AT Chemistry
2012
Chemical Equilibrium
Equilibrium - a condition in which the rates of the forward and reverse reactions are
equal. As a result, the overall reaction appears static (unchanging) but it is not.
Equilibrium is a dynamic process. At equilibrium the concentrations of reactants and
products remain constant. Consider the equilibrium involving the synthesis of ammonia
from its elements:
3H2(g) + N2(g) = 2NH3(g)
All reversible reactions will tend to go to an equilibrium condition because this is
thermodynamically most favorable. Factors that determine the equilibrium position
include the potential energies of the reactants and products, the enthalpy change as well
as the entropy change.
Law of Mass Action
One way to understand the significance of K (the equilibrium constant) is to
assume that the forward reaction involves nA and mB as the rate-determining step, and
the reverse reaction as the rate-determining step pC and qD
rateforward = Kf[A]n[B]m
ratereverse = Kr[C]p[D]q
At equilibrium,
rateforward = ratereverse or Kf[A]n[B]m = Kr[C]p[D]q
Rearranging,
K is the ratio of the forward to reverse rate constants (not the rates)! As with the
kinetic rate-determining step, reactant or product coefficients become exponents when
put in the mass action expression (or "equilibrium expression").
K is a temperature dependent constant which can have different equilibrium
concentrations. Each set of equilibrium concentrations is called an equilibrium position.
For a reaction at a given temperature, there are many equilibrium positions but only one
value for K.
_______________________________________________________________________________
Starting and Equilibrium Concentrations for the Reaction Between Hydrogen and Iodine at 490oC
H2(g) + I2(g) = 2HI(g)
______________________________________________________________
Concentrations
Starting Concentrations
at Equilibrium
________________________________________________
Experiment
[H2]
[I2]
[HI]
[H2]
[I2]
[HI]
_______________________________________________________________
1
1.00
1.00
0
0.228 0.228 1.544
2
0
0
1.00
0.114 0.114 0.772
3
0
0
1.50
0.171 0.171 1.158
4
0.600 0.400 0
0.245 0.045 0.711
5
0.800 1.200 0
0.090 0.490 1.423
_______________________________________________________________
K = [HI]2/[[H2][I2]
Experiment 1:
[1.544]2/[0.228][0.228] = 45.9
Experiment 2:
[0.772]2/[0.114][0.114] = 45.9
Experiment 3:
[1.158]2/[0.171][0.171] = 45.9
Experiment 4:
[0.711]2/[0.245][0.045] = 45.9
Experiment 5:
[1.423]2/[0.090][0.490] = 45.9
Equilibrium Involving Pressures
Equilibria for gases can be expressed either in concentration terms or pressure
terms. We use K when dealing with concentrations and Kp when dealing with pressures.
2AB(g) = A2(g) + B2(g)
For a general reaction, Kp = K(RT)n
(see text for derivation)
n =  coefficients products -  coefficients reactants
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Summary of properties of K (or Kc)






K is temperature dependent
K is independent of original concentrations, container volume or pressure
Pure solids or liquids do not appear in K (do not appear in mass action expression;
they are incorporated into the value of K)
A reverse reaction has a value of Kreverse equal to the recipricol of Kforward
In adding equilibrium reaction steps to get a net reaction, we multiply the equilibrium
constants for each reaction to get the net reaction’s K.
If you multiply any chemical equation through by a factor it raises K to that power
There are several types of equilibrium problems that you will need to solve in this
chapter:



solving for K (knowing the equilibrium concentrations)
solving for equilibrium concentrations (knowing K)
solving for equilibrium concentrations (knowing K and initial concentrations)
Heterogeneous Equilibria
Equilibrium expressions involve concentrations (or pressures) of substances that
change from initial to the equilibrium condition. A pure substance, such as water,
changes amount but not concentration. The concentrations of pure solids and liquids
remains constant. That means that pure solids and liquids can be incorporated into the
equilibrium constant. For example, the equilibrium expression for
CaCO3(s) = CaO(s) + CO2(g)
IS NOT K = [CaO][CO2]/[CaCO3]
Rather, [CaCO3] and [CaO] are constant. The amounts of each will change, but
their concentrations (a derivative term) are constant. Thus, we can incorporate them into
our equilibrium expression.
K = [CO2]
another example:
or
Kp = PCO2
HCl(g) + NH3(g) = NH4Cl(s)
K=
or
Kp =
textbook problems 21, 23, 33
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Applications of the Equilibrium Constant

K does not reflect how fast a reaction goes

the magnitude of K depends on E; the reaction rate depends on Ea

a large value of K means that mostly products will be present at equilibrium

a small value of K means that mostly reactants will be present at equilibrium
KEY IDEA: use the reaction quotient (Q) to predict the direction in which the reaction
will go to reach equilibrium. Calculate Q by using initial concentrations (or pressures).
If Q = K, system at equilibrium
If Q > K, system will shift left to reach equilibrium
If Q < K, system will shift right to reach equilibrium
Problem:
H2(g) + I2(g) = 2HI(g)
K = 710
Predict direction of shift to reach equilibrium given the following conditions:
a) Q = 427
b) Q = 1522
c) [H2]o = 0.18M, [I2]o = 0.44M, [HI]o = 0.58M
Problem:
N2(g) + O2(g) = 2NO(g)
K = 4.1 X 10-4
If 0.50 moles of N2 and 0.86 moles of O2 are put into a 2.0 L container, calculate the
equilibrium concentration of all species.
The most important question you must raise in doing equilibrium problems is "What
does the value of K tell me about the extent of reaction?"
With a large K, reaction will stay far to the right.
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With a small K, reaction will stay far to the left.
N2(g) + O2(g) = 2NO(g)
K = 4.1 X 10-4
Textbook problems: 39, 41, 45, 47, 53, 57 and 59
Le Chatelier's Principle
If a change is imposed on a system at equilibrium, the position of the equilibrium
will shift in a direction that tends to reduce that change.

Concentration - if you add a product or reactant the equilibrium shifts in the direction
that uses it up.

Pressure - an increase in pressure results in a shift toward the side with smaller
volume (smaller number of moles) and vice-versa. If both sides have the same
volume (same number moles of gas) there will be no effect.

Temperature - an increase in temperature favors the endothermic direction to try to
decrease the temperature.

Catalyst - no effect on the equilibrium since catalyst lowers rate for both forward and
reverse reactions.
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
Note that volume changes are like pressure changes. Increasing the volume decreases
the pressure and decreasing the volume increases the pressure.

Adding a species that does not participate in the reaction has no effect on the
equilibrium.
Problem:
N2(g) + O2(g) = 2NO(g)
shift
H = 181 kJ
Kp = 2.3 X 1030
at new equilibrium point
a) N2 added:
b) He added:
c) Container is
made larger:
d) System is cooled:
Problem:
2H2(g) + O2(g) = 2H2O(g)
shift
H = -484 kJ
at new equilibrium point
a) H2O removed:
b) H2 added:
c) system cooled:
d) O2 removed:
e) pressure is increased:
textbook problems 63-69 odd
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Equilibrium Demonstrations
The Effect of Concentration and Temperature
on an Equilibrium System
Given the following system at equilibrium:
(Co(H2O)6)2+(aq) + 4 Cl-1(aq) = (CoCl4)2-(aq) + 6 H2O(l)
PINK
BLUE
Predict the direction of equilibrium shift for each of the following changes:
1.
Addition of concentrated hydrchloric acid:
2.
Addition of distilled water:
3.
Addition of silver nitrate solution (describe and explain what happens):
4.
Addition of Acetone (CH3COCH3) (describe and explain what happens):
5. How can you determine whether the reaction is endothermic or exothermic? Test
your hypothesis and indicate your conclusion.
The Chromate-Dichromate Equilibrium
NIE:
2CrO42-(aq) + 2H+(aq) = Cr2O72-(aq) + H2O(l)
yellow
orange
(a) addition of HCl:
(b) addition of NaOH:
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NIE:
2CrO42-(aq) + 2H+(aq) = Cr2O72-(aq) + H2O(l)
yellow
orange
(c) addition of Ba(NO3)2:
(d) addition of HCl to (c):
Compare the relative solubilities of the barium salts:
The NO2(g) = N2O4(g) System
A sample of copper metal is placed in a flask containing concentrated nitric acid. A
reaction occurs producing a red-brown gas, nitrogen dioxide. In a closed system the
nitrogen dioxide exists in equilibrium with dinitrogen tetroxide:
2NO2(g) = N2O4(g)
red-brown
colorless
1. What is the entropy change involved?
2. What can you do to determine whether the reaction is endothermic or exothermic?
3. Discuss pressure effects on the equilibrium system by changing volume.
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Additional Problems
1. Sulfuryl chloride, SO2Cl2, is a highly reactive gaseous compound. When heated, it
decomposes as follows:
SO2Cl2(g) = SO2(g) + Cl2(g)
This decomposition is endothermic. A sample of 3.509 grams of SO2Cl2 is placed in
an evacuated 1.00 liter bulb and the temperature is raised to 375K.
(a) What would be the pressure in atmospheres in the bulb if no dissociation of the
SO2Cl2(g) occured?
(b) When the system has come to equilibrium at 375K, the total pressure in the bulb is
found to be 1.43 atmospheres. Calculate the partial pressures of SO2, Cl2, and SO2Cl2
at equilibrium at 375K.
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(c) Give the expression for the equilibrium constant (either Kp or Kc) for the
decomposition of SO2Cl2 at 375K.
(d) If the temperature were raised to 500K, what effect would this have on the
equilibrium constant? Explain briefly.
2. Ammonium hydrogen sulfide is a crystalline solid that decomposes as follows:
NH4HS(s)  NH3 (g) + H2S(g)
(a) Some solid NH4HS is placed in an evacuated vessel at 25oC. After equilibrium is
attained, the total pressure inside the vessel is found to be 0.659 atmospheres. Some
solid NH4HS remains in the vessel at equilibrium. For this decomposition, write the
expression for Kp and calculate its numeric value at 25oC.
(b) Some extra NH3 gas is injected into the vessel containing the sample described in
part (a). When equilibrium is reestablished at 25oC, the partial pressure of NH3 in the
vessel is twice the partial pressure of H2S. Calculate the numerical value of the partial
pressure of NH3 and the partial pressure of H2S in the vessel after the NH3 has been
added and the equilibrium has been reestablished.
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NH4HS(s)  NH3 (g) + H2S(g)
(c) In a different experiment, NH3 gas and H2S gas are introduced into an empty 1.00
liter vessel at 25oC. The initial partial pressure of each gas is 0.500 atmospheres.
Calculate the number of moles of solid NH4HS that is present when equilibrium is
established.
3. Given:
2NaHCO3(s) = Na2CO3(s) + H2O(g) + CO2(g)
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Solid sodium hydrogen carbonate, NaHCO3, decomposes on heating to the equation
above.
(a) A sample of 100. grams of solid NaHCO3 was placed in a previously evacuated
rigid 5.00 liter container and heated to 160oC. Some of the original solid remained
and the total pressure in the container was 7.76 atmospheres when rquilibrium was
reached. Calculate the number of moles of H2O(g) present at equilibrium.
(b) How many grams of the original solid remain in the cotainer under the conditions
described in (a)?
(c) Write the equilibrium expression for the equilibrium constant, Kp, and calculate its
value for the reaction under the conditions in (a).
(d) If 110. grams of solid NaHCO3 had been placed in the 5.00 liter container and
heated to 160oC, what would the total pressure have been at equilibrium? Explain.
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4. At elevated temperatures, SbCl5 gas decomposes into SbCl3 gas and Cl2 gas as shown
by the following equation:
SbCl5(g) = SbCl3(g) + Cl2(g)
(a) An 89.7 gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated
15.0 litre container at 182oC.
SbCl5(g) = SbCl3(g) + Cl2(g)
1. What is the concentration in moles per liter of SbCl5 in the container before
any decomposition occurs?
2. What is the pressure in atmospheres of SbCl5 in the container before any
decomposition occurs?
(b) If the SbCl5 is 29.2 percent decomposed when equilibrium is established at
182oC, calculate the value for either Kp or Kc, for this decomposition reaction.
Indicate whether you are calculating Kp or Kc.
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(c) In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first placed in an
empty 2.00 litre container maintained at a temperature different from 182oC. At
this temperature, Kc equals 0.117. How many moles of Cl2 must be added to this
container to reduce the number of moles of SbCl3 to 0.700 mole at equilibrium?
5. Given:
NH4HS(s) = NH3(g) + H2S(g)
Ho = 93 kJ
The equilibrium above is established by placing solid NH4HS in an evacuated
container at 25oC. At equilibrium, some solid NH4HS remains in the ontainer.
Predict and explaineach of the following.
(a) The effect on the equilibrium partial pressure of NH3 gas when additional solid
NH4HS is introduced into the container.
NH4HS(s) = NH3(g) + H2S(g)
Ho = 93 kJ
(b) The effect on the equilibrium partial pressure of NH3 gas when additional H2S gas
is introduced into the container.
(c) The effect on the mass of solid NH4HS present when the volume of the container
is decreased.
(d) The effect on the mass of solid NH4HS present when the temperature is increased.
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