1
Today…
• Turn in:
– Get out POGIL (Day 1)
• Our Plan:
– Test Results/Calendar
– POGIL discussion
– Intro to Equilibrium Notes
• Homework (Write in Planner):
– Begin Homework Problems
– Gizmos Due next class
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
2
Test Results
A
B
C
D
F
Average
High Score
Prentice Hall © 2005
7
8
3
0
2
83.2
98.5
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
3
Chapter Fourteen
Chemical Equilibrium
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
4
What is Equilibrium?
• Reactions don’t always have a
beginning and an end. They find a
balance and have to work through
stress to find it.
• Many reaction never EVER finish.
– They are reversible
– Can go forward or backward without
extra energy used
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
5
What is Equilibrium?
• Equilibrium state is reached
when a reaction’s forward
progress is perfectly
balanced with the reverse
process.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
6
Dynamic Nature of Equilibrium
When a system reaches equilibrium, the forward and reverse
reactions continue to occur … but at equal rates.
We are usually concerned
with the situation after
equilibrium is reached.
After equilibrium the concentrations of reactants and
products remain constant.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
7
Concentration vs. Time
If we begin with only 1
M HI, the [HI]
decreases and both [H2]
and [I2] increase.
Prentice Hall © 2005
Beginning with 1 M H2
and 1 M I2, the [HI]
increases and both [H2]
and [I2] decrease.
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Beginning with 1 M each
of H2, I2, and HI, the [HI]
increases and both [H2]
and [I2] decrease.
Chapter Fourteen
8
Regardless of the
starting concentrations;
once equilibrium is
reached …
Prentice Hall © 2005
… the expression with products
in numerator, reactants in
denominator, where each
concentration is raised to the
power of its coefficient, appears
to give a constant.
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
9
The Equilibrium Constant Expression
For the general reaction:
aA + bB  gG + hH
The equilibrium expression is:
Each concentration is
simply raised to the
power of its coefficient
[G]g[H]h
Kc =
[A]a[B]b
Products in
numerator.
Reactants in
denominator.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
10
The Equilibrium Constant
• The equilibrium constant is constant regardless of the
initial concentrations of reactants and products.
• This constant is denoted by the symbol Kc and is
called the concentration equilibrium constant.
• The constant is only good at a given temperature. If
temperature is changed, so is the constant.
• Concentrations of the products appear in the
numerator and concentrations of the reactants appear
in the denominator.
• The exponents of the concentrations are identical to
the stoichiometric coefficients in the chemical
equation.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
11
Quick Check
• Write the equilibrium constant expression
for:
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
12
Example 14.1
If the equilibrium concentrations of COCl2 and Cl2 are the
same at 395 °C, find the equilibrium concentration of CO
in the reaction:
CO(g) + Cl2(g)
COCl2(g)
Kc = 1.2 x 103 at 395 °C
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
13
Modifying the Chemical Equation
Consider the reaction:
2 NO(g) + O2(g)
2 NO2(g)
[NO2]2
Kc = –––––––––
=
4.67
x 1013 (at 298 K)
[NO]2 [O2]
Now consider the reaction: 2 NO2(g)
2 NO(g) + O2(g)
What will be the equilibrium constant K'c for the new reaction?
[NO]2 [O2]
1
1
1
K'c = –––––––––
=
–––––––––––
=
––
=
–––––––––
=
2.14
x 10–14
[NO2]2
[NO2]2
Kc
4.67 x 1013
–––––––––
[NO]2 [O2]
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
14
Modifying the Chemical Equation (cont’d)
Consider the reaction:
2 NO(g) + O2(g)
2 NO2(g)
[NO2]2
Kc = –––––––––
=
4.67
x 1013 (at 298 K)
[NO]2 [O2]
Now consider the reaction: NO2(g)
NO(g) + ½ O2(g)
What will be the equilibrium constant K"c for the new reaction?
[NO] [O2]1/2
1
K"c = –––––––––
= –––
2
[NO2]
Kc
Prentice Hall © 2005
1/2
= 2.14 x 10–14 = 1.46 x 10–7
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
15
Modifying the Chemical Equation (cont’d)
• For the reverse reaction, K is the reciprocal of K for the
forward reaction.
• When an equation is divided by two, K for the new
reaction is the square root of K for the original reaction.
• General rule:
– When the coefficients of an equation are multiplied by a
common factor n to produce a new equation, we raise
the original Kc value to the power n to obtain the new
equilibrium constant.
• It should be clear that we must write a balanced chemical
equation when citing a value for Kc.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
16
Quick Check
• So…
– If we reverse a reaction, to find the
new Kc we would…
– If we double a reaction, to find the
new Kc we would…
– If we third a reaction, to find the new
Kc we would…
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
17
Example 14.2
The equilibrium constant for the reaction:
½ H2(g) + ½ I2(g)
at 718 K is 7.07.
HI(g)
(a) What is the value of Kc at 718 K for the reaction
HI(g)
½ H2(g) + ½ I2(g)
(b) What is the value of Kc at 718 K for the reaction
H2(g) + I2(g)
2 HI(g)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
18
Try It Out
N2(g) + 3H2(g) ↔2NH3(g)
For the equation above the Kc = 3.6 x 108
What is the value of Kc at the same temp
for:
NH3 ↔ ½ N2 + 3/2 H2
5.3 x
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
-5
10
Chapter Fourteen
19
Try It Out!
N2(g) + 3H2(g) ↔ 2NH3(g)
For the equation above the Kc = 3.6 x 108
Determine the value for Kc for the reaction:
1/3 N2 + H2 ↔2/3 NH3
7.1 x
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
2
10
Chapter Fourteen
20
Equilibria Involving Gases
• In reactions involving gases, it is often convenient to
measure partial pressures rather than molarities.
• In these cases, a partial pressure equilibrium constant, Kp,
is used.
g
h
Kp =
(PG) (PH)
(PA)a(PB)b
Kc and Kp are related by:
Kp = Kc (RT)Δn(gas)
where Dn(gas) is the change in number of moles of gas as the
reaction occurs in the forward direction.
Dn(gas) = mol gaseous products – mol gaseous reactants
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
21
Remember “R”
.0821 atm · L
mol · K
Units in problem have to
be Kelvin and atm!
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
22
An Example
2H2S (g) ↔ 2H2 (g)+ S2 (g)
-2
Kp = 1.2 x 10
Kc = ? (@ 1065 deg C)
1.1 x
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
-4
10
Chapter Fourteen
23
Example 14.3
Consider the equilibrium between dinitrogen tetroxide and
nitrogen dioxide:
N2O4(g)
2 NO2(g)
Kp = 0.660 at 319 K
(a) What is the value of Kc for this reaction? (b) What is the
value of Kp for the reaction 2 NO2(g)
N2O4(g)? (c) If
the equilibrium partial pressure of NO2(g) is 0.332 atm,
what is the equilibrium partial pressure of N2O4(g)?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
24
The Equilibrium Constant
for an Overall Reaction
Suppose we need: N2O(g) + 3/2 O2(g)
2 NO2(g)
Kc(1) = ??
and we’re given:
N2O(g) + ½ O2(g)
2 NO(g) + O2(g)
2 NO(g)
Kc(2) = 1.7 x 10–13
2 NO2(g)
Kc(3) = 4.67 x 1013
• Adding the given equations gives the desired equation.
• Multiplying the given values of K gives the equilibrium
constant for the overall reaction.
• (To see why this is so, write the equilibrium constant
expressions for the two given equations, and multiply
them together. Examine the result …)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
25
Let’s Try One… (#25)
What is the value of Kp at 298 K for the
reaction ½ CH4 (g) + H2O (g) ↔ ½ CO2 (g) +
2H2 (g), given the following data at 298 K?
CO2 (g) + H2 (g) ↔ CO (g) + H2O (g)
Kp = 9.80 x 10-6
CO (g) + 3 H2 (g) ↔ CH4 (g) + H2O (g)
Kp = 8.00 x 1024
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
26
Try It Out! (#27)
• Determine Kc at 298 K for the reaction
2CH4 (g) ↔ C2H2 (g) + 3 H2 (g), given the
following data at 298 K.
CH4 (g) + H2O (g) ↔ CO (g) + 3 H2 (g)
Kp = 1.2 x 10-25
2C2H2 (g) + 3O2 (g) ↔ 4CO (g) + 2H2O (g)
Kp = 1.1 x 102
H2 (g) + ½ O2 (g) ↔ H2O (g)
Kp = 1.1 x 1040
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
27
Equilibria Involving Pure
Solids and Liquids
• The equilibrium constant expression does not include
terms for pure solid and liquid phases because their
concentrations do not change in a reaction.
• Although the amounts of pure solid and liquid phases
change during a reaction, these phases remain pure and
their concentrations do not change.
Example:
CaCO3(s)
CaO(s) + CO2(g)
[CaO] [CO2]
Kc = ––––––––––
[CaCO3]
Prentice Hall © 2005
Kc = [CO2]
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
28
Write the Kc expression for:
C(s) + H2O (g) ↔ CO(g) + H2(g)
MgCO3(s) ↔ MgO(s) + CO2(g)
2H2(g) + O2(g) ↔ 2H2O(l)
H2S(g) + I2(s) ↔ 2HI(g) + S(s)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
29
Example 14.4
The reaction of steam and coke (a form of carbon)
produces a mixture of carbon monoxide and hydrogen,
called water-gas. This reaction has long been used to make
combustible gases from coal:
C(s) + H2O(g)
CO(g) + H2(g)
Write the equilibrium constant expression for Kc for this
reaction.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
30
Equilibrium Constants: When Do We
Need Them and When Do We Not?
• A very large numerical value of Kc or Kp signifies that a
reaction goes (essentially) to completion.
• A very small numerical value of Kc or Kp signifies that the
forward reaction, as written, occurs only to a slight extent.
• An equilibrium constant expression applies only to a
reversible reaction at equilibrium.
• Although a reaction may be thermodynamically favored, it
may be kinetically controlled …
• Thermodynamics tells us “it’s possible (or not)”
• Kinetics tells us “it’s practical (or not)”
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
31
2H2 (g) + O2(g) ↔ 2 H2O(g)
Kp = 1.4 x 10
83
@ 298 K
A term in the denominator of the
equilibrium approaches zero and
makes the value of the equilibrium
constant very large.
Very large Kc or Kp favors forward reaction
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
32
C(s)+ H2O(g) ↔CO(g) + H2(g)
Kp = 1.6 x
-21
10
@ 298 K
To account for a small numerical value
of an equilibrium constant, the
numerator must be very small.
Very small Kc or Kp favors reverse reaction
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
SO…..
33
A reaction is most likely to reach a
state of equilibrium in which
significant quantities of both
reactants and products are present
if the numerical value of Kc and Kp
is neither very large nor very
small, roughly in the range from
10-10 to 1010
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
34
The Reaction Quotient, Q
• For nonequilibrium conditions, the expression
having the same form as Kc or Kp is called the
reaction quotient, Qc or Qp.
• The reaction quotient is not constant for a
reaction, but is useful for predicting the direction
in which a net change must occur to establish
equilibrium.
• To determine the direction of net change, we
compare the magnitude of Qc to that of Kc.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
35
The Reaction Quotient, Q
When Q is smaller
than K, the
denominator of Q is
too big; we have “too
much reactants.”
When Q = K,
equilibrium has been
reached.
When Q is larger
than K, the
numerator of Q is too
big; we have “too
much products.”
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
36
A better graphic
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
37
CO(g) + 2 H2(g) ↔ CH3OH (g)
If the Kc for this reaction = 14.5, then
what direction will the reaction
proceed if the initial concentrations
are
CO = 0.100 M Left to
H2 = 0.100 M right
CH3OH = 0 M
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
38
Try It Out! (#49)
• If a gaseous mixture at 588 K contains 20.0
g each of CO and H2O and 25.0 g each of
CO2 and H2, in what direction will a net
reaction occur to reach equilibrium?
Explain
• CO (g) + H2O (g) ↔ CO2 (g) + H2 (g)
Kc = 31.4 at 588 K
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
39
Stop – Homework Time
• You should be able to
complete problems 20,
22, 24, 26, 28, 30, 32, 50,
& 52
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
40
Today…
• Turn in:
– Gizmos - basket
• Our Plan:
– Review/Quiz
– LeChatelier’s POGIL
– Notes
– Demonstration
• Homework (Write in Planner):
– Investigation 13 Pre-Lab
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
41
Quick Check
• For the reaction 2NO(g) ↔ N2 (g) + O2 (g), Kc =
9.92 at 2273 K.
1. Write the equilibrium expression (Kc) for this
reaction.
2. What is Kc for this reaction:
2N2 (g) + 2O2 (g) ↔ 4NO(g) (0.0102)
3. At equilibrium, the concentration of nitrogen and
oxygen are both 1.5 x 10-3 M. What is the
concentration of NO? (4.76 x 10-4 M)
4. What is Kp for the reaction? (9.92)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
42
Stresses on Equilibrium
• Chemists make a living preventing
equilibrium. They want to “knock
it out of whack” to maximize a
certain product.
• Stresses can be applied to shift
equilibrium and maximize the
amount of product, as you saw in
the POGIL.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
43
Le Châtelier’s Principle
• When any change in concentration, temperature,
pressure, or volume is imposed on a system at
equilibrium, the system responds by attaining a new
equilibrium condition that minimizes the impact of the
imposed change.
• Analogy: Begin with 100 men and 100 women at a dance.
• Assume that there are 70 couples dancing, though not
always the same couples (dynamic equilibrium).
• If 30 more men arrive, what happens?
• The equilibrium will shift, and shortly, more couples will
be dancing … but probably not 30 more couples.
• Real Example – Deer populations
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
44
Changing the Amounts of
Reacting Species (concentration)
• At equilibrium, Q = Kc.
• If the concentration of one of the reactants is
increased, the denominator of the reaction quotient
increases.
• Q is now less than Kc.
• This condition is only temporary, however, because
the concentrations of all species must change in such a
way so as to make Q = Kc again.
• In order to do this, the concentrations of the products
increase; the equilibrium is shifted to the right.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
45
Concentration
•An increase in
concentration will cause
the reaction to shift to the
opposite side until a new
equilibrium is achieved.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
46
Important Note
• Remember that this is a change in
concentration so changing the
amount of a pure solid or liquid
which is already in equilibrium
should not make the reaction
consume the additional solid or
liquid.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
47
Example
• Example: 2HgO (g) ↔2Hg (l) + O2 (g)
∆ H = +43.4 kcal
• Add HgO, what happens to equilibrium?
• Add O2, what happens to equilibrium?
• Add Hg, what happens to equilibrium?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
48
Example 14.7
Water can be removed from an equilibrium mixture in the
reaction of 1-octanol and acetic acid, for example, by
using a solid drying agent that is insoluble in the reaction
mixture. Describe how the removal of a small quantity of
water affects the equilibrium.
CH3(CH2)6CH2OH(aq) + CH3COOH(aq)
H+
CH3(CH2)6CH2OCOCH3(aq) + H2O(soln)
p. 591 – As water is removed, the reverse reaction is slowed and the forward, waterforming reaction is stimulated. Not all of the water that is removed can be replaced,
however, because a new equilibrium could not tolerate a constant concentration of H2O
at the same time that the other three concentrations change. Thus, in the new
equilibrium the amount of water is somewhat less than in the original equilibrium, the
amount of octyl acetate is greater, and the amounts of both 1-octanol and acetic acid are
less.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
49
Changing External Pressure or
Volume in Gaseous Equilibria
• When the external pressure is increased (or system volume
is reduced), an equilibrium shifts in the direction producing
the smaller number of moles of gas.
• When the external pressure is decreased (or the system
volume is increased), an equilibrium shifts in the direction
producing the larger number of moles of gas.
• If there is no change in the number of moles of gas in a
reaction, changes in external pressure (or system volume)
have no effect on an equilibrium.
Example: H2(g) + I2(g)
2 HI (g) What would
happen if pressure is increased?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
50
Example
• Example: 2HgO (g) ↔2Hg (l) + O2 (g)
∆ H = +43.4 kcal
• Add pressure, what happens to equilibrium?
• Increase volume, what happens to
equilibrium?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
51
Try It Out!
Example 14.8
An equilibrium mixture of O2(g), SO2(g),
and SO3(g) is transferred from a 1.00-L
flask to a 2.00-L flask. In which direction
does a net reaction proceed to restore
equilibrium? The balanced equation for the
reaction is
2 SO3(g) ↔ 2 SO2(g) + O2(g)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
52
Temperature Changes
• Raising the temperature of an equilibrium
mixture shifts equilibrium in the direction of
the endothermic reaction; lowering the
temperature shifts equilibrium in the
direction of the exothermic reaction.
– Consider heat as though it is a product of an
exothermic reaction or as a reactant of an
endothermic reaction, and apply Le Châtelier’s
principle.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
53
Example 14.9
Is the amount of NO(g) formed from given amounts of
N2(g) and O2(g),
N2(g) + O2(g)
↔ 2 NO(g)
ΔH° = +180.5 kJ
greater at high or low temperatures?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
54
Catalysts & Equilibrium
• A catalyst lowers the activation
energy … of both the forward and
the reverse reaction.
• Adding a catalyst does not affect
an equilibrium state.
• A catalyst merely causes
equilibrium to be achieved faster.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
55
Demo Time!
• This is going to be similar to
Investigation 13! Pay
attention!
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
56
Today…
• Turn in:
–Nothing
• Our Plan:
–Investigation 13
• Homework (Write in Planner):
–Lab Report Due Monday
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
57
Today…
• Turn in:
– Lab Report (basket) – rubric on top
• Our Plan:
– Review Quiz (use your notes)
– Notes – Equilibrium Calculations
• Homework (Write in Planner):
– Homework Problems (due next Monday)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
58
Determining Values of Equilibrium
Constants Experimentally
• When initial amounts of one or more species, and
equilibrium amounts of one or more species, are given, the
amounts of the remaining species in the equilibrium state
and, therefore, the equilibrium concentrations often can be
established.
• A useful general approach is to tabulate under the chemical
Reaction:
– the concentrations of substances present Initially
– Changes in these concentrations that occur in reaching equilibrium
– the Equilibrium concentrations.
• This sort of table is sometimes called a “RICE” table:
Reaction/Initial/Change/Equilibrium.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
59
Important Note
• The “RICE” table may be
used with moles,
molarity, or Liters (for
gaseous equilibria)…
never grams.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
60
Example
• Equilibrium involving SO2, O2,
and SO3 is important in sulfuric
acid production. When a 0.0200
mol sample of SO3 is introduced
into an evacuated 1.52 L flask @
900 K, 0.0142 mol SO3 is found to
be present at equilibrium. What is
the value of Kp ?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
61
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
64
Example 14.11
In a 10.0-L vessel at 1000 K, 0.250 mol SO2
and 0.200 mol O2 react to form 0.162 mol
SO3 at equilibrium. What is Kc, at 1000 K,
for the reaction that is shown here?
2 SO2(g) + O2(g)
2 SO3(g)
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
65
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
66
Try It Out!
• 14.11 A
A reaction starts with 1.00 mol each of PCl3
and Cl2 in a 1.00 L flask. When equilibrium
is established at 250 degrees C in the reaction
PCl3 (g) + Cl2 (g) ↔ PCl5 (g), the amount of
PCl5 present is 0.82 mol. What is Kc for this
reaction?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
67
Calculating Equilibrium Quantities
from Kc and Kp Values
• When starting with initial reactants and no products and
with the known value of the equilibrium constant, these
data are used to calculate the amount of substances present
at equilibrium.
• Typically, a RICE table is constructed, and the symbol x is
used to identify one of the changes in concentration that
occurs in establishing equilibrium.
• Then, all the other concentration changes are related to x,
the appropriate terms are substituted into the equilibrium
constant expression, and the equation solved for x.
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
68
Example 14.12
Consider the reaction
H2(g) + I2(g) ↔ 2 HI(g) Kc = 54.3 at 698 K
If we start with 0.500 mol I2(g) and 0.500 mol H2(g) in a
5.25-L vessel at 698 K, how many moles of each gas will
be present at equilibrium?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
69
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
Example 14.14
70
Carbon monoxide and chlorine react to form phosgene, COCl2, which is used in the
manufacture of pesticides, herbicides, and plastics:
CO(g) + Cl2(g)
COCl2(g)
Kc = 1.2 x 103 at 668 K
How much of each substance, in moles, will there be at equilibrium in a reaction mixture that
initially has 0.0100 mol CO, 0.0100 mol Cl2, and 0.100 mol COCl2 in a 10.0-L flask?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
71
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
72
Example 14.15
A sample of phosgene, COCl2(g), is introduced into a
constant-volume vessel at 395 °C and observed to exert an
initial pressure of 0.351 atm. When equilibrium is
established for the reaction
CO(g) + Cl2(g)
COCl2(g)
Kp = 22.5
what will be the partial pressure of each gas and the total
gas pressure?
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
73
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
74
Try It Out!
• 14.13 A: Starting with 0.100 mol CO and
0.200 mol Cl2 in a 25.0 L flask, how many
moles of COCl2 will be present at
equilibrium in the following reaction?
CO (g) + Cl2 (g) ↔ COCl2
Kc = 1.2 x 102 at 668 K
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
75
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
76
The Effect of T on Kp
• For an exothermic reaction;
as T increases, Kp decreases
• For an endothermic reaction;
as T decreases, Kp increases
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
77
Today…
• Turn in:
– Nothing
• Our Plan:
– Crash Course Review - Equilibrium
– Practice, Practice, Practice RICE Tables
(guided worksheet)
– Homework
• Homework (Write in Planner):
– Homework due Monday
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
78
Today…
• Turn in:
– Nothing
• Our Plan:
– Crash Course Review – Calculations
– Practice, Practice, Practice RICE Tables
– Quiz
– Work on Homework
• Homework (Write in Planner):
– Homework Due Monday
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
79
Today…
• Turn in:
– Mark Homework Questions on the Board
• Our Plan:
– Homework Discussion
– Station Review
– Test Review
• Homework (Write in Planner):
– Test Next Class
– Breakfast Club Wednesday at 6 am
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen
80
Today…
• Turn in:
–Nothing
• Our Plan:
–Questions on Test Review
–Unit 10 Test
• Homework (Write in Planner):
–Enjoy your long weekend!
Prentice Hall © 2005
General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry
Chapter Fourteen