Chemistry 102(01) Spring 2002
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Instructor: Dr. Upali Siriwardane
e-mail:upali@chem.latech
Office: CTH 311 Phone 257-4941
Office Hours:
8:00-9:00 & 11:00-12:00 a.m., M, W,
8:00-10:00 a.m., Tu,Th, F
March 27,
April 26,
May 15,
May 16,
2002 (Test 1): Chapter 12 &13.
2002(Test 2): Chapter 14 & 15.
2002 (Test 3): Chapter 17 & 18.
2002 (Comprehensive Test):
Chapters 12,13,14,15,17,18
Chapter 14.Chemical Equilibrium
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Introduction to Chemical Equilibrium
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Equilibrium Constants and Expressions
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Calculations Involving Equilibrium
Constants
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Using Le Châtelier’s Principle
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Some Important Equilibria
Dynamic Equilibrium
Equilibrium
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A state where the forward and reverse
conditions occur at the same rate.
I’m in static
equilibrium.
Dynamic
Equilibrium
Chemical Equilibrium
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Branch of chemistry dealing with reactions
where reactants and products coexist in a
dynamic equilibrium
the rates of forward and backward reactions
have comparable rates reaction
Chemical equilibrium
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Different types of arrows are used in chemical
equations associated with equilibria.
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Single arrow
Assumes that the reaction proceeds to completion as
written.
Two single-headed arrows
Used to indicate a system in equilibrium.
Two single-headed arrows of different sizes.
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May be used to indicate when one side of an
equilibrium system is favored.
Chemical Equilibrium
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Equilibrium region.
A point is finally reached where the forward and
reverse reactions occur at the same rate.
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H2 + I2
2HI
There is no net change in the concentration of
any of the species.
Partial Pressure
Chemical Equilibrium
Kinetic
Region
Equilibrium
Region
HI
I2
H2
Time
Equilibrium
Rate
A point is ultimately
reached where the
rates of the forward
and reverse changes
are the same.
At this point,
equilibrium
is reached.
Time
Equilibrium
Equilibrium
Region
Concentration
Kinetic
Region
Time
Terminology
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Initial concentration:
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concentration (M) of reactants and
products before the equilibrium is
reached.
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Equilibrium Concentration
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Concentration (M) of reactants and
products After the equilibrium is
reached.
Law of mass Action
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Defines an equilibrium constant (K) for the
process
j A + k B <==> l C + m D
[C]l[D]m
K = ----------------- ; [A], [B] etc are
[A]j[B]k
Equilibrium
concentrations
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Pure liquid or solid concentrations are not
written in the expression.
Writing an equilibrium expression
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Write a balanced equation for the
equilibrium.
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Put the products in the numerator and
the reactants in the denominator.
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Omit pure solids and liquids from the
expression
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Omit solvents if your solutes are dilute
(<0.1M).
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The exponent of each concentration
should be the same as the coefficient
for the species in the equation.
Equilibrium Expression
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An equilibrium expression could be
written for any reaction
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E.g. 2H2 (g) + O2 (g) ---> 2H2O( g)
[H2O]2
K = --------- = a is equal to (infinity)
[H2]2[O2]
1
K = --------- = a is equal to (infinity)
[H2]2[O2]
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Reaction profile
Potential
Energy
This type of plot
shows the energy
changes during
a reaction.
H
activation
energy
Reaction coordinate
Value of K
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k+
K = ---------------------------------- = --rate of forward Reaction
rate of backward Reaction
K-
K = a (infinity) -> Irreversible reactions
K = 0 -> No reaction
K = between 0 and 1 -> Equilibrium
reactions
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E.g. Formation of NH3 gas.
N2(g) + 3H2(g) <===> 2NH3(g)
[NH3]2
K = ------------ = 6.02 x 10-2 L2/mole2 at 127oC
[N2][H2]3
Types of Equilibria
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Homogenous equilibrium: Chemical
equilibrium where reactants and products are
in same phase.
Heterogeneous equilibrium: Chemical
Equilibrium where at least one phase of a
reactant or product is different from the rest.
What is K (Kc) and Kp
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Kc (K) - equilibrium constant calculated
based on [A]-Concentrations.
Kp- equilibrium constant calculated
based on partial pressure
Kp = K(RT) n
R = universal gas constant
T = Kelvin Temperature,
n = (sum of stoichiometric coefficients
of gaseous products) - (sum of the
stoichiometric coefficients of gaseous
reactants)
Partial pressure & Equilibrium
Constants
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For the following equilibrium, Kc = 1.10 x 107 at
700. oC. What is the Kp?
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2H2 (g) + S2 (g)
2H2S (g)
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Kp
= Kc (RT)ng
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T
= 700 + 273 = 973 K
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R
= 0.08206
atm L
ng = ( 2 ) - ( 2 + 1) = mol
-1 K
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Partial pressure & Equilibrium
Constants
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Kp = Kc (RT)ng
= 1.10 x
107
[ (0.08206
= 1.378 x105
atm L
mol K
) (973 K)
]
-1
Equilibrium Calculations
Hydrogen iodide, HI, decomposes according to
the equation
2 HI(g)
<====> H2(g) + I2(g)
When 4.00 mol of HI placed in a 5.00-L vessel
at 458ºC, the equilibrium mixture was found to
contain 0.442 mol I2. What is the value of Kc for
the reaction?
2 HI(g)
<====> H2(g) + I2(g)
Initial 4.00/5=.80
Change -2x
0
0
x
x
Eqilibrium 0.80-2x
x x=0.442/5
x = 0.0884
Equilibrium concentrations
[HI] = 0.80 - 2x = 0.8 - 2 x 0.0884 = 0.62
[H2] = x = 0.0884
[I2] = x = 0.0884
[H2] [I2]
0.0884 x 0.0884
Kc = ---------------- = ------------------------- = 0.0201
[HI]2
(0.62) 2
What is the reaction quotient, Q
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(Q) is constant in the equilibrium expression
when initial concentration of reactants and
products are used.
SO2(g)+ NO2(g) <===> NO(g) +SO3(g)
[NO][SO3]
Q = ---------------[SO2][NO2]
comparing to K and Q provide the net direction
to achieve equilibrium.
Equilibrium calculations
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We can predict the direction of a reaction by
calculating the reaction quotient.
Reaction quotient, Q
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For the reaction: aA + bB
[E]e [F]f
Q = [A]a [B]b
eE + fF
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Q has the same form as Kc with one important
difference. Q can be for any set of concentrations,
not just at equilibrium.
Reaction quotient
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Any set of concentrations can be given and a Q
calculated. By comparing Q to the Kc value, we
can predict the direction for the reaction.
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Q < Kc Net forward reaction will occur.
Q = Kc No change, at equilibrium.
Q > Kc Net reverse reaction will occur.
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Q Calculation
Consider the following reaction:
SO2(g) + NO2(g) <===> NO(g) + SO3(g)
(Kc = 85.0 at 460oC)
Given 0.040 mole of SO2(g), 0.500 mole of NO2(g),
0.30 mole of NO(g),and 0.020
mole of SO3(g) are mixed in a 5.00 L flask,
determine:
a) The net the reaction quotient, Q .
b) Direction to achieve equilibrium.
Equilibrium Calculation Example
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A sample of COCl2 is allowed to decompose.
The value of Kc for the equilibrium
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COCl2 (g)
CO (g) + Cl2 (g)
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is 2.2 x 10-10 at 100 oC.
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If the initial concentration of COCl2 is 0.095M,
what will be the equilibrium concentrations for
each of the species involved?
Equilibrium Calculation Example
COCl2 (g)
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CO (g)
Cl2 (g)
0.000
0.000
+X
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Initial conc., M 0.095
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+X
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Change
-X
in conc. due to reaction
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Equilibrium
X
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Concentration,
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Kc =
M(0.095 -X)
[ CO ] [ Cl2 ]
[ COCl2 ]
=
X
X2
(0.095 - X)
Equilibrium calculation example
2
X
Kc = 2.2 x 10-10 = (0.095 - X)
Rearrangement gives
X2 + 2.2 x 10-10 X - 2.09 x 10-11 = 0
This is a quadratic equation. Fortunately,
there is a straightforward equation for their
solution
Quadratic equations
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An equation of the form
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a X2 + b X + c = 0
Can be solved by using the following
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-b +
b2 - 4ac
2a
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x=
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Only the positive root is meaningful in
equilibrium problems.
Equilibrium Calculation Example
X2 + 2.2 x 10-10 X - 2.09 x 10-11 = 0
a
b
c
-b + b2 - 4ac
X=
2a
X=
- 2.2 x 10-10 + [(2.2 x 10-10)2 - (4)(1)(- 2.09 x 10-11)]1/2
2
X = 9.1 x 10-6 M
Equilibrium Calculation Example
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Now that we know X, we can solve for the
concentration of all of the species.
COCl2
CO
Cl2
= 0.095 - X
= X
= X
= 0.095 M
= 9.1 x 10-6 M
= 9.1 x 10-6 M
In this case, the change in the concentration of
is COCl2 negligible.
Le Chatelier's Principle
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One of the general principles applicable to
any equilibrium process. It simply states
that: 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.
For chemical equilibrium the changes could
be: Temperature and Pressure changes
Le Chatelier’s principle
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Any stress placed on an equilibrium system
will cause the system to shift to minimize the
effect of the stress.
You can put stress on a system by adding or
removing something from one side of a
reaction.
N2(g) + 3H2 (g)
2NH3 (g)
What effect will there be if you added more
ammonia? How about more nitrogen?
Predicting Shifts in Equilibria
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Equilibrium concentrations are based on:
– The specific equilibrium
– The starting concentrations
– Other factors such as:
• Temperature
• Pressure
• Reaction specific conditions
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Altering conditions will stress a system,
resulting in an equilibrium shift.
Changes in Concentration
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Changes in concentration do not change the
value of the equilibrium constant at constant
temperature.
When a material is added to a system in
equilibrium, the equilibrium will shift away from
that side of the equation.
When a material is removed from a system in
equilibrium, the equilibrium will shift towards
that sid of the equation.
Changes in pressure

In general, increasing the pressure by
decreasing volume shifts equilibria towards
the side that has the smaller number of
moles of gas.
Unaffected by pressure
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H2 (g) + I2 (g)
2HI (g)
Increased pressure, shift to left
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N2O2 (g)
2NO2 (g)
For the following equilibrium reactions:
H2(g) + CO2(g) <===> H2O(g) + CO(g)
H = 40 kJ
Predict the equilibrium shift if:
a) The temperature is increased
b) The pressure is decreased
At 100o C the equilibrium constant (K) for
the reaction:
H2(g) + I2(g) <===> 2HI(g)
is 1.15 x 102. If 0.400 moles of H2 and
0.400 moles of I2 are placed into a 12.0-liter
container and allowed to react at this
temperature, what is the HI concentration
(moles/liter) at equilibrium?
At a certain temperature the value of the
equilibrium constant is 3.24 for the
reaction:
H2(g) + CO2(g) <===> H2O(g) + CO(g)
If 0.400 mol H2 and 0.400 mol CO2 are
placed in a 1.00 L vessel, what is the
concentration of of CO at equilibrium?
Consider the following reaction:
SO2(g) + NO2(g) <===> NO(g) + SO3(g)
(Kc = 85.0 at 460oC)
Given 0.040 mole of SO2(g), 0.500 mole
of NO2(g), 0.30 mole of NO(g),and 0.020
mole of SO3(g) are mixed in a 5.00 L
flask, determine:
a) The net the reaction quotient, Q .
b) Direction to achieve equilibrium.
A 5.0-g sample of solid NH4Cl is heated in a
2.5-L container to 900oC. At equilibrium the
pressure of NH3(g) (reaction 2h in problem
2) is 0.60 atm. Calculate the equilibrium
constant, Kp, for this reaction.