Free Energy, and Equilibrium

Transcription

1 17 Ch a pt e r Thermodynamics: Entropy, Free Energy, and Equilibrium Chemistry 4th Edition McMurry/Fay Dr. Paul Charlesworth Michigan Technological University Spontaneous Processes and Entropy 01 The 2nd Law explains why chemical reactions tend to favor a particular direction. It is important to predict whether a reaction will occur. A reaction that does occur under specific conditions is called a spontaneous reaction. A reaction that does not occur under specific conditions is called a nonspontaneous reaction. Chapter 17 Slide 2 Spontaneous Processes and Entropy 02 A spontaneous reaction will always move a reaction mixture toward equilibrium. Reaction spontaneity is independent of reaction rate. Chapter 17 Slide 3 1

2 Spontaneous Processes and Entropy 03 To predict spontaneity we need to know the energy change and the entropy. Entropy (S) is a measure of the randomness or disorder of a system. The greater the disorder, the greater the entropy. Nature tends to the greatest entropy. S solid < S liquid < S gas Chapter 17 Slide 4 Spontaneous Processes and Entropy 04 Chapter 17 Slide 5 Spontaneous Processes and Entropy 04 Standard Entropy is the absolute entropy of a substance at 1 atm and 25 C. Change in entropy is given by?s = S f S i Chapter 17 Slide 6 2

3 Spontaneous Processes and Entropy 05 The decomposition of N 2 O 4 (O 2 N NO 2 ) is also accompanied by an increase in randomness. Whenever molecules break apart, randomness increases. Chapter 17 Slide 7 Spontaneous Processes and Entropy 06 When NaCl dissolves in water, the crystal breaks up, and the ions are surrounded by water molecules. Chapter 17 Slide 8 Spontaneous Processes and Entropy 07 Consider the gas phase reaction of A 2 molecules (red) with B atoms (blue). (a) Write a balanced equation for the reaction. (b) Predict the sign of?s for the reaction. Chapter 17 Slide 9 3

4 Entropy and Temperature 01 Entropy is associated with molecular motion. As temperature increases, entropy increases. Third Law of Thermodynamics:Entropy of a perfectly ordered crystalline substance at 0 K is zero. At the melting and boiling point there is a discontinuous jump in entropy. Chapter 17 Slide 10 Entropy and Temperature 02 Chapter 17 Slide 11 Standard Molar Entropies 01 Chapter 17 Slide 12 4

5 Standard Molar Entropies 01 Standard Molar Entropy, S : The entropy of 1 mol of the pure substance at 1 atm pressure and a specified temperature, usually 25 C. Standard molar entropies are absolute entropies measured against an absolute reference point. Standard entropy of reaction:?s =? ns (products)? ns (reactants) Chapter 17 Slide 13 Standard Molar Entropies 02 Chapter 17 Slide 14 Standard Molar Entropies 03 Calculate the standard entropy of reaction at 25 C for the synthesis of ammonia: N H 2 2 NH 3 Calculate the standard entropy of reaction at 25 C for the decomposition of calcium carbonate: CaCO 3 (s) CaO(s) + CO 2 Chapter 17 Slide 15 5

6 Standard Molar Entropies 04 Predict the entropy change, and then calculate the standard entropy change for the following reactions at 25 C. a. 2 CO + O 2 2 CO 2 b. 3 O 2 2 O 3 c. 2 NaHCO 3 (s) Na 2 CO 3 (s) + H 2 O(l) + CO 2 Chapter 17 Slide 16 2nd Law of Thermodynamics 01 The total entropy increases in a spontaneous process and remains unchanged in an equilibrium process. Spontaneous:?S total =?S sys +?S sur > 0 Equilibrium:?S total =?S sys +?S sur = 0 The system is what you observe; surroundings are everything else. Chapter 17 Slide 17 2nd Law of Thermodynamics 02 For?S total, we need to know?s sys and?s surr.?s sys is often determined from the standard entropy of reaction,?s rxn :?S rxn =? ns (Products)? ns (Reactants) Chapter 17 Slide 18 6

7 2nd Law of Thermodynamics 03 Entropy Changes to the Surroundings: Exothermic increase?s surr S surr = H sys T Endothermic decrease?s surr Chapter 17 Slide 19 Gibbs Free Energy 01 The 2nd law tells us a process will be spontaneous if?s total > 0 which requires a knowledge of?s surr. Since we only concern ourselves with the system we derive an expression using only?s sys. T?S total =?H sys T?S sys < 0 Chapter 17 Slide 20 Gibbs Free Energy 02 The expression T?S total is equated as Gibbs free energy change (?G), or simply free energy change:?g =?H T?S sys?g < 0 Reaction is spontaneous in forward direction.?g = 0 Reaction is at equilibrium.?g > 0 Reaction is spontaneous in reverse direction. Chapter 17 Slide 21 7

8 Gibbs Free Energy 03 Using?G =?H T?S, we can predict the sign of?g from the sign of?h and?s. If both?h and?s are positive,?g will be negative only when the temperature value is large. If?H is positive and?s is negative,?g will always be positive. If?H is negative and?s is positive,?g will always be negative. If both?h and?s are negative,?g will be negative only when the temperature value is small. Chapter 17 Slide 22 Gibbs Free Energy 04 Chapter 17 Slide 23 Gibbs Free Energy 05 Iron metal can be produced by reducing iron(iii) oxide with hydrogen: Fe 2 O 3 (s) + 3 H 2 2 Fe(s) + 3 H 2 O?H = kj;?s = J/K 1. Is this reaction spontaneous at 25 C? 2. At what temperature will the reaction become spontaneous? Chapter 17 Slide 24 8

9 Gibbs Free Energy 06 What are the signs (+,, or 0) of?h,?s, and?g for the following spontaneous reaction of A atoms (red) and B atoms (blue)? Chapter 17 Slide 25 Gibbs Free Energy 07 Chapter 17 Slide 26 Gibbs Free Energy 07 Standard Free Energy (?G rxn ) is the free energy for a reaction occurring under standard state conditions. Reactants in their standard states are converted to products in their standard states.?g rxn =? n?g ƒ (products)? n?g ƒ (reactants)?g ƒ is the standard free energy of formation of a compound from its elements in their standard states. Chapter 17 Slide 27 9

10 Gibbs Free Energy 08 Chapter 17 Slide 28 Gibbs Free Energy 09 Calculate the standard free energy changes for the following reactions at 25 C: a. H 2 + Br 2 2 HBr b. 2 C 2 H O 2 4 CO H 2 O(l) c. CH O 2 CO H 2 O(l) Chapter 17 Slide 29 Free Energy and Chemical Equilibrium 01 The sign of?g tells the direction of spontaneous reaction when both reactants and products are present at standard state conditions. Under nonstandard conditions,?g becomes?g.?g =?G + RT lnq The reaction quotient is obtained in the same way as an equilibrium expression. Chapter 17 Slide 30 10

11 Free Energy and Chemical Equilibrium 02 If?G is a large negative value, RT lnq will not be positive enough to match?g until a significant amount of product is formed. If?G is a large positive value, RT lnq will only be more negative when very little product has been formed. Chapter 17 Slide 31 Free Energy and Chemical Equilibrium 03 Chapter 17 Slide 32 Free Energy and Chemical Equilibrium 04 Chapter 17 Slide 33 11

12 Free Energy and Chemical Equilibrium 05 At equilibrium?g = 0 and Q = K?G rxn = RT ln K Using free energy data, calculate the equilibrium constant (K p ) for the following at 25 C: a. 2 H 2 O(l) æ 2 H 2 + O 2 b. 2 O 3 æ 3 O 2 Chapter 17 Slide 34 Free Energy and Chemical Equilibrium 06 Using the solubility product of silver chloride at 25 C (1.6 x ), calculate?g for the process: AgCl(s) æ Ag + (aq) + Cl (aq) The?G for the reaction H 2 + I 2 æ 2 HI is 2.60 kj at 25 C. In one experiment, the initial pressures are P H2 = 4.26 atm, P I2 = atm, and P HI = 0.23 atm. Calculate?G for the reaction and predict the reaction direction. Chapter 17 Slide 35 12