Spontaneity Entropy and Free Energy Chapter 10 Email

Spontaneity, Entropy and Free Energy Chapter 10 E-mail: benzene 4 president@gmail Web-site: http: //clas. sa. ucsb. edu/staff/terri/

Spontaneity, Entropy and Free Energy - ch 10 1. Which has the greatest entropy? a. 1 mol of He at 0. 5 atm and 25°C or 1 mol of He at 1 atm and 25°C b. 1 mol of Ne at STP or 1 mol of CH 4 at STP c. 1 mol of Cl 2 at 1 atm and 25°C or 1 mol of Br 2 at 1 atm and 25°C

Spontaneity, Entropy and Free Energy - ch 10 Hierarchy of Entropy (S) Considerations: 1. Phase - S(s) < S(l) << S(g) 2. There is more entropy at higher temperatures and/or larger volumes (lower pressures) 3. The more bonds per molecule the greater the positional probability ex: CH 4 > H 2 4. If there are the same number of atoms in the molecules/elements; then the one with more electrons has the greater the positional probability ex: Ar > He 5. For the same atom but different structures (allotropes) the positional probability is greater in the more disordered structure ex: C (graphite) > C (diamond)

Spontaneity, Entropy and Free Energy - ch 10 2. Predict if ∆Ssys and ∆Ssurr is positive, negative or zero for the following under standard conditions. a. melting ice b. photosynthesis 6 CO 2 (g) + 6 H 2 O (l) → C 6 H 12 O 6 (s) + 6 O 2 (g)

Spontaneity, Entropy and Free Energy - ch 10 3. One mole of an ideal gas at 25°C undergoes a reversible expansion from 125. 0 L to 250. 0 L. Which statement is correct? a. ∆Sgas = 0 b. ∆Ssurr = 0 c. ∆Suniv = 0

Spontaneity, Entropy and Free Energy - ch 10 4. A 100 -m. L sample of water is placed in a coffee cup calorimeter. When 1. 0 g of an ionic solid is added, the temperature decreases from 21. 5°C to 20. 8°C as the solid dissolves. Which of the following is true for the dissolving of the solid? a. ∆H < 0 b. ∆Suniv > 0 c. ∆Ssys < 0 d. ∆Ssurr > 0 e. none of these

Spontaneity, Entropy and Free Energy - ch 10 5. One mole of an ideal gas is compressed reversibly at 607. 4 K from 5. 60 atm to 8. 90 atm. Calculate ∆S for the gas. a. 2. 34 J/K b. – 2. 34 J/K c. – 3. 85 J/K d. 3. 85 J/K e. 0 J/K

Spontaneity, Entropy and Free Energy - ch 10 6. Calculate the change in entropy (in J/K) for a process in which 54 g of ice at – 5 °C is mixed with 112 g of liquid water at 100°C in a perfectly insulated container. The specific heat capacities of ice and liquid water is 2. 03 J/g°C and 4. 18 J/g°C respectively. The heat of fusion for water is 6. 01 k. J/mol.

Spontaneity, Entropy and Free Energy - ch 10 7. Consider the process A (l) at 75°C → A (g) at 155°C which is carried out at constant pressure. The total ΔS for this process is 75. 0 Jmol-1 K-1. For A (l) and A (g) the Cp values are 75. 0 Jmol-1 K-1 and 29. 0 Jmol-1 K-1 respectively. Calculate ΔHcondensation at 125°C (its boiling point).

Spontaneity, Entropy and Free Energy - ch 10 8. Indicate true or false for each of the following statements: a. Spontaneous reactions must have a positive ΔSº for the reaction. b. When the change in free energy is less than zero for a chemical reaction, the reaction must be exothermic. For a spontaneous reaction, if ΔSº < 0 then the reaction must be exothermic.

Spontaneity, Entropy and Free Energy - ch 10 Spontaneous ⇒ wants to go forward on its own K > Q ∆Suniverse > 0 or ∆Gsystem < 0 Non-spontaneous ⇒ wants to go backward on its own K < Q ∆Suniverse < 0 or ∆Gsystem > 0 Equilibrium ⇒ doesn’t prefer one direction over the other K = Q ∆Suniverse = 0 or ∆Gsystem = 0

Spontaneity, Entropy and Free Energy - ch 10 ∆H ∆S ∆G Spontaneous Temperatures ∆H < 0 negative ∆S > 0 positive Always Negative All Temperatures ∆H > 0 positive ∆S < 0 negative Always Positive No Temperatures ∆H > 0 positive ∆S > 0 positive Depends on Temperature ∆H < 0 negative ∆S < 0 negative Depends on Temperature

Spontaneity, Entropy and Free Energy - ch 10 9. The following graph of G° versus temperature (T) corresponds to which of the following situations? a. H° < 0 and S° > 0 b. H° > 0 and S° < 0 c. H° > 0 and S° > 0 d. H° < 0 and S° < 0

Spontaneity, Entropy and Free Energy - ch 10 10. Which of the following is true for the dissociation of fluorine? F 2 (g) → 2 F (g) a. spontaneous at all temperatures b. spontaneous at high temperatures c. spontaneous at low temperatures d. never spontaneous

Spontaneity, Entropy and Free Energy - ch 10 11. At 1 atm, the freezing point of mercury is – 39 °C. Which of the following is true regarding the freezing of mercury at – 30 °C and 1 atm? a. Ssurr > 0, Suniv > 0 b. Ssurr > 0, Suniv = 0 c. Ssurr < 0, Suniv > 0 d. Ssurr < 0, Suniv < 0 e. Ssurr > 0, Suniv < 0

Spontaneity, Entropy and Free Energy - ch 10 12. Given the following data, calculate the normal boiling point formic acid (HCOOH). ∆Hfo (k. J/mol) S° (J/K·mol) HCOOH (l) – 410. 130. HCOOH (g) – 363 251 a. 2. 57 K b. 1730°C c. 388°C d. 82°C e. 115°C

Spontaneity, Entropy and Free Energy - ch 10 13. Consider the following reaction at 25 o. C. CO (g) + H 2 O (g) → H 2 (g) + CO 2 (g) For this reaction ΔHo = – 5. 36 k. J and ΔSo = – 109. 8 J /K. At what temperatures will the reaction be spontaneous? a. T > 48. 8 K b. T < 48. 8 K c. T > 20. 5 K d. T < 20. 5 K e. Spontaneous at all temperatures.

Spontaneity, Entropy and Free Energy - ch 10 14. Consider the following reaction. 2 POCl 3 (g) → 2 PCl 3 (g) + O 2 (g) The free energies of formation at 25 °C are given below. ΔGf° POCl 3 (g) = – 502 k. J/mol, ΔGf° PCl 3 (g) = – 270 k. J/mol Indicate true or false. a. The entropy change for the reaction is positive. b. The enthalpy change for the reaction is positive. c. The reaction is non-spontaneous at standard conditions and 25 °C but will eventually become spontaneous if the temperature is increased. The equilibrium constant for the reaction at 298 K is less than 1. e. Increasing the pressure of POCl 3 will cause an increase in ΔG.

Thermodynamics - ch 10 ΔG° verses Keq ΔG° tells you where the equilibrium lies If ΔG° < 0 the equilibrium favors the products or Keq > 1 If ΔG° > 0 the equilibrium favors the reactants or Keq < 1 ΔG° = –RTln. Keq

Spontaneity, Entropy and Free Energy - ch 10 15. Consider the following reaction at 800 K. 2 NF 3 (g) → N 2 (g) + 3 F 2 (g) At equilibrium, the partial pressures are PN 2= 0. 040 atm, PF 2= 0. 063 atm and PNF 3= 0. 66 atm. Which of the following is true about the value of ∆G°? a. is a positive number b. is a negative number c. is equal to zero d. is independent of the temperature e. can not be predicted from this data

Spontaneity, Entropy and Free Energy - ch 10 16. Use the following reaction to answer the following. 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g) a. Calculate ∆G° b. What is the equilibrium constant at 25 °C? c. At 25 °C the initial pressures for SO 2, O 2 and SO 3 are 0. 001 atm, 0. 002 atm and 40 atm respectively. Will SO 3 be consumed or will SO 3 be formed? Substance ∆Gf° (k. J/mol) SO 2 (g) – 300 SO 3 (g) – 321

Spontaneity, Entropy and Free Energy - ch 10 17. Consider the following reaction and thermodynamic data: 2 NO (g) + O 2 (g) → 2 NO 2 (g) Ho = – 190 k. J Go = – 71 k. J at 600 K Calculate the equilibrium constant (K) for this reaction at 370 K. a. 2. 9 x 1016 b. 7. 1 x 1013 c. 1. 9 x 10– 7 d. 6. 1 x 105 e. 3. 7 x 103

Spontaneity, Entropy and Free Energy - ch 10 You have completed ch 10

Answer key – ch. 10 1. Which has the greatest entropy? a. 1 mol of He at 0. 5 atm and 25°C or 1 mol of He at 1 atm and 25°C as P↓ , V ↑ , S ↑ b. 1 mol of Ne at STP or 1 mol of CH 4 at STP as the number of atoms/molecule ↑ , S ↑ c. 1 mol of Cl 2 at 1 atm and 25°C or 1 mol of Br 2 at 1 atm and 25°C gases have more entropy than liquids

Answer key – ch. 10 2. Predict if ∆Ssys and ∆Ssurr is positive, negative or zero for the following under standard conditions. a. melting ice ⇒ H 2 O (s) H 2 O (l) ⇒ ∆Ssys >0 since Sliquid>Ssolid ⇒ ∆Hsys>0 since bonds are broken ⇒ ∆Ssurr<0 b. photosynthesis => 6 CO 2 (g) + 6 H 2 O (l) C 6 H 12 O 6 (s) + 6 O 2 (g) ∆Ssys<0 since the moles of gas are equal and considering the moles solids ↑ and liquids ↓ ⇒ ∆Hsys>0 because the reaction is combustion in reverse ⇒ ∆Ssurr<0

Answer key – ch. 10 3. One mole of an ideal gas at 25°C is expanded isothermally and reversibly from 125. 0 L to 250. 0 L. Which statement is correct? a. ∆Sgas = 0 b. ∆Ssurr = 0 ∆Sgas > 0 √ c. ∆Suniv = 0 since expanding a gas causes an increase in entropy qrev > 0 ⇒ ∆Ssurr < 0 ∆Suniv = 0 since reversible a. k. a. equilibrium

Answer key – ch. 10 4. A 100 -m. L sample of water is placed in a coffee cup calorimeter. When 1. 0 g of an ionic solid is added, the temperature decreases from 21. 5°C to 20. 8°C as the solid dissolves. Which of the following is true for the dissolving of the solid? a. ∆H < 0 false ⇒ it was observed that the T of the surroundings cooled √ b. ∆Suniv > 0 true ⇒ because the solid dissolved spontaneously c. ∆Ssys < 0 false ⇒ aqueous ions have more S than in a solid d. ∆Ssurr > 0 false ⇒ since the dissolving was endothermic e. none of these

Answer key – ch. 10 5. One mole of an ideal gas is compressed isothermally and reversibly at 607. 4 K from 5. 60 atm to 8. 90 atm. Calculate ∆S for the gas. a. 2. 34 J/K b. -2. 34 J/K ∆S = n. Rln. V 2/V 1 √ c. -3. 85 J/K since P 1 V 1=P 2 V 2 d. 3. 85 J/K ∆S = n. Rln. P 1/P 2 e. 0 J/K ∆S = (1 mol)(8. 314 J/mol. K)(ln(5. 6 atm/8. 9 atm)) ∆S = -3. 85 J/K

Answer key – ch. 10 6. Calculate the change in entropy for a process in which 3. 00 moles of liquid water at 0°C is mixed with 1. 00 mole of liquid water at 100°C in a perfectly insulated container. The molar heat capacity of liquid water is 75. 3 Jmol-l. K-1

Answer key – ch. 10 o

Answer key – ch. 10 o

Answer key – ch. 10 8. Indicate true or false for each of the following statements. a. Spontaneous reactions must have a positive ΔSº for the reaction. False, ΔSº can be either positive or negative in spontaneous reactions b. When the change in free energy is less than zero for a chemical reaction, the reaction must be exothermic. False, ΔHº can be either positive or negative in spontaneous reactions c. For a spontaneous reaction, if ΔSº < 0 then the reaction must be exothermic. True, ΔG = ΔH – TΔS

Answer key – ch. 10 9. The following graph of G° versus temperature (T) corresponds to which of the following situations? a. H° < 0 and S° > 0 ⇒ always spontaneous ( G° < 0 ) b. H° > 0 and S° < 0 ⇒ always non-spontaneous( G° < 0 ) √c. H° > 0 and S° > 0 ⇒ only spontaneous at high T’s d. H° < 0 and S° < 0 ⇒ only spontaneous at low T’s Non-spontaneous at lower T’s Spontaneous at higher T’s

Answer key – ch. 10 10. Which of the following is true for the dissociation of fluorine? F 2 (g) → 2 F (g) a. spontaneous at all temperatures √ b. spontaneous at high temperatures c. spontaneous at low temperatures d. never spontaneous

Answer key – ch. 10 11. At 1 atm, the freezing point of mercury is – 39 °C. Which of the following is true regarding the freezing of mercury at – 30 °C and 1 atm? √ a. Ssurr > 0, Suniv > 0 Ssurr > 0 b. Ssurr > 0, Suniv = 0 since freezing is exothermic due to c. Ssurr < 0, Suniv > 0 bonds being formed d. Ssurr < 0, Suniv < 0 Suniv > 0 since the T< FP and e. Ssurr > 0, Suniv < 0 therefore will spontaneously freeze

Answer key – ch. 10 12. Given the following data, calculate the normal boiling point formic acid (HCOOH). ∆Hfo (k. J/mol) S° (J/K·mol) HCOOH (l) – 410. 130. HCOOH (g) – 363 251 a. 2. 57 K b. 1730°C c. 388°C d. 82°C √ e. 115°C

Answer key – ch. 10 13. Consider the following reaction at 25 o. C. CO (g) + H 2 (g) → H 2 (g) + CO (g) For this reaction ΔHo = – 5. 36 k. J and ΔSo = – 109. 8 J /K. At what temperatures will the reaction be spontaneous? √ a. T > 48. 8 K b. T < 48. 8 K c. T > 20. 5 K d. T < 20. 5 K e. Spontaneous at all temperatures.

Answer key – ch. 10 14. Consider the following reaction. 2 POCl 3 (g) → 2 PCl 3 (g) + O 2 (g) The free energies of formation at 25 °C are given below. ΔGf° POCl 3 (g) = – 502 k. J/mol, ΔGf° PCl 3 (g) = – 270 k. J/mol Indicate true or false. a. The entropy change for the reaction is positive. True ⇒ moles of gas ↑ b. The reaction is not spontaneous at standard conditions and 25 °C but will eventually become spontaneous if the temperature is increased. True ⇒ ΔG° = Σ ∆Gf°(products) - Σ∆Gf°(reactants) ΔG° = (2 mol PCl 3)(-270 k. J/mol) – (2 mol POCl 3)(-502 k. J/mol) = 462 k. J ⇒ positive so nonspontaneous ⇒ how ever since ΔH° and ΔS° are both positive it will be come spontaneous when the temperature becomes high enough c. The equilibrium constant for the reaction at 298 K is less than 1. True – if ΔG° > 0 then reactants are favored or K <1 d. The enthalpy change for the reaction is positive. True – this reaction is a combustion reaction in reverse e. Increasing the pressure of POCl 3 will cause an increase in ΔG. False – more reactants makes the reaction more spontaneous which causes ΔG ↓

Answer key – ch. 10 15. Consider the following reaction at 800 K. 2 NF 3 (g) → N 2 (g) + 3 F 2 (g) At equilibrium, the partial pressures are PN 2= 0. 040 atm, PF 2= 0. 063 atm and PNF 3= 0. 66 atm. Which of the following is true about the value of ∆G°? √ a. is a positive number b. is a negative number c. is equal to zero d. is independent of the temperature e. can not be predicted from this data

Answer key – ch. 10 o

Answer key – ch. 10 17. Consider the following reaction and thermodynamic data: 2 NO (g) + O 2 (g) → 2 NO 2 (g) Ho = – 190 k. J Go = – 71 k. J at 600 K Calculate the equilibrium constant (K) for this reaction at 370 K. √ a. 2. 9 x 1016 b. 7. 1 x 1013 c. 1. 9 x 10– 7 d. 6. 1 x 105 e. 3. 7 x 103