Question
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.
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.
Answers
Al(s)+3/2O2(g)-->Al2O3(s)
delta H=-167.6kj
delta H=-167.6kj
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