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Chemistry 2720, Fall 1996, Test 2 Solutions

Part 1

  1. The vapour pressure is related to the equilibrium constant for the process tex2html_wrap_inline268 by tex2html_wrap_inline270 . For this process,

    displaymath272

    at 298K. The equilibrium constant at this temperature is

    displaymath274

    We want the equilibrium constant at 323K. To calculate this, we need tex2html_wrap_inline276 :

    eqnarray44

    The vapour pressure of sodium at tex2html_wrap_inline278 is therefore tex2html_wrap_inline280 .

  2. The osmotic pressure is

    displaymath282

    Thus

    displaymath284

    or tex2html_wrap_inline286 . The number of moles of the protein is therefore

    displaymath288

    and the molar mass is

    displaymath290

    Molar masses of this magnitude are usually expressed as 193kg/mol.

  3. The overall transport process is

    displaymath292

    Note that tex2html_wrap_inline294 is unaffected because the standard free energies of formation of the glucoses cancel out. The maximum ratio of glucose in to glucose out is obtained when all of the available free energy is begin used up by transport, i.e. when tex2html_wrap_inline296 . This implies that

    displaymath298

    In terms of the activities,

    eqnarray126

    The glucose concentration in the cell can be as high as 71mol/L.

  4. The equilibrium is

    displaymath300

    Let s be the molar solubility. Then

      eqnarray148

    To use the Debye-Hückel equation, we need to express the ionic strength in terms of s.

    displaymath306

    The Debye-Hückel equation becomes

      eqnarray159

    Use tex2html_wrap_inline308 as an initial guess (i.e. assume ideal behaviour) to kick off chain iteration of equations 1 and 2.

    tabular170

    The solubility of lead (II) iodide is tex2html_wrap_inline322 .

Part 2

  1. The key to this kind of problem is to be methodical:

    displaymath320

  2. We expand tex2html_wrap_inline326 near x=0.

    eqnarray192

  3. If the reaction is endothermic, increasing the temperature should favour products. Conversely, we should decrease the temperature if the reaction is exothermic.

    Since the reaction converts two moles of gas to one, Le Chatelier's principle suggests that increasing the pressure on the system should favour the products.

    We can use free energy coupling to increase the yield if we arrange for one of the reactants to be produced by a reaction with a large, negative tex2html_wrap_inline330 .


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Marc Roussel
Fri Nov 22 09:58:15 MST 1996