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Estimate molarity from enthalpy, gibbs energy and entropy of formation |
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| Dec24-12, 06:04 PM | #1 |
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Estimate molarity from enthalpy, gibbs energy and entropy of formation
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## 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. |
| Dec25-12, 09:43 AM | #2 |
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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 :) |
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