Ch. 13: Chemical Equilibrium

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Ch. 13 Chemical Equilibrium

• 13.1 The Equilibrium Condition

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Equilibrium

• dynamic equilibrium
• may seem like no changes are occurring but there
  are changes
• no NET changes
• When did this reaction reach it?

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reactants
products
H2O(g) CO(g) ? H2(g) CO2(g)
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• At equilibrium, forward and reverse reaction
  rates are ____________

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Equilibrium

• equilibrium position of a reaction is determined
  by
• initial _________________
• ____________ of reactants and products
• degree of __________________ of reactants and
  products
• GOAL
• _
• _

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Equilibrium Position

• lies to the left
• more _________
• less ___________
• lies to the right
• less __________
• more _________

• If reactants are mixed and concentrations do not
  change
• could already be at equilibrium
• reaction rates are so ________ that change is too
  difficult to detect

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Ch. 13 Chemical Equilibrium

• 13.2 Equilibrium Constant

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Law of Mass Action

• created in 1864 by Guldberg and Waage (Norweigen)
• For a reaction jA kB ? lC mD
• equilibrium constant K

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Law of Mass Action

• because
• Rateforward
• Ratereverse
• so if Ratef Rater
• kfAjBk krClDm

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Example

• Write the equilibrium expression for
• 4NH3(g) 7O2(g) ? 4NO2(g) 6H2O(g)
• What would it be for the reverse reaction?

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Equilibrium Constant

• will always have the same value at a certain
  temperature
• no matter what amounts are added
• ratio at equilibrium will always be same

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Equilibrium Position

• each set of equilibrium concentrations
• depends on initial concentrations

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Ch. 13 Chemical Equilibrium

• 13.3 Equilibrium Expressions with Pressure

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Equilibrium with Gases

• equilibria involving gases can be described using
  __________ instead of _______________
• N2(g) 3H2(g) ? 2NH3(g)

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Equilibrium with Gases

• N2(g) 3H2(g) ? 2NH3(g)

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Calculating K from KP

• KP KC RT can cancel out if total of
  coefficients are same on each side
• where ?n is the difference in moles of gas on
  either side of the equation
• ?n (lm) (jk)
• N2(g) 3H2(g) ? 2NH3(g)
• ?n

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Example

• Setup the expression for KP in terms of KC, R and
  T
• 2NO(g) Cl2(g) ? 2NOCl(g)

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Ch. 13 Chemical Equilibrium

• 13.4 Heterogeneous Equilibria

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Heterogeneous Equilibria

• involve more than one phase
• position of heterogeneous equilibria does NOT
  depend on amounts of
• _
• _
• because their concentrations stay constant (since
  they are PURE)

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Heterogeneous Equilibria

• do not include liquids or solids in equilibrium
  expression
• only include ________ and ______________

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Example 1

• 2H2O(l) ? 2H2(g) O2(g)
• 2H2O(g) ? 2H2(g) O2(g)

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Ch. 13 Chemical Equilibrium

• 13.5/6 Applications of Equilibrium Constant (K)

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Equilibrium Constant

• if we know the value of K, we can predict
• tendency of a reaction to occur
• if a set of concentrations could be at
  equilibrium
• equilibrium position, given initial
  concentrations

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Equilibrium Constant

• If you start a reaction with only reactants
• concentration of reactants will decrease by a
  certain amount
• concentration of products will increase by a same
  amount

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Example 2

• The following reaction has a K of 16. You are
  starting reaction with 9 O3 molecules and 12 CO
  molecules.
• Find the amount of each species at equilibrium.
• O3(g) CO(g) ? CO2(g) O2(g)

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Example 2
O3(g) CO(g) ? O2(g) CO2(g) O3(g) CO(g) ? O2(g) CO2(g) O3(g) CO(g) ? O2(g) CO2(g) O3(g) CO(g) ? O2(g) CO2(g) O3(g) CO(g) ? O2(g) CO2(g) O3(g) CO(g) ? O2(g) CO2(g)
Initial I
Change C
Equilibrium E
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Example 2
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Example 2
O3(g) CO(g) ? O2(g) CO2(g) O3(g) CO(g) ? O2(g) CO2(g) O3(g) CO(g) ? O2(g) CO2(g) O3(g) CO(g) ? O2(g) CO2(g) O3(g) CO(g) ? O2(g) CO2(g)
I 9 12 0 0
C -x -x x x
E
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Extent of a Reaction

• If _________
• mostly products
• goes essentially to completion
• lies far to right
• If _________
• mostly reactants
• reaction is negligible
• lies far to left

• size of K and time needed to reach equilibrium
  are NOT related
• time required is determined by reaction rate (Ea)

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Reaction Quotient

• Q equal to equilibrium expression but
  ___________ have to be at equilibrium
• used to tell if a reaction is at equilibrium or
  not
• relationship between Q and K tells which way the
  reaction will shift
• _______ at equilibrium, no shift
• _______ too large, forms reactants, shift to
  left
• ______ too small, forms products, shift to right

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Example 3

• For the synthesis of ammonia at 500C, the
  equilibrium constant is 6.0 x 10-2. Predict the
  direction the system will shift to reach
  equilibrium in the following case
• N2(g) 3H2(g) ? 2NH3(g)

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Example 3

• NH30 1.0x10-3 M,
• N201.0x10-5 M
• H202.0x10-3 M
• Q __ K so forms _________, shifts to _____

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Example 4

• In the gas phase, dinitrogen tetroxide decomposes
  to gaseous nitrogen dioxide
• N2O4(g) ? 2NO2(g)
• Consider an experiment in which gaseous N2O4 was
  placed in a flask and allowed to reach
  equilibrium at a T where KP 0.133. At
  equilibrium, the pressure of N2O4 was found to be
  2.71 atm.
• Calculate the equilibrium pressure of NO2.

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Example 4
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Example 5

• At a certain temperature a 1.00 L flask initially
  contained 0.298 mol PCl3(g) and 8.70x10-3 mol
  PCl5(g). After the system had reached
  equilibrium, 2.00x10-3 mol Cl2(g) was found in
  the flask.
• PCl5(g) ? PCl3(g) Cl2(g)
• Calculate the equilibrium concentrations of all
  the species and the value of K.

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Example 5
PCl5(g) ? PCl3(g) Cl2(g) PCl5(g) ? PCl3(g) Cl2(g) PCl5(g) ? PCl3(g) Cl2(g) PCl5(g) ? PCl3(g) Cl2(g)
I
C
E
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Approximations

• If K is very small, we can assume that the change
  (x) is going to be negligible
• can be used to cancel out when adding or
  subtracting from a normal sized number
• to simplify algebra

0
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Example 6

• At 35C, K1.6x10-5 for the reaction
• 2NOCl(g) ? 2NO(g) Cl2(g)
• Calculate the concentration of all species at
  equilibrium for the following mixtures
• 2.0 mol NOCl in 2.0 L flask
• 1.0 mol NOCl and 1.0 mol NO in 1.0 L flask
• 2.0 mol NOCl and 1.0 mol Cl2 in 1.0 L flask

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Example 6

• 2.0 mol NOCl in 2.0 L flask
• NOCl
• NO Cl2

2NOCl(g) ? 2NO(g) Cl2(g) 2NOCl(g) ? 2NO(g) Cl2(g) 2NOCl(g) ? 2NO(g) Cl2(g) 2NOCl(g) ? 2NO(g) Cl2(g)
I
C
E
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Example 6

• 1.0 mol NOCl and 1.0 mol NO in 1.0 L flask
• NOCl
• NO Cl2

2NOCl(g) ? 2NO(g) Cl2(g) 2NOCl(g) ? 2NO(g) Cl2(g) 2NOCl(g) ? 2NO(g) Cl2(g) 2NOCl(g) ? 2NO(g) Cl2(g)
I
C
E
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Example 6

• 2.0 mol NOCl and 1.0 mol Cl2 in 1.0 L flask
• NOCl
• Cl2 NO

2NOCl(g) ? 2NO(g) Cl2(g) 2NOCl(g) ? 2NO(g) Cl2(g) 2NOCl(g) ? 2NO(g) Cl2(g) 2NOCl(g) ? 2NO(g) Cl2(g)
I
C
E
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Ch. 13 Chemical Equilibrium

• 13.7 Le Chatliers Principle

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Le Châtliers Principle

• can predict how certain changes in a reaction
  will affect the position of equilibrium

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Changing Concentration

• system will shift away from the added component
  or towards a removed component
• Ex N2 3H2 ? 2NH3
• if more N2 is added, then equilibrium position
  shifts to right
• if some NH3 is removed, then equilibrium position
  shifts to right

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Change in Pressure

• adding or removing gaseous reactant or product is
  same as changing conc.
• adding inert or uninvolved gas
• increase the ___________________
• ___________effect the equilibrium position

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Change in Pressure

• changing the volume
• decrease V
• decrease in gas molecules
• shifts towards the side of the reaction with
  _____ gas molecules
• increase V
• increase in of gas molecules
• shifts towards the side of the reaction with
  _____ gas molecules

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Change in Temperature

• all other changes alter the concentration at
  equilibrium position but dont actually change
  value of K
• value of K does change with temperature

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Change in Temperature

• if energy is added, the reaction will shift in
  direction that consumes energy
• treat energy as a
• __________ for endothermic reactions
• __________ for exothermic reactions

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As4O6(s) 6C(s) ? As4(g) 6CO(g)

• add CO
• add C
• remove C
• add As4O6

• remove As4O6
• remove As4
• decrease volume
• add Ne gas

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P4(s) 6Cl2(g) ? 4PCl3(l)

• decrease volume
• increase volume
• add P4

• remove Cl2
• add Kr gas
• add PCl3

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energy N2(g) O2(g) ? 2NO(g)

• endo or exo?
• increase temp

• increase volume
• decrease temp

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N2(g) 3H2(g) ? 2NH3(g)
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