, to attain equilibrium.
(Practice problems on old Chemistry worksheet: writing equilibrium
expressions and solving, comparing Q & K)
13.6 Solving Equilibrium Problems (p. 635)
1. write the balanced equation for the reaction.
2. write the equilibrium expression using the law of mass action.
3. list the initial concentrations.
4. calculate Q, and determine the direction of the shift to equilibrium.
5. define the change needed to reach equilibrium, and define the
equilibrium concentrations by applying the change to the initial
concentrations.
6. substitute the equilibrium concentrations into the equilibrium
expression; solve for the unknown.
7. check your calculated equilibrium concentrations by making sure
that they give the correct value of K.
Small values of K and the resulting small shift to the right to reach
equilibrium allow for simplification. This is okay as long as the
resulting change in value is less than 5% of the original value. If not,
then you must use the quadratic formula.
Example: 3.0 mol of iodine and 4.0 mol of
bromine are placed in a 2.0 L reaction vessel
at 150oC. The reaction comes to equilibrium at
which point 3.2 mol of iodine bromide are
present. Determine the equilibrium constant
for the reaction: I2(g) + Br2(g)  2IBr(g)
Set up a rice!!
Kc = 3.0
Example: Consider the equilibrium:
H2(g) + I2(g)  2HI(g)
If 2.0 mol of hydrogen gas and 2.0 mol of iodine
gas are introduced into a 2.0 L reaction vessel
at 400oC, determine the concentrations of all
species at equilibrium. (Kc = 64 @ 400oC)
[H2] = 1.0M - x = 0.20 M
[I2] = 1.0M - x = 0.20 M
[HI] = 2x = 1.60M
Example: Using above reaction, Calculate Kp for this reaction @ 400oC.
Determine the equilibrium partial pressure of HI @ 400oC if initially the
partial pressures of H2 and I2 are 2.0 and 4.0 atm respectively, at
400oC.
Kp = 64
(Kp = Kc whenever the # of moles of gas on each side
of the equation is equal)
PHI = 2x = 2(1.9) = 3.8 atm
13.7 LeChatelier’s Principle- if a change is imposed
on a system in equilibrium, the position of the
equilibrium will shift in a direction that tends to
reduce that change (or, in other words, it will
shift in a direction to alleviate the stress applied
to the system). (p. 641)
• Effect of change in concentration
If a reactant or product is added to a system at
equilibrium, the system will shift away from the
added component to decrease the amount of the
added component. If a reactant or product is
removed, the system will shift toward the
removed component, to make more of the
removed component. (p. 641)
• Effect of a change in pressure (p. 643)
The addition of an inert gas increases the total
pressure but has no effect on the concentrations
or partial pressures of the reactants or products.
When the volume of the container holding a
gaseous system is reduced, the system responds
by reducing its own volume. This is done by
decreasing the total number of gaseous
molecules in the system, so the reaction will shift
in the direction which has the least number of
molecules.
Adding or removing a gas product or reactant (see
Effect in Change of Concentration)
• Effect of a Change in Temperature
It is important to realize that although the changes
discussed so far may alter the equilibrium
position, they do not alter the equilibrium
constant. The value of K is altered with a change
in temperature. If energy is added to an
exothermic system at equilibrium by heating it,
the shift will occur in the direction which will
consume energy, that is, to the left. Opposite
would be true for an endothermic reaction. (In
other words, this acts like concentration, the
equilibrium will shift away if there is an increase
on the side where the energy term is written.)(p.
646)
Example: Consider the equilibrium
CaCO3(s)  CaO(s) + CO2(g)
A 1.0 L reaction vessel contains 10.0 g of CaO at
1000oC. At this temperature, 0.20 atm of carbon
dioxide gas is introduced into the reaction vessel.
Will any calcium carbonate form? Justify your
answer. At 1000oC, Kc = 0.0370 for this reaction.
Kp = Kc(RT)/\n = (0.0370)[(0.0821)(1273)]1
= 3.87
Q = 0.20
Q<K, therefore, system not @ equilibrium,
therefore, no CaCO3 will form, reaction will
continue to proceed to the RHS, forming more
product
For the above reaction, /\Ho = 175 kJ/mol
How will the equilibrium be affected by each of the following?
Adding CaCO3(s)
(no effect; solids never appear in the equilibrium expression)
Removing CO2(g)
(a net shift to the right; system acts to replace the carbon dioxide)
Halving the volume of the reaction vessel
(a net shift to the left; halving the volume doubles all of the partial pressures;
the reaction shifts to the side w/ fewer gas moles)
Adding helium gas to the reaction vessel
(no effect; even though it increases the total pressure, it does not change the partial
pressure of carbon dioxide)
Heating the reaction vessel
(a net shift to the right; endothermic reaction. Think of heat as a reactant term. You
will consume it and send the reaction to the other side)
Adding a catalyst to the reaction vessel
(no effect; changes the rate, equilibrium is attained faster, but it does not change
equilibrium positions)