Ch' 13 Equilibrium

1
Ch. 13 Equilibrium
2
Chemical Equilibrium

• The state where the concentrations of all
  reactants and products remain constant with time.
• On the molecular level, there is frantic
  activity. Equilibrium is not static, but is a
  highly dynamic situation.
• (BDVD)

3
Dynamic Equilibrium

• Reactions continue to take place.
• A B C D ( forward)
• C D A B (reverse)
• Initially there is only A and B so only the
  forward reaction is possible
• As C and D build up, the reverse reaction speeds
  up while the forward reaction slows down.
• Eventually the rates are equal.

4
Forward Reaction
Reaction Rate
Equilibrium
Reverse reaction
Time
5
What is equal at Equilibrium?

• Rates are equal.
• Concentrations are not.
• Rates are determined by concentrations and
  activation energy.
• The concentrations do not change at equilibrium.

6
13.2 Law of Mass Action

• jA kB lC mD
• j, k, l, m are coefficients
• The law of mass action is represented by the
  equilibrium expression
• K ClDm PRODUCTSpower
  AjBk REACTANTSpower
• K is called the equilibrium constant.
• is how we indicate a reversible
  reaction. x represents concentration.

7
Playing with K

• If we write the reaction in reverse.
• lC mD jA kB
• Then the new equilibrium constant is
• K AjBk 1/K ClDm

8
Playing with K

• If we multiply the equation by a constant
• njA nkB nlC nmD
• Then the equilibrium constant is
• K AnjBnk (A jBk)n Kn
  CnlDnm (ClDm)n

9
K is CONSTANT

• At any temperature.
• Temperature affects rate.
• The equilibrium concentrations dont have to be
  the same only K.
• Equilibrium position is a set of concentrations
  at equilibrium.
• One value at each temperature, but there are an
  unlimited number of possibilities.
• Usually written without units.

10
Equilibrium Expression

• 4NH3(g) 7O2(g) 4NO2(g) 6H2O(g)
• What is the equilibrium expression for this
  reaction?
• N2(g) O2(g) 2NO(g)

11
Calculate K

• N2 3H2 2NH3
• Initial At Equilibrium
• N20 1.000 M N2 0.921M
• H20 1.000 M H2 0.763M
• NH30 0 M NH3 0.157M

12
Calculate K

• N2 3H2 2NH3
• Initial At Equilibrium
• N20 0 M N2 0.399 M
• H20 0 M H2 1.197 M
• NH30 1.000 M NH3 0.203M
• K is the same no matter what the amount of
  starting materials (at the same temperature).

13
13.3 Equilibrium and Pressure

• Some reactions are gaseous
• PV nRT
• P (n/V)RT
• P CRT
• C is a concentration in moles/Liter
• C P/RT

14
Equilibrium and Pressure

• 2SO2(g) O2 (g) 2SO3 (g)
• In term of partial pressures
• Kp (PSO3)2 (PSO2)2 (PO2)
• In terms of concentration
• Kc SO32 SO22 O2

15
K v. Kp

• For
• jA kB lC mD
• Kp K(RT)?n
• ?n sum of coefficients of gaseous products
  minus sum of coefficients of gaseous reactants.

16
Homogeneous Equilibria

• So far every example dealt with reactants and
  products where all were in the same phase.
• We can use K in terms of either concentration or
  pressure.
• Units depend on reaction.

17
13.4 Heterogeneous Equilibria

• Are equilibria that involve more than one phase.
• If the reaction involves pure solids or pure
  liquids, the concentration of the solid or the
  liquid doesnt change.
• The position of a heterogeneous equilibrium does
  not depend on the amounts of pure solids or
  liquids present.
• As long as they are not used up we can leave them
  out of the equilibrium expression.

18
For Example

• H2(g) I2(s) 2HI(g)
• K HI2 H2I2
• But the concentration of I2 does not change,
  therefore
• K HI2 H2

19
13.5 Applying the Equilibrium Constant

• Reactions with large K (gtgt1), essentially to
  completion. Large negative ?E.
• Reactions with small K (ltlt1) consist mostly of
  reactants.
• Time to reach equilibrium is related to rate and
  AE. It is not related to size of K.

20
The Reaction Quotient

• Tells you the directing the reaction will go to
  reach equilibrium
• Calculated the same as the equilibrium constant,
  but for a system not at equilibrium by using
  initial concentrations.
• Q Productscoefficient Reactants
  coefficient
• Compare value to equilibrium constant

21
What Q tells us

• If QltK
• Not enough products
• Shift to right
• If QgtK
• Too many products
• Shift to left
• If QK system is at equilibrium

22
Example

• For the reaction
• 2NOCl(g) 2NO(g) Cl2(g)
• K 1.55 x 10-5 M at 35ºC
• In an experiment 0.10 mol NOCl, 0.0010 mol NO(g)
  and 0.00010 mol Cl2 are mixed in 2.0 L flask.
• Which direction will the reaction proceed to
  reach equilibrium?

23
13.6 Solving Equilibrium Problems

• Balance the equation.
• Write the equilibrium expression.
• List the initial concentrations.
• Calculate Q and determine the shift to
  equilibrium.
• Define equilibrium concentrations.
• Substitute equilibrium concentrations into
  equilibrium expression and solve.
• Check calculated concentrations by calculating K.

24
Intro to ICE

• H2 F2 2HF
• The Equilibrium constant for the above reaction
  is 115 at a certain temperature. 3.000 mol of
  each component was added to a 1.500 L flask.
  Calculate the equilibrium concentrations of all
  species.
• (46)

25
What if youre not given equilibrium
concentration?

• The size of K will determine what approach to
  take.
• First lets look at the case of a LARGE value of
  K ( gt100).
• Allows us to make simplifying assumptions.

26
Example

• H2(g) F2(g) 2HF(g)
• K 1.15 x 102 at 25ºC
• Calculate the equilibrium concentrations if a
  3.00 L container initially contains 3.00 mol of
  H2 and 6.00 mol F2 .
• H20 3.00 mol/3.00 L 1.00 M
• F20 6.00 mol/3.00 L 2.00 M
• HF0 0

27

• Q 0ltK so more product will be formed.
• Assumption since K is large reaction will go to
  completion.
• Stoichiometry tells us H2 is LR, it will be
  smallest at equilibrium let it be x
• Set up table of initial, change and equilibrium
  in concentrations.

28

H2(g) F2(g) 2HF(g)
Initial 1.00 M 2.00 M 0 M
Change Equilibrium

• For H2 and F2 the change must be -X
• Using to stoichiometry HI must be 2X
• Equilibrium initial change

29
H2(g) F2(g) 2HF(g) Initial 1.00
M 2.00 M 0 M Change -X -X
2X Equili 1.00 -X 2.00-X 2X

• Therefore, ice chart looks like this.
• Change in HF twice change in H2

30
H2(g) F2(g) 2HF(g) Initial 1.00
M 2.00 M 0 M Change -X -X
2X Equili 1.00 -X 2.00-X 2X

• Now plug these values into the equilibrium
  expression
• K 1.5 x 102 (2X)2
• (1.00-x)(2.00-x)
• Solving this gives us a quadratic equation.
• Quadratic gives us 2.14 mol/L and 0.968 mol/L.
  Only 0.968 is reasonable.

31

• H2 1.00 M - 0.968 M 3.2 x 10-2M
• F2 2.00 M - 0.968 M 1.032 M
• HF 2(0.968 M) 1.936 M
• If substituted into the equilibrium expression we
  get 1.3 x 102 which is very close to given K.

32
Practice

• For the reaction Cl2 O2 2ClO(g) K
  156
• In an experiment 0.100 mol ClO, 1.00 mol O2 and
  0.0100 mol Cl2 are mixed in a 4.00 L flask.
• If the reaction is not at equilibrium, which way
  will it shift?
• Calculate the equilibrium concentrations.

33
Problems with small K

• Klt .01

34
Process is the same

• Set up table of initial, change, and equilibrium
  concentrations.
• Choose X to be small.
• For this case it will be a product.
• For a small K the product concentration is small.
• (52)

35
For example

• For the reaction 2NOCl 2NO
  Cl2
• K 1.6 x 10-5
• If 1.00 mol NOCl is put in a 2.0 L container,
  what are the equilibrium concentrations?
• K NO2Cl2 1.6 x 10-5 NOCl2
• NOCl0 0.50M, NO Cl2 0

36
2NOCl 2NO Cl2 Initial 0.50
0 0 Change -2X 2X X Equil 0.50 -2X
2X X

• K NO2Cl2 (2x)2(x) 1.6 x 10-5
  NOCl2 (0.50 -2x)2
• Since K is so small, we we can make an
  approximation that 0.50-2x 0.50
• This makes the math much easier. X 1.0 x 10-2

37
5 Rule

• Many of the systems we will deal with have very
  small equalibrium conatants.
• When this is the case, there will be very little
  shift to the right to reach equilibrium. Since x
  is so small, we will ignore it. However, the
  final value must be checked against the initial
  concentration. If the difference is less than 5,
  then our assumption is valid.
• In the previous problem X 1.0 x 10-2
• 0.50 -2x 0.50 - 2(1.0 x 10-2) 0.48
• Is 1.0 x 10-2 five percent or less than 0.50?
  0.01/0.50 x 100 2

38
Practice Problem

• For the reaction 2ClO(g) Cl2 (g) O2
  (g)
• K 6.4 x 10-3
• In an experiment 0.100 mol ClO(g), 1.00 mol O2 is
  put into a 4.00 L container.
• What are the equilibrium concentrations?

39
Problems Involving Pressure

• Solved exactly the same, with same rules for
  choosing X depending on KP
• For the reaction N2O4(g) 2NO2(g) KP
  0.133 atm. At equilibrium, the pressure of N2O4
  was found to be 2.71 atm? Calculate the
  equilibrium pressure of NO2(g)

40
Le Chateliers Principle

• If a change is applied to a system at
  equilibrium, the position of the equilibrium will
  shift in a direction that tends to reduce the
  change.
• 3 Types of change Concentration (adding or
  reducing reactant or product), Pressure,
  Temperature.

41
Change amounts of reactants and/or products

• Adding product makes QgtK
• Removing reactant makes QgtK
• Adding reactant makes QltK
• Removing product makes QltK
• Determine the effect on Q, will tell you the
  direction of shift.
• The system will shift away from the added
  component.

42
Change Pressure

• By changing volume.
• When volume is reduced, the system will move in
  the direction that has the least moles of gas.
• When volume is increased, the system will move in
  the direction that has the greatest moles of gas.
• Because partial pressures (and conc.) change a
  new equilibrium must be reached.
• System tries to minimize the moles of gas.

43
Change in Pressure

• By adding an inert gas.
• Partial pressures of reactants and product are
  not changed.
• No effect on equilibrium position.

44
Change in Temperature

• Affects the rates of both the forward and reverse
  reactions.
• Doesnt just change the equilibrium position,
  changes the equilibrium constant.
• The direction of the shift depends on whether it
  is exo - or endothermic.

45
Exothermic

• ?H lt 0
• Releases heat.
• Think of heat as a product.
• Raising temperature push toward reactants.
• Shifts to left.

46
Endothermic

• ?H gt 0
• Produces heat.
• Think of heat as a reactant.
• Raising temperature push toward products.
• Shifts to right.