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Estimate molarity from enthalpy, gibbs energy and entropy of formation

  1. Dec 24, 2012 #1
    Hi there, I have to solve this problem:

    Use the following data to estimate the molarity of a saturated aqueous solution of ##Sr(IO_3)_2##

    attachment.php?attachmentid=54225&stc=1&d=1356393140.png

    So, I think I should use the Van't Hoff equation in some way, but I don't know how.
    I also have:

    ##\Delta_r G=\Delta G^o+RT\ln K##

    ##K## is the equilibrium constant, and ##\Delta G^o## is the Gibbs energy of formation.

    In equilibrium ##\Delta_r G=0## and the equation can be managed to get the Van't Hoff equation, which is:

    ##\ln K_1-\ln K_2=-\displaystyle\frac{\Delta H^o}{R} \left( \displaystyle\frac{1}{T_2}-\displaystyle\frac{1}{T_1} \right)##

    I think that I should handle this equations to get the equilibrium constant in some way, and then the molarity. Another equation that may be useful is the definition of the Gibbs energy:

    ##\Delta G^o=\Delta H^o-T\Delta S^o##

    The chemical equation involved I think should be:
    ##Sr(IO_3)_2(s)+H_2O(l) \rightleftharpoons Sr^{2+}(aq)+2IO_3^{-}##

    And from it: ##K'=\displaystyle\frac{[Sr^{2+}][IO_3^{-}]^2}{[Sr(IO_3)_2]}##

    The solid concentration remains constant, and then: ##K=[Sr^{2+}][IO_3^{-}]^2##

    Can anybody help me to work this out?

    Thanks.
     

    Attached Files:

    Last edited: Dec 24, 2012
  2. jcsd
  3. Dec 25, 2012 #2
    Ok. The chemical equation I set before was wrong. I wrote it hurried, because of christmass I had to dinner with my family and all that stuff.

    Here is the correct chemical equation as I think it should be:
    ##Sr(IO_3)_2(s) \rightleftharpoons Sr^{2+}(aq)+2IO_3^{-}(aq)##

    Alright, so I tried to solve this in the following manner. I am trying to find the equilibrium constant, I think that if I find it, then I will find the asked molarity.

    So I thought of using that at equilibrium:
    ##K_c=e^{\displaystyle\frac{\Delta G^o}{RT}}##

    So, I have to find the temperature at first. And for that I thought of using

    ##\Delta G=\Delta H-T\Delta S\rightarrow T=\displaystyle\frac{\Delta H-\Delta G}{\Delta S}## (1)

    And then for the reaction I have:

    ##\Delta S=\sum \nu S^o(products)-\sum \nu S^o (reactants)##

    nu stands for the stoichiometric coefficients. From the data in the table I get:

    ##\Delta S=-0.0.0298\frac{kJ}{mol K}##

    Similarly: ##\Delta G=\sum \nu \Delta G_f^o(products)-\sum \nu \Delta G_f^o (reactants)=-0.4\frac{kJ}{mol}##

    And: ##\Delta H=\sum \nu \Delta H_f^o(products)-\sum \nu \Delta H_f^o (reactants)=30.8\frac{kJ}{mol}##

    Then, back to (1) I get:

    ##T=\frac{252.1-127.6}{-0.1482}K=-1020.13K##

    And there is the problem, I'm getting a negative temperature. I don't know what I did wrong. Besides, at first I found a negative entropy, which implies not spontaneous reaction. And the enthalpy is positive, with means endothermic reaction, I think that is consistent. But I don't know why I get this negative temperature, which is obviously wrong.

    Thanks for your attention :)
     
    Last edited: Dec 25, 2012
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