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Entropy and Free Energy

Modified 02/02/12 by Laura Peck to accompany Silberberg: Principles of General Chemistry Chapter 20. Entropy and Free Energy. How to predict if a reaction can occur, given enough time? THERMODYNAMICS. How to predict if a reaction can occur at a reasonable rate? KINETICS. Thermodynamics.

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Entropy and Free Energy

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  1. Modified 02/02/12 by Laura Peck to accompany Silberberg: Principles of General Chemistry Chapter 20 Entropy and Free Energy How to predict if a reaction can occur, given enough time? THERMODYNAMICS How to predict if a reaction can occur at a reasonable rate? KINETICS

  2. Thermodynamics • If the state of a chemical system is such that a rearrangement of its atoms and molecules would decrease the energy of the system--- • AND the K is greater than 1, • then this is a product-favoredsystem. • Most product-favored reactions are exothermic—but this is not the only criterion

  3. Thermodynamics • The sign of ΔH cannot predict spontaneous change. • CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) ΔH°rxn = - 802kJ K>1 • N2O5(s)  2NO2(g) + 1/2O2(g) ΔH°rxn = +109.5 kJ K>1 • *****Both are product favored*** • Both product- and reactant-favored reactions can proceed to equilibrium in a spontaneousprocess. • AgCl(s) Ag+(aq) + Cl–(aq) K<1 Reactant favored Reaction is not product-favored, but it moves spontaneously toward equilibrium. • Spontaneous does not imply anything about time for reaction to occur.

  4. Thermodynamics and Kinetics Diamond is thermodynamically favored to convert to graphite, but not kinetically favored. Paper burns — a product-favored reaction. Also kinetically favored once reaction is begun.

  5. Spontaneous Reactions In general, spontaneous reactions are exothermic. Fe2O3(s) + 2 Al(s) ---> 2 Fe(s) + Al2O3(s) ∆H = - 848 kJ

  6. Spontaneous Reactions But many spontaneous reactions or processes are endothermic or even have ∆H = 0. NH4NO3(s) + heat ---> NH4NO3(aq) Discuss your labs using urea & water or vinegar & baking soda!

  7. Entropy, S One property common to spontaneous processes is that the final state is more DISORDERED or RANDOM than the original. Spontaneity is related to an increase in randomness. The thermodynamic property related to randomness is ENTROPY, S. Reaction of K with water

  8. The entropy of liquid water is greater than the entropy of solid water (ice) at 0˚ C.

  9. Entropy and the Number of Microstates • Ludwig Boltzmann defined the entropy (S) of a system in terms of W: • S= klnW • K is the universal gas constant R divided by Avogadro’s number. It equals 1.38x10-23J/K • A system with fewer microstates (smaller W) among which to spread its energy has lower entropy (lower S) • A system with more microstate (larger W) among which to spread its energy has higher entropy (higher S) • Smore microstates > Sless microstates

  10. Directionality of Reactions How probable is it that reactant molecules will react? PROBABILITY suggests that a spontaneous reaction will result in the dispersal * ofenergy * or ofmatter * or of energy & matter.

  11. Directionality of Reactions Probability suggests that a spontaneous reaction will result in the dispersal of energy or of matter or both. Matter Dispersal

  12. Directionality of Reactions Probability suggests that a spontaneous reaction will result in the dispersal of energy or of matter or both. Energy Dispersal

  13. Directionality of ReactionsEnergy Dispersal Exothermic reactions involve a release of stored chemical potential energy to the surroundings. The stored potential energy starts out in a few molecules but is finally dispersed over a great many molecules. The final state—with energy dispersed—is more probable and makes a reaction spontaneous.

  14. Entropy, S (Δssys = Sfinal – Sinitial) So (J/K•mol) H2O(liq) 69.95 H2O(gas) 188.8 S (gases) > S (liquids) > S (solids) ΔSsys = SliquidH2O – SgasH2O <0 = 69.95J/K*mol – 188.8J/K*mol = -118.9 J/K*mol Δssys = SgasH2O – SliquidH2O >0 = 188.8J/K*mol – 69.95J/K*mol = +118.9J/K*mol (discuss in light of microstates)

  15. Entropy and States of Matter S˚(Br2 liq) < S˚(Br2 gas) S˚(H2O sol) < S˚(H2O liq)

  16. Entropy, S Entropy of a substance increases with temperature. Molecular motions of heptane at different temps. Molecular motions of heptane, C7H16

  17. Entropy, S Increase in molecular complexity generally leads to increase in S.

  18. Entropy, S Entropies of ionic solids depend on coulombic attractions. So (J/K•mol) MgO 26.9 NaF 51.5 Mg2+ & O2- Na+ & F-

  19. Entropy, S Entropy usually increases when a pure liquid or solid dissolves in a solvent.

  20. Practice problems • Choose the member with the higher entropy in each of the following pairs, and justify your choice (assume constant temperature except E) • A) 1 mol of SO2(g) or 1 mol of SO3(g) • B) 1 mol of CO2(s) or 1 mol of CO2(g) • C) 3 mol of O2(g) or 2 mol of O3(g) • D) 1 mol of KBr(s) or 1 mol of KBr(aq) • E) Seawater at 2°C or at 23°C • F) 1 mol of CF4(g) or 1 mol of CCl4(g) SO3(g) >atoms CO2(g) s<l<g 3 mol O2(g) >#mols KBr(aq) s<l<g 23°C >T CCl4(g) >molar mass

  21. Where does the entropy go? • The second law of thermodynamics states that entropy of the universe will increase. • An exothermic reaction will release energy (E) to the environment and thus, entropy within the system will decrease. The system becomes more ordered. • An endothermic reaction will absorb energy (E) from the environment and thus, entropy within the system will INCREASE. The system becomes more disordered. • In both cases, the total S will be greater than zero. • ΔSuniv = ΔSsys + ΔSsurr >0 • (discuss combustion & urea/water reactions)

  22. Standard Molar Entropies

  23. Entropy Changes for Phase Changes For a phase change, ∆S = q/T where q = heat transferred in phase change For H2O (liq) ---> H2O(g) ∆H = q = +40,700 J/mol

  24. S increases slightly with T S increases a large amount with phase changes Entropy and Temperature

  25. Calculating ∆S for a Reaction ∆So =  So (products) -  So (reactants) Consider 2 H2(g) + O2(g) ---> 2 H2O(liq) ∆So = 2 So (H2O) - [2 So (H2) + So (O2)] ∆So = 2 mol (69.9 J/K•mol) - [2 mol (130.7 J/K•mol) + 1 mol (205.3 J/K•mol)] ∆So = -326.9 J/K Note that there is a decrease in S because 3 mol of gas give 2 mol of liquid.

  26. Practice Problem • Calculate ΔS°rxn for the combustion of 1 mol of propane (C3H8) at 25°C Write a balanced reaction. Since its combustion we know The products have to be CO2(g) and H2O(l), the other reactant Has to be O2(g) *hint: since there’s 6 mols of gas reactants And only 3 mols of gas products – what do You think entropy should do? C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) ΔS°rxn = [(3mol CO2)(S°CO2) + (4mol H2O)(S°H2O)] – [(1mol C3H8)(S°C3H8) + (5mol O2)(S°O2)] ΔS°rxn = [(3mol)(213.7J/mol*K) + (4mol)(69.9J/mol*K)] – [(1mol)(269.9J/mol*K) + (5mol)(205.0J/mol*K)] ΔS°rxn = -374 J/K

  27. LE CHATELIER'S PRINCIPLE Le Chatelier's Principle:If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. • 1) Using Le Chatelier's Principle with a change of concentration • Suppose you have an equilibrium established between four substances A, B, C and D. • Increase concentration of A: Increase concentration of B:

  28. Using Le Chatelier's Principle with a change of pressure (This only applies to reactions involving gases):

  29. Using Le Chatelier's Principle with a change of temperature • For this, you need to know whether heat is given out or absorbed during the reaction. Assume that our forward reaction is exothermic (heat is evolved):

  30. What would happen if you changed the conditions by increasing the temperature? • According to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the temperature is reduced again. • Suppose the system is in equilibrium at 300°C, and you increase the temperature to 500°C. How can the reaction counteract the change you have made? How can it cool itself down again? • To cool down, it needs to absorb the extra heat that you have just put in. In the case we are looking at, the back reaction absorbs heat. The position of equilibrium therefore moves to the left. The new equilibrium mixture contains more A and B, and less C and D.

  31. What would happen if you changed the conditions by decreasing the temperature? • The equilibrium will move in such a way that the temperature increases again. • Suppose the system is in equilibrium at 500°C and you reduce the temperature to 400°C. The reaction will tend to heat itself up again to return to the original temperature. It can do that by favoring the exothermic reaction.

  32. Le Chateliers principle

  33. 2nd Law of Thermodynamics A reaction is spontaneous if ∆S for the universe is positive. ∆Suniverse = ∆Ssystem + ∆Ssurroundings ∆Suniverse > 0 for spontaneous process First calc. entropy created by matter dispersal (∆Ssystem) Next, calc. entropy created by energy dispersal (∆Ssurround)

  34. 2nd Law of Thermodynamics Dissolving NH4NO3 in water—an entropy driven process. ∆Suniverse = ∆Ssystem + ∆Ssurroundings

  35. 2nd Law of Thermodynamics 2 H2(g) + O2(g) ---> 2 H2O(liq) ∆Sosystem = -326.9 J/K Can calc. that ∆Horxn = ∆Hosystem = -571.7 kJ ∆Sosurroundings = +1917 J/K

  36. Practice problem:Determining Reaction Spontaneity • At 298K, the formation of ammonia has a negative ΔS°sys. Calculate ΔSuniv, and state whether the reaction occurs spontaneously at this temperature. • N2(g) + 3H2(g)  2NH3(g) ΔS°sys = -197J/K ΔH°sys = ΔH°rxn = [(2molNH3)(-45.9kJ/mol)] – [(3molH2)(0 kJ/mol) + (1mol N2)(0 kJ/mol)] = -91.8kJ ΔSsurr = - ΔH°sys = - (-91.8kJ)(1000 J) T (298K)(1kJ) = 308 J/K ΔSuniv = ΔS°sys + ΔSsurr = -197 J/K + 308 J/K = 111 J/K ΔSuniv > 0, so the rxn is spontaneous at 298K

  37. 2 H2(g) + O2(g) ---> 2 H2O(liq) ∆Sosystem = -326.9 J/K ∆Sosurroundings = +1917 J/K ∆Souniverse = +1590. J/K The entropy of the universe is increasing, so the reaction is product-favored. 2nd Law of Thermodynamics

  38. Spontaneous or Not? Remember that –∆H˚sys is proportional to ∆S˚surr An exothermic process has ∆S˚surr > 0.

  39. Entropy and Gibbs free energy, ΔG = ΔH - TΔS What happens when one of the potential driving forces behind a chemical reaction is favorable and the other is not? We can answer this question by defining a new quantity known as the Gibbs free energy (G) of the system, which reflects the balance between these forces. Favorable Unfavorable The entropy term is ΔH° < 0 ΔH° > 0 therefore subtracted ΔS° > 0 ΔS° < 0 from the enthalpy term when calculating ΔG° ΔG°< 0 ΔG° > 0 for a reaction

  40. When to use Gibbs

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