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[Need Help] Finding Gibbs Energy of a reaction

Hello!

Can anyone help me if i want to find the Gibbs free energy of the following reaction at different temperatures? The H2O in the reactant is the superheated steam at 3barG and 380oC.

CaCl2.2H2O(l) + 2H2O(g) Ca(OH)2 (s)+ 2HCl(g) +2 H2O(g)

:smile:
Reply 1
Original post by dalila_amnani
Hello!

Can anyone help me if i want to find the Gibbs free energy of the following reaction at different temperatures? The H2O in the reactant is the superheated steam at 3barG and 380oC.

CaCl2.2H2O(l) + 2H2O(g) Ca(OH)2 (s)+ 2HCl(g) +2 H2O(g)

:smile:

[delta]Gr(T,P)= [delta]Hr(T) - T*[delta]Sr(T) + R*T*ln((Ppartial(HCl)/Ptotal)^2)

You must find [delta]Hr(T) and [delta]Sr(T) All data are in a handbook about thermodynamic values of substances.

But you must remember that [delta]Hr(T) = [delta]Hr(298) +{integral}([delta]Cp(298)*dT) and
[delta]Sr(T) = [delta]Sr(298) +{integral}([delta]Cp(298)*dT/T)

It is very easy.
Reply 2
how about the H2O(g) in the product? do we need to consider the pressure also?
Reply 3
Original post by dalila_amnani
how about the H2O(g) in the product? do we need to consider the pressure also?

No. The pressures of gaseous H2O are reduced in logarithm. And in general you calculate to the equation:

CaCl2*2H2O Ca(OH)2 (s)+ 2HCl(g)

2 molecules of H2O (g) is as in reagents and in products. You can reduce their mentally.

And I think, that CaCl2*2H2O not (l) but (s))))) this is a solid substance))
(edited 10 years ago)
Reply 4
Or if you're working with a real life issue I can assume you have access to data-books or the Internet for data. Then find the enthalpies of formation of each of your products and reactants, find the standard enthalpy change of the reaction; molar entropies for each of your products and reactants and find the standard entropy change; calculate from ΔG(stn)=ΔH(stn)-TΔS(stn) the standard Gibbs' free energy; and then you will have to use ΔG=ΔG(stn)+R*T*ln(Q) where Q=P(HCl)^2=P(total)^2 * x(HCl)^2 where the likelihood that HCl is not the only gas contributing to your total pressure is treated. All this assumes that enthalpy change and entropy change are independent of temperature; that can be fixed by using heat capacities to modify them for the desired temperature.
(edited 10 years ago)
Reply 5
Original post by Big-Daddy
Write out the problem properly for God's sake.

Or if you're working with a real life issue I can assume you have access to data-books or the Internet for data. Then find the enthalpies of formation of each of your products and reactants, find the standard enthalpy change of the reaction; molar entropies for each of your products and reactants and find the standard entropy change; calculate from ΔG(stn)=ΔH(stn)-TΔS(stn) the standard Gibbs' free energy; and then you will have to use ΔG=ΔG(stn)+R*T*ln(Q) where Q=P(HCl)^2=P(total)^2 * x(HCl)^2 where the likelihood that HCl is not the only gas contributing to your total pressure is treated. All this assumes that enthalpy change and entropy change are independent of temperature; that can be fixed by using heat capacities to modify them for the desired temperature.


I already wrote this. More in detail.
Reply 6
thank you for helping me out with a detailed explanation. appreciate that very much :biggrin:

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