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Spontaneity Practice Problems:
  1. Which has higher positional entropy?
    1. Gas H2O or Water H2O
    2. 100 atm  gas or 1 atm gas
  2. Calculate DSsurr for: (assume 1 atm and 298 K)
    1. Cgraphite (s) + O2 ----> CO2 (s)     DH = -394 kJ
    2. SbO 6 (s) + 6C (s) ----> 4Sb (s)+ 6CO (g)       DH = 778 kJ
  3. Calculate DSo and DHo for the following reaction: (hint, use the thermodynamic data)

    Cl2 (aq) ---> Cl2 (g)
  4. Calculate what temperature the following reaction becomes spontaneous: Cl2 (aq) ---> Cl2 (g)
  5. Estimate DSo:
    1. 2C10H22(l) + 31O2 ----> 20CO2 + 22H2O
    2. HCl (g) + NH3(g) ----> NH4Cl(s)
  6. Calculate DG: N2(g) + 3H2 (g) <---> 2NH3 (g)
    PNH3 = 3.2 atm PN2 = 4.0 atm PH2 = 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.) DSsurr: -DH / T = -(-394/298) = 1.32

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

3.)
DSo: 223-121 = 102 J/K*mol
DHo: 0-(-231) = 23 kJ/mol

4.) DGo = DHo - TDSo

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

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