1. Spontaneity, Entropy and Free Energy
Chapter 10
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2. Spontaneity, Entropy and Free Energy - ch 10
1. Which has the greatest entropy?
a. 1 mol of He at STP or 1 mol of He at 25°C
b. 1 mol of Ne at STP or 1 mol of CH4 at STP
c. 1 mol of Cl2 at STP or 1 mol of F2 at STP

3. Spontaneity, Entropy and Free Energy - ch 10
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: CH4 > H2
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)

4. Spontaneity, Entropy and Free Energy - ch 10
2. Predict if ∆Ssys and ∆Ssurr is positive or negative for the following under standard conditions.
a. melting ice
b. photosynthesis ⇒
6 CO2 (g) H2O (l) → C6H12O6 (s) + 6 O2 (g)
c. precipitation of AgCl

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

6. Spontaneity, Entropy and Free Energy - ch 10
4. One mole of an ideal gas is compressed isothermally and reversibly at K from 5.60 atm to 8.90 atm. Calculate ∆S for the gas.
a J/K
b. – 2.34 J/K
c. – 3.85 J/K
d J/K
e. 0 J/K

7. Spontaneity, Entropy and Free Energy - ch 10
4. In a certain reversible expansion, a system at 300. K absorbs exactly 6.00 × 102 J of heat. In the irreversible recompression to the original state of the system, twice as much work is done on the system as is performed on the surroundings in the expansion. What is the entropy change of the system in the recompression step?
a) – 4.00 J/K
b) – 2.00 J/K
c) 0.00 J/K
d) 2.00 J/K
e) 4.00 J/K

8. Spontaneity, Entropy and Free Energy - ch 10
5. A system composed of a ideal gas expands spontaneously in one step from an initial volume of 1.00 L to a final volume of 2.00 L at a constant temperature of 200 K. During the process the gas does 200 J of work. What conclusion can be reached about the value of the entropy change, ΔS, for this process?
a) ΔS = J/K
b) ΔS = – 1.00 J/K
c) ΔS is less than J/K
d) ΔS is greater than J/K
e) None of the above

9. Spontaneity, Entropy and Free Energy - 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-lK-1

10. Spontaneity, Entropy and Free Energy - ch 10
7. Calculate ∆S when 54 g of water is heated from – 22°C to 156 °C at a constant pressure of 1 atm. The heat capacities for solid, liquid and gaseous water are 2.03 J/g°C, 4.18 J/g°C and 2.02 J/g°C respectively. The enthalpies of fusion and vaporization are 6.01 kJ/mol and 40.7 kJ/mol respectively.

11. Spontaneity, Entropy and Free Energy - ch 10
8. 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-1K-1. For A (l) and A (g) the Cp values are 75.0 Jmol-1K-1 and 29.0 Jmol-1K-1 respectively. Calculate ΔHvap at 125°C (its boiling point).

12. Spontaneity, Entropy and Free Energy - ch 10
9. 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.
c. For a spontaneous reaction, if ΔSº < 0 then the reaction must be exothermic.

13. 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

14. Spontaneity, Entropy and Free Energy - ch 10
Spontaneous Temperatures
∆H < 0
negative
∆S > 0
positive
All
Temperatures
∆H > 0
∆S < 0 No
“High” Temperatures
Tspont > ∆H/ ∆S or Tspont > Teq**
“Low” Temperatures
Tspont< ∆H/ ∆S or Tspont < Teq**
** Note ⇒ If ∆H and ∆S are the same sign Teq = ∆H /∆S

15. Spontaneity, Entropy and Free Energy - ch 10
10. A 100-mL 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

16. Spontaneity, Entropy and Free Energy - ch 10
11. At what temperatures would the dissociation of hydrogen be spontaneous?
H2 (g)  H (g)
a. all temperatures
b. high temperatures
c. low temperatures
d. it’ll never be spontaneous

17. Spontaneity, Entropy and Free Energy - ch 10
13. For the freezing of water at – 10 °C and 1 atm, predict whether ∆H, ∆S, and ∆G should be positive, negative or zero.

18. Spontaneity, Entropy and Free Energy - ch 10
14. Given the following data, calculate the normal boiling point for formic acid (HCOOH).
∆Hfo (kJ/mol) S° (J/K·mol)
HCOOH (l) –
HCOOH (g) –
a K
b. 1730°C
c. 388°C
d. 82°C
e. 115°C

19. Spontaneity, Entropy and Free Energy - ch 10
15. Consider the following reaction at 25 oC.
CO (g) + H2O (g) → H2 (g) + CO2 (g)
For this reaction ΔHo = –5.36 kJ 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.

20. Spontaneity, Entropy and Free Energy - ch 10
16. Consider the following reaction.
2 POCl3 (g) → 2 PCl3 (g) + O2 (g)
The free energies of formation at 25 °C are given below. ΔGf° POCl3 (g) = –502 kJ/mol, ΔGf° PCl3 (g) = –270 kJ/mol
Indicate true or false.
a. The entropy change for the reaction is positive.
b. The reaction is not spontaneous at standard conditions and 25 °C but will eventually become spontaneous if the temperature is increased.
c. The equilibrium constant for the reaction at 298 K is less than 1.
d. The enthalpy change for the reaction is positive.

21. 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° = –RTlnKeq

22. Spontaneity, Entropy and Free Energy - ch 10
17. Consider the following reaction at 800 K.
2 NF3 (g) → N2 (g) + 3 F2 (g)
At equilibrium, the partial pressures are PN2= atm, PF2= atm and PNF3= 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

23. Spontaneity, Entropy and Free Energy - ch 10
18. Use the following reaction to answer the following SO2 (g) O2 (g) → 2 SO3 (g)
a. Calculate ∆G°
b. What is the equilibrium constant at 25 °C?
c. At 25 °C the initial pressures for SO2, O2 and SO3 are atm, atm and 40 atm respectively. Will SO3 be consumed or will SO3 be formed?
Substance
∆Gf° (kJ/mol)
SO2 (g)
–300
SO3 (g)
–321

24. Spontaneity, Entropy and Free Energy - ch 10
19. For a certain reaction the Kp is 50 at 25 °C and 9 x 103 at 110 °C.
a. Is the reaction endothermic or exothermic?
b. Calculate Kp at 275 °C?

25. Spontaneity, Entropy and Free Energy - ch 10
20. For a particular reaction a graph was made by plotting ln(Keq) as a function of 1000/K. The x and y intercepts were found to be 3.0 and 40. respectively. Determine the values of ΔH° and ΔS° for this reaction.

26. Answer key – ch. 10 1. Which has the greatest entropy?
a. 1 mol of He at STP or 1 mol of He at 25°C
as T ↑ S ↑
b. 1 mol of Ne at STP or 1 mol of CH4 at STP
as #atoms/molecule ↑ S ↑
c. 1 mol of Cl2 at STP or 1 mol of F2 at STP
as #subatomic particles/molecule ↑ S ↑

27. Answer key – ch. 10
2. Predict if ∆Ssys and ∆Ssurr is positive or negative for the following under standard conditions.
a. melting ice => H2O (s)  H2O (l) => ∆Ssys >0 since Sliquid>Ssolid => ∆Hsys>0 since bonds are broken => ∆Ssurr<0
b. photosynthesis =>
6 CO2 (g) H2O (l)  C6H12O6 (s) + 6 O2 (g)
∆Ssys<0 since the moles of gas are equal and considering the moles solids ↑ and liquids ↓ => ∆Hsys>0 because the reaction is combustion flipped => ∆Ssurr<0
c. precipitation of AgCl => Ag+ (aq) + Cl-(aq)  AgCl(s)
∆Ssys<0 since Ssolid<Saqueous => ∆Hsys<0 since bonds are forming => ∆Ssurr>0

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

29. ∆S = (1mol)(8.314J/molK)(ln(5.6atm/8.9atm))
Answer key – ch. 10
4. One mole of an ideal gas is compressed isothermally and reversibly at K from 5.60 atm to 8.90 atm. Calculate ∆S for the gas.
a J/K
b J/K
c J/K
d J/K
e. 0 J/K
∆S = nRlnV2/V1
since P1V1=P2V2
∆S = nRlnP1/P2
∆S = (1mol)(8.314J/molK)(ln(5.6atm/8.9atm))
∆S = -3.85J/K

30. therefore for the recompression
Answer key – ch. 10
5. In a certain reversible expansion, a system at 300. K absorbs exactly 6.00 × 102 J of heat. In the irreversible recompression to the original state of the system, twice as much work is done on the system as is performed on the surroundings in the expansion. What is the entropy change of the system in the recompression step?
a) J/K
b) J/K
c) 0.00 J/K
d) 2.00 J/K
e) 4.00 J/K
∆S = qrev/T
for the expansion
∆S = (600J)/(300K) = 2 J/K
therefore for the recompression
∆S = -2 J/K

31. Answer key – ch. 10
6. A system composed of a ideal gas expands spontaneously in one step from an initial volume of 1.00 L to a final volume of 2.00 L at a constant temperature of 200 K. During the process the gas does 200 J of work. What conclusion can be reached about the value of the entropy change, ΔS, for this process?
a) ΔS = J/K
b) ΔS = – 1.00 J/K
c) ΔS is less than J/K
d) ΔS is greater than J/K
e) None of the above
wirrev = -200J
∆E = 0 since isothermal
qirrev = 200J (qirrev <qrev)
∆S = qrev/T
qrev > qirrev
∆S > qirrev/T
∆S > (200J)/(200K) > 1J/K

32. Answer key – ch. 10
7. 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-lK-1
∆Stotal = ∆Scold + ∆Shot
temperature is changing => ∆S = nClnT2/T1
∆Stotal = nClnT2/T1 + nClnT2/T1
T2 is unknown
-nC∆T = nC∆T where C cancels out
-(1mol)(T2-100°C) = (3mol)(T2-0°C) => T2 = 25 °C
∆Stotal = (3mol)(75.3J/molK)(ln(298K)/(273K)) +
(1mol)(75.3J/molK)(ln(298K)/(373K)) = 2.89 J/K
2.89 J/K

33. Answer key – ch. 10
8. Calculate ∆S when 54 g of water is heated from -22°C to 156 °C at a constant pressure of 1 atm. The heat capacities for solid, liquid and gaseous water are 2.03 J/g°C, 4.18 J/g°C and 2.02 J/g°C respectively. The enthalpies of fusion and vaporization are 6.01 kJ/mol and 40.7 kJ/mol respectively.
Water will undergo two phase changes within this temperature range – so we need to break it down into steps
Step 1 => water will be a solid from -22°C to 0°C => ∆S1 = mClnT2/T1
∆S1 = (54g)(2.03J/g°C)(ln(273K/251K)) = 9.2 J/K
Step 2 => water will melt at 0°C => ∆S2 = qrev/T
∆S2 = (54g/18g/mol)(6010J/mol)/(273K) = 66 J/K
continue to next slide…

34. Answer key – ch. 10 ∆S5 = (54g)(2.02J/g°C)(ln(429K/373K) = 15 J/K
8. …continued
Step 3 => water will be a liquid from 0°C to 100°C => ∆S3 = mClnT2/T1
∆S3 = (54g)(4.18J/g°C)(ln(373K/273K)) = 70.4 J/K
Step 4 => water will boil at 100°C => ∆S4 = qrev/T
∆S4 = (54g/18g/mol)(40,700J/mol)/(373K) = 327J/K
Step 5 => water will be a gas from 100°C to 156°C => ∆S5 =mClnT2/T1
∆S5 = (54g)(2.02J/g°C)(ln(429K/373K) = 15 J/K
∆Stotal = 9.2 J/K + 66 J/K J/K J/K J/K = 473 J/K

35. Answer key – ch. 10 9. 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-1K-1. For A (l) and A (g) the Cp values are 75.0 Jmol-1K-1 and 29.0 Jmol-1K-1 respectively. Calculate ΔHvap at 125°C (its boiling point).
Since a phase change will occur within this temperature range => break it into steps
Step 1 => Substance A is a liquid from 75°C to 125°C
∆S1 = ClnT2/T1
continue to next slide…

36. Answer key – ch. 10 Step 2 => Substance A will boil at 125°C
∆S2 = qrev/T
Step 3 => Substance A will be a gas from 125°C to 155°C
∆S3 = ClnT2/T1
∆Stotal = ∆S1 + ∆S2 + ∆S3
75J/molK = (75J/molK)(ln(398K/348K)) + qrev/398K + (29J/molK)(ln(428K/398K))
qrev = 25 kJ/mol

37. Answer key – ch. 10
10. 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

38. Answer key – ch. 10
11. A 100-mL 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

39. Answer key – ch. 10
12. At what temperatures would the dissociation of hydrogen be spontaneous?
H2 (g)  H (g)
∆H > 0 since bonds are broken
∆S > 0 since 2 moles of gas have more entropy than one
∆G <0 only at “high” temperatures

40. Answer key – ch. 10
13. For the freezing of water at -10 °C and 1 atm, predict whether ∆H, ∆S, and ∆G should be positive, negative or zero.
∆H < 0 since bonds are formed
∆S < 0 since liquids have more entropy than solids
∆G < 0 only at “low” temperatures => T < ∆H/∆S => We’re not given ∆H or ∆S values but we know the freezing point of water is 0°C => ∆H/∆S = 0°C => so if the temperature is below 0°C it’s spontaneous

41. Answer key – ch. 10
14. Given the following data, calculate the normal boiling point for formic acid (HCOOH).
∆Hfo (kJ/mol) S° (J/K·mol)
HCOOH (l) –
HCOOH (g) –
a K
b. 1730°C
c. 388°C
d. 82°C
e. 115°C
Boiling points (and freezing points) are
temperatures at equilibrium (∆G=0)
T = ∆H/∆S
T = [(-363kJ/mol)-(-410kJ/mol)]/[(0.251J/molK) –
(0.13J/molK)] = 388K or 115 °C

42. Answer key – ch. 10 15. Consider the following reaction at 25 oC.
CO (g) + H2 (g) → H2 (g) + CO (g)
For this reaction ΔHo = –5.36 kJ 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.
Since both ΔHo and ΔS° are negative
the reaction will be spontaneous at
T > ΔH°/ΔS°
T > (-5.36 kJ)/( kJ/K)
T > 48.8 K

43. Answer key – ch. 10 16. Consider the following reaction.
2 POCl3 (g) → 2 PCl3 (g) + O2 (g)
The free energies of formation at 25 °C are given below. ΔGf° POCl3 (g) = –502 kJ/mol, ΔGf° PCl3 (g) = –270 kJ/mol
Indicate true or false.
a. The entropy change for the reaction is positive. True – moles of gas increased
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)
Continue to next slide…

44. Answer key – ch. 10 ΔG° = (2 mol PCl3)(-270 kJ/mol) –
(2 mol POCl3)(-502 kJ/mol) = 462 kJ =>
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 flipped

45. Answer key – ch. 10 17. Consider the following reaction at 800 K.
2 NF3 (g) → N2 (g) + 3 F2 (g)
At equilibrium, the partial pressures are PN2= atm, PF2= atm and PNF3= 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
since
K = (0.04)(0.063)3/(0.66)2
K = 2.3 x 10-5
when K <1 the reaction
favors the reactants or
∆G° > 0

46. continue to next slide…
Answer key – ch. 10
18. Use the following reaction to answer the following SO2 (g) O2 (g)  2 SO3 (g)
a. Calculate ∆G°
b. What is the equilibrium constant at 25 °C?
c. At 25 °C the initial pressures for SO2, O2 and SO3 are atm, atm and 40 atm respectively. Will SO3 be consumed or will SO3 be formed?
Substance
∆Gf° (kJ/mol)
SO2 (g)
-300
SO3 (g)
-321
continue to next slide…

47. Answer key – ch. 10 18. …continued
a. ΔG° = Σ ∆Gf°(products) - Σ∆Gf°(reactants)
ΔG° = (2 mol SO3)(-321kJ/mol) – (2 mol SO2)(-300kJ/mol)
ΔG° = -42 kJ
b. ΔG° = -RTlnK
-42 kJ = -(8.314x10-3kJ/molK)(298K)(lnK)
K = 2.3x107
c. There’s 2 ways to do this =>
option 1 => ΔG = ΔG° + RTlnQ = -42kJ + (8.314x10-3kJ/molK)
(298K)(ln(40)2/(0.001)2(0.002)) = 26 kJ or nonspont.
option 2 => compare K to Q => Q = (40)2/(0.001)2(0.002) = 8x1011 => since K < Q reaction wants to go backward

48. Answer key – ch. 10
19. For a certain reaction the Kp is 50 at 25 °C and Kp is
9 x 103 at 110 °C.
a. Is the reaction endothermic or exothermic? if K↑ and T↑
b. Calculate Kp at 275 °C? ln(K2/K1) = (ΔH/R)(1/T1 – 1/T2)
since we don’t have ΔH =>
ln(9x103/50) = (ΔH/(8.314 J/mol))(1/298K – 1/383K)
ΔH = 57,970 J/mol
now that we have ΔH lets make K2 be unknown
ln(K2/50) = ((57,970 J/mol)/(8.314 J/mol))(1/298K – 1/548K)
K2 = 2.16 x 106

49. Answer key – ch. 10
20. For a particular reaction a graph was made by plotting ln(Keq) as a function of 1000/K. The x and y intercepts were found to be 3.0 and 40. respectively. Determine the values of ΔH° and ΔS° for this reaction.
lnK = (-∆H °/R)(1/T) + ΔS°/R =>
equation of a line => y = mx + b
so y intercept = ΔS°/R => 40 = ΔS°/8.314 J/molK => ΔS°= 333 J/molK
x intercept means y = 0 or lnK = 0 and 1000/T = 3 =>
1/T = 0.003/K
0 = (- ∆H°/8.314J/molK)(0.003/K) + 40 => ∆H°= 111kJ/mol