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March 29, 2015

March 29, 2015

Posted by **Christian** on Wednesday, May 16, 2007 at 2:32pm.

delta H=-196.0 kJ and delta S=125.7J/K

Is the decomposition of hydrogen peroxide spontaneous at 25 degrees?

My answer: delta G= (-196.0kJ)-25(125.7J/K)

so the free energy change at 25 degrees is:-2946.5 but I'm not sure what units to put on it.Then the decomposition of hydrogen peroxide at 25 degrees would be spontaneous because the free energy is negative.

First, the T must be put into the equation in Kelvin and note that delta H is listed in kJ (I presume for the rxn, as in 2 mols) while delta S is listrd in J; you can't add those two together without changing one or the other.

To put 25 degrees into Kelvin all I do is add 25 to 273 right? So them plug 298 into my equation instead of 25 but then how do I change delta H or delta S so I can add them together?

Was the first answer (the one this one is to be compared with) in kJ. I think so; therefore, let's change this to kJ. Delta H is already in kJ. S is in J/K, to change that to kJ, just divide S by 1000 and it will have the units kJ/K. Yes, add 273 to 25 for 298.

I used 273.16 (since S has that many s.f.) and delta G = -233.5 kJ from the data you have. I don't know how to compare this with the other since I don't know what they problem means "enthalpy change" and "entropy change"' that is, I don't know if those changes are for 1 mol or 2 mols. Anyway, I think the answer is that the reaction is spontaneous. Check my thinking. I don't think it will change the answer but you might want to check with the prof or another student and see how they interpret the 1 mol vs 2 mol thing.

Ok so my new equation is: delta G=(-196.0)-(298)(0.1257)

then the free energy change at 25 degrees is -233.4586 kJ.So compared with the previous equation the free energy change is the same because in my previous post the answer was -233.4kJ.And the reaction is still spontaneous. Can you check this for me? Thanks

- Chemistry -
**ben**, Sunday, January 29, 2012 at 5:21pmRight

firstly entropy is 125.7j/molk. Thus it is 0.1257kj/molk. Now try using the legendre relationship again an you should get an answer of -233.5KJ/mol, which is the same as you get if you calculate it with gibbs free energies of formation. Hope this helps - have fun with thermodynamics

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