Transcript C - UCSB CLAS

Spontaneity, Entropy and
Free Energy
Chapter 10
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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 1atm and 25°C
b. 1 mol of Ne at STP or 1 mol of CH4 at STP
c. 1 mol of Cl2 at 1 atm and 25°C or 1 mol of Br2 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: 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)
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 CO2 (g) + 6 H2O (l) → C6H12O6 (s) + 6 O2 (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-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
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 kJ/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-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 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.
c. 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
“High” Temperatures
∆H
Tspont >
or Tspont > Teq**
∆S
∆H < 0
negative
∆S < 0
negative
Depends on
Temperature
“Low” Temperatures
∆H
Tspont >
or Tspont < Teq**
∆S
** Note ⇒ If ∆H and ∆S are the same sign Teq =
∆H
∆S
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?
F2 (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 for formic acid
(HCOOH).
∆Hfo (kJ/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 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.
Spontaneity, Entropy and Free Energy - ch 10
14. 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 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.
d. The equilibrium constant for the reaction at 298 K is less than 1.
e. Increasing the pressure of POCl3 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° = –RTlnKeq
Spontaneity, Entropy and Free Energy - ch 10
15. Consider the following reaction at 800 K.
2 NF3 (g) → N2 (g) + 3 F2 (g)
At equilibrium, the partial pressures are PN2= 0.040 atm, PF2= 0.063 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
Spontaneity, Entropy and Free Energy - ch 10
16. Use the following reaction to answer the following.
2 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 0.001 atm,
0.002 atm and 40 atm respectively. Will SO3 be consumed or will
SO3 be formed?
Substance
∆Gf° (kJ/mol)
SO2 (g)
–300
SO3 (g)
–321
Spontaneity, Entropy and Free Energy - ch 10
17. Consider the following reaction and thermodynamic data:
2 NO (g) + O2 (g) → 2 NO2 (g)
Ho = –190 kJ Go = –71 kJ at 600 K
Calculate the equilibrium constant (K) for this reaction at 370 K.
a. 2.9x1016
b. 7.1x1013
c. 1.9x10–7
d. 6.1x105
e. 3.7x103
Spontaneity, Entropy and Free Energy - ch 10
You have completed ch10
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 1atm and 25°C
as P↓ , V ↑ , S ↑
b. 1 mol of Ne at STP or 1 mol of CH4 at STP
as the number of atoms/molecule ↑ , S ↑
c. 1 mol of Cl2 at 1 atm and 25°C or 1 mol of Br2 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 ⇒ H2O (s)  H2O (l) ⇒ ∆Ssys >0 since Sliquid>Ssolid ⇒
∆Hsys>0 since bonds are broken ⇒ ∆Ssurr<0
b. photosynthesis =>
6 CO2 (g) + 6 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 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-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
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 = nRlnV2/V1
√ c. -3.85 J/K
since P1V1=P2V2
d. 3.85 J/K
∆S = nRlnP1/P2
e. 0 J/K
∆S = (1mol)(8.314J/molK)(ln(5.6atm/8.9atm))
∆S = -3.85J/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-lK-1
∆Stotal = ∆Scold + ∆Shot
T
temperature is changing ⇒ ∆S = nCln 2
T1
T
T
∆Stotal = nCln 2 + nCln 2
T1
T1
T2 is unknown
-nC∆T = nC∆T where C cancels out
-(1 mol)(T2 - 100°C) = (3 mol)(T2 - 0°C) ⇒ T2 = 25 °C
298K
298K
∆Stotal = (3mol)(75.3J/molK)ln
+ (1mol)(75.3J/molK)ln
= 2.89 J/K
273K
373K
Answer key – 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-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
T
Step 1 ⇒ Substance A is a liquid from 75°C to 125°C ⇒ ∆S1 = Cln 2
T1
q
ΔHvap
Step 2 ⇒ Substance A will boil at 125°C ⇒ ∆S2 = rev =
T
T
T
Step 3 ⇒ Substance A is a gas from 125°C to 155°C ⇒ ∆S3 = Cln 2
T1
T ΔHvap
T
∆Stotal = ∆S1 + ∆S2 + ∆S3 = Cln 2 +
+ Cln 2
T1
T
T1
continue to next slide…
Answer key – ch. 10
7. …continued
75J/molK = (75J/molK)ln
ΔHvap = 25 kJ/mol
398K ΔHvap
428K
+
+ (29J/molK)ln
348K 398K
398K
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?
F2 (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 for formic acid
(HCOOH).
∆Hfo (kJ/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
Boiling points (and freezing points) are
temperatures at equilibrium (∆G=0)
T=
ΔH ΔH°f 𝑝𝑟𝑜𝑑 − ΔH°f (𝑟𝑒𝑎𝑐𝑡)
=
ΔS
S° 𝑝𝑟𝑜𝑑 − S°(𝑟𝑒𝑎𝑐𝑡)
T=
(−363kJ/mol)−(−410kJ/mol)
(0.251J/molK) – (0.13J/molK)
T = 388K or 115 °C
Answer key – ch. 10
13. 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
Since both ΔHo and ΔS° are negative
b. T < 48.8 K
the reaction will be spontaneous at
c. T > 20.5 K
ΔH
T<
d. T < 20.5 K
ΔS
e. Spontaneous at all temperatures. T < (−5.36 kJ)
(−0.1098 kJ/K)
T < 48.8 K
Answer key – ch. 10
14. 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 ↑
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 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
reaction in reverse
e. Increasing the pressure of POCl3 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 NF3 (g) → N2 (g) + 3 F2 (g)
At equilibrium, the partial pressures are PN2= 0.040 atm, PF2= 0.063 atm and
PNF3= 0.66 atm. Which of the following is true about the value of ∆G°?
√ a. is a positive number
∆G°= -RTlnK
b. is a negative number
since
c. is equal to zero
(0.04)(0.063)3
K=
d. is independent of the temperature
(0.66)2
e. can not be predicted from this data
K = 2.3 x 10-5
when K <1 the reaction
favors the reactants or
∆G° > 0
Answer key – ch. 10
16. Use the following reaction to answer the following.
2 SO2 (g) + O2 (g)  2 SO3 (g)
a. Calculate ∆G°
ΔG° = Σ ∆Gf°(products) - Σ∆Gf°(reactants)
ΔG° = (2 mol SO3)(-321kJ/mol) – (2 mol SO2)(– 300kJ/mol)
ΔG° = – 42 kJ
b. What is the equilibrium constant at 25 °C?
ΔG° = -RTlnK
-42 kJ = -(8.314x10-3kJ/molK)(298K)(lnK)
K = 2.3x107
c. At 25 °C the initial pressures for SO2, O2 and SO3 are 0.001 atm, 0.002 atm and
40 atm respectively. Will SO3 be consumed or will SO3 be formed?
ΔG = ΔG° + RTlnQ
2
(40)
11
Q = (0.001)
2(0.002) = 8x10
ΔG = – 42 kJ + (0.008314 kJ/molK)(298K)(ln8x1011) = 26 kJ
Since ΔG > 0 the reaction is non-spontaneous or will go backward
Answer key – ch. 10
17. Consider the following reaction and thermodynamic data:
2 NO (g) + O2 (g) → 2 NO2 (g)
Ho = –190 kJ Go = –71 kJ at 600 K
Calculate the equilibrium constant (K) for this reaction at 370 K.
√ a. 2.9x1016
The equilibrium constant depends on T
b. 7.1x1013
𝐾2
− ∆𝐻° 1
1
–7
ln
=
−
c. 1.9x10
𝐾1
𝑅
𝑇2
𝑇1
1st we need to get the equilibrium constant at 600K
d. 6.1x105
ΔG° = -RTlnK
e. 3.7x103
–71kJ = –(0.0083145kJ/molK)(600K)lnK
K = 1.52 x 106
− –190 kJ
𝐾
ln 1.52 x2 106 =
0.0083145kJ/molK
K2 = 2.9x1016
1
370𝐾
−
1
600𝐾