Chapter 16: Thermodynamics

1
Chapter 16 Thermodynamics

• We are interested in the changes in thermodynamic
  properties that occur during a chemical reaction.
• Question How do we know when a reaction will
  occur spontaneously (no outside intervention)?
• We have already seen that a negative ?H tends to
  be favorable.

2

• Ex Consider the melting of ice at 25C.
• This is observed to be a spontaneous process.
• What do we know about the sign of ?H for this
  process (is it endothermic or exothermic)?
• The reaction requires that heat be absorbed,
  i.e., it is an endothermic process.
• Since the process is spontaneous, there must be
  other factors to consider besides ?H.

3

• Entropy, S another state function (Clausius,
  1854)
• Definition S k ln w
• k Boltzmanns constant 1.38 x 1023 J/K
• R /No (8.314 J/mol K) / (6.022 x 1023
  mol1)
• w probability of a given state occurring
• ex consider 2 coins being tossed
• HH 0.25 probability
• HT or TH 0.50 probability
• TT 0.25 probability
• S is related to disorder or randomness.
  Disorder is more probable than order.

4
Number of Equivalent Combinations for Various
Types of Poker Hands

• Hand
  W
  ln W
• Royal flush (AKQJ10 in one suit)
  4 1.39
• Straight flush (five cards in sequence in one
  suit) 36 3.58
• Four of a kind
  624 6.44
• Full house (three of a kind plus a pair
  3,744 8.23
• Flush (five cards in the same suit)
  5,108 8.54
• Straight (five cards in sequence)
  10,200 9.23
• Three of a kind
  54,912 10.91
• Two pairs
  123,552 11.72
• One pair
  1,098,240 13.91
• No pairs
  1,302,540 14.08
• 
  
• Total
  2,598,960

5
Figure 16.5Entropy
6

• The Second Law of Thermodynamics
• In a spontaneous process, there is an increase in
  the entropy of the universe.
• ?Suniverse ?Ssystem ?Ssurroundings
• After a spontaneous process, it is impossible to
  restore the system to its initial state without
  the input of energy.
• In any energy transaction, there is a decrease in
  the amount of useful energy available for further
  transactions.

7

• Many times, we can qualitatively recognize what
  the sign of ?S will be. Some examples
• Solid ? Liquid or Liquid ? Gas
• Pure substances ? Mixture
• The absolute temperature increases.
• The volume of a gas increases.
• Molecular complexity increases (H2 vs. C8H18).
• A larger number of particles is formed.
• ex 4 C3H5(NO3)3 ? 6 N2 10 H2O 12 CO2 O2

8
Figure 16.6H2O Molecule
9

• Now, how can we calculate ?S?
• For reversible (equilibrium) reactions
• ?S qrev / T in J/K
• Ex A reversible phase change at constant T P
• H2O(s) ? H2O(l) at 273K and 1
  atm
• ?Hfusion 6.00 kJ/mol qp
• ?S (6.00 x 103 J/mol) / 273 K
• 22.0 J/K
• The positive ?S value is favorable.

10

• What if it is an irreversible (spontaneous)
  process?
• Then ?S gt qirrev / T
• Are there any other methods for calculating ?S?
• The Third Law of Thermodynamics
• The entropy of a perfect crystal is zero at 0 K.
• This allows us to establish a scale of absolute
  entropies. Standard entropies of substances at
  298 K have been calculated and are found in
  tables (Appendix 4).

11

• Ex for I2(s), S 116.1 J/mol K
• So, how do we make use of standard entropies?
• Recall that ?H for a reaction can be calculated
  from the standard enthalpies of formation of the
  reactants and products.
• The ?S for a reaction is calculated in a similar
  manner

12

• Ex 2 H2(g) O2(g) ? 2 H2O(l)
  at 298 K
• S, J/mol K 130.6 205.0
  69.91
• ?S (2 mol)(69.91 J/mol K)
• (2 mol)(130.6 J/mol K) (1 mol)(205.0 J/mol
  K)
• 326.4 J/K
• Why is ?S negative?
• 1) 3 moles ? 2 moles
• 2) gases ? liquid
• The negative ?S is unfavorable, yet this is a
  spontaneous reaction.