CHEMICAL EQUILIBRIUM

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CHEMICAL EQUILIBRIUM

• .occurs when the forward and reverse reactions
  progress at the same rate in a reversible
  reaction, i.e. dynamic equilibrium.
• ..is the state where concentrations of all
  reactants and products remain constant with time.
• Factors influencing equilibrium
• forward and reverse rates
• partial pressures
• concentrations
• temperature

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Rules of reversible first order reaction
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EQUILIBRIUM CONSTANTS

• Use rate constants for forward and reverse
  reactions to calculate equilibrium constants.
• Ratef kf Butane
• Rater kr iButane
• At equilibrium
• Ratef rater and
• kf Butaneequil kr iButaneequil
• so
• Where Kc equilibrium constant in M, mole/L.

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EQUILIBRIUM CONSTANTCALCULATION OF Kc

• LAW OF MASS ACTION At a given T, the same Kc,
  state of equilibrium, will be reached regardless
  of initial conditions.
• For the reaction,
• jA kB ?????lC mD
• The standard state is 1 Molar.

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• When the reversible reaction
• 2 A B ??? 2 C
• Reached equilibrium, the following concentrations
  were measured A 0.40 M, B 0.30 M,
  and C 0.55 M. What is the value of Kc for
  this reaction?

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EQUILIBRIUM CONSTANTCALCULATION, Kp

• Kp uses partial pressures of each species for
  gas
• phase reactions.
• so
• jA(g) kB(g) ??? lC(g) mD(g)
• The standard state is 1 atmosphere
• At a specific temperature, the equilibrium
  constant,
• Kp , remains a constant regardless of the initial
  conditions.

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• At 427oC, a 1.0-L flask contains 20.0 mol H2,
  18.0 mol CO2, 12.0 mol H2O, and 5.9 mol CO at
  equilibrium. Calculate K for the reaction
• CO2(g) H2(g) ??? CO(g) H2O(g)

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Kp, Kc and PV nRT

• P nRT/V but n/V moles/L M
• so
• P MTR, P/RT M.
• Kp Kc (RT) ?n
• If ?n 0, Kp Kc

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• At 127oC, K 2.6 x 10-5 mol2/L2 for the reaction
• 2NH3(g) ??? N2(g) 2H2(g)
• Calculate Kp at this temperature

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REVERSE EQUILIBRIA

• Kc 1/Kc
• And
• Kp 1/Kp
• When reverse reactions are written
• When a reaction is multiplied through by a value
  of n, then
• Knew (Koriginal)n

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• At a given temperature, K 278 for the reaction
• 2SO2(g) O2(g) ??? 2SO3(g)
• Calculate values of K for the following reactions
  at this temperature.
• a. SO2(g) ½ O2(g) ??? SO3(g)
• b. 2SO3(g) ??? 2SO2(g) O2(g)
• c. SO3(g) ??? SO2(g) ½ O2(g)
• d. 4SO2(g) 2O2(g) ??? 4SO3(g)

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• Write the equilibrium expression (for Kp) for
  each of the following gas-phase reactions, which
  occur in the atmosphere
• a. NO(g) O3(g) ???NO2(g) O2(g)
• b. O3(g) ??? O2(g) O(g)
• c. Cl(g) O3(g) ??? ClO(g) O2(g)
• d. 2O3(g) ??? 3O2(g)

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HETEROGENEOUS EQUILIBRIA

• For equilibria involving more than one state,
  (g), (l), (s), (aq), the
• (s) and (l) 1 and are omitted (drop out)
• (g), (aq) use M, moles/L for Kc.
• (g), (s) and (l) use either Kp or Kc

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• Write expressions for K for the following
  reactions
• P4(s) 5O2(g) ??? P4O10(s)
• NH4NO3(s) ??? N2O(g) 2H2O(g)
• CO2(g) NaOH(s) ??? NaHCO3(s)
• S8(s) 8O2(g) ??? 8SO2(g)

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Heterogeneous Equilibria

• 2NaHCO3(s) ? Na2CO3(s) H2O(g) CO2(g)
• Kp PCO2 PH2O
• Kc CO2 H2O
• 2FeCl3(aq) 3H2O(l) ? 2Fe(OH)3(s) 6HCl(aq)
• Kc ?
• But HCl(aq) ?? H(aq) Cl-(aq)
• Kc ?

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• An equilibrium mixture contains 0.60 g solid
  carbon and the gases carbon dioxide and carbon
  monoxide at partial pressures of 2.9 atm and 2.6
  atm, respectively. Calculate Kp for the
  reaction.
• C(s) CO2(g) ??? 2CO(g)
• A sample of gaseous PCl5 was introduced into an
  evacuated flask so that the pressure of pure PCl5
  would be 0.50 atm at 523 K. However PCl5
  decomposes to gaseous PCl3 and Cl2, and the
  actual pressure in the flask was found to be 0.84
  atm. Calculate Kp for the decomposition
  reaction.
• PCl5(g) ??? PCl3(g) Cl2(g)
• At 523 K. Calculate Kp at this temperature

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EQUILIBRIUM CONSTANTSAND SPONTANEITY

• Knowing Kp or Kc for a reaction and the reaction
  mixture concentrations or pressures, predict
• 1. Will there be any net reaction?
• 2. If so, in which direction will it go?
• Calculate Q, then compare it to Kp or Kc.
• Q the reaction Quotient, calculated with
  concentrations or pressures just as the Kp or Kc.

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Kp, Kc and SPONTANEITYwith the Reaction
Quotient, Q

• K Q Equilibrium
• K gt Q Spontaneous
• as written.
• K lt Q Spontaneous
• in reverse.

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• At a particular temperature, Kp 0.133 atm for
  the reaction
• N2O4(g) ??? 2NO2(g)
• Which of the following conditions correspond to
  equilibrium positions?
• a. PNO2 0.144 atm, PN2O4 0.156 atm
• b. PNO2 0.175 atm, PN2O4 0.102 atm
• c. PNO2 0.056 atm, PN2O4 0.048 atm
• d. PNO2 0.064 atm, PN2O4 0.0308 atm

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LeChateliers Principle

• A system at equilibrium will shift to a new
  equilibrium to partially remove any external
  stress. Consider the effects of
• -- Concentration changes
• -- Pressure changes
• -- Temperature
• -- Catalysts

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LeChateliers Principle

• Concentration changes
• K constant but look at Q
• K gt Q causes a shift to increase Q
• K lt Q causes a shift to decrease Q
• H2(g) Cl2(g) ??? 2HCl(g)
• So as HCl goes up, Q goes up, K lt Q

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LeChateliers Principle

• Pressure changes
• A system under increasing pressure will shift
  equilibrium to reduce the pressure.
• 3CuO(s) 2NH3(g) ???3Cu(s) N2(g) 3H2O(g)
• What happens to Cu produced as gas pressures
  change?
• PCl5(g) ??? PCl3(g) Cl2(g)
• What happens with addition of ? Gas?

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The ammonia synthesis equilibrium
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The Effect of decreased volume on the ammonia
synthesis equilibrium
N2
H2
NH3
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LeChateliers Principle

• Temperature changes
• At equilibrium,
• Look at heat as another reactant or product.
• ? Ho Enthalpy.
• Heat on the left, ?Ho gt 0, or it is ,
  (Endothermic)
• Heat on the right, ?Ho lt 0, or it is -,
  (Exothermic)
• If T goes up, K goes up when ? Ho gt 0
• T goes up, K goes down when ? Ho lt 0
• At two temperatures,

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LeChateliers Principle

• Catalysts
• A catalyst DOES NOT change the equilibrium state,
  only the rate of reaching equilibrium.

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• Hydrogen for use in ammonia production is
  produced by the reaction
• CH4(g) H2O(g) ??? CO(g) 3H2(g)
• What will happen to a reaction mixture at
  equilibrium if
• H2O(g) is removed.
• The temperature is increased.
• An inert gas is added.
• CO(g) is removed.
• The Ni catalyst is removed

Ni catalyst
750oC
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Equilibrium position in the Haber process (part 1)
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Equilibrium position in the Haber process (part 2)
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EQUILIBRIUM CONSTANTCALCULATIONS

• Use
• - a table of concentrations and changes
• - substitution into the Kc or Kp
• - algebraic manipulation including
• perfect squares
• dropping small components
• quadratic equations

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• 1. Write a balanced chemical equation.
• 2. Select one of the concentration changes and
  call it x.
• 3. Use the stoichiometry to determine all the
  concentration changes in terms of x.
• 4. Make a table containing the substances, their
  initial concentrations, their changes in
  concentration, and their equilibrium
  concentrations (calculated from the initial
  concentrations and the concentration changes).
• 5. Write the equilibrium-constant expression.
• 6. Insert the equilibrium concentrations from the
  table into the equilibrium-constant expression.
• 7. Solve the equation for x.
• 8. Check any simplifications for validity.
• 9. Substitute the value of x into the expressions
  for the equilibrium concentrations and determine
  their values.
• 10. Use the equilibrium concentrations to
  calculate the reaction quotient, and compare it
  with the equilibrium constant to verify the
  accuracy of the answers.
• The same product works for partial pressures
  simply substitute pressure for concentration in
  this list of steps.

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• The equilibrium constant, Kc, for the water-gas
  shift reaction has a value of 0.227 at 2000 K.
• CO(g) H2O(g) ??? CO2(g) H2(g)
• Suppose 0.0500 moles of CO and 0.0500 mole of
  H2O(g) are placed in a 2.00 liter flask at 2000
  K. What are the equilibrium concentrations of
  all species?

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• Solving with the Quadratic Equation
• The equilibrium constant, Kc, at 55 degrees C for
  the dissociation of dinitrogen tetroxide is
  0.0245.
• N2O4(g) ??? 2NO2(g)
• Suppose 1.50 mole of N2O4(g) is place in a 10.0
  liter container. What are the equilibrium
  concentrations of the species involved?

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Solving with a substance added

• The equilibrium constant, Kc, at 55oC for the
  dissociation of dinitrogen tetroxide is 0.0245.
• N2O4(g) ??? 2NO2(g)
• Suppose 1.50 mole of N2O4(g) is placed in a 10.0
  liter container. The equilibrium concentrations
  of the species are N2O4(g) 0.123 M and
  NO2(g) 0.0548 M. Now 0.0200 mole/L NO2(g) is
  added to the mixture. What are the new
  equilibrium concentrations of the species
  involved?

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• At a particular temperature, assume that K 1.0
  x 102 for the reaction
• H2(g) F2(g) ??? 2HF(g)
• In an experiment, 2.0 mol H2 and 2.0 mol F2 are
  introduced into a 1.0-L flask. Calculate the
  concentrations of all species when equilibrium is
  reached.
• To the equilibrium mixture in part a, an
  additional 0.40 mol H2 is added. Calculate the
  new equilibrium concentrations of H2, F2, and HF.