TOPIC E: ENTROPY

• 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.

• 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.

• We might also assume that a reaction with a the reactants. + ΔH would be spontaneous since the products have a higher enthalpy than • • 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.

• 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

• 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

• 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

• • • Practice: • 1. Predict the sign of ΔS in these reactions. (a) MgO(s) + H 2 O(l)  (b) Na 2 CO 3 (s)  Na 2 Mg(OH) O(s) + CO 2 2 (s) (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.

• The change in entropy (ΔS) in a reaction can be found by subtracting the entropy of the reactants from the entropies of the products å

(

products)-

(

reactants) =

o • 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)

• 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  N 2 (g) = 192 J / K  H 2 (g) = 131 J / K  mol mol 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)

• Gibbs Free Energy • • Enthalpy and entropy are brought together in the Gibbs free energy equation. D Gº = D Hº - T D 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.

• • 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 at a measureable rate. ΔG o may not necessarily occur • When a reaction is thermodynamically favored, lots of products will form, and K (the equilibrium constant) will be greater than 1.

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

• A value for ΔG of a reaction can be calculated by using the following equation. å

(

f o

products)-

(

f o

reactants) =

o • 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.

• 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)

• 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 .

• • • Biochemistry Example: • The addition of phosphate group to a glucose molecule.

It can be summarized: PO 4 3 ΔG o + glucose   = +13.8 kJ / mol glucose-6-phosphate (reaction 1) 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 favored. and is thermodynamically

• 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 .

• 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)

• 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 on a new ΔG value under new conditions value, will take • 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

• 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.