Thermodynamics Practice ANSWERS.pdf - Thermodynamics...

Info icon This preview shows pages 1–18. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
Image of page 11

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 12
Image of page 13

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 14
Image of page 15

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 16
Image of page 17

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 18
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern