CPT 4. CHEMICAL EQUILIBRIA

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CPT 4. CHEMICAL EQUILIBRIA

• Objectives
• Describe and explain the quantitative relations
  between the species involved in a chemical
  reaction equilibrium.
• Calculate and apply the principles associates
  with chemical equilibria

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CHEMICAL EQUILIBRIAThought teaser

• Examine the graphic on the previous slide showing
  the following reaction
• H2 I2 -gt 2HI
• At what point can you tell that the reaction has
  reached equilibrium?
• Hint watch the HI molecules. At what point do
  you notice that the proportion of products in the
  mixture does not changer any more?

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4.1. Introduction Equilibrium and equilibrium
constant

• DEMO NO2 lt-gt N2O4 CD E\Chapter_14\Present\Media\
  AADZZSE0.html
• E\Chapter_14\Present\Media\EquilibriumConstant\Eq
  uilibriumConstant.html
• Equilibrium state of a reaction process in
  steady state, when the concentrations of
  reactants and products no longer change.
• Law of mass action regulates relation between
  species at equilibrium.
• General case aA bB lt-gt cC dD
• K CcDd / (AaBb)

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Law of mass action (Continued)

• K CcDd / (AaBb)
• K Equilibrium constant mathematical form of
  the law of mass action
• If K gt 1, products are more abundant than
  reactants, reaction favors products.
• If K lt 1, reactants is more abundant than
  products, reaction favors reactants.
• Demo Tro/Media 14.10
• Conclusion K constant for each reaction at a
  given temp.

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RQ2-23B Previous Material Review

• The equilibrium constant for the following
  reaction 2C2H6 7O2 -gt 4CO2 6H2O is K
• a. C2H62O27/ CO24H2O6
• b. CO24O27/ H2O6 C2H62
• c. CO24H2O6 / C2H62O27

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(RQ2-23B2) Previous Materal Review

• If the value of K in the previous reaction is 3,
  the reaction favors a, reactants b, products
  c, both
• If you find more C2H6 and O2 than CO2 and H2O
  after the reaction has reached equilibrium, then
  the value of K is a, lt1 b, gt1 c, lt0

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Law of mass action (Continued 2)

• K CcDd / (AaBb)
• Equilibrium positions specific concentrations of
  reactants and products.
• Positions may change K stays constant

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Law of mass action (RQ2-24)

• The graphics on the previous slide show the
  reaction C2H4 X2 lt-gt C2H4X2 at equilibrium. X
  is Cl, Br, or I. Rank the equilibrium constants
  of the reactions from largest to smallest.
  Explain.
• a. Green rxn K gt Red rxn K gt Purple rxn K. The
  amount of X2 in the rxn mixture decreases in the
  same order.
• b. Purple rxn K gt Green rxn K gt Red rxn K. The
  amount of X2 in the rxn mixture decreases in the
  same order.
• c. Green rxn K gt Red rxn K gt Purple rxn K. The
  amount of X2 in the rxn mixture increases in the
  same order.

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Law of mass action (Illustrations)

• Example 1 4NH3 7O2 lt-gt 4NO2 6H2O
• K NO24H2O6 / NH34O27
• Example 2
• Info provided 2SO2 O2 lt-gt 2SO3
• SO2 3.77E-3 mol/L O2 4.30E-3 mol/L
  SO3 4.13E-3 mol/L
• Info requested K ?
• Extra exercise 25, pg 656

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Equilibrium Constant Determination (RQ2-24a)

• Consider the following equilibrium constant K
  CO23H2O4 / C3H8O25 What reaction does it
  correspond to?
• 3CO2 4H2O -gt C3H8 5O2
• C3H8 5O2 -gt 3CO2 4H2O
• C3H8 4H2O -gt 3CO2 5O2

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Equilibrium Constant Calculation (Illustrations
2)

• Example 3
• Info provided H2 CO2 lt-gt H2O CO
• Reaction container volume 1.00 L
• Amounts of CO and H2O at equilibrium 0.11 mol
• Amounts of H2 and CO2 at equilibrium 0.087 mol
• Info requested (a) Calculate K for the reaction

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Equilibrium Constant Calculation (Illustrations 2
Solution)

• K H2OCO / H2CO2
• H2O CO 0.11 mol/L (how come?)
• H2 CO2 0.087 mol/L
• K 0.11 M2 / 0.087 M2 ?

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Equilibrium Constant Calculation (Illustration 2
Continued)

• Problem continued
• Info provided (b) Reaction container volume
  2.00 L
• amounts of H2 and CO2 at beginning of reaction
  0.050 mol
• Info requested (b) amounts of CO and H2O at
  equilibrium.
• Extra example 37, pg 657

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Equilibrium Constant Calculation (Illustration 2
Continued)

• Assume K 1.60
• K H2OCO / H2CO2 CO2 / CO22
• If at equilibrium x mol/L of CO are produced,
  then the initial concentration of CO2 is reduced
  by an amount equal to x. CO2 CO20 x
• Then 1.60 x2 / (CO20 x)2

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Equilibrium Constant Calculation (Illustration 2
Continued 2)

• CO20 0.050 mol / 2.00 L 0.025 M
• 1.60 x2 /(0.025 x)2 which leads to
• x / 0.025 x 1.26. Solve for x
• Ultimately x 0.0251.26 / (1 1.26) ? CO
  H2O
• CO2 0.025 M CO ?

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Reaction Positions Reaction Quotient(Q)

• Applies to reactions outside equilibrium
  conditions
• Q Pn / Rm Pn , Rm concentrations of
  products and reactants before or after
  equilibrium.
• Q enables to calculate the reaction orientation.

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Reaction Quotient (Continued)

• If Q K, reaction system is at equilibrium
• If Q lt K, there still are more reactants than at
  equilibrium. The system shifts toward consumption
  of more reactants and formation of products
• If Q gt K there are more products than at
  equilibrium. The system shifts toward consumption
  of products and formation of reactants.

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Reaction Quotient (Illustration)

• Example
• Info provided N2 O2 lt-gt2NO
• K 4.0E-4
• At 2000 K, N2 0.50 M O2 0.25 M NO
  4.2E-3 M.
• Info requested Is the reaction at equilibrium?
  If not which way will the reaction shift to reach
  equilibrium?

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Reaction Quotient (Illustration Solution)

• Expression of Q NO2/N2O2. Plug in
  numbers. Q ?
• Compare Q to K 4.0E-4.
• What conclusion do you draw from the value of Q ?
  See RQ2-25
• Extra Exercise 45, pg 657

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Reaction Quotient (RQ2-25)

• Based on the value of Q, what do you conclude?
• a. The reaction shifts toward the formation of
  more products because Q lt K
• b. The reaction shifts toward the formation of
  more reactants because Q gt K
• c. The reaction shifts toward the formation of
  more products because Q lt 1

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4.2 Equilibrium expression for gases

• For gases PV nRT lt-gt P nRT / V CRT,
  Therefore C P/RT. C molarity concentration.
• Case study N2 3H2 lt-gt 2NH3
• Kc K C2NH3 / (CN2C3H2) concentration-based
  equilibrium constant
• Replace C by P/RT in expression of Kc. After
  regroupment and simplification of RT factors
• Kc (RT)2P2NH3 / (PN2P3H2). Move RT term to
  constant side. Kc becomes Kp
• Final result Kp P2NH3 / (PN2P3H2)

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Equilibrium expression for gases (Continued)

• Relation between Kc and Kp
• Case of N2 3H2 lt-gt 2NH3
• Kp Kc/(RT)2 Kc(RT)-2.
• -2 NH3 coefficient - sum of N2 and H2
  coefficients
• General Rule Kp Kc(RT)Dn
• Dn change in mol from reactants to products
  sum of coefficients of products sum of
  coefficients of reactants

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RQ2-25b

• Consider the relation Kp Kc(RT)Dn In what
  conditions is Kp gt Kc ? Justify your answer
• When the number of moles of products is gt the
  number of moles of reactants, because Dn is
  negative.
• When the number of moles of products is lt the
  number of moles of reactants, because Dn is
  negative.
• When the number of moles of products is gt the
  number of moles of reactants, because Dn is
  positive.

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Equilibrium expression for gases (RQ2-26)

• Consider the following reaction
• 4NH3 (g) 5O2 (g) lt-gt 4NO (g) 6H2O (g)
• Express Kc of the reaction in terms of Kp
• Kc Kp/(RT)Dn Kp/(RT)(46-4-5) Kp/RT
  Kp(RT)-1
• Kc Kp(RT)Dn Kp (RT)(46-4-5) KpRT
• Kc Kp/(RT)Dn Kp/(RT)(45-4-6) Kp/RT-1
  Kp(RT)

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Equilibrium Constants for Gases (Illustration)

• Example Decomposition of N2O4 to NO2
• Info provided N2O4 lt-gt 2NO2
• Pressure of N2O4 at equilibrium 0.85 atm
• Kp 0.15 (at 25 deg C)
• Info requested Total pressure of the gas mixture
  in the container
• How do you proceed to solve the problem? See
  RQ2-25C

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RQ2-25C

• How do you proceed to solve the problem?
• a. Set up the expression of the final answer to
  Ptotal PNO2 PN2O4. PN204 is known. Figure out
  PNO2 using Kp P2NO2 / PN2O4
• b. Set up the expression of the final answer to
  Ptotal PNO2 - PN2O4. PN204 is known. Figure out
  PNO2 using Kp P2NO2 / PN2O4
• c. Set up the expression of the final answer to
  Ptotal PNO2 PN2O4. PN204 is known. Figure out
  PNO2 using Kp PN2O4 / P2NO2

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Kp (45, pg 793 Solution)

• Kp P2NO2 / PN2O4 0.15
• Plug in value of PN2O4
• 0.15 P2NO2 / 0.85
• P2NO2 ?
• PNO2 ?
• Total pressure PNO2 PN2O4 ?
• Extra Exercise 39, PG 657

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4.3. Heterogeneous equilibria

• Context when components of a reaction are in
  different physical states.
• CaCO3(s) lt-gt CaO(s) CO2(g)
• K CaOCO2/CaCO3
• CaO and CaCO3 Constant, because mol/V of
  solid does not change. Conclusion incorporate
  CaO and CaCO3 into K. Same consideration
  holds for pure liquids.
• Result K CO2
• Rule Concentrations of pure liquids and solids
  are not included in the expression of the
  equilibrium constants of reactions.

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Heterogeneous equilibria (RQ2-27)

• The concentrations of pure solids and liquids are
  considered constant in a chemical reaction. What
  about the amounts of solids and liquids? Explain
• a. They change. If the mol increases, V
  decreases, which keeps mol/V constant.
• b. They do not change. Otherwise the
  concentrations would change too.
• c. They change. However the mol and V change
  proportionately, which keeps mol/V constant.

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Heterogeneous equilibria (Illustration)

• Example Decomposition of NH4HS
• Info provided NH4HS(s) lt-gt NH3(g) H2S(g).
• Kp 0.11 at 25 deg C
• Info requested Total Pressure in the rxn
  container
• Kp P2NH3 0.11 PNH3 ?
• Extra exercise 53, pg 657

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RQ2-27B Previous Material Review

• When you have a reaction equilibrium involving
  substances in different physical states, you are
  dealing with a a, heteroclite b, heterogeneous
  c, composite equilibrium.
• When expressing the equilibrium constant of a
  heterogeneous equilibrium, dont take into
  account a, pure solids and gases b, pure
  liquids and gases c, solids and pure liquids.

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4.4. Combination of Equilibrium constants

• Study Case Reactions resulting from combination
  of other reactions with known Ks
• General case A B lt-gt D E
• K
• Equation 1 A B lt-gt C K1
• Equation 2 C lt-gt D E K2
• Eq 1 Eq 2 A B C lt-gt C D E
• K CDE / (ABC) K1K2

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Combination of Equilibrium constants (Continued)

• Additional equilibrium constant relations
• If for the equation A lt-gt B, K1 B / A,
  then
• For the equation B lt-gt A,
• K2 A / B 1/K1
• For 2A lt-gt 2B,
• K3 B2 / A2 (K1)2

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Combination of Equilibrium constants (RQ2-28)

• Consider the following reactions
• a. 2A lt-gt B 2C
• b. 3D lt-gt B 2C
• c. 2A lt-gt 3D
• How should you process equations a and b to
  produce equation c?

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Combination of Equilibrium constants (RQ2-28
Answer)

• a. Reverse equation a and add it to equation b to
  get equation c.
• b. Reverse equation b and add it to equation a to
  get equation c.
• c. Simply add equation a to equation b to get
  equation c.

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Combination of Equilibrium constants
(Illustration)

• Info provided Fe(s) H2O(g) lt-gt FeO(s)
  H2(g)
• H2O(g) CO(g) lt-gt H2(g) CO2(g)
• K1 1.6
• FeO(s) CO(g) lt-gt Fe(s) CO2(g)
• K2 0.67
• Info requested Find K for the 1st reaction.
• Extra exercise 27, 29, pg 656

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Combination of Equilibrium constants (Solution)

• Expression of K H2/H2O
• Expression of K1 H2CO2/(H2OCO)
• Expression of K2 CO2/CO
• What is needed in order to transform K1 into K?
  See RQ2-28b

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RQ2-28b

• Given the expressions of K, K1 and K2 n the
  previous slide, how can you change K1 to obtain
  K?
• a. Cancel H2 and H2O by multiplying K1 by
  1/K2
• b. Cancel CO2 and CO by multiplying K1 by K2
• c. Cancel CO2 and CO by multiplying K1 by
  1/K2

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• Multiply the expressions of K1 and 1/K2. What
  happens to CO2 and CO?
• K ?
• Extra ex 27, pg 656

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4.5. Applications of the Equilibrium Constants

• Equilibrium constants can be used to determine
  the amounts of reactants and products in a
  reaction.
• Tools needed
• Chemical Equation
• Changes in concentrations of reactants caused
  by formation of products

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a. Calculating equilibrium concentrations and
pressures

• Procedure
• Write the chemical equation
• Set up the Initial-Change-Equilibrium (ICE)
  quantitative relationships between changes in
  concentrations of reactants due to formation of
  products.
• Use the ICE relations in to set the expression
  of the equilibrium constant.
• Figure out the concentrations of the
  components.

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Calculating equilibrium concentrations
(Illustration)

• Example Reaction of H2 and I2 to make HI.
• Info available
• H2 I2 lt-gt 2HI K 55.64 at 425 deg. C.
• Amounts of reactants before the reaction
  starts 1 mol of H2 and I2
• Volume of the reaction container 0.500 L
• Info requested concentrations if each component
  at equilibrium.
• Extra exercise 51, pg 658

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Calcultating equilibrium concentrations (Solution)

• ICE RELATIONS
• Concentrations (mol / V(Ls)
• Initial Change At Equilibrium
• H2 2 -x 2 x
• I2 2 -x 2 x
• HI 0 2x() 2x
• () According to the mole ratios as shown in the
  equation of the reaction

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Solution (Continued)

• K (2x)2 / (2-x)2 ?
• (K)1/2 2x / 2-x (55.64)1/2 7.46
• This leads to 7.46(2-x) 2x 14.9 - 7.46x
• Which leads to x(27.46) 14.9
• Therefore x ?
• Conclusion at equilibrium
• HI 2x ?
• H2 I2 2 x ?

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RQ2-28C Previous Material Review

• If you know K1 and K2 for reactions A lt-gt B and
  Blt-gt C respectively, all you need to do to find K
  for reaction Alt-gt C is to a, add b, multiply
  c, divide K1 and K2
• In order to find the equilibrium concentration of
  a reactant for which you know the initial
  concentration in a reaction a, add b, multiply
  c, subtract the change in concentration from the
  initial concentration.

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Calcultating equilibrium concentrations (RQ2-29)

• Consider the reaction A lt-gt 3C
• If you start with A 5 mol/L, and form product
  with a 3n mol/L concentration, what are A and
  C at equilibrium?
• a. A (5 n) mol/L and C 3n mol/L because
  A increases and C decreases during the
  reaction
• b. A (3n) mol/L and C 5 - n mol/L because
  A decreases and C increases during the
  reaction
• c. A (5 n) mol/L and C 3n mol/L because
  A decreases and C increases during the
  reaction

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b. Equilibria with small constants

• General equation A Blt-gt C.
• Simplifications can be done to avoid use of
  complex math processes. Condition A0 gt 100 K
  or A0 / K gt 100
• Simplification A0 almost Aeq because very
  little products are formed and therefore changes
  in initial concentrations of reactants are
  negligible.

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Small equilibrium constants (Illustration)

• Example Reaction of N2 and O2 to form NO
• Info available
• N2 O2 ? 2NO K 1E-5 at 1500K
• N20 0.80 M
• O20 0.20M
• Info requested equilibrium concentrations of the
  components
• Extra exercise 57, pg 658

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Small equilibrium constants (Solution)

• ICE RELATIONS between components
• Concentrations (mol/l)
• Initial Change Equilibrium
• N2 0.8 -x 0.8 - x
• O2 0.2 -x 0.2 x
• NO 0 2x() 2x
• () According to the mole ratios as shown in the
  equation of the reaction.

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Small equilibrium constants (Solution, continued)

• K NO2 / N2O2 (2x)2 / (0.8 - x)(0.2 -
  x)
• Compare K to the initial concentrations of
  reactants N20 / K 0.8 / 1E-5 0.8E5 gtgt 100
  and O20 / K 0.2E5 gtgt 100. Therefore the
  simplifications are justified very little
  products are formed
• N2 x 0.8 x approximately 0.8
• O2 x 0.2 x approximately 0.2

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Small equilibrium constants (RQ2-30)

• Given the simplifications on the previous slide,
  what is the next step in the problem solving
  process?
• a. Simplify the expression which becomes K
  (2x)2 / (x2)
• b. Simplify the expression of K which becomes K
  (2x)2 / (0.80.2)
• c. Simplify the expression which becomes K
  (2x)2 / (0.80.2 1.0x x2)

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Small equilibrium constants (Solution, continued
2)

• Expression of K (2x)2 / (0.8 x 0.2)
• (K)1/2 2x / (0.16)1/2 (1E-5)1/2 3.16E-3
  2x
• Result x ?
• Conclusion at equilibrium
• NO 2x ?
• N2 0.8 x ?
• O2 0.2 x ?

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4.7. Disturbing a Chemical Equilibrium 
LeChateliers Principle

• If a change is made to an equilibrium, the
  equilibrium shifts in the direction that consumes
  the change

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LeChatelier Principle

• Case 1 Changing the amounts of reactants /
  products. The equilibrium shifts in a direction
  that consumes the change.
• Example N2(g) 3H2(g) ?? 2NH3(g)
• Adding N2 or H2 results in the consumption of the
  addition. The system consumes the reagents et
  forms products.
• The same effect result if a product (NH3) is
  removed instead of adding reactants.
• Extra example 63, pg 658 (hint heterogeneous)

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LeChatelier Principle (RQ2-31)

• Consider this reaction
• NH3 H2O lt-gt NH4 OH-
• What would be the effect of removing water from
  the reaction?
• Justify your answer
• The equilibrium would shift to the left. It would
  seek to replace the removed H2O by consuming NH4
  and OH-
• b. The equilibrium would shift to the right. It
  would seek to
• replace the removed H2O by consuming NH4 and OH-
• c. The equilibrium would shift to the left. It
  would seek to
• replace the removed H2O by consuming NH3 and OH-

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LeChatelier Principle (Continued)

• Case 2 Changing the volume by changing pressure.
  
• Decreasing the volume equilibrium reacts by
  shifting toward the side that produces less
  amounts of substances (Production of NH3 in the
  example)
• Example running the previous reaction in a
  container half the size of the container in the
  original reaction.

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LeChatelier Principle (Continued)

• Increasing the volume equilibrium reacts by
  shifting toward the side that produces more
  amounts of substances
• Example running the reaction in a container with
  a volume twice the size of the original volume.
• Extra example 69, pg 659

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LeChatelier Principle (Continued)

• Case 3 Changing the temperature. The equilibrium
  shifts in a direction that consumes the change.
  Note Contrary to previous cases, changing T
  leads to changing the value of K.
• Exothermic reactions heat is produced by the
  reaction. Heat is treated as a product. Adding
  more heat pushes the system toward consumption of
  products formation of reactants. K is decreased
• Extra example 69, pg 659

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LeChatelier Principle (Continued)

• Endothermic reactions heat is consumed by the
  reaction. Heat is treated as a reactant. Adding
  more heat pushes the system toward consumption of
  reactants formation of products. The value of
  K is increased
• Extra example 69, pg 659

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LeChatelier Principle (RQ2-32)

• Consider this reaction endothermic reaction
• NH3 H2O lt-gt NH4 OH-
• How would you use temperature to make it produce
  more products?
• a. Lower the temperature. It is like removing the
  heat as a reactant. It pushes the equilibrium to
  the right
• b. Raise the temperature. It is like adding the
  heat as a reactant. It pushes the equilibrium to
  the right
• c. Raise the temperature. It is like removing the
  heat as a reactant. It pushes the equilibrium to
  the right