Thermodynamics: Free Energy and Entropy. Suggested Reading: Chapter 19

Transcription

1 Thermodynamics: Free Energy and Entropy Suggested Reading: Chapter 19

2 System and Surroundings System: An object or collection of objects being studied. Surroundings: Everything outside of the system. the surroundings the system

3 Direction of Energy Transfer When E is transferred as heat between a system and its surroundings, the direction is described as endo or exothermic. Exothermic: Energy is transferred from the system to its surroundings. E of the system decreases, E of the surroundings increases. Endothermic: Energy is transferred from the surroundings to the system. E of the system increases, E of the surroundings decreases.

4 First Law of Thermodynamics The energy change for a system (ΔU) is the sum of the E transferred as heat (q) and the E transferred as work (w). Work: Energy transfer that occurs as a mass is moved against an opposing force. If a system does work on its surroundings, E sys will decrease. If work is done by the surroundings, E system will increase. heat energy transferred internal energy change ΔU = q + w work energy transferred

5 Enthalpy Heat energy transferred at constant pressure. ΔU = q p + w p ΔU = q p - PΔV q p = ΔU + PΔV ΔH ΔH = q p = ΔU + PΔV ΔH (-) E transfer from the system ΔH (+) E transfer to the system ΔH and ΔU differ by PΔV. ΔV is typically very small therefore work is small. ΔV is large when gases are formed or consumed.

6 State Functions A state function depends on the initial and final state of the system but is independent of the path taken. (Doesn t matter how you got to the end product!) No matter how you go from reactants to products in a reaction, the values of ΔH and ΔU are always the same. P, V and T are also state functions q and w are not state functions

7 Standard Reaction Enthalpy The state of matter effects the magnitude of the enthalpy. H 2 (g) + ½ O 2 (g) H 2 O (g) Δ r H = kj/mol H 2 (g) + ½ O 2 (g) H 2 O (l) Δ r H = kj/mol The difference is equal to the enthalpy change for the condensation of 1 mol of H 2 O vapor to 1 mol of liquid H 2 O. Keep in Mind: 1. ΔH is specific to each reaction; it depends on the reactants, products and their states of matter. 2. ΔH depends on the number of moles of reactant. 3. ΔH (-) = exothermic reaction ΔH (+) = endothermic reaction

8 Standard Molar Enthalpy of Formation The enthalpy change for the formation of 1 mol of a cmpd directly from its elements in their standard state. ΔH f does not always describe a rxn that can be done in lab For an element, ΔH f = 0 If ΔH f is negative, formation of the compound is exothermic is positive, formation of the compound is endothermic The more negative the ΔH f, the more stable the compound Which of the hydrogen halides is most stable? Compound ΔH f (kj/mol) HF (g) HCl (g) HBr (g) HI (g)

9 Standard Molar Enthalpy of Formation Which of the following chemical equations does not correspond to a standard molar enthalpy of formation? a. Ca (s) + C (s) + 3/2 O 2 (g) CaCO 3 (s) b. C (s) + O 2 (g) CO 2 (g) c. NO (g) + ½ O 2 (g) NO 2 (g) d. N 2 (g) + 2 O 2 (g) N 2 O 4 (g) e. H 2 (g) + ½ O 2 (g) H 2 O (l) Have to form ONLY 1 mol of product Elements must be in standard states (No compounds as reactants!)

10 Standard Reaction Enthalpy Enthalpy changes accompany all chemical reactions. The standard reaction enthalpy (Δ r H ) is used when all reactants and products are in their standard state [ie O 2 (g), graphite (s)]. H 2 O (g) H 2 (g) + ½ O 2 (g) Δ r H = kj/mol Positive value for enthalpy indicates an endothermic rxn. 2 H 2 (g) + O 2 (g) 2 H 2 O (g) Δ r H = kj/mol Reverse reaction has an enthalpy with an opposite sign; If two moles are used in the rxn, the enthalpy is doubled.

11 Is a Reaction Spontaneous? Spontaneous: a change that occurs without intervention The change occurs to lead to equilibrium Has nothing to do with rate of reaction! All chemical reactions proceed in the direction to achieve equilibrium.

12 Spontaneous Reactions Many spontaneous reactions are exothermic, such as combustion, formation of NaCl, strong acid/base reactions Exothermicity is not a requirement of spontaneity. NH 4 NO 3 dissolves spontaneously in water and ΔH rxn = kj/mol. Expansion of gas into a vacuum is energy neutral (or endothermic for real gases). Phase changes (s à l à g) require E input (and depend on Temperature!) These examples illustrate spontaneity cannot be equated to whether a reaction is endo/exothermic.

13 Entropy (S) Energy is conserved, so it is not a good indicator of spontaneity. In a spontaneous process, energy goes from being concentrated to being more disperse. Example: a book falling from a table The book will fall spontaneously, but will not jump back up! How do we predict directionality and quantify a process? Entropy (S): state function Second Law of Thermodynamics: A spontaneous process results in an increase in the entropy of the universe (i.e ΔS universe > 0)

14 Calculating Entropy Thermal energy is caused by the random motion of particles, so potential energy is dispersed when it is converted to thermal energy (i.e. heat). Therefore heat (q) is related to entropy. Energy dispersal due to heat is inversely proportional to T. ΔS = q rev T ΔS units = J/K q rev = heat transferred under reversible conditions (adding energy by heating an object slowly) T = Temperature (in Kelvin)

15 Entropy on the Microscopic Level Why does entropy increase in a spontaneous process? Why must energy be more dispersed? E transferred as heat between hot and cold gaseous atoms Initial Equilibrium

16 Entropy on the Microscopic Level Why does energy disperse? Consider a statistical approach. Imagine 4 atoms, 1 which has 2 units of energy Overall energy of all 4 atoms = 2 units Over time, 10 different ways of distributing the energy between particles can be observed

17 Entropy on the Microscopic Level Each of the different configurations is called a microstate. In only one state do both energy units remain on atom 1 4 ways to distribute 2 units on a single atom 6 ways to distribute the units between different atoms There is a preference that energy by dispersed over a greater number of atoms (60 % chance).

18 Entropy on the Microscopic Level As number of atoms increases, probability of energy dispersing increases dramatically. For a 6 atom system: 6 microstates with the energy concentrated on one atom 15 microstates with the energy present on two atoms Statistics dictate a better chance of having the energy dispersed between more atoms (71.4 % chance).

19 Entropy on the Microscopic Level Instead of increasing # atoms, what about more E units? Consider 6 energy units, dispersed over four atoms Initially two atoms have 3 quanta of E and two atoms have no E. Through collisions, E is transferred to achieve different distributions of E. Now there are 84 possible microstates.

20 Entropy on the on Microscopic the Microscopic Level Level There are 4 different ways to have four particles such that one particle has 3 quanta of energy and the other three each have one quantum of energy. Increasing # energy units also increases the # microstates

21 Boltzmann s Equation Entropy of a system is determined by # of microstates available Entropy is proportional to the number of microstates S = k lnw W = # of microstates k = Boltzmann s constant = x J/K

22 Dispersal of Energy: Gas Diffusion Why does a gas diffuse throughout a container? Energy is quantized Molecules have a distribution of energies Think of each energy as a microstate As the gas disperses across the two flasks, there are more spaces for the molecules, and therefore more microstates available!

23 Dispersal of Energy: Gas Diffusion Why does a gas diffuse throughout a container? The energy is conserved, so the microstates are closer.

24 A Chemist s Perspective on Entropy Chemists are interested in the change in entropy (ΔS) ΔS = S final S initial = k ln(w final /W initial ) ΔS depends on the number of microstates available before and after (a state function). We are interested in determining a numerical value for ΔS Third Law of Thermodynamics: A perfect crystal at 0 K has S = 0. This is used as a reference point. The entropy for any other set of conditions is measured relative to the absolute zero point. Entropy (S) is always positive (but ΔS can be negative).

25 Standard Entropy (S ) Standard entropy (S ) of a substance: Entropy gained by converting 1 mol of a substance from a perfect crystal at 0 K to standard conditions (1 bar, 1 m). Units = J/K mol Generally find in tables (Appendix L) Entropy increases with increase in T Changes of state accompany large entropy increases

26 Standard Entropy (S ) General Trends: Large increase in S accompanies a phase change Larger molecules generally have larger S Big molecules have more ways to orient therefore more microstates S gas > S liquid > S solid Compound S (J/K mol) I 2 (s) 116 Br 2 (l) 152 Cl 2 (g) 223

27 range the substances on the left in order of increasing entropy. Assume 1 mole of each at standard conditions. Entropy Practice Which substance has the higher entropy? NO 2 (g) or N 2 O 4 (g) N 2 O 4 is a larger, more complex molecule. In the same state of matter (g), it will have a greater S. I 2 (g) or I 2 (l) For the same substance, the gas phase will have the higher S. Arrange in order of increasing entropy, assuming 1 mol CH 3 COOH (l) CO 2 (g) HCOOH (l) Al (s) (1) lowest entropy

28 range the substances on the left in order of increasing entropy. Assume 1 mole of each at standard conditions. Entropy Practice Predict whether ΔS for each reaction is greater than zero, less than zero or too close to determine. Consider which side of the reaction has more disorder (count moles!) H 2 (g) + F 2 (g) 2 HF (g) Too close 4 HCl (g) + O 2 (g) 2 H 2 O (g) + 2 Cl 2 (g) ΔS < 0 2 H 2 O (l) 2 H 2 (g) + O 2 (g) ΔS > 0 2 NOBr (g) 2 NO (g) + Br 2 (g) ΔS > 0 2 HBr (g) + Cl 2 (g) 2 HCl (g) + Br 2 (g) Too close

29 Calculating ΔS rxn ΔS rxn = ΣS (products) - ΣS (reactants) Calculate entropy change for a system in which reactants are completely converted to products. What is ΔS rxn for the formation of NO 2? 2 NO (g) + O 2 (g) 2 NO 2 (g) Compound S (J/K mol) NO O NO ΔS rxn = [2 mol NO 2 x 240.0] [2 mol NO x mol O 2 x 205.1] = J/K mol

30 Calculating ΔS rxn Calculate ΔS rxn for the formation of ammonia. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Predict the sign of ΔS rxn Less gas moles in product therefore entropy must decrease ΔS rxn = ΣS (products) - ΣS (reactants) Compound S (J/K mol) N H NH ΔS rxn = [2 mol NH 3 x ] [1 mol N 2 x mol H 2 x 130.7] = J/K mol ΔS rxn for a system can be positive or negative

31 Entropy Changes & Spontaneous Rxns ΔS (universe) > 0 if the reaction is spontaneous ΔS (universe) = 0 at equilibrium ΔS (universe) < 0 if the reaction is not spontaneous ΔS (universe) must increase in a spontaneous process ΔS for the system can be positive or negative ΔS (universe) = ΔS (system) + ΔS (surroundings) ΔS (system) can be negative if the magnitude of ΔS (surrounding) is larger so that ΔS (universe) is positive.

32 Calculating ΔS (universe) Calculate ΔS (universe) for the following reaction at 25 C CO (g) + 2 H 2 (g) CH 3 OH (g) ΔS (universe) = ΔS (system) + ΔS (surroundings) ΔS rxn = ΣS (products) - ΣS (reactants) ΔS rxn = [1 mol CH 3 OH x 127.2] [1 mol CO x mol H 2 x 130.7] = J/K mol ΔS (surroundings) = q rev T = ΔH(surroundings) T = - ΔH sys = ΣΔH f (products) - ΣΔH f (reactants) ΔH sys = [1 mol CH 3 OH x ] [1 mol CO x mol H 2 x 0] = kj/mol ΔH(system) T

33 Calculating ΔS (universe) ΔS (surroundings) = - ΔH(system) T = J/K mol = 127,900 J/mol 298 K ΔS (universe) = ΔS (system) + ΔS (surroundings) ΔS (universe) = J/K mol J/K mol = 97.3 J/K mol ΔS (universe) > 0 therefore the reaction is spontaneous

34 Calculating ΔS (universe) Calculate ΔS (universe) when moles of NO (g) react under standard conditions at K. ΔH = kj and ΔS = J/K 2 NO (g) + 2 H 2 (g) N 2 (g) + 2 H 2 O (l) ΔS (universe) = ΔS (system) + ΔS (surroundings) ΔS (surroundings) = - ΔH(system) T = = J/K 752,200 J K ΔS (universe) = J/K J/K = J/K moles NO J/K x = 2134 J/K 2 moles NO The reaction is product favored and enthalpy favored.

35 Spontaneous Reactions: A Summary Four possible outcomes of ΔH (system) and ΔS (system) 2 C 4 H 10 (g) + 13 O 2 (g) 8 CO 2 (g) + 10 H 2 O (g) ΔH = kj/mol, ΔS = J/K mol N 2 (g) + 2 H 2 (g) N 2 H 4 (l) ΔH = kj/mol, ΔS = J/K mol

36 Gibbs Free Energy (G) A single equation to predict reaction spontaneity. Considers only the system, not the surroundings. ΔG = ΔH - TΔS Gibbs free energy is a state function.

37 Where Does ΔG Come From? ΔS (universe) = ΔS (surroundings) + ΔS (system) ΔS (universe) = - ΔH(system) T -TΔS (universe) = ΔH(system) TΔS(system) ΔG = -TΔS (universe) Under standard conditions: + ΔS (system) ΔG < 0 ΔG = 0 ΔG > 0 ΔG = ΔH - TΔS reaction spontaneous in direction written reaction at equilibrium reaction not spontaneous

38 Interpreting ΔG Reactant Favored Product Favored Lowest possible free energy at equilibrium.

39 Converting between ΔG rxn and ΔG rxn For a reaction under nonstandard conditions ΔG rxn = ΔG rxn + RT lnq R = ideal gas constant Q = reaction quotient At equilibrium: (i.e. ΔG rxn = 0) 0 = ΔG rxn + RT lnk ΔG rxn = -RT lnk If K >1 (products favored) then ΔG rxn must be negative (spontaneous reaction) If K <1 (reactants favored) the ΔG rxn must be positive (not spontaneous reaction)

40 Gibbs Free Energy: A Summary At equilibrium, Gibbs free energy is at a minimum When ΔG rxn is negative (Q < K), the rxn is spontaneous Reactions always proceed toward equilibrium When ΔG rxn is positive (K > Q), the reaction is not spontaneous in that direction When ΔG rxn = 0, the reaction is at equilibrium (Q = K) For ΔG rxn (under standard conditions): ΔG rxn = 0 rxn is at equilibrium (K = 1) ΔG rxn < 0 rxn spontaneous in direction written (K > 1) ΔG rxn > 0 rxn not spontaneous (K < 1)

41 Calculating Free Energy for a Reaction Standard free energy of formation (ΔG f ): The free energy change when one mole of compound is formed from elements in their standard states ΔG f = 0 for an element in its standard state ΔG f can be used to calculate ΔG rxn ΔG rxn = ΣΔG f (products) - ΣΔG f (reactants) free energy products < free energy reactants: spontaneous reaction (ΔG rxn < 0) free energy products > free energy reactants not a spontaneous reaction (ΔG rxn > 0)

42 Problems Involving Gibbs Free Energy Three types of problems: 1. Using ΔH f and ΔS values, calculate ΔG rxn ΔG = ΔH - TΔS 2. Using ΔG f values, calculate ΔG rxn ΔG rxn = ΣΔG f (products) - ΣΔG f (reactants) 3. Knowing K or Q, calculate ΔG rxn or ΔG rxn ΔG rxn = ΔG rxn + RT lnq ΔG rxn = -RT lnk

43 Calculating Free Energy Calculate ΔG rxn for the following reaction at 298 K, using values of ΔH f and ΔS. Is the reaction spontaneous? Does the reaction favor products or reactants? C (graphite) + 2 H 2 (g) CH 4 (g) C (graphite) H 2 (g) CH 4 (g) ΔH f (kj/mol) S (J/Kmol) Remember units of ΔH f and ΔS and are different! (kj versus J) ΔH rxn = ΣΔH f (products) - ΣΔH f (reactants) ΔH rxn = [1 mol CH 4 x -74.9] [1 mol C x mol H 2 x 0] = kj/mol

44 Calculating Free Energy ΔS rxn = ΣS (products) - ΣS (reactants) ΔS rxn = [1 mol CH 4 x 186.3] [1 mol C x mol H 2 x 130.7] = J/K mol ΔG rxn = ΔH rxn - TΔS rxn ΔG rxn = kj/mol [(298 K)( kj/k mol)] = kj/mol ΔG rxn is negative therefore the reaction is spontaneous Product favored (K > 1) We can actually calculate K! ΔG rxn = -RT lnk K = 8.36 x 10 8

45 Calculating Free Energy Calculate ΔG rxn for the following reaction, using ΔG f for reactants and products. Is the reaction spontaneous? Does the reaction favor products or reactants? CH 4 (g) + 2 O 2 (g) 2 H 2 O (g) + CO 2 (g) CH 4 (g) O 2 (g) H 2 O (g) CO 2 (g) ΔG f (kj/mol) ΔG rxn = ΣΔG f (products) - ΣΔG f (reactants) ΔG rxn = [2 mol H 2 O x mol CO 2 x ] [1 mol CH 4 x mol O 2 x 0] = kj/mol Large negative ΔG rxn means spontaneous reaction Product favored

46 Relating ΔG rxn and K Calculate ΔG rxn for the following reaction and then determine the equilibrium constant at 25 C. ½ N 2 (g) + 3 / 2 H 2 (g) NH 3 (g) ΔG f (NH 3 (g)) = kj/mol (from table or calculate from reaction equation) ΔG rxn = -RT lnk p kj/mol = -( kj/k mol)(298 K)lnK p kj/mol = (-2.48 kj/mol)lnk p lnk p =6.64 K p = 7.65 x 10 2 Convert R and ΔG rxn to same units!

47 Calculating K What is the equilibrium constant for the following reaction at 276 K? ΔH = 41.2 kj, ΔS = 42.1 J CO 2 (g) + H 2 (g) CO (g) + H 2 O (g) ΔG rxn = ΔH rxn - TΔS rxn ΔG rxn = 41.2 kj (276 K)( kj/k) = kj ΔG rxn = -RT lnk kj = -( kj/k mol)(276 K) lnk = lnk K = 2.52 x 10-6

48 Relating ΔG rxn and K The value of K sp for AgCl (s) at 25 C is 1.8 x Calculate ΔG rxn for the following reaction at 25 C Ag + (aq) + Cl - (aq) AgCl (s) K sp is the equilibrium constant for the reverse reaction, so the K for the given reaction is: K = 1/K sp = 5.6 x 10 9 ΔG rxn = -RT lnk ΔG rxn = -( J/K mol)( K) ln(5.6 x 10 9 ) ΔG rxn = kj/mol The reaction is product favored at equilibrium (i.e. favors AgCl precipitation)

49 Free Energy and Temperature Can a reactant favored reaction turn to product favored by a change of temperature? Yes. This is vital to chemical industry. Where is the line between reactant and product favored? ΔG rxn = 0 marks the line between product or reactant favored Solve for when ΔG rxn = 0 and find T

50 Free Energy and Temperature Free energy is a function of temperature. Is the following reaction spontaneous at 298 K? CaCO 3 (s) CaO (s) + CO 2 (g) at 298 K ΔG f (kj/mol) ΔH f (kj/mol) S (J/K mol) ΔH rxn = 179 kj ΔS rxn = J/K ΔG rxn = ΔH rxn - TΔS rxn Entropically favored (more moles of product than reactant) Enthalpy disfavored (endothermic) ΔG rxn > 0 until the temperature goes above 1118 K Reaction spontaneity depends on temperature!

51 Qualitative Fun! Without doing any calculations, match the following thermodynamic properties with their appropriate numerical sign for the following exothermic reaction. 4 NH 3 (g) + 5 O 2 (g) 4 NO (g) + 6 H 2 O (g) ΔH rxn ΔS rxn ΔG rxn ΔS universe 1. > 0 2. < 0 3. = 0 4. > 0 low T, < 0 high T 5. < 0 low T, > 0 high T Count the number of moles.

52 Summary ΔS = q rev T S = k lnw ΔS (universe) = ΔS (system) + ΔS (surroundings) ΔG rxn = ΣΔG f (products) - ΣΔG f (reactants) ΔH rxn = ΣΔH f (products) - ΣΔH f (reactants) ΔS rxn = ΣS (products) - ΣS (reactants) ΔG = ΔH - TΔS ΔG rxn = ΔG rxn + RT lnq ΔG rxn = -RT lnk p