Download Principles of Reactivity: Chemical Equilibria

Principles of Reactivity:
Chemical Equilibria
Chapter 16
16.1 Nature of the Equilibrium
State
All chemical reactions are reversible.
 When both the forward and reverse
reactions continue to occur at equal rates
but no net change is observed.
 Use a double arrow to show this situation.
 Outside forces can affect the equilibrium!

16.2 The Reaction Quotient and
Equilibrium Constant
For the reaction H2 + I2  2HI the
relationships among the substances has
been experimentally established as
[HI]2/[H2][I2]
 We can make an “ICE” table to summarize
this relationship…

The Reaction Quotient and
Equilibrium Constant
I2

Initial conc. 0.0175
(M)
0.0175
0
Change in
-0.0138
conc. as rxn
proceeds (M)
-0.0138
+0.0276
Equilibrium 0.0037
conc. (M)
0.0037
0.0276
Equation
H2
+
2HI
The Reaction Quotient and
Equilibrium Constant

Putting the equilibrium concentrations into
the expression …
[HI]2/[H2][I2] = (0.0276)2/(0.0037)(0.0037)
= 56
This quotient is always the same for all
experiments under these conditions no
matter what the initial concentrations.
The Reaction Quotient and
Equilibrium Constant
The concentrations of reactants and
products are always related by an
expression to a value called the reaction
quotient.
 Reaction quotient = Q = [C]c[D]d/[A]a[B]b
 When a reaction has reached equilibrium,
the reaction quotient has a constant value
referred to as the equilibrium constant, K.

The Reaction Quotient and
Equilibrium Constant

The expression is commonly referred to as the
equilibrium constant expression.




Product concentrations always appear in the
numerator.
Reactant concentrations always appear in the
denominator.
Each concentration is always raised to the power of
its stoichiometric coefficient in the balanced equation.
When the reaction has reached equilibrium, the value
of the constant K depends on the particular reaction
and on the temperature. Units are never given for K!
Writing Equilibrium Constant
Expressions

Reactions involving solids, water, and pure
liquids.
Concentrations of any solid reactants and
products are omitted from the equilibrium
constant expression.
 The molar concentration of water (or of any
liquid reactant or product) is omitted from the
equilibrium constant expression.

Writing Equilibrium Constant
Expressions

Reactions involving gases…
If reactant and product quantities are given in
partial pressures, then K is given the subscript
p, as in Kp.
H2(g) + I2(g)  2HI(g) Kp = P2HI/(PH2PI2)

Writing Equilibrium Constant
Expressions
For reactions concerned with substances
in aqueous solution, the symbol K
sometimes has the subscript c for
“concentration”, as in Kc.
 In some cases Kc and Kp may be the
same, but they are often different.

Practice Problem
Write the equilibrium constant expression
for each of the following reactions in terms
of concentrations:
 PCl5(g)  PCl3(g) + Cl2(g)
 Cu(OH)2(s)  Cu2+(aq) + 2OH-(aq)
 Cu(NH3)42+(aq)  Cu2+(aq) + 4NH3(aq)


CH3CO2H(aq) + H2O(l)  CH3CO2-(aq) + H3O+(aq)
Meaning of the Equilibrium
Constant, K
K >> 1: Reaction is product-favored.
Equilibrium concentrations of products are
greater than equilibrium concentrations of
reactants. “The reaction has gone to
completion.”
 K<< 1: Reaction is reactant-favored.
Equilibrium concentrations of reactants are
greater than equilibrium concentrations of
products.

Meaning of the Reaction Quotient,
Q
If Q < K, the system is not at equilibrium
and some reactants will be converted to
products.
 If Q > K, the system is not at equilibrium
and some products will be converted to
reactants.
 If Q = K, the system is at equilibrium.

Practice Problem

Answer the following questions regarding the
butane  isobutane equilibrium (K = 2.50 at
298K).


Is the system at equilibrium when [butane] = 0.97 M
and [isobutane] = 2.18 M? If it is not at equilibrium, in
which direction will the reaction proceed in order to
achieve equilibrium?
Is the system at equilibrium when [butane] = 0.75 M
and [isobutane] = 2.60 M? If it is not at equilibrium, in
which direction will the reaction proceed in order to
achieve equilibrium?
Practice Problem

At 2000K the equilibrium constant, K, for
the formation of NO(g)…
N2(g) + O2(g)  2NO(g) is 4.0 x 10-4. If,
at 2000K, the concentration of N2 is 0.50
M, that of O2 is 0.25 M, and that of NO is
4.2 x 10-3 M, is the system at equilibrium?
If not, predict which way the reaction will
proceed to achieve equilibrium.
16.3 Determining an Equilibrium
Constant

If the experimental values of the
concentrations of all the reactants and
products at equilibrium are known, an
equilibrium constant can be calculated by
substituting the data into the equilibrium
constant expression.
Determining an Equilibrium
Constant

If not all concentrations are known, use an
ICE table to show the initial
concentrations, how they change on
proceeding to equilibrium, and the
concentrations at equilibrium.
Practice Problem

a.
b.
A solution is prepared by dissolving 0.050 mol
of diiodocyclohexane, C6H10I2, in the solvent
CCl4. The total solution volume is 1.00 L.
When the reaction
C6H10I2  C6H10 + I2 has come to
equilibrium at 35oC, the concentration of I2 is
0.035 M.
What are the concentrations of C6H10I2 and
C6H10 at equilibrium?
Calculate Kc, the equilibrium constant.
Homework
After reading sections 16.1-16.3, you
should be able to do the following…
 P. 752 (1-10)

16.4 Using Equilibrium Constants

The value of K and the initial amounts of
reactants are often known, and you are
asked to calculate the quantities present at
equilibrium.
Practice Problem

At some temperature, K = 33 for the
reaction H2(g) + I2(g)  2HI(g)
Assume the initial concentration of both H2
and I2 are 6.00 x 10-3 mol/L. Find the
concentration of each reactant and
product at equilibrium.
Using Equilibrium Constants in
Calculations





If the solution is a
quadratic
expression…
K= 1.20 = x2/(1.60-x)
x2 + 1.20x -1.92 = 0
Use appdx A to find
roots to equation
Negative values
aren’t chemically
meaningful! So,
answer is x = 0.91.
PCl5  PCl3 + Cl2
I
1.60
0
0
C
-x
+x
+x
E
1.60-x
x
x
Practice Problem

Graphite and carbon dioxide are kept at
1000 K until the reaction
C(graphite) + CO2(g)  2CO(g)
has come to equilibrium. At this
temperature, K = 0.021. The initial
concentration of CO2 is 0.012 mol/L.
Calculate the equilibrium concentration of
CO.
Is there an easier way?
If K is very small, then you can use
another expression…
 If 100K < [A]o the approximate expression
will give acceptable values of equilibrium
concentrations.
 In general, when K is about 1 or greater,
the approximation cannot be made.
 This means you can drop the “-x” off of
your expression!

Practice Problem

The reaction N2 + O2  2NO contributes
to air pollution whenever a fuel is burned
in air at a high temperature, as in a
gasoline engine. At 1500K, K=1.0x10-5.
Suppose a sample of air has [N2] = 0.80M
and [O2] = 0.20M before any reaction
occurs. Calculate the equilibrium
concentrations of reactants and products
after the mixture has been heated to
1500K.
16.5 Balanced Equations and
Equilibrium Constants



When the stoichiometric coefficients of a
balanced equation are multiplied by some factor,
the equilibrium constant for the new equation
(Knew) is the old equilibrium constant (Kold) raised
to the power of the multiplication factor.
The equilibrium constants for a reaction and its
reverse are the reciprocals of each other.
When two or more chemical equations are
added to produce a net equation, the equilibrium
constant for the net equation is the product of
the equilibrium constants for the added
equations.
Practice Problem

a.
b.
The conversion of oxygen to ozone has a
very small equilibrium constant.
3/ O (g)  O (g)
-29
K
=
2.5
x
10
2
2
3
What is the value of K when the equation
is written using whole number
coefficients?
What is the value of K for the conversion
of ozone to oxygen?
Practice Problem

The following equilibrium constants are
given at 500 K:
H2(g) + Br2(g)  2HBr(g) Kp = 7.9x1011
H2(g)  2H(g)
Kp = 4.8x10-41
Br2(g)  2Br(g)
Kp = 2.2x10-15
Calculate Kp for the reaction of H and Br
atoms to give HBr.
H(g) + Br(g)  HBr(g)
Kp = ?
16.6 Disturbing a Chemical
Equilibrium

A change in any factor that determines the
equilibrium conditions of a system will
cause the system to change to counteract
the effect of the change. (Le Chatelier’s
principle)
Change temperature
 Change concentrations of reactants or
products
 Change volume (for gas system)

Disturbing a Chemical Equilibrium
When temp of a system at equilibrium
increases, rxn shifts in direction that
absorbs heat energy (endothermic
direction)
 If temp decreases, rxn shifts in direction
that releases heat energy (exothermic
direction)
 Either way, equilibrium composition will
change and therefore K will change

Practice Problem
Consider the effect of temperature
changes on the following equilibria.
 Does the concentration of NOCl increase
or decrease at equilibrium as the
temperature of the system is increased?



2NOCl(g)  2NO(g) + Cl2(g) ΔH = +77.1kJ
Does the concentration of SO3 increase or
decrease when the temperature
increases?

2SO2(g) + O2(g)  2SO3(g) ΔH = -198 kJ
Effect of Addition or Removal of
Reactant/Product

If the concentration of a reactant or
product is changed from its equilibrium
value at a given temperature, equilibrium
will be reestablished eventually. The value
of K will still be the same!
Practice Problem

Equilibrium exists between butane and
isobutane when [butane] = 0.20M and
[isobutane] = 0.50M. What are the
concentrations of butane and isobutane,
after equilibrium has been established, if
2.00 mol/L of isobutane is added to the
original mixture? K = 2.50
Effect of Volume Changes on GasPhase Equilibria



The stress of a volume decrease (pressure
increase) will be counterbalanced by a change
in the equilibrium composition to one having a
smaller number of gas molecules.
For a volume increase (pressure decrease), the
equilibrium will shift toward the side of rxn with
the greater number of gas molecules.
If there is no change in the number of gas
molecules in the reaction, then a volume change
will have no effect.
Practice Problem



The formation of ammonia from its elements is
an important industrial process.
3H2(g) + N2(g)  2NH3(g)
How does the equilibrium composition change
when extra H2 is added? When extra NH3 is
added?
What is the effect on the equilibrium when the
volume of the system is increased? Does the
equilibrium composition change or is the system
unchanged?
Homework
After reading sections 16.4-16.6, you
should be able to do the following…
 P. 753 (15-20,25-28)
