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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.

### 1 Answer

- 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.