Give the enthalpy change and entropy change values shown below calculate the free energy change for this reaction at 25 degrees C using Gibbs free energy equation. How does this value compare to the free energy change using the standard free energy of formation data?

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

Right

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

To calculate the free energy change (ΔG) for a reaction using the Gibbs free energy equation, you need to use the formula: ΔG = ΔH - TΔS, where ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in Kelvin.

In this case, the given values are ΔH = -196.0 kJ and ΔS = 125.7 J/K. However, to ensure consistency in units, you need to convert ΔS to kJ/K by dividing it by 1000. So, ΔS = 0.1257 kJ/K.

To convert the temperature from degrees Celsius to Kelvin, add 273 to 25, giving T = 298 K.

Now you can substitute the values into the equation: ΔG = -196.0 kJ - 298 K * 0.1257 kJ/K = -196.0 kJ - 37.4766 kJ = -233.4766 kJ.

Therefore, the free energy change for this reaction at 25 degrees Celsius is -233.4766 kJ. The negative value indicates that the reaction is spontaneous at this temperature because the free energy change is negative.

Regarding the comparison to the standard free energy of formation data, you haven't provided any information about the standard free energy of formation values for the reactants and products of the reaction. Therefore, it is not possible to compare the calculated ΔG to the standard free energy values without knowing the specific compounds involved in the reaction.

Please note that while I've explained how to calculate the free energy change, it's always a good idea to double-check your calculations and consider any additional information provided in your course material or by your instructor.