Lecture Problem 10'45

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Lecture Problem 10.45

• Matthew Weiner
• 4/11/07

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Problem Statement

• From the phase diagram for the MgO-Al2O3 system,
  it may be noted that the spinel solid solution
  (MgAl2O4) exists over a range of compositions,
  which means that it is nonstoichiometric at
  compositions other than 50 mol MgO-50 mol
  Al2O3.
• The maximum nonstoichiometry on the Al2O3-rich
  side of the spinel phase field exists at about
  2000 degrees C corresponding to approximately 82
  mol (92 wt) Al2O3. Determine the type of
  vacancy defect that is produced and the
  percentage of vacancies that exist at this
  composition.
• The maximum nonstoichiometry on the MgO rich side
  of the spinel phase field exists at about 2000
  degrees C corresponding to approximately 39 mol
  (62 wt) Al2O3. Determine the type of vacancy
  defect that is produced and the percentage of
  vacancies that exist at this composition.

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Important Information

• For the first part of the problem, we will be
  interested in the region of the phase diagram at
  82 mol Al2O3
• For the second part, we will be interested in the
  39 mol Al2O3 region

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MgO Al2O3 Phase Diagram
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Part A

• Type of Vacancy Created
• Since the compound is Al2O3-rich, some Mg2 must
  be removed so that there can be room for more
  Al3 (Al3 substitutes for Mg2)
• The overall charge must be kept neutral, so for
  every 2 Al3 added, 3 Mg2 must be removed
• ? There are Mg2 vacancies (cation vacancy)

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Part A (cont)

• Choose a basis to get an expression for mole
  fraction
• Use 100 mol of MgO-Al2O3, which means there are
  100 mol MgO and 100 mol Al2O3
• If 3 Mg2 need to be removed for every 2 Al3
  added, then for every mol of Al2O3 formed, 3
  moles of MgO need to be removed (the O2- lattice
  stays the same)

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Part A (cont)

• Write the mole fraction of Al2O3 in terms of the
  number of moles of Al2O3 created past 50 mol
  Al2O3-50 mol MgO

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Part A (cont)

• Solve the Al2O3 mole fraction equation for a mole
  fraction of 82 (given in problem statement)
• Since three moles of MgO are removed for every
  mole of Al2O3 created, 72.72 mols of Mg2 were
  removed
• This means a total of 72.72-48.48 24.24 mols of
  vacancies were created

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Part A (concluded)

• The fraction of vacancies is equal to the number
  of vacancies over the total number of sites that
  Mg2 could occupy
• 100 mol sites from the 100 mol of MgO
• 200 mol sites from the 100 mol of Al2O3
• So, the percentage of vacancies is

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Part B

• Now, we are looking at the MgO-rich side of the
  spinel phase diagram
• The Mg2 will be replacing the Al3
• If two Al3 are removed to allow 2 Mg2 to come
  in, there is still a net -2 charge, so one O2-
  must be removed as well
• ?There are O2- vacancies (anion vacancy)

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Part B (cont)

• For every 2 Mg2 added, 1 O2- will be removed, so
  we can write the equation for the mole fraction
  of MgO

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Part B (cont)

• Since we are looking at 39 mol Al2O3, the spinel
  is 61 mol MgO
• Setting the above equation equal to 0.61
• Since the amount of O2- removed is half the
  amount of Mg2 added, 15.8 moles of O2- were
  removed

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Part B (concluded)

• Follow the same steps as in Part A to find the
  percentage of vacancies
• There are 100 mol sites for O2- in MgO and 300
  mol sites for O2- in Al2O3
• The percentage of vacancies is

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Applications

• Can analyze the phase diagram for a ceramic
  system like for a metal-metal system
• With an increase in the number of vacancies, the
  diffusivity increases
• Conductivity increases (undesired effect for an
  insulator)