Equilibrium

1
Equilibrium
2
Reactions are reversible

• 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

3
Forward Reaction
Reaction Rate
Equilibrium
Reverse reaction
Time
4
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.
• Or if the reaction is verrrry slooooow.

5
Law of Mass Action

• For any reaction
• jA kB lC mD
• K ClDm PRODUCTSpower
  AjBk REACTANTSpower
• K is called the equilibrium constant.
• is how we indicate a reversible
  reaction
• Only gases and aqueous solutions are included in
  the law of mass action

6
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

7
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

8
Manipulating K
From a practical point of view, here is how K
gets manipulated If a reaction gets doubled,
the value of K is squared. If it is tripled, K
is raised to the third power. If the reaction
gets cut in ½ then K is found by taking the
square root of K.
9
The units for K

• This one is simple.
• K has no units.

10
K is CONSTANT

• Temperature affects the equilibrium constant.
• The equilibrium concentrations dont have to
  be the same only K.
• Equilibrium position is a set of
  concentrations at equilibrium.
• There are an unlimited number of K values so
  the temperature needs to be stated.

11
Equilibrium Constant

• One for each Temperature

12
Calculate K

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

13
Calculate K

• N2 3 H2 2 NH3 (all gases)
• Initial At Equilibrium
• N20 0 M N2 0.399 M
• H20 0 M H2 1.197 M
• NH30 1.000 M NH3 0.157M
• K is the same no matter what the amount of
  starting materials

14
Equilibrium and Pressure

• Dont forget your gas laws, they are often
  needed to get some information for the problem.
  Especially PV nRT
• Some reactions are gaseous
• P (n/V)RT
• P CRT
• C is a concentration in moles/Liter
• C P/RT

15
Equilibrium and Pressure

• 2 SO2(g) O2(g) 2 SO3(g)
• Kp (PSO3)2
• (PSO2)2 (PO2)
• K SO32
  SO22 O2

16
Equilibrium and Pressure For those that are a
glutton for punishment and want to see the full
treatment.

• K (PSO3/RT)2 (PSO2/RT)2(PO2/RT)
• K (PSO3)2 (1/RT)2
  (PSO2)2(PO2) (1/RT)3
• K Kp (1/RT)2 Kp RT (1/RT)3

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General Equation

• jA kB lC mD
• Kp (PC)l (PD)m (CC x RT)l (CD x RT)m (PA)j
  (PB)k (CA x RT)j(CB x RT)k
• Kp (CC)l (CD)m x (RT)lm (CA)j(CB)k x
  (RT)jk
• Kp K (RT)(lm)-(jk) K (RT)Dn
• Dn(lm)-(jk) Change in moles of gas

18
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.

19
Heterogeneous Equilibria

• If the reaction involves pure solids or pure
  liquids the concentration of the solid or the
  liquid doesnt change.
• As long as they are not used up we can leave
  them out of the equilibrium expression.
• For an example, check out the next slide.

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For Example

• H2(g) I2(s) 2 HI(g)
• K HI2 H2I2
• But the concentration of I2 does not change, so
  it factors out of the equilibrium expression and
  looks like this.
• K HI2 H2

21
Solving Equilibrium Problems

• Type 1

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The Reaction Quotient

• Tells you the direction the reaction will go to
  reach equilibrium
• Calculated the same as the equilibrium
  constant, but for a system not at equilibrium
• Q Productscoefficient Reactants
  coefficient
• Compare value to equilibrium constant

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What Q tells us

• If Q lt K
• Not enough products
• Shift to right
• If Q gt K
• Too many products
• Shift to left
• If Q K system is at equilibrium. We like
  these ones there is nothing to do !!

24
Example

• For the reaction
• 2 NOCl(g) 2 NO(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?

25
Solving Equilibrium Problems

• Given the starting concentrations and one
  equilibrium concentration.
• Use stoichiometry to figure out other
  concentrations and K.
• Learn to create a table of initial and final
  conditions.
• Check out the next slide for an actual problem.

26

• Consider the following reaction at 600ºC
• 2 SO2(g) O2(g) 2 SO3(g)
• In a certain experiment 2.00 mol of SO2, 1.50 mol
  of O2 and 3.00 mol of SO3 were placed in a 1.00 L
  flask. At equilibrium 3.50 mol of SO3 were found
  to be present. Calculate the equilibrium
  concentrations of O2 and SO2, K and KP

27
The ICE BOX

• The table we create is called the ICE Box.
• 2 SO2(g) O2(g) 2 SO3(g)
• Here is the information given in the problem.
  The blank spaces are what we have to determine.
  Some are determined and others are calculated.
• Initial 2 1.5 3
• Change
• Equilibrium 3.5

28
The ICE BOX

• The table we create is called the ICE Box.
• 2 SO2(g) O2(g) 2 SO3(g)
• Here is the information given in the problem.
  The blank spaces are what we have to determine.
  Some are determined and others are calculated.
• Initial 2 1.5 3
• Change - 2x - x 2x
• Equilibrium 3.5

29
Where are we at ?

• The values in the CHANGE row are a combination of
  stoichiometry and equilibrium.
• Notice that the numbers are the same as the
  coefficients. We cant forget stoichiometry.
• The SO3 increases so it has to be a 2x.
• This tells us it shifts to the right and that the
  value of the reactants has to decrease.
• We combine the or sign along with the
  stoichiometry to set up the equation.

30
The ICE BOX

• The table we create is called the ICE Box.
• 2 SO2(g) O2(g) 2 SO3(g)
• Here is the information given in the problem.
  The blank spaces are what we have to determine.
  Some are determined and others are calculated.
• Initial 2 1.5 3
• Change - 2x - x 2x
• Equilibrium 2 2x 1.5 x 3.5

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Where are we at ? Part 2

• The values of the equilibrium are known
  algebraically. Here is our Law of Mass Action.
  We can solve that x .25
• K SO3 2 (3.5) 2 O2
  SO22 (1.25) (1.5)2
• K 4.36

32
A New Problem

• Consider the same reaction at 600ºC
• 2 SO2(g) O2(g) 2 SO3(g)
• In a different experiment .500 mol SO2 and .350
  mol SO3 were placed in a 1.000 L container. When
  the system reaches equilibrium 0.045 mol of O2
  are present.
• Calculate the final concentrations of SO2 and SO3
  and K.

33
Lets set up the ICE Box

• 2 SO2(g) O2(g) 2 SO3(g)
• Initial .5 0 .35
• Change 2x x - 2x
• Notice the signs. Since there is no oxygen
  present at the start, the reaction must shift to
  the left and that means products consumed and
  reactants created.
• Equilibrium .5 2x .045 .35 2x
• This tells us that x .045 and therefore the
  concentrations must be as follows.
• SO2 .5 2x or .5 .09 .59
• SO3 .35 2x or .35 - .09 .26
• K (.26)2 / (.59)2 (.045) 4.32

34
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).
• We can make simplifying assumptions.

35
5 rule

• Now is as good a time as any to introduce the 5
  rule.
• Equilibrium concentrations are often a bit of an
  approximation. Therefore its allowable to
  simplify the math to make it more workable.
• The 5 rule states that if x is small in
  relation to the stated amount, and the K, then
  it can be ignored in the calculation.
• Generally works if the Keq is less than 10-3 or
  greater than 103

36
Example

• H2 (g) I2 (g) ? 2 HI (g) K 7.1 x 102 at 25ºC
• Calculate the equilibrium concentrations if a
  5.00 L container initially contains 15.9 g of H2
  and 294 g I2 .
• H2 0 (15.7g/2.02)/5.00 L 1.56 M
• I2 0 (294g/253.8)/5.00L 0.232 M
• HI 0 0

37
Lets set up the Ice Box !!

• Q gt K so more product will be formed.
• Since K is large, rxn. will go to completion.
• Another great clue is that when there is a
  component with a concentration of zero, the
  reaction must shift in that direction.
• Stoichiometry tells us I2 is LR, it will be
  smallest at equilibrium let it be x
• Set up table of initial, final and change in
  concentrations.

38
This equation needs manipulating

H2(g) I2(g) ? 2 HI(g)
initial 1.56 M 0.232 M 0 M
change - x - x 2xequilibrium
1.56 - x .232 - x 2x

• Choose x so it is small.
• Since we know the I2 is going to lose most of its
  concentration, x is actually going to be a very
  high percentage of the amount of I2
• There is a way around this dilemma.

39
Here we go.
In the work below, we are going to assume that
the reaction goes to completion.

• H2(g) I2(g) ? 2 HI(g) initial
  1.56 M .232 M 0 M change -
  .232 - .232 .464 equilibrium 1.328
  0 .464

Lets combine the following facts and ideas and
solve the problem.
40
Piecing it together

• 1. We didnt change the conditions of the rxn.
  (ie. Temp. ) so the Keq remains unchanged.
• 2. The Keq is large so we know there will be more
  product than reactant at equilibrium.
• 3. Since we assumed the reaction went to
  completion we now have no Iodine and the reaction
  must move to the left.
• 4. X must be small because there needs to be
  more product to satisfy Keq. X will be
  negligible compared to the amounts given.
• 5. However, x cant be ignored for the I2
  because it is at zero and x is 100 of the total
  amount of I2

41
H2(g) I2(g) ? 2 HI(g) initial
1.328 0 .464 change x
x -2x equilibrium 1.328 x x .464
2x

• Here is our new Law of Mass Action after we
  remove the x according to the 5 rule.
• Remember we cant remove the x in the I2 because
  that is all that there is of it.
• 710 (.464)2 / (1.328) ( x ) 2.28 x 10-4

42
Checking the assumption

• The rule of thumb is that if the value of X is
  less than 5 of all the other concentrations, our
  assumption was valid.
• If not we would have had to use the quadratic
  equation
• More on this later.
• Our assumption was valid.

43
Practice

• For the reaction Cl2 O2 ? 2 ClO(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? Q ?
• Calculate the equilibrium concentrations of all
  components.

44
Problems with small K

• Klt .01

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Process is the same

• Set up table of initial, change, and final
  concentrations. Your ICE Box
• Choose X to be small.
• Sounds familiar so far, I hope !!
• For this case it will be a product.
• For a small K the product concentration is
  small.

46

Lets start the set up.

• For the reaction 2 NOCl ? 2 NO Cl2
• K 1.6 x 10-5 Make note the small K favors
  more reactant
• If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol
  Cl2 are mixed in a 1 L container
• What are the equilibrium concentrations ?
• Q NO2Cl2 (0.45)2(0.87) 0.15
  NOCl2 (1.20)2
• Q gt K so it shifts to the left.

47
Here is the ICE Box set up
Once again we can manipulate the components to
make the math easier and we do this by assuming
the reaction goes to completion according to
stoichiometry.
2 NOCl ? 2 NO Cl2 Initial 1.20
0.45 0.87 Change .45 -.45
-.225 Equilibrium 1.65 0 .645

• We can now use the Equilibrium concentrations to
  rework the problem. Remember that the K does not
  change. K favors the reactants. Having NO
  0 means that the reaction will shift to the
  right. Combining the two means that the change
  will be small so x will be small and we can use
  the 5 simplification rule. This is a Happy Day
  !

48
Where are we at ?

• We can now use the Equilibrium concentrations to
  rework the problem.
• Remember that the K does not change.
• K favors the reactants.
• Having NO 0 means that the reaction will
  shift to the right. Combining the two means that
  the change will be small so x will be small and
  we can use the 5 simplification rule.
• This is truly a Happy Day !

49
Here is the new ICE Box
2 NOCl ? 2 NO Cl2 Initial 1.65 0
.645 Change - 2x 2x x Equilibrium 1.65
2x 2x .645 x Equilibrium (5) 1.65
2x .645

• Solving 1.6 x 10-5 (.645) (2x)2
• (1.65)2
• The algebra is trickier but doable x .0042

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Always verify the 5 rule

• If x .0042, we compare that to the
  concentration. .0042 / .87 .48
• We are good !!

51
Practice Problem

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

52
Mid-range Ks

• .01ltKlt10

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No Simplification

• Choose X to be small.
• Cant simplify so we will have to solve the
  quadratic formula
• H2 (g) I2 (g) ? 2 HI (g) K 38.6
• What is the equilibrium concentrations if 1.800
  mol H2, 1.600 mol I2 and 2.600 mol HI are mixed
  in a 2.000 L container?

54
Here is the set - up

• Q 2.34 so Q lt K it shifts to the right.
• H2 (g) I2 (g) ? 2 HI (g)
• Initial .9 .8 1.3
• Change -x -x 2x
• Equilibrium .9 x .8 x 1.3 2x
• K 38.6 (1.3 2x)2 / (.9 x) (.8 x) ??
• Fortunately, we dont see them this type often.
• You are encouraged to discover the many uses of
  your TI 84 or equivalent.

55
Problems Involving Pressure

• Solved exactly the same, with same rules for
  choosing X but use pressures due to Kp
• For the reaction N2O4 (g) ? 2NO2 (g)
• Kp .131 atm. What are the equilibrium
  pressures if a flask initially contains 1.000 atm
  N2O4?
• N2O4 (g) ? 2 NO2 (g)
• Initial 1 0
• Change - x 2x
• Equilibrium 1-x 1 2x

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Le Chateliers Principle

• If a stress is applied to a system at
  equilibrium, the position of the equilibrium will
  shift to reduce the stress, and establish a new
  equilibrium.
• 3 Types of stress
• Temperature changes
• Pressure changes
• Concentration changes

57
The Basic Steps

• Here is the best way to always get LeChateliers
  problems correct.
• Sample Problem A B ? C D
• How is the system going to change if more B is
  added ?
• 1. Identify the stress The addition of B
• 2. What is the opposite The removal of B
• 3. Which way must the reaction shift to achieve
  step 2 It must shift to the right

58
Change amounts of reactants and/or products

• Adding product makes Q gt K
• Removing reactant makes Q gt K
• Adding reactant makes Q lt K
• Removing product makes Q lt K
• Determine the effect on Q, will tell you the
  direction of shift remember that it is important
  to know what Q is equal to.

59
Change in Pressure 1

• This is usually done by changing volume.
• Decreasing the volume
• increases the pressure
• System shifts in the direction that has the less
  moles of gas less gas less pressure
• Because partial pressures (and concentrations)
  change a new equilibrium must be reached and
  established.
• System tries to minimize the moles of gas.

60
Change in Pressure 2

• This is usually done by changing volume.
• Increasing the volume
• decreases the pressure
• System shifts in the direction that has more
  moles of gas more gas more pressure
• Because partial pressures (and concentrations)
  change a new equilibrium must be reached and
  established.
• System tries to maximize the moles of gas.

61
Change in Pressure 3

• By the addition of an inert gas.
• Partial pressures of reactants and product are
  not changed .
• No effect on equilibrium position.
• This is a common question on the AP exam so be
  aware of it. It should be easy points.
• It might phrase it as an inert gas or it might
  use one of the Noble Gases
• He Ne Ar Kr Xe Rn

62
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 magnitude of the shift depends on
  whether the reaction is exothermic or
  endothermic.
• On many questions, it states a temperature that K
  is measured in. This is more to make the question
  technically correct. Look back on our solved
  problems. Temperature was not calculated in.

63
Change in Temperature Exothermic

• DH lt 0 It is a negative number.
• The reaction releases heat.
• Think of heat as a product since you know an
  equation can be written with a heat term
• A B ? C D Heat
• Raising temperature (adding heat) effectively
  increases the amount of heat (product)
• Shifts to left.
• Remember the steps to solving LeChateliers
  problems.

64
Change in Temperature Endothermic

• DH gt 0 It is a positive number.
• The reaction absorbs heat.
• Think of heat as a reactant since you know an
  equation can be written with a heat term.
• A B Heat ? C D
• Lowering temperature push toward reactants.
• Shifts to left.
• Remember the steps to solving LeChateliers
  problems.

65
Changes in Temperature Summary

• Place the heat term in the reaction.
• Determine if heat is being added or taken away
  from the reaction.
• If heat is being added to an exothermic reaction,
  the reaction shifts to the left.
• If heat is being added to an endothermic
  reaction, the reaction shifts to the right.
• If heat is being taken away from an exothermic
  reaction, the reaction shifts to the right
• If heat is being taken away from an endothermic
  reaction, the reaction shifts to the left.