1. TOPIC E: ENTROPY

2. Does a reaction with a – ΔH always proceed spontaneously since the products have a lower enthalpy than the reactants and are more stable? If a reaction has a very high E a (energy of activation), then it will not occur and is described as kinetically stable. In these circumstances it’s possible to predict this reaction as highly likely, yet it produces little or no products.

3. For example: The oxidation of ammonia by oxygen to produce nitrogen monoxide and water. NH 3(g) + O 2(g)  NO (g) + H 2 (g) The ΔH for this reaction is - 909 kJ / mol showing ammonia and oxygen to be very unstable. However the reaction does not proceed due to the high E a Reactants are kinetically stable Reaction does not occur.

4. We might also assume that a reaction with a + ΔH would be spontaneous since the products have a higher enthalpy than the reactants. Some endothermic reactions do proceed like dissoving KCl (s) in water: KCl(s)  KCl(aq) ΔH = + 19.2 kJ/mol There is also an Ea that must be obtained before the reaction will proceed. We can say that the ΔH and Ea are not the only factors that determine whether a chemical reaction will (or won’t), take place.

5. For the process of KCl solid dissolving: there is a change from an ordered to a less ordered state solid (s)  solution (aq) The degree of disorder in a reaction is called the entropy (S) In the reaction above there has been an increase in entropy, +ΔS

6. A pure, perfect crystal at 0 K (absolute zero) is assigned an absolute entropy = 0 It is completely ‘organized’ or completely ‘ordered’. Entropies of substances not at 0 K are measured relative to that All have values that are greater than zero. Important Factors in Entropy a. the state b. number of particles c. volumes of gases d. temperature

7. A. entropy of solids < entropy of liquids <<< entropy of gases B. entropies of small numbers of particles are less than entropies of large number of particles (This is called positional entropy – there are a greater number of possible positions for molecules) more particles = greater entropy C. entropies of gases with smaller volumes are lower than the entropies of gases with larger volumes since the ones with larger volumes have more space to disperse and move around in D. entropy increases with increasing temperature since the particles can move around more and are more dispersed

8. Practice: 1. Predict the sign of ΔS in these reactions. (a) MgO(s) + H 2 O(l)  Mg(OH) 2 (s) (b) Na 2 CO 3 (s)  Na 2 O(s) + CO 2 (g) 2. Which of the following has the greater entropy? (a) A metal at 273 K, or the same metal at 400 K (b) A flexible soft metal like lead, or diamond, a rigid solid. (c) Two samples of the same gas, at the same temperature, but one with at a pressure of 1 atm and the other at a pressure of 0.5 atm.

9. The change in entropy (ΔS) in a reaction can be found by subtracting the entropy of the reactants from the entropies of the products For example, calculate the entropy change, ΔS, for the formation of ammonia below. The absolute entropies of the chemicals involved are; 193 J/K mol-1 for NH3(g) 192 J/K mol-1 for N2(g) 131 J/K mol-1 for H2(g)

10. For example, calculate the entropy change, ΔS, for the formation of ammonia below. The absolute entropies of the chemicals involved are: NH 3 (g) = 193 J / K  mol N 2 (g) = 192 J / K  mol H 2 (g) = 131 J / K  mol N 2 (g) + 3H 2 (g)  2NH 3 (g) ΔS rxn = [ 2(193) ] – [ 192 + 3(131) ] = -199 J/K This makes sense since the negative value implies that the system has become more ordered (four gas molecules are converted to two gas molecules)

11. Gibbs Free Energy Enthalpy and entropy are brought together in the Gibbs free energy equation.  Gº =  Hº - T  Sº Entropy values tend to be given in units that involve J Enthalpy values tend to be given in units that involve kJ. When performing calculations that involve both entropy and enthalpy, remember to convert one unit to match the other.

12. All thermodynamically favored chemical reactions have a - ΔG o. The equation on the previous slide indicates: - ΔG o is favored by a - ΔH o and a + ΔS o. A thermodynamically favored reaction may still be a very slow one if the Ea is very high So a reaction that has a - ΔG o may not necessarily occur at a measureable rate. When a reaction is thermodynamically favored, lots of products will form, and K (the equilibrium constant) will be greater than 1.

13. Analysis of the possible sign combinations of ΔH o, ΔS o and ΔG o.

14. A value for ΔG of a reaction can be calculated by using the following equation. If ΔG o is (+) positive the reaction is not thermodynamically favored K<1 reactants are favored If ΔG o is (0) zero the reaction is favored equally in both the forward and backward directions reaction is at equilibrium.

15. Forcing non-thermodynamically favored reactions to occur A reaction with a positive ΔG o can be forced to occur by applying energy from an external source. Three such examples: 1. Using electricity in the process of electrolysis 2. Using light to overcome a highly endothermic ionization energy or light initiated photosynthesis in the equation below that has a ΔG o = +2880 kJ / mol 6CO 2 (g) + 6H 2 O(l)  C 6 H 12 O 6 (aq) + 6O 2 (g)

16. 3. The coupling of a thermodynamically unfavorable reaction ( + ΔG o ) to a favorable one ( - ΔG o ) taken together the reactions combine to form an overall reaction with a favorable, - ΔG o.

17. Biochemistry Example: The addition of phosphate group to a glucose molecule. It can be summarized: PO 4 3- + glucose   glucose-6-phosphate (reaction 1) ΔG o = +13.8 kJ / mol This reaction has a +ΔG o value and is not thermodynamically favored. Another reaction of cells is the conversion of ATP to ADP ATP   ADP + PO 4 3- (reaction 2) ΔG o = -30.5 kJ / mol This reaction has a – ΔG o and is thermodynamically favored.

18. If a cell can simultaneously: convert ATP to ADP (- ΔG o ) and use the PO 4 3- generated to cause ( +ΔG o ) reaction 1 to go as long as the - ΔG o (rxn 2) exceeds the + ΔG o (rxn 1), then the overall reaction will have a – ΔG. ATP + glucose   glucose-6-phosphate + ADP (overall reaction) ΔG o = +13.8 + (-30.5) = -16.7 kJ / mol The process of combining non-thermodynamically favored and thermodynamically favored reactions to produce an overall thermodynamically favored reaction, is called coupling.

19. Consider the relationship between ΔG o and K. ΔG o = -RT ln K This shows us that a large – ΔG o will lead to large + K (In unit 6, we will see how large, positive values of K show that reactions are very likely to occur)

20. You should understand that conditions not at standard ones, may cause: a favored reaction to produce very few products a non-favored one to produce products ΔG o assumes standard conditions of 1 atm for gases and 1 M for solutions So, a previously positive (or negative) ΔG o value, will take on a new ΔG value under new conditions In some cases (where ΔG o is close to 0) the sign will change too, causing a previously non-favored reaction to be come favorable or vice-versa

21. For reactions not at standard state use: ΔG = ΔG o + RT ln Q This is no longer on the equation sheet, so might not be on the new Exam.