Chemistry 121. Name _______________________________

Mid-Term Examination #2 [32 Marks] Friday, 26th March, 1999.

[2]1. What is the role of the "Second Law of Thermodynamics" in chemistry?

The Second Law of Thermodynamics states that "in any spontaneous process the entropy of the Universe must increase." This means that for a chemical process to be spontaneous, the entropy of the Universe must increase as it takes place. Calculating the change in the entropy of the Universe for a process can be used to find whether it will be spontaneous or not under the given conditions.

[2]2. The reaction between ethyne and oxygen is used in oxyacetylene welding:

2 C_{2}H_{2}(g) + 5 O_{2}(g) Ž 4 CO_{2}(g) + 2 H_{2}O(g) DHº = -2510 kJ.

a. Is the reaction exothermic or endothermic? Explain your answer.

The enthalpy of the system is decreasing. Since energy cannot be created or destroyed, this must be lost to the surroundings. The process is exothermic.

b. Is the sign of the entropy change for the process +, -, or neither

(ie DSº Ÿ 0)? Explain your answer.

The entropy change will be negative, since the final state of the system is more ordered than the initial.

[3]3. PbI_{2} dissolves in water: PbI_{2}(s) ¾ Pb^{++}(aq) + 2 I^{-}(aq).

The K value for this equilibrium at 298 K is 8.3 x 10^{ -9}. What is [I^{-}] at equilibrium?

PbI_{2}(s) |
Pb^{2+}(aq) |
+2 I^{-}(aq) | ||

I | some | 0 | 0 | |

C | -x | x | 2x | |

E | some | x | 2x |

K = a(Pb^{2+}) * a(I^{-})^{2} / a(PbI_{2}) = (x)*(2x)^{2} / 1 = 8.3 x 10^{-9}

So x = 1.28 x 10^{-3}

And [I^{-}] = 2.55 x 10^{-3} M.

[18]4. Consider the equilibrium and the given data:

COCl_{2}(g) |
¾ | CO(g) + | Cl_{2}(g) | |

DHº_{f} (kJ/mol) |
? | -110.5 | 0 | |

Sº_{298} (J/mol.K) |
? | 198 | 223 |

_{}By measuring the total pressure in the flask at two different temperatures, you are to calculate the missing entries in
the above table for COCl_{2}(g).

[1]a. Why is DHº_{f}(Cl_{2}) = 0 kJ/mol?

By definition, the enthalpy of formation is the change in enthalpy when 1 mole of the compound is made from its
elements in their standard states. So converting Cl_{2}(g)into Cl_{2}(g) involves no change.

[1]b. Why is Sº_{298}(Cl_{2}) > Sº_{298}(CO)?

Cl_{2} is a more complex molecule in that it contains many more electrons, which must be ordered about the two nuclei
than does CO.

[1]c. 100 kPa of COCl_{2} is placed in a 4.00 L flask at 300 K and heated to 800 K. What is the pressure of COCl_{2} at
800 K?

P_{1} / T_{1} = P_{2} / T_{2}. So P_{2} = 267 kPa.

[2]d. Define K and K_{p} for this reaction.

K = a(CO)*a(Cl_{2}) / a(COCl_{2})

K_{p} = P(CO)*P(Cl_{2}) / P(COCl_{2}) kPa.

[3]e. At this temperature COCl_{2} dissociates according to the equation given. At equilibrium, the TOTAL pressure in
the flask is found to be 381.8 kPa. Calculate K_{p} and then K_{800} at this temperature.

COCl_{2} |
CO | Cl_{2} | ||

I (kPa) | 267 | 0 | 0 | |

C | (-1)x | (+1)x | (+1)x | |

E | 267 - x | x | x |

K_{p} = x^{2} / (267 - x)

Need K_{p} and x! Two unknowns, need another equation:

Total pressure at equilibrium = 381.8 kPa = 267 -x +x +x = 267 + x

So x = 115.1

And K_{p} = 87.22 kPa.

From which K = 0.8722.

[1]f. The reaction flask is heated to 900 K. If the COCl_{2} were not dissociated, what would the pressure of it be at this
temperature?

P_{1} / T_{1} = P_{2} / T_{2}. So P_{2} = 300 kPa.

[3]g. In fact the COCl_{2} does dissociate and the TOTAL pressure at this temperature is found to be 515.8 kPa.
Calculate K_{p} at this temperature and K_{900}.

COCl_{2} |
CO | Cl_{2} | ||

I (kPa) | 300 | 0 | 0 | |

C | (-1)x | (+1)x | (+1)x | |

E | 300 - x | x | x |

K_{p} = x^{2} / (300 - x)

Need K_{p} and x! Two unknowns, need another equation:

Total pressure at equilibrium = 515.8 kPa = 300 -x +x +x = 300 + x

So x = 215.8

And K_{p} = 553 kPa.

From which K = 5.53.

[3]h. From the two values of K at different temperatures (K_{800} and K_{900}), calculate DHº and DSº for the reaction. (Note:
**If you do not** have calculated values for K_{800} and K_{900} use K_{800} = 1.00 and K_{900} = 6.00 in this part. **These are not the
calculated answers!)**

R = 8.314 J. mol^{-1}. K^{-1}.

ln K = DS/R - DH/RT

or

ln (K_{1} / K_{2}) = (DH/R)(1/T_{2} - 1/T_{1})

from which DH = ln (K_{1} / K_{2})*R / (1/T_{2} - 1/T_{1}) = 110,600 J

then using ln (5.53) = DS/R - 110,600/RT gives DS = 137 J/K

[3]i. Now calculate the missing table entries for COCl_{2}, DHº_{f} and Sº_{298}.

DHº = Sm_{i}DHº_{f}(i) = (-1)(X) + (+1)(-110.5) + (+1)(0) = 110.6 kJ (remember UNITS)

from which DHº_{f}(COCl_{2}) = X = -221 kJ

DSº = Sm_{i}Sº_{298}(i) = (-1)(X) + (+1)(198) + (+1)(223) = 137 J/K

from which Sº_{298}(COCl_{2}) = X = 284 J/K

[3]5. Predict whether BF_{3} will react with water or not under standard conditions:

BF_{3}(g) + 3 H_{2}O Ž H_{3}BO_{3}(s) + 3 HF(l)

DHº_{f} -1137 -286 -1094 -271 kJ/mol

Sº_{298} 254 70 89 174 J/mol.K

This reaction will be spontaneous of DGº_{298} < 0.

DHº = Sm_{i}DHº_{f}(i) = (-1)(-1137) + (-3)(-286) + (+1)(-1094) + (+3)(-271) = 88 kJ

DSº = Sm_{i}Sº_{298}(i) = (-1)(254) + (-3)(70) + (+1)(89) + (+3)(174) = 147 J/K

from which DGº_{298} =DHº - TDSº = 88 - 298*147/1000 = 44.2 kJ (watch UNITS).

Since DGº_{298} is positive, the reaction is not spontaneous under standard conditions at 298 K.

[7]6. Consider the process: N_{2}O_{4}(g) ¾ 2 NO_{2}(g). (endothermic)

[4]a. At a certain temperature, P(N_{2}O_{4}) = 30 kPa, and P(NO_{2}) = 120 kPa. at equilibrium. If the volume of the
container is doubled at constant temperature, calculate the **new** equilibrium partial pressures.

N_{2}O_{4} |
2 NO_{2} | |

I (after the volume change) | 15 | 60 |

C | (-1)x | (+2)x |

E | 15 - x | 60 + 2x |

K_{P} = P^{2}(NO_{2}) / P(N_{2}O_{4}) kPa = (60 + 2x)^{2} / (15 - x)

At this point there are two unknowns K_{P} and x. However since we know that before the volume change the system
was at equilibrium, we can calculate K_{P}:

K_{p} = (120)^{2} / 30 kPa = 480 kPa.

Now the quadratic can be solved giving x = 4.88

and P(NO_{2}) = 69.8 kPa and P(N_{2}O_{4}) = 10.1 kPa

[3]b. Complete the following table for this equilibrium. (Use **UP, DOWN, NONE**)

Change | Affect on P(N_{2}O_{4}) |
Affect on K. |

Addition of more N_{2}O_{4} at constant
volume and temperature. |
UP | NONE |

Addition of a catalyst | NONE | NONE |

Heating the equilibrium mixture | DOWN | UP |