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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 l-mole 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 kJ-mole'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)(8.314J-I=<§“)(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 “ *f-W- 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.Jr-I<;“‘=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—SiEfﬁrat-thesamﬁemperatureandfpressurew—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