1. Which has higher positional entropy?
   1. Gas H₂O or Water H₂O
   2. 100 atm  gas or 1 atm gas
2. Calculate DS_surr for: (assume 1 atm and 298 K)
   1. C_{graphite (s)} + O₂ ----> CO_{2 (s)}                      DH = -394 kJ
   2. SbO _{6 (s)} + 6C _(s) ----> 4Sb _(s)+ 6CO (g)       DH = 778 kJ
3. Calculate DS^o and DH^o for the following reaction: (hint, use the thermodynamic data)
   
   Cl_{2 (aq)} ---> Cl_{2 (g)}
4. Calculate what temperature the following reaction becomes spontaneous: Cl_{2 (aq)} ---> Cl_{2 (g)}
5. Estimate DS^o:
   1. 2C₁₀H_22(l) + 31O₂ ----> 20CO₂ + 22H₂O
   2. HCl _(g) + NH_3(g) ----> NH₄Cl_(s)
6. Calculate DG: N_2(g) + 3H_{2 (g)} <---> 2NH_{3 (g)}
   P_NH3 = 3.2 atm    P_N2 = 4.0 atm   P_H2 = 1.2 atm

Answers:

1a.) Gas water has a higher positional entropy because under constant temperature and pressure, a gas takes more volume than a solid. The gas molecules, therefore, have more places they could be, which increases the positional entropy.

1b.) 1 atm. At lower pressure, there is an increase in volume (PV = nRT), and with an increase of volume, positional entropy increases.

2a.) DS_surr: -DH / T = -(-394/298) = 1.32

2b.) DS_surr: -DH / T = -(778/298) = -2.61

3.)
DS^o: 223-121 = 102 J/K*mol
DH^o: 0-(-231) = 23 kJ/mol

4.) DG^o = DH^o - TDS^o

0 = 23000 - T(102)

230000 = T(102)

T = 225 K

At temperatures greater than 225 K, the reaction will be spontaneous.

5a.) DS^o will be positive, because more gas is created.

5b.) DS^o will be negative, because a solid is being formed from two gas molecules.

6.) Use the equation:

DG = DG^o + RT ln (Q).

Q = (3.2)² / ((1.2)³ * (4.0)) = 1.48

DG = -33300 + 8.3145 * 298 ln (1.48)

DG = -32 kJ/mol