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HELP PLEASE!!! I don't know how to do this Gibbs energy problem: Use the thermodynamic data provided below to determine ΔG (in kJ/mol)?
Use the thermodynamic data provided below to determine ΔG (in kJ/mol) for the condensation of H2O2 at 2.08 °C if the initial partial pressure of H2O2 is 2.53 atm.
Substance ΔH°f (kJ/mol) S° (J mol-1K-1)
H2O2 (l) -187.8 109.6
H2O2 (g) -136.3 232.7
Last answer: -17.6 kJ/mol. This answer is wrong, and I don't know why.
I was trying to lead up to the equation Delta(G)= Delta(G^0) +RTln(K)
First I converted C to K.
Then I found the Delta(H) and Delta(S) for each,
then found Delta(G^0) using Delta(G^0)= H -T(S)
and found Q = P(H2O2)(l)/P(H2O2)(g)
And plugged in the numbers.
Please help, and thank you in advance.
- hcbiochemLv 75 years agoFavorite Answer
The reaction under question is H2O2(g) <--> H2O2(l)
I don't know whether you did this or not, but because H2O2(g) is the only gaseous component, Q = 1/P(H2O2)
Calculate DeltaGo for each component under standard conditions:
DGof = DHo = TDSo
DGof(H2O2(l)) = -187.8 kJ/mol - 209.15 (0.1096 kJ/mol K) = - 220.5 kJ/mo
DGof(H2O2(g)) = -136.3 - 298.15(0.2327) = -205.7 kJ/mol
DGo(rxn) = -205.7 - (-220.5) = +14.8 kJ/mol
Now, use the equation you tried to get DG:
DG = DGo - RT ln Q
DG = +14.8 kJ/mol - (8.314 J/molK)(275.23K) ln(1/2.53)
DG = +14.8 kJ/mol + 2.2 kJ/mol = +17.0 kJ
Not certain that's correct, but it might be. I think the approach is right.