The Concept of Equilibrium

1
The Concept of Equilibrium

• As the reaction progresses
• A decreases to a constant,
• B increases from zero to a constant.
• When A and B are constant, equilibrium is
  achieved.

2
Start with NO2
Start with N2O4
Start with NO2 N2O4
14.1
3
14.1
4
The Equilibrium Constant

• No matter the starting composition of reactants
  and products, the same ratio of concentrations is
  achieved at equilibrium.
• For a general reaction
• the equilibrium constant expression is
• where Kc is the equilibrium constant.

5
4.63 x 10-3
Law of Mass Action
Equilibrium Will
K gtgt 1
Lie to the right
Favor products
K ltlt 1
Lie to the left
Favor reactants
14.1
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Homogenous equilibrium applies to reactions in
which all reacting species are in the same phase.
Kp Kc(RT)Dn
Dn moles of gaseous products moles of gaseous
reactants
(c d) (a b)
14.2
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Homogeneous Equilibrium
H2O constant
Kc
14.2
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The Equilibrium Expression

• Write the equilibrium expression for the
  following reaction

9
The equilibrium concentrations for the reaction
between carbon monoxide and molecular chlorine to
form COCl2 (g) at 740C are CO 0.012 M, Cl2
0.054 M, and COCl2 0.14 M. Calculate the
equilibrium constants Kc and Kp.
Kc
220
Kp Kc(RT)Dn
Dn 1 2 -1
R 0.0821
T 273 74 347 K
Kp 220 x (0.0821 x 347)-1 7.7
14.2
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The equilibrium constant Kp for the reaction is
158 at 1000K. What is the equilibrium pressure
of O2 if the PNO 0.400 atm and PNO 0.270 atm?
2
347 atm
14.2
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Heterogenous equilibrium applies to reactions in
which reactants and products are in different
phases.
CaCO3 constant CaO constant
Kc CO2
The concentration of solids and pure liquids are
not included in the expression for the
equilibrium constant.
14.2
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Kp
14.2
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Writing Equilibrium Constant Expressions

• The concentrations of the reacting species in the
  condensed phase are expressed in M. In the
  gaseous phase, the concentrations can be
  expressed in M or in atm.
• The concentrations of pure solids, pure liquids
  and solvents do not appear in the equilibrium
  constant expressions.
• The equilibrium constant is a dimensionless
  quantity.
• In quoting a value for the equilibrium constant,
  you must specify the balanced equation and the
  temperature.
• If a reaction can be expressed as a sum of two or
  more reactions, the equilibrium constant for the
  overall reaction is given by the product of the
  equilibrium constants of the individual reactions.

14.2
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Calculating Equilibrium Concentrations

1. Express the equilibrium concentrations of all
   species in terms of the initial concentrations
   and a single unknown x, which represents the
   change in concentration.
2. Write the equilibrium constant expression in
   terms of the equilibrium concentrations. Knowing
   the value of the equilibrium constant, solve for
   x.
3. Having solved for x, calculate the equilibrium
   concentrations of all species.

14.4
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At 12800C the equilibrium constant (Kc) for the
reaction Is 1.1 x 10-3. If the initial
concentrations are Br2 0.063 M and Br
0.012 M, calculate the concentrations of these
species at equilibrium.
Let x be the change in concentration of Br2
Initial (M)
0.063
0.012
ICE
Change (M)
-x
2x
Equilibrium (M)
0.063 - x
0.012 2x
Solve for x
14.4
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4x2 0.048x 0.000144 0.0000693 0.0011x
4x2 0.0491x 0.0000747 0
ax2 bx c 0
x -0.00178
x -0.0105
At equilibrium, Br 0.012 2x -0.009 M
or 0.00844 M
At equilibrium, Br2 0.062 x 0.0648 M
14.4
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Example Problem Calculate Concentration
Note the moles into a 10.32 L vessel stuff ...
calculate molarity. Starting concentration of HI
2.5 mol/10.32 L 0.242 M
2 HI H2 I2
0.242 M 0 0
Initial Change Equil
-2x x x
0.242-2x x x
What we are asked for here is the equilibrium
concentration of H2 ... ... otherwise known as
x. So, we need to solve this beast for x.
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Example Problem Calculate Concentration
And yes, its a quadratic equation. Doing a bit
of rearranging
x 0.00802 or 0.00925 Since we are using this
to model a real, physical system, we reject the
negative root. The H2 at equil. is 0.00802 M.
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Example Problem Calculate Keq
This type of problem is typically tackled using
the three line approach 2 NO O2 2 NO2

Initial
Change
Equilibrium
20
Approximating

• If Keq is really small the reaction will not
  proceed to the right very far, meaning the
  equilibrium concentrations will be nearly the
  same as the initial concentrations of your
  reactants.
• 0.20 x is just about 0.20 is x is really
  dinky.
• If the difference between Keq and initial
  concentrations is around 3 orders of magnitude or
  more, go for it. Otherwise, you have to use the
  quadratic.

21
Example
Initial Concentration of I2 0.50 mol/2.5L 0.20
M I2 2 I
More than 3 orders of mag. between these numbers.
The simplification will work here.
0.20 0 -x 2x 0.20-x 2x
Initial change equil
With an equilibrium constant that small, whatever
x is, its near dink, and 0.20 minus dink is 0.20
(like a million dollars minus a nickel is still
a million dollars). 0.20 x is the same as 0.20
x 3.83 x 10-6 M
22
Example
Initial Concentration of I2 0.50 mol/2.5L 0.20
M I2 2 I
These are too close to each other ... 0.20-x
will not be trivially close to 0.20 here.
0.20 0 -x 2x 0.20-x 2x
Initial change equil
Looks like this one has to proceed through the
quadratic ...
23
Kc
14.2
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216
14.2
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The reaction quotient (Qc) is calculated by
substituting the initial concentrations of the
reactants and products into the equilibrium
constant (Kc) expression.

• IF
• Qc gt Kc system proceeds from right to left to
  reach equilibrium
• Qc Kc the system is at equilibrium
• Qc lt Kc system proceeds from left to right to
  reach equilibrium

14.4
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Example
27
Chemistry In Action The Haber Process
28
Le Châteliers Principle
Change
Shift Equilibrium
Concentration
yes
no
Pressure
yes
no
Volume
yes
no
Temperature
yes
yes
Catalyst
no
no
14.5