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Unformatted text preview: Thermodynamics Practice 1. Given the following standard molar entropies measured at 25 °C and 1 atm pressure,
calculate AS° in (J/K) for the reaction 2 Al(s) + 3 MgO(s) ——> 3 Mg(s) 4 M20309) A10) = 28.0 J/K MgO(s) = 27.0 MC [39% 5,9,me ﬁéwaei
Mg(s) = 33.0 J/K A0030) = 51.0 J/K :[3{% 0? a” (5% e g
7. ”‘ ‘7“ . 429.01/K
—13.0 J/K
13.0 J/K
69.0 J/K
139 J/K 2. For the given reaction and the following information, caiculate AG“ at 25°C. 2 Pb0(s) + 2 SOg(g) ——> 2 PbS(s) + 3 02(9) Species AH° (Id/mole) AS° U/mole  K)
at 25°C and 1 atm at 25°C and 1 atm PbOG)
502(9) '
PbSG) A. 273.0 k]
438.0 M
634.0 M ; 782.0kJ E g
E. 830.0kJ Agni”: [1(@%@*“5(2050§j [ng W: : 70H d/moi zoj'” 3. Given the information that follows, calculate the standard ffee energy change, 136°, for the reaction CH4(g) + 2 02(g) a 2 H2002) + C02(g) AG° (Id/mole) at AH“ (kl/mole) at
25°C and 1 atm 25°C and 1 atm . ~919.00 kJ/mole
~817.00 kJ/mole
c. —408.50 kJ/mole D. 459.50 kJ/mole
E. 919.00 kJ/mole 4. Calculate the approximate standard free energy change for the ionization of hydrofluoric
acid, HF (Kn = 1.0 x 10—3), at 25°C. A. —9.0kJ [My :2 a RT M K I 5% B. —4.0 kJ , ,.. (@ﬁEQd/Q§<[email protected] if} EEE (H3 KEVE’E: C. 0.050 10 E ‘43; ‘3: ”MA E K 5. Arrange the following reactions according to increasing ASUHn values. 1. H20(g) —> 1120(9) , be
2' _ 2 HCI(g) a H2(g) + (312(5)) 5% VQN 5%“ (ME Swag ‘90:?“ éedaﬁ Wm
3. Si02(s) ——> Si(s) + 02(g) + My ngsgs‘a’“ ’7 mo 33:!» ”by, g lowest a highest
AS°(1) < AS°(2) < AS°(3)
AS°(2) < AS°(3) < AS°(1) .
AS°(3) < AS°(1) < AS°(2) 
AS°(1) < AS°(3) < AS°(2) dﬁB) < AS°(2) < AS°(1) v—a Hg(g) that AH” = 63.0 k]  mole‘1 and
culate the normal boiling point of Hg. 6. Given for the reaction HgGZ)
AS" = 100. J . K“ . mole", cal A. 6.30 K 6.30 x 103 K ' 3W E. cannot be determined from the information provided 7. Given the following data: Fe203(s) + 3 C0(g) —> 2 Fe(s) + 3‘C01(g) AH° = —27 kJ/mole 3 Fe203(s) + C0(g) —> 2 Fe304(s) + C02(g) AH” '
Fe304(s) + C0(g) —> 3 FeO(s) 1 C02(g) = ~61 kJ/mole
AH° = 38 kJ/mole Calculate the approximate AG” (at 25°C) for the reaction FeO(s) + CO(g) —+ Fe(s) + C02(g)
A. —26 kJ/mole
B. —13 kJ/mole
C. 13 kJ/mole
D. 26 kJ/mole
E. 39 kJ/mole Energyvand Spontaneity r7. Given the following data: Fe203(s) ,+ 37C0(g) —> 2Fe(s)ri+73 C02(g) AH0 = ~217kJ/mole , 3 Fe203(s) + C0(g) —> 2 Fe304(s) + C02(g) AH° = —61 kJ/mole Fe304(s) + C0(g) —> 3 FeO(s) + C02(g) ' ‘ AH° = 38 kJ/mole i Species ' AS" (I  K”  mole") “ Fe203(5) 87.0 3 CO(g) 190.0 1 Fe($) 27.0 i 2‘
F COz(g) 214.0 I l Fe304ts) 146.0 1 Calculate the approximate AG° (at 25°C) for the reaction spew? ~26 kJ/mole
~13 kJ/mole
l3 kJ/mole
26 kJ/mole
39 kJ/mole FeO($) 61.0 : FeO(s) + CO(g) —a Fe(s) + C02(g) ‘ ' l :3; Answer: B To solve this problem, use the Gibbs—Helmholtz equation: «1 3: Step I .' Solve for AH". Realize that you will have to use Hess’s law to determine AH°. Be sure
to multiply through the stepwise equations to achieve the lowest common denominator (6), and
reverse equations where necessary. ! WWCOQ) —>6Fe(s)
Wmoﬁg) agFezQaéS§+CO(g) 61kJ/mole
6FeO(s) +2C02(g)—+2Fe—an€£%+2C0(g)  —76kJ/mole_ AG° = AH" _— TAS" +9C02(g) —81kJ/mole 6Fe0(s) +6co(g) —»6Fe(s) AH°=ﬂklin9l§=—16kJ/moie 6 + 6C02(g) —96 kJ/mole 175 Part II: S ecific To ics 5 Step 2: Solve for AS°. FeO(s) + CO(g) a Fe(s) + C02(g) ’ ASO = ZASoproducts _ Z:A‘Soreactants
= (27.0 + 214.0)  (61.0 + 190.0) = ~10.0 J  K—l  mole'l Step 3: Substitute AS° and AH° into the Gibbs—Helmholtz equation. AG° = AH“ — TAS° .
= ~16 kJ/mole  298 K(—0.0100 kJ  K“1  mole—1)
z —13 kJ/mole ' i; changed from the; standard? atmgo the following :11 . ”@593;me umsam y “M (.3 {Q‘mh‘pviﬂ‘m é Mg) = 2.00 atm ‘4 HFQg) .= 1.00 atm H2(g) = 0.50 atm A. 1090 kJ/mole . ~546 kJ/mole
c. —273 kJ/mole .
D. 546 kJ/mole
E. 1090 kJ/mole : GELEW® Jr {@5335} (mm “a? m {$95) : «gaivaagwﬁ/W
: ’QLW i/Jéfmﬁé 9. If AH0 and AS“ are both negative, then AG° is A. always negative B. always positive C. positive at low temperatures and negative at high temperatures \ negative at low temperatures and positive at high temperatures Answer: D 1O 10. Determine the entropy change that takes place when 500 grams of compound x are
heated from 50°C to 2,957°C. It is found that 290.7 kilojoules of heat are absorbed. A. ~461 J/K
B. 0.00 J/K (C. i} 230 J/K D. 461 J/K
E. 921 J/K Answer: C Use the equation 2:;
T1' where Cp represents the heat capacity AS = 2.3030,; log (q, = AH/AT = 290700 J / 2907 K = 100.0 J/K)’. ' Substituting into the equation yields AS: 2.303Cplog%
=2.303<100.0%>10g 33223;); = 230 J/K 11 Free Response 1 1. Given the equation N204(g) ——> 2 N02(g) and the following data. (a) Calculate AG°. (b) Calculate AH".
Calculate the equilibtium constant K, at 298 K and 1 atm.
Calculate K at 500°C and 1 atm.
Calculate AS" at 298 K and 1 atm. Calculate the temperature at which AG" IS equal to zero at 1 atm, assuming that AH3
and AS” do not change significantly as the temperature increases. 12 Free Response 2 Deﬁne the concept of entropy. From each of the pairs of substances listed, and assuming lmole of each substance,
choose the one that would be expected to have the lower absolute entropy. Explain
your choice in each case. ‘ (1) H2099) or SiC(s)~at the same temperature and pressure (2) 02(g) at 3.0 atm or 02(g) at 1.0 atm, both at the same temperature '
(3) NH3(Q) or CGHGGZ) at the same temperature and pressure (4) Na(s) or Si02(s) ' 13 Part II: Speciﬁc Topics . ,. .,_.______________________.____._..———————————————'~—'—“ 1. Given the equation N204(g) —> 2 N02(g) and the follo'Wing‘ data: _ (a) Calculate AG°. (b) Calculate AH° . (c) Calculate the equilibrium constant K, at 298 K and 1 atm.
(d) CalCulate”K‘at*500"€"andl'atm’. (6) Calculate AS° at 298 K and 1 atm. (f) Calculate the temperature at which AG° is equal to zero at 1 atm assuming that AH°
and AS° do not change signiﬁcantly as the temperature increases. Answer 1. Given: AHf° and AGf° information for the equation
N204(g) —> 2 N02(g) . (a) Restatement: Calculate KG°. AG° =ZAGf° products — EAGI" reactants
= 2(51. 30)— (97 82): 4. 78 kJmole'1 (b) Restatement: Calculate AH° . AH“: ElAHfo products ~2AH}° reactants
= 2(33. 2)— 9. 16:57. 2 k] mole (c) Restatement: Calculate the equilibrium constant Kp at 298 K and 1 atm. 2 K, = 1133:”; (where P represents the partial pressure of a gas in atmospheres) AG° = '—2 303 RTlog K (R: 8. 314 J K")
AG° 4.78161" rnole’l :,_____——————*—"”"‘
10ng —2.303RT —2.303(0.0083141e1 K )(298K) ::.—0 838
KP: 0.145 (at standard temperature of 298 K) 178  Ener and S ontanei
M (d) Restatement: Calculate K at 500°C and 1 atm. AH°(T2—T1)__1 KT:
“2.303RT1T2 ‘ 0g KT.
57, 2001(77314— 298K) Km (2.303)(8.314JI=<§“)(77314;)(298142)zlog K298
6.16 = log~K773 log K773 — log K298 = 6.16 log K298 = — 0.838 from part (c)
log K773 = 6.16 + (—O.838) = 5.32
K = 2.09 ><105 (e) Restatement: Calculate AS" at 298 K and 1 atm.
AG° = AH“ — TAS°
o_AH°AG°_57,200J—4,780J_ . 1
AS “ *fW 176] K (f) Restatement: Calculate the temperature at which AG° is equal to zero at 1 atm, assuming
that AH° and AS° do not change signiﬁcantly as the temperature increases. AG° = AH° — TAS°
0 = 57,200 J — T(176 J  K“) T_ 57,200; ‘176.JrI<;“‘=3'25K 2. (3) Deﬁne the concept of entropy. (b) From each of the pairs of substances listed, and assuming 1 mole of each substance, choose the one that would be expeCted to have the lower absolute entropy. Explain
your choieeinweach case. ‘ (~1)—H79,(S)or—SiEfﬁratthesamﬁemperatureandfpressurew—mﬂ“"'
(2) 02(g) at 3.0 atm or 02(g) at 1.0 atm, both at the same temperature
(3) NH3(Q) br C6H6(Q) at the same temperature and pressure (4) Na(s) or Si02(s) Answer 2. (a) Restatement: Deﬁne entropy. Entropy, which has the symbol S, is a thermodynamic function that is a measure of
the disorder of a system. Entropy, like enthalpy, is a state function. State functions
arethosequantities wh’oserhan’ged values are determined by their initiala’nd'ﬁnal
values. The quantity of entropy of a system depends on the temperature and pressure
of the system. The units of entropy are commonly J  K‘l  mole". If S has a ° (3°) 7 179 ...
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 Spring '17
 jane doe

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