1. Use standard entropies and heats of formation to calculate delta-G of formation at 25° C for
a) cadmium(II) chloride (s)
b) methyl alcohol, CH3OH (l)
c) copper(I) sulfide (s)

My problem is that I don't know how to calculate delta-G of formation. I know that the Gibbs-Helmholtz equation can take the form
delta-G= delta-H -T(delta-S), but that's not really the same thing, is it? Could I modify the equation so it was all in terms of formation of a substance?

Yes, you can do that but you will need to calculate delta S from Cd, Cl2, and CdCl2. Delta S = S(products)-S(reactants).
Cd + Cl2 ==> CdCl2

I'm still having trouble getting the correct answers. I was able to get the correct one for CdCl2 (344 kJ) but not for the others. The back of my book says CH3OH should be -166.3 kJ, and Cu2S should be -53.6 kJ

Here's my work... maybe you can find where I'm making an error?

CO + 2H2 --> CH3OH
calculating delta-s:
standard molar entropy CH3OH: 126.8 J/K
" " CO: 197.6 J/k
" " H2: 130.6 J/K
-126.8 J/K - (197.6 J/K + 2(130.6 J/K)) = -585.6 J/k
-238.7 kJ (delta-h formation) - 298.2(-.5856 kJ/K) = -154.4 kJ [not correct]

2Cu+1 + S-2 = Cu2S
calculating delta-s
standard molar entropy Cu2S: 120.9 J/k
" " Cu+1: 40.6 J/K
" " S-2: -14.6 J/K

120.9 J/K - (2(20.6 J/K)+ -14.6 J/K)= 54.3 J/K

-79.5 kJ (heat of formation Cu2S) - 298.2(.0543 kJ/K)= -95.69 kJ [not correct]

Where did I go wrong??

My tables don't list CH3OH or inorganic salts so I can't verify any of the other numbers. However, I did verify S for CO and H2. You switched signs from the initial listing of CH3OH delta H as 126.8 from + at the initial listing to - in the calculation. I tried that change and it didn't get 166 either. Have you checked to see if CH3OH is a liquid or a gas? Just as H2O is listed in the tables as a gas and as a liquid, I suspect CH3OH may be too. Look at that. Sorry my tables don't allow me to do any more.

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asked by Chris
  1. Al(s)+3/2O2(g)-->Al2O3(s)
    delta H=-167.6kj

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    posted by Jillian

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