Chapter 18 Thermodynamics: Directionality of Chemical Reactions Ludwig Boltzmann Famous for his equation statistically defining entropy. Josian W. Gibbs Pioneered concepts of chemical thermodynamics and free energy.

So far, we have tried to answer the following questions: (1) What are the energetics (heat) of a reaction? Is it exothermic (  H= -) or endothermic (  H= +)? (2) How fast (kinetics) and how (mechanism) does the reaction go? (3) To what extent does it go? (equilibrium) (4) Does it go, i.e., is it spontaneous? This is the subject of this chapter. And finally now ……..

Spontaneous processes: defined descriptively as a process that occurs by itself (and the reverse does not occur by itself) hotcold The opposite: hotcold is not spontaneous, but, it is possible (how does a refrigerator work?). Other spontaneous processes (ask yourself: does reverse ever occur by itself?) nail rusting eggs breaking (Humpty Dumpty) paper burning water freezing at -10 o C ice melting at +10 o C gases mix heat is spontaneous,

All of these spontaneous processes are also described as: irreversible There are reversible processes, but the systems must be at equilibrium. heat + ice → water at +10 o C Spontaneous, irreversible heat + ice ← water at -10 o C Spontaneous, irreversible heat + ice  water at 0 o C Reversible; equilibrium Both ice and water coexist at 0 o C Either process, → or ← can occur at equilibrium Irreversibility ═ Spontaneity

What makes a process spontaneous (irreversible)? Exothermic reactions tend to be spontaneous (exception, dissolving ammonium salts), and increasing entropy (randomness) tends to cause processes to be spontaneous; but overall Gibbs Free Energy must decrease in order for a process to be spontaneous. A B  G=(-)  G =  H - T  S Free energy Enthalpy Entropy

Expansion of an Ideal Gas FG19_004.JPG This is a statistical explanation of why increasing entropy is spontaneous. Consider an initial state: two flasks connected by a closed stopcock. One flask is evacuated and the other contains 1 atm of gas. The final state: two flasks connected by an open stopcock, and the gas distributes itself equally in both flasks; each flask Now contains gas at 0.5 atm. Why does the gas expand?

How many ways? A,B,C,DW=1 way How many ways? A,B,C D A,B,D C A,C,D B B,C,D A W=4 ways (none) How to distribute 4 particles (A,B,C,D) in two vessels?? Let W = number of ways A configuration is possible

How many ways? A, B C,D A,C B,D A,D B,C B,C A,D B,D A,C C,D A,B W=6 ways W=4 ways W=1 way

Recap: way ways ways ways way The point here is that there are more ways to have an even distribution of particles. With more than 4 particles (say Avogadro’s number), the number of ways have an even distribution is enormous!! (That’s why air on one side of a room doesn’t suddenly rush into the other side, asphyxiating everyone in the airless side). for 4 particles Left Right An equal distribution (2 and 2) has the greatest number of ways of distributing particles (6 ways).

So, again, why does the gas expand? Because there are many more ways of having gas molecules evenly divided in number between two chambers than having them all in one chamber. There is a greater and natural tendency toward randomness and disorder. Boltzmann: S = k ln W

Other examples: Ice melting – similar to gas expansion – more randomness and disorder, even though process is endothermic Ink drops in water – ink becomes evenly distributed in water; increase in randomness and disorder. Decay of biological organisms – increase in randomness and disorder. Dissolving of salts in water – increase in randomness and disorder To Summarize : what contributes to spontaneity? 1. Exothermic processes (heat is evolved). 2. Any process which increases randomness and disorder.

The thermodynamic quantity which describes randomness and disorder is called ENTROPY and denoted as S The SECOND LAW OF THERMODYNAMICS postulates the existence of entropy; it also states that the entropy of the universe is constantly increasing. It is not a conserved quantity. 1.Gases have more entropy than liquids, which have more entropy than solids. 2.Corollary: Melting, or vaporization, increases entropy. 3.Corollary: In a chemical rx., increasing the number of moles of a gas, increases the entropy (e.g., H 2 O(g)  H 2 (g) + ½O 2 (g)). 4. Dissolving or mixing increases entropy. 5. Corollary: precipitation decreases entropy. 6. Increasing the temperature increases entropy.

Thermodynamic Calculations What is ΔG for the oxidation of SO 2 to SO 3 at 25°C? Is the reaction spontaneous? exothermic? SO 2 (g) + ½O 2 (g)  SO 3 (g) ΔH f ° ΔH = kJ ΔS° ½(0.205) ΔS = kJ ΔG = ΔH – TΔS = – (298)( ) = = kJ The reaction is spontaneous. The reaction is exothermic. Note: Be sure you convert ΔS values from J to kJ

What is ΔG for the decarboxylation of limestone at 25°C? Is the reaction spontaneous? exothermic? CaCO 3 (s)  CaO(s) + CO 2 (g) ΔH f ° ΔH = kJ ΔS° ΔS = kJ ΔG = ΔH – TΔS = – (298)( ) = – = The reaction is not spontaneous. The reaction is endothermic. Note: Be sure you convert ΔS values from J to kJ

How do we make the decarboxylation of limestone spontaneous? Set ΔG = 0, the the crossing over point where the reaction converts from nonspontaneous to spontaneous. ΔG = 0 = ΔH – TΔS 0 = –T( ) T = 1109°K = 837°C When the temperature falls below 837°C, CO 2 begins spontaneously to react with CaO to form CaCO 3 : CaO(s) + CO 2 (g)  CaCO 3 (s) At room temperature, ΔG = kJ

Calculate the boiling point of methanol. CH 3 OH(l)  CH 3 OH(g) ΔH f ° ΔH = kJ ΔS ΔS = kJ At equilibrium, ΔG is always 0. ΔG = 0 = ΔH – TΔS =+38.0 –T(+0.113) T b = 336°K = 63.3°C Note: Be sure you convert ΔS values from J to kJ

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Additional aspects of Free Energy Even though a reaction has a negative  G it may occur too slowly to be observed (i.e. combustion). Thermodynamics gives us the direction of a spontaneous process, it does not give us the rate of the process. A nonspontaneous process can be driven if coupled with a spontaneous process – this is very important in life processes (i.e., respiration to form ATP), and can be used in industrial processes, such as smelting. To calculate K values, use ΔG° = -RT ln K eq. This refers to the ΔG difference of the standard states of compounds, before equilibrium is attained.

Calculating K eq from ΔG values R = J/mol-K Let’s calculate K eq from the following reaction, which we previously studied in the Equilibrium chapter: N 2 O 4 (g)  2NO 2 (g) ΔG f ° (51.8) ΔG = 5.3 kJ ΔG = -RT ln K 5300 = (407) ln K ln K = K = very close to the experimental value

Let’s calculate K sp of AgCl. AgCl(s)  Ag + (aq) + Cl - (aq) ΔG° = ΔG = ΔG = -RT ln K = -(8.314)(298) ln K = ln K K sp = 1.7 x – very close to the experimental value

Let’s take a look at the dissolution of NH 4 Cl. NH 4 Cl(s)  NH 4 + (aq) + Cl - (aq) ΔH f ° ΔH = ΔS° ΔS = ΔG = ΔH – T ΔS = 14.7 – (298)(0.0754) = 14.7 – = kJ The reaction is endothermic, but is spontaneous! Hence, ammonium chloride is excellent for cold packs.

Laws of Thermodynamics 1 st Law. Energy is neither created nor destroyed. In chemistry, chemical energy can be converted into heat and vice versa. 2 nd Law. Entropy increases spontaneous; i.e., the natural tendency is for randomization. 3 rd Law. The entropy of a perfect crystal at 0°K is zero (it is impossible to attain 0°K).