Chapter 17. Spontaneity, Entropy, and Free Energy

Chapter 17 Spontaneity, Entropy, and Free Energy

Thermodynamics Thermodynamics is the study of the relationship between heat and other forms of energy in a chemical or physical process. Thermodynamics can be used to predict if a reaction will occur. Kinetics tell us how fast the reaction will occur. 2

Thermodynamics vs. Kinetics Domain of Kinetics Rate of a reaction depends on the pathway from reactants to products. Domain of Thermodynamics Using the properties of reactants and products to predict whether a reaction is spontaneous or not. 3

First Law of Thermodynamics Energy can be converted from on form to another, but it cannot be created or destroyed. Enthalpy, H, is defined as H = U + PV, where U is internal energy of the system P is pressure of the system V is volume of the system Change in enthalpy, DH, equals the heat of reaction at constant pressure, q P. 4

Enthalpy and Internal Energy Many reactions take place at constant pressure, so the change in enthalpy can be given by: DH = DU + PDV The first law of thermodynamics can now be expressed as follows: DU = DH PDV The term ( PDV) is the energy needed to change volume against the atmospheric pressure, P. It is called pressure-volume work. 5

Enthalpy and Internal Energy For the reaction: 2 Na(s) + 2 H 2 O(l) 2 NaOH(aq) + H 2 (g) PDV The H 2 gas does pressure-volume work to raise the piston: At 1 atm, PDV = 2.5 kj In addition, 368.6 kj of heat are evolved. Internal energy of the system decreases: DU = DH PDV = 368.6 kj 2.5 kj = 371.1 kj 6

Spontaneous Processes To understand why a chemical reaction goes in a particular direction, we need to study spontaneous processes. A spontaneous process is a physical or chemical change that occurs by itself. These processes occur without requiring an outside force and continue until equilibrium is reached. 7

Spontaneous Processes 8

Spontaneous Processes spontaneous nonspontaneous 9

Spontaneous Physical and Chemical Processes 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 10

Does a decrease in enthalpy mean a reaction proceeds spontaneously? Spontaneous reactions CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (l) DH 0 = -890.4 kj H + (aq) + OH - (aq) H 2 O (l) DH 0 = -56.2 kj H 2 O (s) H 2 O (l) DH 0 = 6.01 kj NH 4 NO 3 (s) H 2 O NH 4 + (aq) + NO 3 - (aq) DH 0 = 25 kj 11

Does a decrease in enthalpy mean a reaction proceeds spontaneously? We cannot decide whether or not a chemical reaction will occur spontaneously solely on the basis of enthalpy changes in the system. To make predictions of spontaneity we need another thermodynamic quantity; ENTROPY 12

Entropy A measure of ecular randomness or disorder. Thermodynamic function that describes the number of arrangements that are available to a system existing in a given state. 13

Entropy Now consider a gas in a flask connected to an equal sized flask that is evacuated: When the stopcock is open, the gas will flow into the evacuated flask to give a uniform distribution because the expanded state has the highest number of arrangements available to the system. The kinetic energy has spread out, and the entropy of the system has increased. 14

Entropy spontaneous entropy increases nonspontaneous entropy decreases 15

Positional Entropy For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state. Therefore: S solid < S liquid << S gas 16

Spontaneous Processes Spontaneous processes lead to an increase in entropy. If the change from initial to final state results in an increase in randomness: DS = S f - S i S f > S i DS > 0 17

How does the entropy of a system change for each of the following processes? (a) Condensing water vapor Randomness decreases Entropy decreases (DS < 0) (b) Forming sucrose crystals from a supersaturated solution Randomness decreases Entropy decreases (DS < 0) (c) Heating hydrogen gas from 60 0 C to 80 0 C Randomness increases Entropy increases (DS > 0) (d) Subliming dry ice Randomness increases Entropy increases (DS > 0) 18

Second Law of Thermodynamics The entropy (S) of the universe always increases for a spontaneous process increases in a spontaneous process and remains unchanged in an equilibrium process. Helps explain why chemical reactions favor a particular direction. The entropy is, like enthalpy, is a state function. 19

Entropy Is a State Function State functions are properties that are determined by the state of the system, regardless of how that condition was achieved. For example: energy, enthalpy, entropy, pressure, volume, temperature Potential energy of hiker 1 and hiker 2 is the same even though they took different paths. 20

Third Law of Thermodynamics The third law of thermodynamics states: the entropy of a perfect crystalline substance is zero at absolute 0 K When temperature is raised, however, the substance becomes more disordered as it absorbs heat. The entropy of a substance is determined by measuring how much heat (thermal energy) is required to change its temperature per Kelvin degree. This allows us to determine the absolute entropies of substances. 21

Standard Entropies and the Third Law of Thermodynamics The standard entropy of a substance or ion, also called its absolute entropy, S o, is the entropy value for the standard state of the species. The symbol S o, rather than DS o, is used for standard entropies to emphasize that they are absolute, rather than relative to some arbitrary standard like enthalpy. Standard state implies 25 o C, 1 atm pressure, and 1 M concentration for dissolved substances. The SI unit of entropy is joules per Kelvin (J/K). Note that the elements have nonzero values, unlike standard enthalpies of formation, DH fo, which by convention, are zero. 22

Entropy Change for a Reaction The standard entropy of reaction ( S o rxn ) is the entropy change for a reaction carried out at 1 atm and 25 0 C. It is calculated using a summation law, similar to the way you obtained DH o : So rxn = n S o (products) n S o (reactants) rxn a A + b B c C + d D So rxn = [c S o C + d S o (D)] [a S o A + b S o (B)] 23

Entropy Change for a Reaction Even without knowing the values for the entropies of substances, you can sometimes predict the sign of S o for a reaction. The entropy usually increases in the following situations: 1. A reaction in which a ecule is broken into two or more smaller ecules. 2. A reaction in which there is an increase in the es of gases. 3. A process in which a solid changes to liquid or gas or a liquid changes to gas. 24

A Problem To Consider 17.1. Calculate the change in entropy, DS o, at 25 o C for the reaction in which urea is formed from NH 3 and CO 2. The standard entropies of the reactants and the products are given below their formulas. 2 NH 3(g) + CO 2(g) NH 2 CONH 2(aq) + H 2 O (l) S o : 193 J K 214 J K 174 J K 70 J K es: 2 1 1 1 25

A Problem To Consider 17.1. Continued. 2 NH 3(g) + CO 2(g) NH 2 CONH 2(aq) + H 2 O (l) S o : 193 J K 214 J K 174 J K 70 J K es: 2 1 1 1 S o = n S o (products) n S o (reactants) S o = 1 70 + 1 174 2 193 + 1 214 = 356 J/K 26

A Problem To Consider 17.2. What is the standard entropy change for the following reaction at 25 0 C? The standard entropies of the reactants and the products are given below their formulas. 2 CO (g) + O 2(g) 2CO 2(g) S o : 197.9 J K 205.0 J K 213.6 J K S o = 2 213.6 2 197.9 + 1 205.0 = 173.6 J/K 27

Entropy Changes of Universe It is convenient to divide the universe into a system and its surroundings. Thus the change in the entropy of the universe can be represented as: S univ = S sys + S surr 28

ΔS surr The magnitude of ΔS surr depends on the temperature. 30

ΔS surr The sign of ΔS surr depends on the direction of the heat flow. 31

ΔS surr At constant pressure change in enthalpy, DH, equals the heat flow: H = q p Therefore: S surr = H T 32

Spontaneous Reactions To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE 33

Free Energy Concept The American physicist J. Willard Gibbs introduced the concept of free energy (sometimes called the Gibbs free energy), G, which is a thermodynamic quantity defined by the equation: G = H TS For a process that occurs at constant temperature the change in free energy is given by the equation: G = H T S 34

Free Energy and Spontaneity G = H sys T S sys Divide both sides by T: G T = H sys T + S sys At constant pressure and temperature: S surr = H sys T Therefore: G T = S surr + S sys S univ = S surr + S sys S univ = G T 35

Free Energy and Spontaneity At constant T and P : S univ = G T Spontaneous process: Equilibrium process: DS univ > 0 DS univ = 0 DG < 0 DG > 0 DG = 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. 36

Standard Free-Energy Change The standard free energy change, DG o, is the free energy change that occurs when reactants and products are in their standard states: G o = H o T S o Changes in H an S during a reaction result in a change in free energy, DG, given by the equation Thus, if you can show that DG is negative at a given temperature and pressure, you can predict that the reaction will be spontaneous. 37

Effect of ΔH and ΔS on Spontaneity G o = H o T S o 38

DG = DH - TDS 39

A Problem To Consider 17.3. What is the standard free energy change, DG o, for the following reaction at 25 o C? Use values of DH fo and S o, from Tables in book. N 2(g) + 3 H 2(g) 2 NH 3(g) DH fo : 0 0 45.9 kj S o : 191.5 J K 130.6 J K 193 J K 40

17.3. (Continued). A Problem To Consider N 2(g) + 3 H 2(g) 2 NH 3(g) DH fo : 0 0 45.9 kj H o = n H o (products) n H o (reactants) H o = 2 45.9 kj = 91.8 kj 41

17.3. (Continued). A Problem To Consider N 2(g) + 3 H 2(g) 2 NH 3(g) S o : 191.5 J K 130.6 J K 193 J K S o = ns o (products) ns o (reactants) S o = 2 193 3 130.6 + 1 191.5 = 197 J K = 0.197 kj K 42

17.3. (Continued). A Problem To Consider N 2(g) + 3 H 2(g) 2 NH 3(g) H o = 91.8 kj S o = 0.197 kj K T = 298 K G o = H o T S o = 91.8 kj 298 K 0.197 kj K = 33.1 kj 43

Standard Free Energies of Formation The standard free energy of formation, DG fo, of a substance is the free energy change that occurs when 1 of a substance is formed from its elements in their standard states at 1 atm pressure and 25 o C. By tabulating DG fo for substances, you can calculate the DG o for a reaction by using a summation law: G o = n G f o m G f o products reactants 44

The standard free-energy of reaction (DG 0 rxn) is the freeenergy change for a reaction when it occurs under standardstate conditions: a A + b B c C + d D DG o rxn = [ cdg o f (C) + ddg o (D)] - [ adg o f (A) + bdg o f (B)] f DG o of any element in its most f stable form is zero. 45

A Problem To Consider 17.4. a) Calculate the standard free energy change, DG o, for the following reaction at 25 o C: 2 C 6 H 6(l) + 15 O 2(g) 12 CO 2(g) + 6 H 2 O (l) DG fo : 124. 5 kj 0 394.4 kj DG fo values are given below each formula in the equation. b) Is the reaction spontaneous at 25 o C? 237.2 kj 46

A Problem To Consider 17.4. a) Calculate the standard free energy change, DG o, for the following reaction at 25 o C: 2 C 6 H 6(l) + 15 O 2(g) 12 CO 2(g) + 6 H 2 O (l) DG fo : 124. 5 kj 0 394.4 kj 237.2 kj G o = n G f o m G f o products reactants G o = CO 2 G f o CO 2 + H 2 O G f o H 2 O C 6 H 6 G f o C 6 H 6 G o = 12 394.4 kj + 6 237.2 kj 2 124.5 kj G o = 4732. 8 + 1423.2 249.0 = 6405 kj 47

A Problem To Consider 17.5. a) Calculate the standard free energy change, DG o, for the following reaction at 25 o C: C 2 H 5 OH (l) + 3 O 2(g) DG fo : 174. 8 kj 2 CO 2(g) + 3 H 2 O (g) 0 394.4 kj DG fo values are given below each formula in the equation. b) Is the reaction spontaneous at 25 o C? 228.6 kj 49

A Problem To Consider 17.5. a) Calculate the standard free energy change, DG o, for the following reaction at 25 o C: C 2 H 5 OH (l) + 3 O 2(g) DG fo : 174. 8 kj 2 CO 2(g) + 3 H 2 O (g) 0 394.4 kj 228.6 kj G o = n G f o m G f o products reactants G o = CO 2 G f o CO 2 + H 2 O G f o H 2 O C 2 H 5 OH G f o C 2 H 5 OH G o = 2 394.4 kj + 3 228.6 kj 1 174.8 kj G o = 788.8 + 685.8 + 174.8 = 1299.8 kj 50

DG o as a Criteria for Spontaneity When DG o is a large negative number (DG o < 10 kj), the reaction is spontaneous as written, and the reactants transform almost entirely to products when equilibrium is reached. When DG o is a large positive number (DG o > +10 kj), the reaction is nonspontaneous as written, and reactants do not give significant amounts of product at equilibrium. When DG o is a small negative or positive value ( 10 kj < DG o < + 10 kj), the reaction gives an equilibrium mixture with significant amounts of both reactants and products. 52

Gibbs Free Energy and Chemical Equilibrium The free energy change when reactants are in nonstandard states (other than 1 atm pressure or 1 M concentration) is related to the standard free energy change, DG o, by the following equation: G = G o + RT lnq R is the gas constant (8.314 J/( K)) T is the absolute temperature (K) Q is the reaction quotient 53

Gibbs Free Energy and Chemical Equilibrium The free energy change when reactants are in nonstandard states (other than 1 atm pressure or 1 M concentration) is related to the standard free energy change, DG o, by the following equation: G = G o + RT lnq At equilibrium DG = 0 and the reaction quotient Q becomes the equilibrium constant K: 0 = G o + RT lnk or K = e Go RT 54

Gibbs Free Energy and Chemical Equilibrium K = e Go RT When DG o < 0 kj, then K > 1, and the reaction mixture at equilibrium contains mostly the products. When DG o > 0 kj, then K < 1, and the reaction mixture at equilibrium contains mostly the reactants. When DG o = 0 kj, then K = 1, and the reaction mixture at equilibrium will contain similar quantities of both the reactants and the products. 55

A Problem To Consider 17.6. Find the value for the equilibrium constant, K, at 25 o C (298 K) for the following reaction: 2 NH 3(g) + CO 2(g) NH 2 CONH 2(aq) + H 2 O (l) The standard free-energy change for this reaction G o = 13.6 kj. 56

A Problem To Consider 17.7. PCl 5(g) PCl 3(g) + Cl 2(g) DG fo : 325 kj 286 kj 0 kj a) Calculate DG o and K p for the reaction at 25 o C. b) Calculate G for the reaction if the partial pressures of the initial mixture are P PCl5 = 0.0029 atm, P PCl3 = 0.27 atm, and P Cl2 = 0.40 atm. Which way is the direction of spontaneous change for this system? 58

A Problem To Consider 17.7. PCl 5(g) PCl 3(g) + Cl 2(g) DG fo : 325 kj 286 kj 0 kj a) Calculate DG o and K p for the reaction at 25 o C: G o = n G o (products) n G o (reactants) G o = 1 286 kj 1 325 kj = 39 kj = 3.9 104 J K p = e Go RT = e 3.9 10 4 J J 8.314 K 298 K = 1 10 7 59

A Problem To Consider 17.7. PCl 5(g) PCl 3(g) + Cl 2(g), DG rxno = 3.9 10 4 J b) P PCl5 = 0.0029 atm P PCl3 = 0.27 atm P Cl2 = 0.40 atm Q p = P PCl 3 P Cl2 P PCl5 = 0.27 0.40 0.0029 = 37 G = Go rxn G = 3.9 10 4 + RTlnQ p J + 8.314 J K 298 K ln37 = 4.8 104 J G increased, so the equilibrium will shift to the left. 60

Calculation of DG o at Various Temperatures G T o = H o T S o In this method you assume that DH o and DS o are essentially constant with respect to temperature. Value of free energy change of a reaction, G T o, at any temperature T can be calculated by using standard enthalpy change, H o and entropy change, S o of a reaction. 61

A Problem To Consider 17.8. a) Find the DG o for the following reaction at 25 o C: CaCO 3(s) CaO (s) + CO 2(g) DH fo : 1206.9 kj 635.1 kj 393.5 kj S o : 92.9 J K 38.2 J K 213.7 J K b) What is the temperature range ( ) in which this reaction occurs spontaneously as written? Assume DH o and DS o do not depend on temperature. 62

A Problem To Consider 17.8. (Continued). part a) CaCO 3(s) CaO (s) + CO 2(g) DH fo : 1206.9 kj 635.1 kj 393.5 kj H o = n H o (products) n H o (reactants) H o = 1 635.1 + 1 393.5 1 1206.9 H o = 178.3 kj 63

A Problem To Consider 17.8. (Continued). part a) CaCO 3(s) CaO (s) + CO 2(g) S o : 92.9 J K 38.2 J K 213.7 J K S o = ns o (products) ns o (reactants) S o = 1 213.7 + 1 38.2 1 92.9 = 159.0 J K = 0.1590 kj K 64

A Problem To Consider 17.8. (Continued). part a) CaCO 3(s) CaO (s) + CO 2(g) H o = 178.3 kj S o = 0.1590 kj K T = 298 K G o = H o T S o = 178.3kJ 298 K 0.1590 kj K = 130.9 kj Reaction is not spontaneous at 25. 65

A Problem To Consider 17.8. (Continued). part b) CaCO 3(s) CaO (s) + CO 2(g) To determine the minimum temperature for spontaneity, we can set DG fo = 0 and solve for T: G o = H o T S o = 0 T = Ho S o T = 178.3 kj 0.1590 kj K = 1121 K = 848 Reaction will be spontaneous at temperatures above 848. 66