Chemistry 1011

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Chemistry 1011

• TOPIC
• Gaseous Chemical Equilibrium
• TEXT REFERENCE
• Masterton and Hurley Chapter 12

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12.2 The Equilibrium Constant

• YOU ARE EXPECTED TO BE ABLE TO
• Write an expression for the equilibrium constant,
  K, for a gaseous reaction
• Recognize that the expression for K depends on
  the form of the balanced chemical equation for
  the reaction.
• Write an expression for the equilibrium constant,
  K, for a gaseous reaction that includes a
  substance in the solid or liquid phase.

3
The Equilibrium Constant

• For a gaseous reaction, the equilibrium constant
  can be written in terms of the partial pressures
  (concentrations) of reactants and products
• For aA(g) bB (g) cC (g) dD (g)
• The equilibrium constant, Kp , is
• K (PC)c x (PD)d
• (PA)a x (PB)b

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The Equilibrium Constant

• Equilibrium partial pressures of the products are
  in the numerator (top)
• Equilibrium partial pressures of the reactants
  are in the denominator (bottom)
• Each partial pressure is raises to a power equal
  to its coefficient in the balanced equation

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Equilibrium Constant Example

• Ammonia is made industrially by the Haber
  Process
• N2(g) 3H2(g) 2NH3(g)
• The equilibrium constant, K, is
• Kp (PNH3)2
• PN2 x (PH2)3

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Equilibrium Constant Example

• Sulfuric acid is a very important industrial
  chemical. It is manufactured from sulfur dioxide
  and oxygen
• 2SO2(g) O2(g) 2SO3(g)
• The equilibrium constant, K, is
• Kp (PSO3)2
• (PSO2 )2 x PO2

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Dependence of K on Equation Stoichiometry

• The expression for K, and its value will depend
  on how the equation is written
• For N2(g) 3H2(g) 2NH3(g)
• Kp (PNH3)2
• PN2 x (PH2)3
• For 1/2N2(g) 3/2H2(g) NH3(g)
• Kp (PNH3)
• (PN2 )1/2 x (PH2)3/2

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Dependence of K on Equation Stoichiometry

• The Coefficient Rule
• If coefficients in a balanced equation are
  multiplied by a factor, n, then
• The equilibrium constant is raised to the nth
  power
• K Kn
• The Reciprocal Rule
• If the equation is written in reverse, then
• K 1/K

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Adding Chemical Equations

• The Rule of Multiple Equilibria
• If a reaction can be expressed as the sum of two
  or more reactions, K for the overall reaction is
  equal to the PRODUCT of the equilibrium constants
  for the individual reactions
• If Reaction 3 Reaction 1 Reaction 2
• Then K(Reaction 3) K(reaction 1) x
  K(Reaction 2)

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Multiple Equilibria Example

• Reaction 1 SO2(g) 1/2O2(g) SO3(g)
• Kp 2.2 (PSO3)
• (PSO2 ) x (PO2 )1/2
• Reaction 2 NO2(g) NO(g) 1/2O2(g)
• Kp 4.0 (PNO) x (PO2 )1/2
• (PNO2 )
• Adding the equations
• SO2(g) NO2(g) SO3(g) NO(g)

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Multiple Equilibria Example

• The equilibrium constant expression for the total
  reaction is
• Kp (PSO3) x (PNO)
• (PSO2 ) x
  (PNO2 )
• This is obtained by multiplying together the
  equilibrium constant expressions for the two
  individual reactions
• (PSO3) x (PNO) x
  (PO2 )1/2
• (PSO2 ) x (PO2 )1/2 (PNO2
  )
• Kp 2.2 x 4.0 8.8

12
Heterogeneous Equilibria

• Up to now, all of the reactions considered have
  been homogeneous gas reactions
• In some cases, one or more of the substances
  involved may be a liquid or solid
• This would be a heterogeneous system
• Example
• I2(s) I2(g)

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Heterogeneous Equilibria the Sublimation of
Iodine

• I2(s) I2(g)
• The rate of the forward process depends only on
  temperature. Sublimation will occur at constant
  rate as long as some solid iodine remains
• The rate of the reverse reaction will depend on
  the concentration (partial pressure) of iodine
  gas
• At equilibrium (in a closed system), this rate
  will become constant
• Kp PI2(g)

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Heterogeneous Equilibria

• For heterogeneous equilibria, it is found that
• The position of equilibrium is independent of the
  amount of solid or liquid component, as long as
  some still remains
• Concentrations of solids and liquids are constant
• Terms for solid or liquid components do not
  appear in the expression for K

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Heterogeneous Equilibria the Decomposition of
Calcium Carbonate

• CaCO3(s) CaO(s) CO2(g)
• Written in terms of concentrations, the
  equilibrium constant expression is
• K CaO(s)CO2(g)
• CaCO3(s)
• But the concentrations of the solids are
  constant, so
• K CO2(g), or Kp PCO2

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Heterogeneous Equilibria with a Liquid Component

• Example
• CO2(g) H2(g) CO(g) H2O(l)
• In this case, the amount of water vapour in the
  system is constant, since it will be in
  equilibrium with the liquid water
• So,
• Kp (PCO)
• (PCO3) (PH3)