The Richard Stockton College of New Jersey
Chemistry Program, School of Natural Sciences and Mathematics
PO Box 195, Pomoma, NJ
CHEM 3410: Physical Chemistry I — Fall 2008
Homework 6— Solutions
1. At constant temperature, we can write dG as:
dG =V dP
Integrating this by applying the ideal gas law:
∆G=Z0.500
10.5
V dP =Z0.500
10.5
nRT
PdP = (2.5mol)(8.314 J/molK )(350 K) ln 0.500
10.5
∆G=−22.1 kJ
2. The reaction of interest is:
CH4+ 2 O2→CO2+ 2 H2O
The maximum other work possible (in this case, electrical work) is equal to ∆Grxn.
We can do this problem several ways. We can find ∆Hrxn and ∆Srxn and use ∆Grxn = ∆Hrxn +
T∆Srxn. However, most data tables also contain ∆Gffor the compounds in this reaction. We
could therefore take ∆Gfof the products minus ∆Gfof the reactants.
∆Grxn = [∆Gf(CO2) + 2∆Gf(H2O)] −∆Gf(C H4)
∆Grxn = (−394.4+2∗ −237.1) −(−50.5) kJ/mol
Max Electrical Work = ∆Grxn =−818.1 kJ/mol
3. Assuming everything behaves ideally:
∆Gmix =RT (Xaln Xa+Xbln Xb)
∆Smix =−R(Xaln Xa+Xbln Xb)
(a) Forming one mole of air
∆Gmix = (8.314 J/molK )(298.15 K)(0.8 ln 0.8+0.2ln 0.2)
∆Gmix =−1240 J=−1.24 kJ
∆Smix =−(8.314 J/molK )(0.8 ln 0.8+0.2ln 0.2)
∆Smix = 4.16 J
(b) Mixing a total of 3 moles of gas
∆Gmix = (8.314 J/molK )(298.15 K)(2
3ln 2
3+1
3ln 1
3)
∆Gmix =−1578 J/mol ×3mol =−4.74 kJ
∆Smix =−(8.314 J/molK )(2
3ln 2
3+1
3ln 1
3)
