14:
Chemical Chapter Equilibrium
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Lecture Outline
Mini-exam 1 results
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LON-CAPA
Lab
Equilibrium
Expectations for next lecture
2
Lab Experiment Introduction:
Chemical Equilibrium
Fe3+ +
SCN-
yellow
colorless
FeSCN2+
red
+
3
Fe3+ + SCN-
FeSCN2+
DeterminationofEquilibriumConstant
Keq =
[FeSCN2+]
[Fe3+][ SCN-]
Whataretheinitialconcentrations?
Howcanweexpresstheconcentrationsatthe
equilibrium?
Youwilldeterminetheequilibrium
concentrationofFeSCN+2Experimentally
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Fe3+ + SCN-
FeSCN2+
DeterminationofEquilibriumConstant
Keq =
[FeSCN2+]
[Fe3+][ SCN-]
[Fe3+]i, [SCN-]i
Whataretheinitialconcentrations?
only
Howcanweexpresstheconcentrationsatthe
equilibrium?
Youwilldeterminetheequilibrium
concentrationofFeSCN+2Experimentally
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Fe3+ + SCN-
[Fe3+]i - X
[SCN-]i - X
FeSCN2+
X
DeterminationofEquilibriumConstant
Keq =
[FeSCN2+]
[Fe3+][ SCN-]
Whataretheinitialconcentrations?
Howcanweexpresstheconcentrationsatthe
equilibrium?
Youwilldeterminetheequilibrium
concentrationofFeSCN2+Experimentally
X = [FeSCN2+]
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Fe3+ + SCN-
FeSCN2+
Wewillneedto:
Determineoperating .
EstablishBeerslawplot(Calibrationcurve).
Determineeq.Conc.FeSCN+2fromplot.
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Use of the Calibration Plot
Measured Absorbance of
an Equilibrium Solution
o
A
o
Calibration
Points
o
Ferricyanate Ion
Concentration in
the Solution
o
o
[FeSCN2+]
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Safety Concerns:
SafetyGogglesarerequiredtobewornatalltimes
whileyouareinthelabroomeven,whencompleting
calculationsandwritingupthelab.
AllsolutionscontainHNO3.
NitricAcidiscorrosivetotheskin,eyes,and
clothing.Wearyoursafetygoggles.Immediately
flushanyareaofskinorclothingthatmayhavecome
incontactwiththesesolutions.
FerricNitratemaystainsomeclothing.
AllFeSCN2+wasteproducedinthisexerciseshould
bediscardedinappropriatewastecontainerslocated
inhoods.
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Chemical Equilibrium
When compounds react, they eventually form a
mixture of products and unreacted reactants, in a
dynamic equilibrium.
A dynamic equilibrium consists of a forward
reaction, in which substances react to give products,
and a reverse reaction, in which products react to give
the original reactants.
Chemical equilibrium is the state reached by a
reaction mixture when the rates of the forward
and reverse reactions have become equal.
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Chemical Equilibrium
reactants
Much like water in a U-shaped tube, there is
constant mixing back and forth through the
lower portion of the tube.
products
Its as if the forward and
reverse reactions are
occurring at the same rate.
The system appears to be
static (stationary) when, in
reality, it is dynamic (in
constant motion).
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Chemical Equilibrium
For example, the Haber process for
producing ammonia from N2 and H2 does
not go to completion.
N 2 ( g ) + 3H 2 ( g )
2NH 3 (g)
It establishes an equilibrium state where all
three species are present.
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A Problem to Consider
Applying Stoichiometry to an Equilibrium
Mixture.
Suppose we place 1.000 mol N2 and 3.000 mol H2
in a reaction vessel at 450 oC and 10.0
atmospheres of pressure. The reaction is
N 2 ( g ) + 3H 2 ( g )
2NH 3 (g)
What is the composition of the equilibrium mixture
if it contains 0.080 mol NH3?
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A Problem to Consider
Using the information given, set up the following
table.
N 2 ( g ) + 3H 2 ( g )
Starting
1.000
3.000
Change
-x
-3x
Equilibrium 1.000 - x 3.000 - 3x
2NH 3 (g)
0
+2x
2x = 0.080 mol
The equilibrium amount of NH3 was given as
0.080 mol. Therefore, 2x = 0.080 mol NH3 (x =
0.040 mol).
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A Problem to Consider
Using the information given, set up the following
table.
N 2 ( g ) + 3H 2 ( g )
Starting
1.000
3.000
Change
-x
-3x
Equilibrium 1.000 - x 3.000 - 3x
2NH 3 (g)
0
+2x
2x = 0.080 mol
Equilibrium amount of N2 = 1.000 - 0.040 = 0.960 mol N2
Equilibrium amount of H2 = 3.000 - (3 x 0.040) = 2.880 mol H2
Equilibrium amount of NH3 = 2x = 0.080 mol NH3
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The Equilibrium Constant
Every reversible system has its own
position of equilibrium under any given
set of conditions.
The ratio of products produced to unreacted
reactants for any given reversible reaction
remains constant under constant conditions of
pressure and temperature.
The numerical value of this ratio is called the
equilibrium constant for the given reaction.
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The Equilibrium Constant
The equilibrium-constant expression for a reaction is
obtained by multiplying the concentrations of products,
dividing by the concentrations of reactants, and raising
each concentration to a power equal to its coefficient in
the balanced chemical equation.
aA + bB
cC + dD
For the general equation above, the
equilibrium-constant expression would
be:
c
d
[C] [D]
Kc =
a
b
[ A ] [B ]
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The Equilibrium Constant
The equilibrium-constant expression for a reaction is
obtained by multiplying the concentrations of products,
dividing by the concentrations of reactants, and raising
each concentration to a power equal to its coefficient in
the balanced chemical equation.
aA + bB
cC + dD
The molar concentration of a substance
is denoted by writing its formula in
square brackets.
c
d
[C] [D]
Kc =
a
b
[ A ] [B ]
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The Equilibrium Constant
The equilibrium constant, Kc, is the value
obtained for the equilibrium-constant
expression when equilibrium
concentrations are substituted.
A large Kc indicates large concentrations of products
at equilibrium.
A small Kc indicates large concentrations of
unreacted reactants at equilibrium.
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The Equilibrium Constant
The law of mass action states that the value of
the equilibrium constant expression Kc is constant
for a particular reaction at a given temperature,
whatever equilibrium concentrations are
substituted.
Consider the equilibrium established in the Haber
process.
N 2 ( g ) + 3H 2 ( g )
2NH 3 (g)
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The Equilibrium Constant
The equilibrium-constant expression would
2
be
[NH ]
Kc =
3
3
[N 2 ][H 2 ]
Note that the stoichiometric coefficients in the
balanced equation have become the powers to
which the concentrations are raised.
N 2 ( g ) + 3H 2 ( g )
2NH 3 (g)
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Equilibrium: A Kinetics
Argument
If the forward and reverse reaction rates in a
system at equilibrium are equal, then it follows
that their rate laws would be equal.
Consider the decomposition of N2O4, dinitrogen
tetroxide.
N 2O 4 ( g )
2NO 2 (g)
If we start with some dinitrogen tetroxide and heat
it, it begins to decompose to produce NO2.
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Equilibrium: A Kinetics
Argument
If the forward and reverse reaction rates in a
system at equilibrium are equal, then it follows
that their rate laws would be equal.
Consider the decomposition of N2O4, dinitrogen
tetroxide.
N 2O 4 ( g )
2NO 2 (g)
However, once some NO2 is produced it can recombine
to form N2O4.
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Equilibrium: Kinetics
Argument
k
N A 2O 4 ( g )
f
kr
2NO 2 (g)
Call the decomposition of N2O4 the forward reaction
and the formation of N2O4 the reverse reaction.
These are elementary reactions, and you can
immediately write the rate law for each.
Rate(forward) = k f [N 2O 4 ]
2
Rate(reverse) = k r [NO 2 ]
Here kf and kr represent
the forward and reverse
rate constants.
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Equilibrium: A Kinetics
Argument
N 2O 4 ( g )
kf
kr
2NO 2 (g)
Ultimately, this reaction reaches an
equilibrium state where the rates of the
forward and reverse reactions are equal.
Therefore,
2
k f [N 2O 4 ] = k r [NO 2 ]
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Equilibrium: A Kinetics
Argument
N 2O 4 ( g )
kf
kr
2NO 2 (g)
Combining the constants you can identify the
equilibrium constant, Kc, as the ratio of the
forward and reverse rate constants.
2
kf
[NO 2 ]
Kc =
=
kr
[ N 2O 4 ]
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Obtaining Equilibrium
Constants for Reactions
Equilibrium concentrations for a reaction
must be obtained experimentally and then
substituted into the equilibrium-constant
expression in order to calculate Kc.
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Obtaining Equilibrium
Constants for Reactions
Consider the reaction below:
CO(g ) + 3 H 2 (g)
CH 4 (g) + H 2O(g)
Suppose we started with initial concentrations of
CO and H2 of 0.100 M and 0.300 M, respectively.
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Obtaining Equilibrium
Constants for Reactions
Consider the reaction below:
CO(g ) + 3 H 2 (g)
CH 4 (g) + H 2O(g)
When the system finally settled into equilibrium we
determined the equilibrium concentrations to be as
follows.
Reactants
[CO] = 0.0613 M
[H2] = 0.1893 M
Products
[CH4] = 0.0387 M
[H2O] = 0.0387 M
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Obtaining Equilibrium
Constants for Reactions
Consider the reaction below:
CO(g ) + 3 H 2 (g)
CH 4 (g) + H 2O(g)
The equilibrium-constant expression for this
reaction is:
[CH 4 ][H 2O]
Kc =
3
[CO][H 2 ]
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Obtaining Equilibrium
Constants for Reactions
Consider the reaction below:
CO(g ) + 3 H 2 (g)
CH 4 (g) + H 2O(g)
If we substitute the equilibrium concentrations,
we obtain:
(0.0387 M )(0.0387 M )
Kc =
= 3.93
3
(0.0613M )(0.1893M )
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Obtaining Equilibrium
Constants for Reactions
Consider the reaction below:
CO(g ) + 3 H 2 (g)
CH 4 (g) + H 2O(g)
Regardless of the initial concentrations (whether
considering reactants or products), the law of
mass action dictates that the reaction will always
settle into an equilibrium where the equilibriumconstant expression equals Kc.
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Obtaining Equilibrium
Constants for Reactions
Consider the reaction below:
CO(g ) + 3 H 2 (g)
CH 4 (g) + H 2O(g)
As an example, lets repeat the previous
experiment, only this time starting with initial
concentrations of products:
[CH4]initial = 0.1000 M and [H2O]initial = 0.1000 M
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Obtaining Equilibrium
Constants for Reactions
Consider the reaction below:
CO(g ) + 3 H 2 (g)
CH 4 (g) + H 2O(g)
We find that these initial concentrations result in
the following equilibrium concentrations.
Reactants
Products
[CO] = 0.0613 M
[CH4] = 0.0387 M
[H2] = 0.1893 M
[H2O] = 0.0387 M
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Obtaining Equilibrium
Constants for Reactions
Consider the reaction below:
CO(g ) + 3 H 2 (g)
CH 4 (g) + H 2O(g)
Substituting these values into the equilibriumconstant expression, we obtain the same result.
(0.0387 M )(0.0387 M )
Kc =
= 3.93
3
(0.0613M )(0.1893M )
Whether we start with reactants or products, the
system establishes the same ratio.
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The Equilibrium Constant, Kp
In discussing gas-phase equilibria, it is often more
convenient to express concentrations in terms of
partial pressures rather than molarities
It can be seen from the ideal gas equation that the
partial pressure of a gas is proportional to its
molarity.
n
P = ( )RT = MRT
V
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The Equilibrium Constant, Kp
If we express a gas-phase equilibria in
terms of partial pressures, we obtain Kp.
Consider the reaction below.
CO(g ) + 3 H 2 (g)
CH 4 (g) + H 2O(g)
The equilibrium-constant expression in terms of
partial pressures becomes:
Kp =
PCH PH O
4
PCO PH
2
3
2
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The Equilibrium Constant, Kp
In general, the numerical value of Kp
differs from that of Kc.
From the relationship n/V=P/RT, we can show that
K p = K c ( RT)
n
where n is the sum of the moles of gaseous
products in a reaction minus the sum of the moles
of gaseous reactants.
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A Problem to Consider
Consider the reaction
2SO 2 (g ) + O 2 (g )
2 SO 3 (g)
Kc for the reaction is 2.8 x 102 at 1000 K. Calculate
Kp for the reaction at this temperature.
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A Problem to Consider
Consider the reaction
2SO 2 (g ) + O 2 (g )
2 SO 3 (g)
We know that
K p = K c ( RT)
n
From the equation we see that Dn = -1. We can
simply substitute the given reaction temperature and
the value of R (0.08206 L.atm/mol.K) to obtain Kp.
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A Problem to Consider
Consider the reaction
2SO 2 (g ) + O 2 (g )
Since
K p = 2.8 10
2 SO 3 (g)
K p = K c ( RT)
2
Latm
(0.08206 molK
n
-1
1000 K) = 3.4
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Equilibrium Constant for the
Sum of Reactions
Similar to the method of combining reactions that
we saw using Hesss law in Chapter 6, we can
combine equilibrium reactions whose Kc values
are known to obtain Kc for the overall reaction.
With Hesss law, when we reversed reactions or
multiplied them prior to adding them together, we had
to manipulate the H values to reflect what we had
done.
The rules are a bit different for manipulating Kc.
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Equilibrium Constant for the
Sum of Reactions
1.
If you reverse a reaction, invert the value of Kc.
2.
If you multiply each of the coefficients in an equation by
the same factor (2, 3, ), raise Kc to the same power
(2, 3, ).
If you divide each coefficient in an equation by the same
factor (2, 3, ), take the corresponding root of Kc
(i.e., square root, cube root, ).
When you finally combine (that is, add) the individual
equations together, take the product of the equilibrium
constants to obtain the overall Kc.
3.
4.
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Equilibrium Constant for the
Sum of Reactions
For example, nitrogen and oxygen can combine
to form either NO(g) or N2O (g) according to the
following equilibria.
Kc = 4.1 x 10-31
N 2 (g ) + O 2 (g )
2 NO(g)
(1)
(2)
N 2 (g ) + 1 O 2 (g )
2
N 2O(g)
Kc = 2.4 x 10-18
Using these two equations, we can obtain Kc for
the formation of NO(g) from N2O(g):
(3)
N 2 O( g ) + 1 O 2 ( g )
2
2 NO(g)
Kc = ?
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(1)
(2)
Equilibrium Constant for the
Sum of Reactions
To combine equations (1) and (2) to obtain
equation (3), we must first reverse equation (2).
When we do that, we must also take the
reciprocal of its Kc value.
Kc = 4.1 x 10-31
N 2 (g ) + O 2 (g )
2 NO(g)
N 2O(g)
N 2 (g) +
1 O (g )
22
Kc =
1
2.4 10-18
N 2 O( g ) + 1 O 2 ( g )
2
2 NO(g)
1
31
K c (overall ) = (4.1 10 ) (
) = 1.7 1013
2.4 1018
(3)
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