CHAPTER 12: Thermodynamics Why Chemical Reactions Happen

CHAPTER 12: Thermodynamics Why Chemical Reactions Happen Useful energy is being "degraded" in the form of unusable heat, light, etc. A tiny fraction of the sun's energy is used to produce complicated, ordered, highenergy systems such as life Our observation is that natural processes proceed from ordered, high-energy systems to disordered, lower energy states. In addition, once the energy has been "degraded", it is no longer available to perform useful work. It may not appear to be so locally (earth), but globally it is true (sun, universe as a whole). 1

Thermodynamics - quantitative description of the factors that drive chemical reactions, i.e. temperature, enthalpy, entropy, free energy. Answers questions such as- will two or more substances react when they are mixed under specified conditions? if a reaction occurs, what energy changes are associated with it? to what extent does a reaction occur to? Thermodynamics does NOT tell us the RATE of a reaction Chapter Outline 12.1 Spontaneous Processes 12.2 Entropy 12.3 Absolute Entropy and Molecular Structure 12.4 Applications of the Second Law 12.5 Calculating Entropy Changes 12.6 Free Energy 12.7 Temperature and Spontaneity 12.8 Coupled Reactions 4 2

Spontaneous Processes A spontaneous process is one that is capable of proceeding in a given direction without an external driving force A waterfall runs downhill A lump of sugar dissolves in a cup of coffee At 1 atm, water freezes below 0 0 C and ice melts above 0 0 C Heat flows from a hotter object to a colder object A gas expands in an evacuated bulb Iron exposed to oxygen and water forms rust Spontaneous chemical and physical changes are frequently accompanied by a release of heat (exothermic H < 0) - C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) H o = -2200 kj 3

But sometimes a spontaneous process can be endothermic H > 0 - Some processes are accompanied by no change in enthalpy at all ( H o = 0), as is the case for an ideal gas spontaneously expanding: spontaneous nonspontaneous 4

There's another factor promoting spontaneity in these processes, and that's the increasing randomness or disorder of the system (this is a qualitative description only quantitative coming up): 1. propane combustion: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(l) H o = -2200 kj 2. water melting: H 2 O(s) H 2 O(l) H o = 6.01 kj 3. gas expansion: Chapter Outline 12.1 Spontaneous Processes 12.2 Entropy 12.3 Absolute Entropy and Molecular Structure 12.4 Applications of the Second Law 12.5 Calculating Entropy Changes 12.6 Free Energy 12.7 Temperature and Spontaneity 12.8 Coupled Reactions 5

Thermodynamics: Entropy Second Law of Thermodynamics: The entropy of the universe increases in any spontaneous process S univ = S sys + S surr 0 Entropy (S): A measure of the amount of disorder (qualitative), or unusable energy in a system at a specific temperature (quantitative). Entropy is affected by molecular motion, or disorder from volume changes (e.g. the previous gas expansion example). Types of Molecular Motion Three types of motion: Translational: Movement through space Rotational: Spinning motion around axis perpendicular to bond Vibrational: Movement of atoms toward/away from each other As temperature increases, the amount of motion increases. 6

Third Law of Thermodynamics The entropy of a perfect crystal is zero at absolute zero Provides a point of reference or baseline for quantitating entropy (placing a numerical value on it) Heat plays a role in the amount of entropy a system has 18-13 Chapter Outline 12.1 Spontaneous Processes 12.2 Entropy 12.3 Absolute Entropy and Molecular Structure 12.4 Applications of the Second Law 12.5 Calculating Entropy Changes 12.6 Free Energy 12.7 Temperature and Spontaneity 12.8 Coupled Reactions 7

Standard Molar Entropy, S o (absolute entropy content, J/mol K The entropy of one mole of a substance in its standard state at 1 atm and 298 K. S o solid < S o liquid < S o gas Trends in S: Phase Changes S solid < S liquid < S gas S = S final - S initial Units: J/mol K 8

The Effect of Molecular Structure on S Summary: S is expected to INCREASE for these types of processes ( S > 0) : 9

Sample Exercise 12.1: Predicting the Sign of S Predict whether or not an increase or decrease in entropy accompanies each of these processes when they occur at constant temperature: (a)h 2 O(l) H 2 O(g) (b)nh 3 (g) + HCl(g) NH 4 Cl(s) (c)c 12 H 22 O 11 (s) C 12 H 22 O 11 (aq) Sample Exercise 12.2: Comparing Standard Molar Entropy Changes Without consulting any standard reference sources, select the component in each of the following pairs that has the greatest standard molar entropy at 298 K. Assume that there is one mole of each component in its standard state (the pressure of each gas is 1 bar and the concentration of each solution is 1 M). (a)hcl(g), HCl(aq) (b)ch 3 OH(l), CH 3 CH 2 OH(l) 10

The Second Law of Thermodynamics: The total entropy of the universe increases in any spontaneous process Spontaneous process: sys = system surr = surroundings S univ = S sys + S surr 0 One can be negative but the other will be even more positive Entropy Changes in the Surroundings ( Ssurr) Exothermic Process S surr > 0 Endothermic Process S surr < 0 12

The change in entropy of the surroundings can be calculated: S surr - H sys S surr 1 T surr if H sys < 0 (exothermic), then S surr > 0 (entropy of the surroundings increases) if H sys > 0 (endothermic), then S surr < 0 (entropy of the surroundings decreases) If the temperature of the surroundings is already high, then pumping heat in or out causes less change in disorder than at lower temperatures Combining the two: S surr - H sys and S surr 1 T surr T so S surr = - H sys (T surr usually = T sys ) e.g. N 2 (g) + 3H 2 (g) 2NH 3 (g) S sys = -198.3 J/K H sys = -92.6 kj * The two main driving forces are in opposition to each other - the release of heat favors a spontaneous reaction while the decrease in entropy does not. Calculating S univ will decide the issue (next slide). Remember: for a spontaneous reaction the entropy of the universe increases. *from Ch 9: H 0 rxn = Sn H 0 f (products) - Sm Hf 0 (reactants) 13

Is the reaction spontaneous at 25 o C? S univ = S sys + S surr The previous example with ammonia illustrated that maybe entropy will decrease in the system, but this will always be accompanied by a greater increase in the entropy of the surroundings such that S univ > 0. S univ = S sys + S surr 0 Another way of stating the 2nd Law is that "You Can't Win!" 14

Chapter Outline 12.1 Spontaneous Processes 12.2 Entropy 12.3 Absolute Entropy and Molecular Structure 12.4 Applications of the Second Law 12.5 Calculating Entropy Changes 12.6 Free Energy 12.7 Temperature and Spontaneity 12.8 Coupled Reactions Standard Entropy of Reaction ( Srxn) The standard entropy of reaction ( S 0 rxn) is the entropy change for a reaction carried out at 1 atm and 25 0 C. aa + bb cc + dd S 0 rxn = [ cs 0 (C) + ds 0 (D)] - [ as 0 (A) + bs 0 (B)] S 0 rxn = SnS 0 (products) - SmS 0 (reactants) 15

Sample Exercise 12.3: Calculating S o Values Given the following standard molar entropy values at 298 K (found in Appendix 4, Table A4.3), what is the of S o rxn for the dissolution of ammonium nitrate under standard conditions? NH 4 NO 3 (s) NH 4+ (aq) + NO 3- (aq) So [J/mol K] 151.1 113.4 146.4 Chapter Outline 12.1 Spontaneous Processes 12.2 Entropy 12.3 Absolute Entropy and Molecular Structure 12.4 Applications of the Second Law 12.5 Calculating Entropy Changes 12.6 Free Energy 12.7 Temperature and Spontaneity 12.8 Coupled Reactions 16

Gibbs Free Energy S univ = S sys + S surr Could use this relation to calculate reaction spontaneity, but not always easy to calculate S surr - so the expression is rearranged to only include terms relating to the system: Under constant temperature and pressure: Gibbs free energy (G) G = H sys -T S sys G < 0 G > 0 G = 0 The reaction is spontaneous in the forward direction. The reaction is nonspontaneous as written. The reaction is spontaneous in the reverse direction. The reaction is at equilibrium. 17

Calculating Free-Energy Changes Using G = H sys -T S sys aa + bb cc + dd G = H sys -T S sys o H rxn = Σn prod H f o prod Σn react H f o react o S rxn = Σn prod S f o prod Σn react S f o react Sample Exercise 12.4: Predicting Reaction Spontaneity under Standard Conditions Consider the reaction of nitrogen gas and hydrogen gas at 298 K to make ammonia at the same temperature: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) (a) Before doing any calculations, predict the sign of So rxn (b) What is the actual value of S o rxn? (c) What is the value of H o rxn? (d) What is the value of G o rxn? (e) Is the reaction spontaneous at 298 K and 1 atm? 18

Calculating Free-Energy Changes Using o G rxn = Σn prod G f o prod Σn react G f o react Standard free energy of formation ( G 0 ) is the free-energy f change that occurs when 1 mole of the compound is formed from its elements in their standard states. G 0 of any element in its most f stable allotropic form is zero, e.g. graphite and not diamond o Sample Exercise 12.5: Calculating G rxn Appropriate G o f Values Using Use the appropriate standard free energy of formation values in App. 4 to calculate the change in free energy as ethanol burns under standard conditions. Assume the reaction proceeds as described by the following chemical equation: CH 3 CH 2 OH(l) + 3 O 2 (g) 2 CO 2 (g) + 3 H 2 O(l) o G rxn = Σn prod G f o prod Σn react G f o react 19

Chapter Outline 12.1 Spontaneous Processes 12.2 Entropy 12.3 Absolute Entropy and Molecular Structure 12.4 Applications of the Second Law 12.5 Calculating Entropy Changes 12.6 Free Energy 12.7 Temperature and Spontaneity 12.8 Coupled Reactions Temperature and Spontaneity G = H T S G = T S + H rearranging terms to match the general formula for a straight line - G = S T + H y = m x + b m = S b = H At what temperature does the reaction spontaneity change? 20

G At what temperature does the reaction spontaneity change? G = H T S When the reaction spontaneity changes, the sign of G changes from to, passing through on the way. Therefore G = 0 at the temperature that spontaneity changes, so - = H T S H = T S and and therefore T = G = H - T S Always Spontaneous: H < 0 and S > 0 b = H (negative) m = - S (overall negative) 0.0 G < 0 spontaneous at all temperatures b = H m = - S T 21

G G G = H - T S Never Spontaneous: H > 0 and S < 0 b = H (positive) m = - S (overall positive) G > 0 not spontaneous at all temperatures m = - S b = H 0.0 T Enthalpy-driven: spontaneous at low T H < 0 and S < 0 G = H - T S b = H (negative) m = - S (overall positive) m = - S G < 0 spontaneous only at low T 0.0 G < 0 b = H T = H S T 22

G G = H - T S Entropy-driven: spontaneous at high T H > 0 and S > 0 b = H (positive) m = - S (overall negative) b = H m = - S G < 0 spontaneous only at high T 0.0 G < 0 T = H S T G = H - T S 23

G Homework Problem #80 The element H 2 is not abundant in nature, but it is a useful reagent in, for example, the potential synthesis of the liquid fuel methanol from gaseous carbon monoxide. Under what temperature conditions if this reaction spontaneous? 2 H 2 (g) + CO(g) CH 3 OH(l) G = 0 T 24