Spontaneity Practice Problems:
- Which has higher positional entropy?
- Gas H2O or Water H2O
- 100 atm gas or 1 atm gas
- Calculate DSsurr for: (assume 1 atm and 298 K)
- Cgraphite (s) + O2 ----> CO2 (s)
DH = -394 kJ
- SbO 6 (s) + 6C (s) ----> 4Sb (s)+ 6CO (g)
DH = 778 kJ
- Calculate DSo and DHo
for the following reaction: (hint, use the thermodynamic
Cl2 (aq) ---> Cl2 (g)
- Calculate what temperature the following reaction becomes spontaneous: Cl2 (aq)
---> Cl2 (g)
- Estimate DSo:
- 2C10H22(l) + 31O2 ----> 20CO2 +
- HCl (g) + NH3(g) ----> NH4Cl(s)
- Calculate DG: N2(g) + 3H2 (g)
<---> 2NH3 (g)
PNH3 = 3.2 atm PN2 = 4.0
atm PH2 = 1.2 atm
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.) DSsurr: -DH / T = -(-394/298) = 1.32
2b.) DSsurr: -DH / T = -(778/298) = -2.61
DSo: 223-121 = 102 J/K*mol
DHo: 0-(-231) = 23 kJ/mol
4.) DGo = DHo
0 = 23000 - T(102)
230000 = T(102)
T = 225 K
At temperatures greater than 225 K, the reaction will be spontaneous.
5a.) DSo will be positive,
because more gas is created.
5b.) DSo will be negative,
because a solid is being formed from two gas molecules.
6.) Use the equation:
DG = DGo + 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