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Is I-(aq) oxidized by air oxigen?

  1. Mar 3, 2008 #1
    [tex]4I^-(aq) + O_2 + 2H_2O \rightarrow 2I_2 + 4OH^-(aq)[/tex]

    According to Atkins - Physical chemistry - fifth edition:

    [tex]\Delta G^0_f(I^-aq)\ =\ -51.57\ kJmol^{-1}[/tex]

    [tex]\Delta G^0_f(OH^-aq)\ =\ -157.24\ kJmol^{-1}[/tex]

    So:

    [tex]\Delta G^0_{reaction}\ =\ -105.67\ kJmol^{-1}[/tex]

    and the reaction should really happen.
     
  2. jcsd
  3. Mar 3, 2008 #2

    mrjeffy321

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    Well, I have worked it again, and one of us still must be making a mistake somewhere since I have come up with a different answer.

    What state of matter did you assume the Iodine to be in on the product side of the reaction? I have, previously, been assuming it to be in the vapor phase since this theoretical reaction called for the Iodide salt to be mildly heating causing the Iodine to sublime (also, Iodine vapor would have a higher entropy and would help drive the reaction forward if it did proceed).
    For the below calculations, I assumed the Iodine to be in the solid phase.

    Using my thero-table, I calculate that,
    ΔH = -127.54 kJ
    ΔS = -600.84 J
    ΔG = +51.58 kJ

    If, if I calculate ΔG manually as,
    ΔG = ΔH – T * ΔS, for T = 298.15 Kelvin (25 °C), I get,
    ΔG = +51.6 kJ,
    which is awfully close to the value I get from the standard values in the table.

    Assuming my calculations are correct, the reaction is not spontaneous at room temperature, however it does become spontaneous upon lowering the temperature to 212 Kelvin (about -60 °C).
     
  4. Mar 4, 2008 #3
    How do you get that value for ΔS?
    Remember that for substances in solution, entropies are not given as absolute values, but only as relative values, with respect to

    [tex]H^+(aq)[/tex]

    to which is conventionally given the value 0 for entropy .
     
  5. Mar 4, 2008 #4
    I believed that the reaction does really depends on the concentration of both sides..
    anyway, this reaction should not be a general way to make I2...
     
  6. Mar 4, 2008 #5

    chemisttree

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    I agree but your reaction should show the acid as a reactant and the water it forms with OH-. I2 is not formed but I3- is. Your calculations are therefore a bit off. Anyone who has done an iodine titration knows from experience that acidic solutions of I- slowly produce I3- at room temperature... no calculus required.
     
  7. Mar 4, 2008 #6
    Yes, you're right, the correct equation, in solution, should be written as:

    [tex]6I^-(aq) + 4H^+(aq) + O_2\rightarrow 2I_3^- + 2H_2O[/tex]
    Certainly. My intention was to discuss, as in the previous thread ( Making iodine from potassium iodide and...), the possibility to get iodine vapours by heating a wet, acidified alcaline iodide in presence of air oxigen, so I3- would give off I2 molecules and I- would stay in the solid phase.
    But my computation was quite simplicistic, the reaction ΔG is clearly more difficult to evaluate than what I previously thought.
     
    Last edited: Mar 4, 2008
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