17.1 Liquid

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17.1 Liquid vapour equilibrium

• 17.1.1 Describe the equilibrium established
  between a liquid and its own vapour and how it is
  affected by temperature changes.
• 17.1.2 Sketch graphs showing the relationship
  between vapour pressure and temperature and
  explain them in terms of the kinetic theory.
• 17.1.3 State and explain the relationship between
  the enthalpy of vaporization, boiling point and
  intermolecular forces.

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Physical equilibrium

• Rate of vapourisation rate of condensation
• Pressure exerted by the particles in the vapour
  phase is known as vapour pressure of the liquid.
• Changing the liquids surface area affects both
  rates equally no overall effect on VP.
• It can affect how long it takes to reach
  equilibrium. More surface area faster rate.

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Enthalpy of vapourisation

• Vapourisation is an endothermic reaction, so
  there is a minimum amount of energy that is
  required for the liquid molecules to escape as
  vapour molecules
• They must overcome the attractive forces between
  molecules (review bonding Unit intermolecular
  forces ppt)
• ?Hvap energy needed to convert 1 mole of liquid
  to gas for a specific substance.

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• At higher temperatures, more molecules have the
  minimum Ek to escape the vapour phase and the
  rate of vapourisation will increase.
• Causing an increase in pressure
• A liquid boils when its surface pressure its
  vapour pressure
• Normal boiling point is the temp where VP is
  equal to standard atmospheric pressure (101.3
  kPa)
• As pressure is altered, so is the liquids BP

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VP and BP
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• Homogeneous equilibrium reactants and products
  are in the same phase
• CO2(g) H2(g) lt - - gt CO(g) H2O(g)
• Heterogeneous equilibirum reactants and
  products are in two or more phases.
• NH4HS(s) lt - - gt NH3(g) H2S(g)
• Kc NH3 H2S ignore the solid, a
  complicated but interesting reason why is found
  here. Essentially (s) and (l) are considered a
  value of 1, so can be ignored.

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Reaction Quotient, Qc

• Identical expression to Kc, but its value is
  calculated using concentrations that arent
  necessarily found at equilibrium.
• If Qcgt Kc shifts to reactants
• If Qclt Kc shifts to products
• If Qc Kc equilibrium

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Practice Problem

• Calculate the Q for the following reaction using
  these equilibrium concentrations N2 0.10 M
  H2 0.30 M and NH3 0.20 M. Is the system
  at equilibrium if Kc of the reaction is 0.40?
• N2 3H2 lt -- gt2NH3
• Answer Not at equilibrium, QgtK, so shift to
  left (reactants)

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Not on our topic 7 test

• E.12.1 Solve problems relating to the removal of
  heavy-metal ions, phosphates and nitrates from
  water by chemical precipitation

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• Solubility equilibrium is any chemical
  equilibrium between solid and dissolved states of
  a compound at saturation.
• The equilibrium expresses the degree to which the
  solid is soluble in water
• Called solubility product or Ksp

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AgCl (s) ltgt Ag (aq) Cl (aq)

• Writing the equilibrium expression for Ksp is
  similar to Kc.
• You only include (aq), not (s). Anything (s) or
  (l) is denoted as worth 1.
• KspAg Cl
• The concentrations of the two ions are EQUAL
  because of 11 ratio, so if we know its Ksp value
  we can solve for the concentration of ions easily.

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You treat the coefficients still as exponents

• Sn(OH)2 (s) ltgt Sn2 (aq) 2 OH (aq)
• Ksp Sn2 OH2

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Solubility Rules (know these)

• All CO32- and PO43- compounds, except Li, Na,
  K and NH4 are insoluble.
• All S2- compounds, except Group 1,2 or NH4 are
  insoluble.
• All OH- and O2- compounds, except Sr2, Ca2 and
  Ba2 are insoluble.
• All Cl-, Br- and I- compounds, except Ag, Hg22,
  and Pb2 are soluble.
• All SO42- compounds, except Sr2, Ca2 , Pb2 and
  Ba2 are soluble.
• Most nitrates (NO3-) are soluble
• Most salts of Na, K and NH4 are soluble

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Write the chemical equation showing how the
substance dissociates and write the Ksp expression

• 1) Na3PO4
• 2) K2SO43) Na2S4) (NH4)3PO45) Sr(OH)2

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Calculations

• The coefficient does two things
• 1) it puts a power on the 'x' which represents
  that particular ion and2) it puts a coefficient
  in front of the 'x'
• Example Problem
• Calculate the molar solubility (in mol/L) of a
  saturated solution of Sn(OH)2 knowing its Ksp is
  5.45 x 1027

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Solve

• Sn(OH)2 ltgt Sn2 2 OH
• Ksp Sn2 OH2
• One Sn2 makes two OH. That means that if 'x'
  Sn2 dissolves, then '2x' of the OH had to have
  dissolved, so
• 5.45 x 1027 (x) (2x)2
• 4x3 5.45 x 1027
• x1.11 x 109 mol/L

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Solve for Fe(OH)3 knowing its Ksp is 2.64 x 1039

• Iron(III) hydroxide
• Fe(OH)3 ltgt Fe3 3OH
• Ksp Fe3 OH3
• 2.64 x 1039 (x) (3x)3
• 27 x4 2.64 x 1039
• x 9.94 x 1011 M

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Solubility and Ksp

• Solubility of a substance is the quantity that
  dissolves to form a saturated solution
• It can be measured in g/L or mol/L
• Ksp is the constant which describes the
  equilibrium between the ionic solid and its ions
  in a saturated solution.

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Flowchart
Solubility of compound (g/L)
Molar solubility of compound (mol/L)
Molar conc. of ions (mol/L)
Ksp
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Precipitation and Separation of Ions

• When two solutions are mixed to form a
  precipitate, we can use Ksp to see if the
  precipitate will actually form, or if it is
  slightly soluble enough in the particular
  scenario to dissolve.
• If Q Ksp, equilibrium exists
• If Q gt Ksp, precipitate will occur
• If Q lt Ksp, solid will dissolve

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Practice

• Will a precipitate form if equal volumes of 3.0 x
  10-3 M Pb(NO3)2 is added 5.0 x10-3 M Na2SO4 ?
• Figure out the double displacement reaction, then
  decide which is the insoluble/slightly soluble
  compound made.
• Determine the solubility product for that
  insoluble compound (Q)
• Compare it with the accepted value of Ksp

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Answer to 1

• Pb(NO3)2 Na2SO4 ? 2NaNO3 PbSO4
• The insoluble compound is PbSO4, so
• PbSO4 Pb2 SO4-2
• Q Pb2SO4-2
• Q 3.0 x 10-3 5.0 x10-3
• Q 1.5 x 10-5 so, QgtKsp, so precipitate
• Ksp 1.8 x 10-8 occurs

Look it up on chart!
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Practice 2

• Will a precipitate form if 0.10 L of 3.0 x 10-3 M
  Pb(NO3)2 is added to 0.40L of 5.0 x10-3 M Na2SO4?
• PbSO4 Pb2 SO4-2
• Total volume of solution is 0.10 0.40 0.50 L
• Find concentrations involved

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• 0.10 L x 3.0 x 10-3 mol/L 3.0 x 10-4 mol
• 3.0 x 10-4 mol/ 0.50 L 6.0 x 10-4 M
• 0.40 L x 5.0 x10-3 mol/L 2.0 x 10-3 mol
• 2.0 x 10-3 mol/ 0.50 L 4.0 x 10-3 M
• Q 6.0 x 10-4 4.0 x 10-3 2.4 x 10-6
• Q gt Ksp so precipitate will form