Chemistry 201: General Chemistry II - Lecture

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1 Chemistry 201: General Chemistry II - Lecture Dr. Namphol Sinkaset Chapter 19 Study Guide Concepts 1. The First Law of Thermodynamics is also known as the Law of Conservation of Energy. 2. In any process that involves an energy transaction, a heat tax must be paid. 3. Thermodynamics allows predictions of which processes will occur spontaneously. 4. A spontaneous process occurs without ongoing outside intervention. 5. For mechanical systems, the direction of spontaneity is towards lower potential energy. 6. For chemical systems, we need an analagous concept. 7. Note that a spontaneous process is not equivalent to a fast process. Thermodynamics and kinetics study different aspects of a chemical reaction. 8. Nonspontaneous processes can and do occur with outside assistance. 9. Examination of spontaneous endothermic processes leads to the importance of increasing disorder in chemical processes. 10. Disorder or randomness is the qualitative description of entropy. 11. entropy (S): a thermodynamic function that increases with the number of energetically equivalent ways to arrange the components of a system to achieve a particular state. 12. The most common units for entropy are J/K. 13. Entropy is a state function, facilitating calculations of changes in entropy. 14. Entropy determines the direction of physical and chemical change. 15. A macrostate is what the system looks like from outside; a microstate is what the system looks like from inside. 16. It is quite possible and common to have many different microstates that represent the same macrostate. 17. Systems tend towards the most probable state which coincides with the state with greatest entropy. 1

2 18. The state with the highest entropy is the one that disperses energy the most. 19. States with low energy concentration have higher entropy than states with high energy concentration (as long as the amount of energy is constant between states). 20. The Second Law of Thermodynamics states that for any spontaneous process, the entropy of the universe increases. 21. As a substance goes from solid to gas, its entropy increases due to the greater number of ways it can hold energy. 22. Reactions resulting in an increase in the number of moles of gas increase entropy. 23. Since the 2nd Law of Thermodynamics requires that the entropy of the universe increase, it is possible for the entropy of a system to decrease as long as the entropy of the surroundings increases enough to offset it. 24. If a system releases energy to the surroundings, the entropy of the surroundings increases. 25. Entropy is temperature dependent. For a constant amount of energy dispersed, S decreases as temperature increases. 26. Gibbs free energy, (G), is formally defined as G = H T S. 27. G can be used to determine spontaneity using only information about the system. 28. If G is negative, the process is spontaneous; if G is positive, the process is nonspontaneous. 29. Since G involves H, S, and T, different combinations of enthalpy, entropy, and temperature determine the sign on G and hence the spontaneity. 30. The standard entropy change for a reaction, S rxn, is the change in entropy for a process in which all reactants and products are in their standard states. 31. The Third Law of Thermodynamics states that the entropy of a perfect crystal at absolute zero (0 K) is zero. 32. The Third Law of Thermodynamics provides a reference point for all other entropies. 33. Entropy increases as a substance goes from solid to liquid to gas. 34. The greater the molar mass, the higher the entropy. 35. Less constrained allotropes have more entropy than more constrained ones. 36. Complex compounds have more entropy than simpler compounds. 37. Dissolved ionic solids have more entropy than the undissolved solid. 38. Changes in standard entropies can be calculated via a Hess s Law type of calculation. 2

3 39. G rxn can be calculated by separately calculating the standard enthalpy change and the standard entropy change and using those values in the Gibbs free energy equation. 40. Just like a standard enthalpy of formation, there is a standard free energy of formation ( G f). The standard free energy of formation is the change in free energy when 1 mole of a compound forms from its constituent elements in their standard states. 41. Reactions with associated free energies can be manipulated like thermochemical equations: (1) If an equation is multiplied by a factor, then the associated free enegy change is multiplied by the same factor; (2) If an equation is reversed, then the free energy change switches sign; (3) If a series of reactions adds up to an overall reaction, the free energy change for the overall reaction is simply the sum of the free energy changes for each step. 42. Free energy is called free because it is the energy that is theoretically available to do work. 43. The only way to get the theoretical maximum work from a chemical system is to have the system change infinitesimally slowly such that the reaction is reversible. 44. Free energies can also be calculated under nonstandard conditions. 45. If there is no change in free energy, i.e. G = 0, then the system is at equilibrium. 46. Equilibrium constants and changes in free energy are intimately related. 47. Equilibrium constants are temperature dependent. Data of K s obtained at different temperature can be used to find the standard enthalpy, standard entropy, and standard free energy changes for a system. Equations 1. S = k ln W (Entropy equation) 2. S = S final S initial (Change in entropy equation) 3. S univ > 0 (2nd Law of Thermodynamics) 4. S univ = S sys + S surr (Components of entropy of the universe) 5. S surr = Hsys T (Equation to calculate entropy change of surroundings) 6. G = H sys T S sys (Gibbs free energy equation) 7. G = H T S (Definition of Gibbs free energy) 8. S rxn = S products S reactants (Standard entropy change for a reaction) 9. S rxn = Σn p S (products) Σn r S (reactants) (Calculation of S rxn) 3

4 10. G rxn = Σn p G f(products) Σn r G f(reactants) (Hess s Law type calculation for G rxn) 11. G rxn = G rxn + RT ln Q (Calculating nonstandard free energy) 12. G rxn = RT ln K (Relationship between G rxn and K) ( ) 13. ln K = H rxn 1 R T + Srxn (Relationship between equilibrium constant and temperature) R Representative Problems R75. Given the reactions and their G values shown below, calculate G for the reaction: CO(NH 2 ) 2(s) + H 2 O (l) CO 2(g) + 2NH 3(g). COCl 2(g) + 4NH 3(g) CO(NH 2 ) 2(s) + 2NH 4 Cl (s) G = kj COCl 2(g) + H 2 O (l) CO 2(g) + 2HCl (g) G = kj NH 3(g) + HCl (g) NH 4 Cl (s) G = kj This is another thermochemical equation manipulation except now we re dealing with G. Always look at what you re trying to get (the equation you want) before you start flipping equations around. We see we need to flip the first equation around to get CO(NH 2 ) 2(s) on the reactant side. We do this and then see what needs to be done next. CO(NH 2 ) 2(s) + 2NH 4 Cl (s) COCl 2(g) + 4NH 3(g) G = kj COCl 2(g) + H 2 O (l) CO 2(g) + 2HCl (g) G = kj NH 3(g) + HCl (g) NH 4 Cl (s) G = kj We don t want NH 4 Cl (s) on the reactant side, so we can get rid of it if we multiply the last equation by 2. Turns out that this is the last thing we need to do. CO(NH 2 ) 2(s) + 2NH 4 Cl (s) COCl 2(g) + 4NH 3(g) G = kj COCl 2(g) + H 2 O (l) CO 2(g) + 2HCl (g) G = kj 2NH 3(g) + 2HCl (g) 2NH 4 Cl (s) 2 G = 2( 91.96) kj CO(NH 2 ) 2(s) + H 2 O (l) CO 2(g) + 2NH 3(g) G = 6.3 kj R83. Which of the following reactions would be expected to be spontaneous at 25 C and 1 atm? (a) PbO (s) + NH 3(g) Pb (s) + N 2(g) + H 2 O (g) (b) Al 2 O 3(s) + Fe (s) Fe 2 O 3(s) + Al (s) To determine spontaneity, we need to know G for the reactions. We will use Hess s Law type calculations, but first we must balance the equations. The balanced equations are: (a) 3PbO (s) + 2NH 3(g) 3Pb (s) + N 2(g) + 3H 2 O (g) (b) Al 2 O 3(s) + 2Fe (s) Fe 2 O 3(s) + 2Al (s) 4

5 We get the G f values we need from a reference table. For (a): = (3 G f,p b (s) + G f,n 2(g) + 3 G f,h 2 O (g) ) (3 G f,p bo (s) + 2 G f,nh 3(g) ) = [3(0 kj/mole) + 0 kj/mole + 3( kj/mole)] [3( kj/mole) + 2(-16.7 kj/mole)] = kj/mole Seeing the negative value for G, we can predict that reaction (a) would be spontaneous. For the equation in (b): = ( G f,f e 2 O 3(s) + 2 G f,al (s) ) ( G f,al 2 O 3(s) + 2 G f,f e (s) ) = [ kj/mole + 2(0 kj/mole)] [ kj/mole) + 2(0 kj/mole)] = kj/mole Seeing the positive value of G, we can predict that reaction (b) would be nonspontaneous. R86. Consider the following reaction: CO 2(g) + CCl 4(g) 2COCl 2(g). Calculate G for this reaction at 25 C under the following conditions: P CO2 = atm, P CCl4 = atm, P COCl2 = atm. To solve this problem, we need to use the equation for finding nonstandard G. Note that just because we re at 25 C does not mean we are at standard conditions. To be at standard conditions, the pressures of each gas would need to be atm as well. G = G + RT ln Q Thus, we need to find the standard change in free energy and the value of Q based on the given conditions. We can calculate G from free energies of formation via a Hess s Law type of calculation. Referring to the Appendix, we find G f s of CO 2(g), CCl 4(g), and COCl 2(g) to be kj/mole, kj/mole, and kj/mole, respectively. = ( ) ( ) 2 G f,cocl 2(g) 1 G f,co2(g) + 1 G f,ccl 4(g) = [2( kj/mole)] [1( kj/mole) + 1(-62.3 kj/mole)] = 46.9 kj/mole Next, we calculate the value of Q. 5

6 P 2 COCl 2 Q = P CO2 P CCl4 = (0.112)(0.174) = Finally, we solve for the value of G under these nonstandard conditions. G = G + RT ln Q = 46.9 kj/mole + (8.314 J/mole K)( K) ln = 46.9 kj/mole J/mole = kj/mole = 55.2 kj/mole R92. Nitrogen dioxide, a pollutant in the atmosphere, can combine with water to form nitric acid. One of the possible reactions is shown below. Calculate G and K p for this reaction at 25 C and comment on the spontaneity of the reaction. 3NO 2(g) + H 2 O (l) 2HNO 3(aq) + NO (g) In this problem, we need to calculate the G for the reaction in order to find the pressure equilibrium constant. We begin by looking up free energies of formation for NO 2(g), H 2 O (l), HNO 3(aq), and NO (g) and find 51.3 kj/mole, kj/mole, kj/mole, and 87.6 kj/mole, respectively. = ( ) ( ) 1 G f,no(g) + 2 G f,hno 3(aq) 3 G f,no2(g) + 1 G f,h 2 O (l) = [1(87.6 kj/mole) + 2( kj/mole)] [3(51.3 kj/mole) + 1( kj/mole)] = kj/mole Since the change in free energy is negative, we know the reaction is spontaneous. Now we find the K p. G = RT ln K p J/mole = (8.314 J/mole K)( K) ln K p = ln K p K p = K p =

4/19/2016. Chapter 17 Free Energy and Thermodynamics. First Law of Thermodynamics. First Law of Thermodynamics. The Energy Tax.

4/19/2016. Chapter 17 Free Energy and Thermodynamics. First Law of Thermodynamics. First Law of Thermodynamics. The Energy Tax. Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro First Law of Thermodynamics Chapter 17 Free Energy and Thermodynamics You can t win! First Law of Thermodynamics: Energy cannot be created or destroyed