1 Chemistry 123/125 Ch. 14 In-Class Exercise Many physical and chemical processes proceed naturally in one direction, but not in the other. In other words, these processes are spontaneous in the direction in which they proceed naturally. For example, water flows downhill, not uphill; ice melts on a warm day, water does not freeze on a warm day. This exercise explores the factors that determine the direction in which reactions proceed spontaneously. 1. a) Write a chemical equation that describes the melting of ice. b) Is the enthalpy change, ΔH, for this process positive negative zero c) Under what temperature conditions does this process proceed spontaneously? 2. a) Write a chemical equation that describes the freezing of water. b) Is the enthalpy change, ΔH, for this process positive negative zero c) Under what temperature conditions does this process proceed spontaneously?
2 3. a) Is it possible to determine whether a process will occur spontaneously solely by examining the sign of ΔH for the process? Explain. b) What other factor(s) must be considered to determine whether or not a process will occur spontaneously under a given set of conditions? Many spontaneously-occurring processes tend to be exothermic, but this is not a requirement. The temperature at which a process occurs also plays a role in determining in which direction a process will proceed spontaneously. Thus there appear to be two important factors in this determination: ΔH of the reaction and another factor whose impact is influenced by the temperature. This second factor, known a entropy, S, is a measure of disorder or randomness. The more disorder, the higher (more positive) the entropy. Consider the following gaseous molecules in a box. (a) (b) (c) Sa Sb Sc 4. In which picture do the molecules have the greatest entropy? (a) (b) (c) 5. Which sequence is most likely the naturally occurring process? (a) to (b) to (c) or (c) to (b) to (a)
3 6. Which of the following inequalities is correct? Sa > Sc or Sc > Sa Figure 2. The dissolution of sodium chloride in water. 1 mole Na + (aq) + 1 mole Cl - (aq) H ΔHº = + 3.86 kj 1 mole NaCl(s) The dissolution of sodium chloride in water is endothermic, yet it proceeds spontaneously at room temperature. Clearly, the driving force for this reaction must be the entropy change. If we did not know that this process is spontaneous, could we predict the direction of entropy change during the process? The following are some generalizations that can often be used in considering the entropy associated with a chemical species, or the change in entropy (ΔS) associated with a chemical reaction: As the number of particles in the system increases, the amount of disorder increases (ΔS is positive). As the volume in which particles can move increases, the amount of disorder increases (ΔS is positive). As the temperature of a system increases, the motion of the particles and the amount of disorder increases (ΔS is positive).
4 7. For the process illustrated in Figure 2 (the dissolution of sodium chloride) is ΔS positive or negative? Explain. 8. For the process described by the following chemical reaction is ΔS positive or negative? Explain. C4H8(g) 2 C2H4(g) 9. For the process described by the following chemical reaction is ΔS positive or negative? Explain. C4H8(g; 298K; 1 atm) C4H8(g; 298K; 0.5 atm) 10. For the process described by the following chemical reaction is ΔS positive or negative? Explain. C4H8(g; 298K; 1 atm) C4H8(g; 398K; 1 atm) 11. Rank the following in order of increasing entropy. H2O(g); H2O(l); H2O(s);
5 12. For each of the following processes, predict whether ΔS 0 is expected to be positive or negative. Explain your reasoning. a) N2(g) + 3 H2(g) 2 NH3(g) b) CO2(g) CO2(s) c) CaCO3(s) CaO(s) + CO2(g) d) The air in a balloon escapes out a hole and the balloon flies wildly around the room. (Consider ΔS for the air molecules originally in the balloon.) e) A precipitate of Pb(OH)2 forms when solutions of lead(ii) nitrate and sodium hydroxide are mixed.
6 The change in entropy, ΔS 0, for a process is the difference between the entropy of the products and that of the reactants at 1 atmosphere of pressure and 25 C. S increasing entropy ½ mole N2(g) + 1 mole O2(g) 1 mole NO2(g) ΔS = -60.3 J/K entropy of half a mole of N 2 (g) and one mole of O 2 (g) entropy of one mole of NO 2 (g) ΔS = +240.45 J/K - [½ (191.5 J/K) + (205.0 J/K)] = -60.3 J/K Table 1. Standard entropy values, Sº, in J/K Substance Sº (J/K) Substance Sº (J/K) Substance Sº (J/K) N2(g) 191.5 H2O(g) 188.83 C6H6(g) 269.2 O2(g) 205.0 H2O(l) 69.91 C6H6(l) 172.8 NO2(g) 240.45 CCl4(g) 309.4 H2(g) 130.58 N2O4(g) 304.3 CCl4(l) 214.4 NH3(g) 192.5 13. Do you expect ΔSº for the following reaction to be positive or negative? Explain. ½ N2(g) + O2(g) NO2(g) 14. Why are all of the values for entropies positive?
7 15. Why do the entropies generally become larger as the number of atoms in the molecule increases? 16. Why are the entropies smaller for liquids than the corresponding entropies for gases? 17. Do you expect ΔSº for the following reaction to be positive or negative? Explain. N2O4(g) 2 NO2(g) 18. Calculate ΔSº for the following reaction using the data in Table 1. N2O4(g) 2 NO2(g) 19. Calculate ΔSº for the following reaction using the data in Table 1. N2(g) + 3 H2(g) 2 NH3(g) Adapted from CHEMISTRY, A Guided Inquiry by Richard S. Moog and John J. Farrell, preliminary edition.