Estimate molarity from enthalpy, gibbs energy and entropy of formation


by Telemachus
Tags: energy, enthalpy, entropy, estimate, formation, gibbs, molarity
Telemachus
Telemachus is offline
#1
Dec24-12, 06:04 PM
P: 533
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.
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Telemachus
Telemachus is offline
#2
Dec25-12, 09:43 AM
P: 533
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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