Self-Test Questions: Elements of Physical Chemistry PDF

Skip to main content A formula found in cheatsheet Return to Elements of Physical Chemistry 7e (NA) student resources Self-test questions: Focus 01...

Skip to main content A formula found in cheatsheet Return to Elements of Physical Chemistry 7e (NA) student resources Self-test questions: Focus 01 12 sor 1. A cylinder of volume 1810 dm3 contains gas at a pressure of 197 atm and temperature of 25°C. Assuming that the gas behaves as a perfect gas, calculate the amount of gas contained in the cylinder. 298 mol pX npT = 1720 mol 19fatm(1010am3) x(0.00ntm/molk) (25+273.1) = x 17547mol = 14600 mol ↓ 1.46 x 106 mol it Silly 39.9u8g/mo) Ar 32.12g/mol ↑ 2. A pressure vessel contains a gaseous mixture made up of 2.34 kg silane, SiH4, and 55.4 kg argon, Ar. Determine the mole fraction of silane. 0.050 mole traction, X Imolecules = 0.041 Foral molecules 0.095 0.4991030293 = 0.064 3. The partial pressure of ethanol in the air above a sample of liquid ethanol is 1400 Pa. Assuming the air to be at normal pressure, calculate the mole fraction of ethanol in the air. 0.0788 ↑Ipanial pressure a pressure 0.0798 P5 xp = 1105Pa ( + 0.0140 100pa x5(1alm X = latm 0.0138 X5 0.0138 = 4. Calculate the root-mean-squared speed of methane, CH4, molecules in a sample of at 25 °C. Vrms * Noles! (3pT7 N 2 6.24 m s-1 = 10.2 m s-1 15 1Pq.m3 = 21.5 m s-1 681 m s-1 = (3(0.3175pa.m3/md(5)(290)3 ,ba 1kgm 52 + 10.049x 0 = 650.715m5' = 5. Calculate the mean speed of oxygen molecules, O2, in air at 25 °C. ↑ 482 m s-1 Umean=(54) Vrs, so I mean M. ms 444 m s-1 Urms (3pT( 1g i = = vi.45)(2av)(* (3(0.31758a.xp, n02.1ms" = 31.20 e https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-01?previousFilter=tag_focus-01 4/18/24, 8:49 AM Page 1 of 3  14.0 m s-1 129 m s-1 90423 atm (bar 10. - 5290.15k, Boltzmann'sconstant, I " 6. Calculate the mean free path of carbon dioxide molecules, CO2, in a sample of gas at standard ambient temperature and pressure. The collision cross section of a carbon dioxide = x(1x10-m) molecule is 0.52 nm2. 0.52nm 5.2x10" me k Boltzmann Mx = constant, J/ 79.1 nm kT = Op x m PamB 1 O = collision cross 791 mm = )(298.15*) k(5* section 4.69 nm 5.2 x 10-9 (0.906923atm 2 -> pa) free path x mean = 7.91 mm = 7.91x10-0m or 79.1nm 7. Calculate the compression factor for carbon dioxide at its critical point given that Tc = 31 °C, pc = 72.9 atm and Vc = 94.0 cm3 mol-1. (103m)x(10am) 3 0.275 z p(/ = anamsx 0.094am3 or0.on e = 1 RT 2.69 = 72.99m (0.004mol) -. AIM (30HK) 26.6 molK 0.2747 = 8. The second virial coefficient of ammonia, NH3, is –40.4 cm3 mol-1 at 600 K. What does this imply about the nature of the interactions between the molecules at this temperature? The interactions between molecules are attractive. The interactions between molecules are negligible. Nothing. The value of the virial coefficient is unrelated to the molecular interactions. The interactions between molecules are repulsive. 3 (iii) as1ram( 0.022M amsL P22.Shasmolx = 9. Use the virial equation to calculate the pressure exerted by exactly 1 mol of hydrogen held in a cylinder of volume 0.250 dm3 at a temperature of 1000 K. The value of the second virial coefficient of hydrogen at this temperature is +22.54 cm3 mol-1. 33.3 kPa *pX/m r (1 Bm vmz 7 2 I atmlbar 1x105Pd T = = + + 333 kPa m 0.022SU p(0.250X) 0.000atm (1000) (1 7 = 33.3 MPa 36.3 MPa mu. π 0.250X p 358a + m 0x 36275215.72Pao 36.3 MPa = -> = 10. Use the van der Waals equation of state to calculate the pressure exerted by exactly1 mol of gaseous ammonia, NH3, held at a temperature of 1000 K in a vessel of volume 2.50 dm3. The values of the van der Waals parameters for ammonia are a = 422.5 kPa dm6 mol-2 and b = 3.71 x 10-2 dm3 mol-1. # a 3.33 kPa p T = - 3.31 MPa Vim Limit it U22.5kPax/orfa-latm = 0.002/atm (1000k) mol - 2.SO2 3.71x10* - m 1.5022 https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-01?previousFilter=tag_focus-01 4/18/24, 8:49 AM Page 2 of 3 32.00752042 atm pa 3310037Pam 3.31kPd = -> =  1970 Pa 224 kPa Submit Quiz Printed from https://learninglink.oup.com/access/content/echem7e-na-student-resources/self-test-questions-focus-01 , all rights reserved. © Oxford University Press, 2024 https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-01?previousFilter=tag_focus-01 4/18/24, 8:49 AM Page 3 of 3 1bar 0.986923 = aN + Skip to main content Return to Elements of Physical Chemistry 7e (NA) student resources 10am3 x (,)m)" 0.01m3 = Self-test questions: Focus 02 ~bar-10.900923 Iam3 (,m)" 1x10m3 x = convertim atmepa 1. Calculate the work done on the system when 1.00 mol of gas held behind a piston expands irreversibly from a volume of 1.00 dm3 to a volume of 10.0 dm3 against an external pressure of 1.00 bar. W PexdV cuLaCONm3 -110 = –5710 J - –900 J =- (arx0.90 +1000 J =-899.999 Pa.msnJ +9120 J Pa.ms pV nRT = 2. 1.00 mol of gas in a cylinder is compressed reversibly by increasing the pressure from 1.00 bar to 10.0 bar at a constant temperature of 500 K. Calculate the work doneC on the gas by the compression. is othermal extensible expansion +9570 J +3740 J npT(n(B) w Bar - = In +37.4 kJ =- 10(0.31455OT) (500) 9572.725, but since work done gas, convertto + 9572.425 - +5710 J = - m 3. The temperature of a copper block of mass 423 g rises by 10.1 °C. Calculate the heat transferred, given that the specific heat capacity of copper is 385 J K–1 kg–1. 385 J a mCIT = 1.63 kJ 4239(3855k-kg 1) (10.1() - = 1.10 kJ 25.6 kJ 1677.8 5 or 1.07k5 / = & 1L 1k9 10009 = = 4. Raising the temperature of 1.00 dm3 of water from a temperature of 25°C to 100°C at constant pressure requires 312 kJ of heat. Calculate the molar heat capacity of water at constant pressure. a mc = 4.16 J K–1 mol–1 3/20005 100090 (75K) = 5.62 J K–1 mol–1 0 4.145g" k x 18.01g10 " = 75 J K–1 mol–1 ImolB0 77.99375K"mot 422 J K–1 mol–1 = https://learninglink.oup.com/access/content/echem7e-na-stud…rces/self-test-questions-focus-02?previousFilter=tag_focus-02 4/18/24, 10:53 AM Page 1 of 5  (91.75 c(2.41(4) ⑨ = 5 pAxV7S 2955k 1 0 - = = 5. A calorimeter was calibrated by passing an electrical current through a heater and measuring the rise in temperature that resulted. When a current of 113 mA from a 24.1 V source was passed through the heater for 254 s, the temperature of the calorimeter rose by 2.61 °C. Determine the heat capacity of the calorimeter. a 0AT = 1.04 J K–1 (113mAX27.1Vx257s) C(2.81 273.15K) + = 4.42 J K–1 - c 20551 692 J K–1 ↓ = 265 J K–1 lomA>,000mA 26. A perfect gas expands reversibly at a constant temperature of 298 K so that its volume doubles. What is the change in the molar internal energy of the gas? +2.27 kJ mol–1 In is othermal extensible expansion, a W, 0 J mol–1 w, since AUm + a = - = +1.72 kJ mol–1 Anm 0 = –2.27 kJ mol–1 7. Which of the following statements is always true for a reaction in which there is no non- expansion work? DU = qp non-expansion work a = 0 = DH = qp DU = 0 a DH = 0 8. Calculate the difference between the molar internal energy and the molar enthalpy for a perfect gas at 298.15 K. 35.859 J mol–1 AU AU An,gRT + = 0 J mol–1 An -AU AngRTERT = 2.4790 kJ mol–1 = 2470.905m2.79k5 8.3145 J mol–1 9. For many substances, the variation with temperature of the molar heat capacity at constant had sa megs pressure of is given by the expression For copper, a = 22.64 J K–1 mol–1, nunu sofer lisod b = 6.28 ´ 10–3 J K–2 mol–1 with the value of c being negligible. Calculate the change in the molar enthalpy of copper when it is heated from 293 to 323 K. +29.7 kJ mol–1 1nm +1.23 kJ mol–1 +737 J mol–1 +1.13 kJ mol–1 f · https://learninglink.oup.com/access/content/echem7e-na-stud…rces/self-test-questions-focus-02?previousFilter=tag_focus-02 4/18/24, 10:53 AM Page 2 of 5  ⑤ 10. The molar heat capacity at constant volume of argon, Ar, is 12.47 J K mol–1. What is the value of the molar heat capacity at constant pressure? 4.15 J K–1 mol–1 Cr-Cv nR = 0(0.3155 N(*) 8.31 J K–1 mol–1 (p-12.775KmOH = mot 15' Cp 20.7555 12.47 J K–1 mol–1 = 20.78 J K–1 mol–1 -11. In an experiment to determine its enthalpy of vaporization, liquid tetrachloromethane, CCl4, was placed in an open boiler that was equipped with a resistive heating coil and brought to the boil at a constant temperature of 350 K and pressure of 1 bar. The passage of a current of 0.933 A from a 24.0 V supply for 30.0 s was found to result in the vaporization of 3.45 g of tetrachloromethane. Calculate the standard enthalpy of vaporization of tetrachloromethane at 350 K. ANS = 671.765 e 0.671 kJ mol–1 3.75gCdux molca 30.0 kJ mol–1 195 J mol–1 ArapU0 299505 mol can = 2.32 kJ mol–1 - 12. For iodine, I2, at 114°C, the standard enthalpy of fusion is 16.1 kJ mol–1 and the standard enthalpy of vaporization is 45.0 kJ mol–1. Calculate the standard enthalpy of sublimation at this temperature. 61.1 kJ mol–1 16.1K5mol" 15.0k5 mrt + 16.1 kJ mol–1 AsubH* 61.1k5mot= 25.0 kJ mol–1 a 28.9 kJ mol–1 calorimeter constant indicates An volume 13. The energy released as heat when liquid propanone, CH3COCH3, is burned in a bomb calorimeter at 298.15 K is 1788 kJ mol–1. Calculate the enthalpy of combustion of propanone. +1790 kJ mol–1 AU = AUAngRT =( 1788x10005mF HMN)(8.3145mol+(298.15) + =-1790478.9485 moltor 1790k5molt +1786 kJ mol–1 Balanced ea'n: - 3LOR –1790 kJ mol–1 CUCOW,;702, 3COct –1786 kJ mol–1 An=products reactants - ⑨ = 3 - 1 = 1 NOTproducts reactants 14. Estimate the&add everything, standard enthalpy change for the processF2(g) + 2 e–(g) ® 2 F–(g) The F–F O 328 kJ mol–1. Fc(g) 2e (g) rF + (g) = o bond enthalpy is +155 kJ mol–1 and the electron gain enthalpy of elemental fluorine, F, is – separate into two processes, then add to getANA –18 kJ mol–1 c dissociation=+155k5mot 118=-328(2) 155 + 05k5 mol" =-320k8mi x2 e-gain enthalpyofF =- –173 kJ mol–1 501k5mol =- - +155kJmol" –501 kJ mol–1 + 501k5 molt https://learninglink.oup.com/access/content/echem7e-na-stud…rces/self-test-questions-focus-02?previousFilter=tag_focus-02 4/18/24, 10:53 AM Page 3 of 5  3 1 –811 kJ mol–1 7 0 1 n 1 xc = use bonds broken- - mean bond enthalpliesgiven ⑤ X bonds formed formula n -u 15. Estimate the standard enthalpy change for the reactionN2(g) + 3 H2(g) ® 2 NH3(g) given the following mean bond enthalpies: H–H: +436 kJ mol–1, N–N: 945 kJ mol–1 and H– N: 388 kJ mol–1. Anobands broken-bands formed +933 kJ mol–1 = (au5 3(734) + - (44308) –75 kJ mol–1 = - 75k5mol" –2690 kJ mol–1 +605 kJ mol–1 enthalpyof atomization: Saddeenything 16. Use the following data to calculate the mean B–Cl bond enthalpy in boron trichloride, BCl3.DHʅ(298 K) Enthalpy of atomization of boronB(s) ® B(g) +590 kJ mol–1Enthalpy of energy change accompa atomization of chlorineCl (g) ® 2 Cl(g) +242 kJ mol–1Enthalpy of formation of boron 2 uying total separation of trichlorideB(s) + 3/2 Cl2(g) ® BCl3(g) –418 kJ mol–1 o all loss in a reaction mean bond enthalpy ofB enthal ofatomization of BAS +975 kJ mol–1 by DO NOT change in +1371 kJ mol–1 so, Aat BAs productsreactants?add atakian - = backwards reachios Bd (g) 10 +418 since needed is bond enthalpy +325 kJ mol–1 breaking down reaction, BCS G (q) - - = or energy when broken +457 kJ mol–1 B(g) B(s): 590 - + 2a - 42 : - M2 - (3/2xM2) 30,410 + 590 1371 +457k5mot + = = 17. Calculate the standard enthalpy of combustion of phenol, C6H5OH, at 298.15 K given that, at this temperature, the standard enthalpy of formation of phenol is –165.0 kJ mol–1, of liquid sa solution, and water, H2O is –285.8 kJ mol–1 and gaseous carbon dioxide, CO2, is –393.51 kJ mol–1. so Giken.Atho, so formula productsreactants balanced is 300... –2202.9 kJ mol–1 Balanced all CallOU 70L = 600L 31cO + –514.3 kJ mol–1 so, AcHO, phenol [6(393.51) 3(205.8)?- IC5105)] = + –844.3 kJ mol–1 =3383.44k5mOH - –1872.9 kJ mol–1 not the above in website:137393.5) 5....] 2202.93k5mol - = - 18. Estimate the standard enthalpy of formation of liquid benzene, C6H6, at 298.15 K. At this temperature, the standard enthalpy of atomization of carbon C(s, graphite) ® C(g) is +717 kJ mol–1 and the standard enthalpy of vaporization of benzene is 34 kJ mol–1, whilst the standard bond enthalpy of hydrogen, H2, is 436 kJ mol–1. The mean bond enthalpy of a C6H5–H bond is 469 kJ mol–1 and of an aromatic C–C bond is 452 kJ mol–1. Givenmean band enthalpies, so formula is bondsbroken bonds formed +198 kJ mol–1 Balanced equation +1102 kJ mol–1 = Callu (9) = CoMuC1 3/2 G((g) 3 U2(a) + +2310 kJ mol–1 GCcs,graphise) = + +50 kJ mol–1 30,6(77) + 3(734) 6(909) 6(452) 34 50k5mor - - = - formation If enthalpyis gixe: use A =products reactants BUT! 19. Itbond enthalpyis given, Use (no bonds broken- https://learninglink.oup.com/access/content/echem7e-na-stud…rces/self-test-questions-focus-02?previousFilter=tag_focus-02 4/18/24, 10:53 AM bonds formed Page 4 of 5 Gluca) RCG) Cello ACHO ofCMoto A += +1560k5mol' since it is formed 19. Calculate the standard enthalpy change for the hydrogenation Back to top  reactionC2H4(g) + H2(g) ® C2H6(g) at 298.15 K. At this temperature, the standard enthalpy of combustion of ethene, C2H4, is –1409 kJ mol–1, of hydrogen, H2, is –286 kJ mol–1 and of ethane, C2H6 is –1560 kJ mol–1. –151 kJ mol–1 AHO = GVAcH0-GVAcH0 –437 kJ mol–1 reactants products –3255 kJ mol–1 (1560k5mol) [(-1)( 1409) ( 1 )( 20u)] - + - - - = - –135 kJ mol–1 = - 135k5mol 20. The standard enthalpy of combustion of propane, C3H8, is –2.220 ´ 103 kJ mol–1 at 400 K. Use the data below, and Kirchhoff's law, to calculate the standard enthalpy of combustion at 600 K. Cʅp,m / J K–1 mol–1C3H8(g) 73.6O2(g) 29.4H2O(g) 33.6CO2(g) 37.1 –2205 kJ mol–1 3C02 1/30 (340 502 = + + –445 kJ mol–1 Kirchoff's law:AfU*(T2) A + U0(T1) - ArCp*CTz-T1) (2001) = –2195 kJ mol–1 2.200x1535mol"x10005 25.15mol ** + = - –2129 kJ mol–1 #5 1 2199.90k5mol - Cp,M =- 21979805mot or - C348:73.0 02:29.4 Submit Quiz 10:33.8 CO2:37.1 Printed from https://learninglink.oup.com/access/content/echem7e-na-student-resources/self-test-questions-focus-02 , all rights reserved. © Oxford University Press, 2024 Comleactants ArCp Com products) - = [(3x37.1) (1733.0)] [73.4 (5429.4)7 = + + + = 25.15mol"K- https://learninglink.oup.com/access/content/echem7e-na-stud…rces/self-test-questions-focus-02?previousFilter=tag_focus-02 4/18/24, 10:53 AM Page 5 of 5  Skip to main content Return to Elements of Physical Chemistry 7e (NA) student resources Self-test questions: Focus 03 a ogas when it is compressed 1. Calculate the change in the molar entropy of a perfect isothermally from a pressure of 1 bar to 10 bar. As accompanying ① +83.1 J K–1 mol–1 ASm nRln(Pt/pi) is othermal expansion = +7.6 J K–1 mol–1 = 0.31455mol"1" (10bar). +19.1 J K –1 mol–1 = 19.17 TmdH15 +183 kJ K–1 mol–1 - I ALWAYS CONVERT TO K 2. Calculate the change in the standard molar entropy of hydrogen chloride, HCl, gas when it is heated from a temperature of 20°C to 30°C. The standard molar constant-pressure heat capacity of hydrogen chloride is 29.1 J K–1 mol–1. AS WITH +11.8 J K–1 mol–1 As a temperature: TEMPERAMPEO +0.976 J K–1 mol–1 * A Sm(Tf) ASmCTi) ((n (Tf/Ti) = + +0.424 J K–1 mol–1 ASmCTA)-ASmCTi) CpIn (TfITi) 303.15K (293.15k) = +29.3 J K–1 mol–1 = 29.15m0 15 = ⑤ 3. The standard enthalpy of sublimation of carbon dioxide is 26.2 kJ K–1 mol–1 at 298.15 K. Calculate the standard entropy of sublimation at this temperature. PUASE O 87.9 J K –1 mol–1 TRANSIONOF –1 AtrsS Ats/ = 78.1 J K mol–1 #S 26.2 J K–1 mol–1 T 1048 J K–1 mol–1 87.95K"mol ⑤ = 4. The boiling temperature of ethylbenzene is 136 °C. Use Trouton's rule to estimate the enthalpy of vaporization of ethylbenzene at this temperature. ·TROUTON'SRULE C 35 kJ mol–1 Trulo's rulemostliquids have O5JKmo Araps · ASUFPHASE 12 kJ mol–1 TRANSIONS 23 kJ mol–1< SO.A pSt A pl + = + 4.8 kJ mol–1 TtA 1ty 1 37.78k5mot' = 5. https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-03?previousFilter=tag_focus-03 4/18/24, 5:07 PM Page 1 of 3  ⑤5. Calculate the change in entropy when exactly 1 mol of solid iodine, I2, at a temperature of 360 K is heated at constant pressure to produce liquid iodine at a temperature of 410 K. The constant pressure molar heat capacity of solid iodine is 54.44 J K–1 mol–1 and of liquid iodine o is 80.67 J K–1 mol–1. The melting temperature of iodine is 387 K, and the molar enthalpy of fusion of iodine is 7.87 kJ mol–1. +8.6 J K–1 mol–1 ASCT2) ASCTD Cp In = (If (i) (from 360 1307) (from 3870410) o +28.9 J K–1 mol–1 AtrS ArU melling = –1 –1 +20.3 J K mol +11.7 J K–1 mol–1 1x 5 3.937 20.3359 4.057 28.93 = + + = - + 6. At low temperatures, the heat capacity of solid chlorine, Cl2, follows the Debye T3 law, Cp,m = a T3, with a = 1.24 ´ 10–3 J K–4 mol–1. Estimate the molar entropy of solid chlorine at Smatlow aI,3 10 K. a 2 0.1m + (nm = = = +1.24 J K mol–1 temperatures o +0.413 J K –1 mol–1 OK) 5CamTCasTapproaches Sm to +6.36 J K–1 mol–1 = +4.13 J K mol–1 0.413 5mol 15 " = 7. Which of the following conditions is necessary for a reaction to be spontaneous? DSsur > 0 Itotu Dsys sur = >O DSsys > 0 DSsur + DSsys < 0 I should always be positive in Spontaneous reactious I 8.possible A hypothetical system consists of 5 molecules and 2 quanta. What is the number of arrangements? 21 3 o5 2 -9. For a spontaneous change in a system at constant temperature and pressure, which of the following statements is true? oDG < 0 For spontaneous changeatconstantT and p. -IC DG = 0 DG > 0 There is no restriction on the value of DG 10. https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-03?previousFilter=tag_focus-03 4/18/24, 5:07 PM Page 2 of 3  - 10. Which of the following statements is always true for a system at equilibrium under conditions of constant temperature and pressure? (aNadaman dt) DS > 0 ↑max nn expansion work, = DG = 0 at constantandp o dG = dw¢max DG > 0 Submit Quiz Printed from https://learninglink.oup.com/access/content/echem7e-na-student-resources/self-test-questions-focus-03 , all rights reserved. © Oxford University Press, 2024 https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-03?previousFilter=tag_focus-03 4/18/24, 5:07 PM Page 3 of 3  Skip to main content Return to Elements of Physical Chemistry 7e (NA) student resources densin M = Self-test questions: Focus 04 V ↑ /m 200.0gmol" 17.75an" molt = = 13.0gan-3 1.475x105m3 molt G = VARIANON OF 1. Mercury has a molar mass of 200.6 g mol–1 and a density of 13.6 g liquid. cm–3 when WITH PRESSURE Calculate the change in the molar Gibbs energy of liquid mercury when the pressure increases by 10.0 bar. Gm(p+) Gm(pi) (Pf pi)Um atm pa aqaaa apa = 10bar x 0.900923 - + = 14.8 ´ 10–3 J mol–1 AGm (Pf Pi)Vm Ibar - = (1.975x105ms most 0 14.8 J mol –1 = (qaa, aaapa) 14.8 kJ mol–1 = 17.75 pam3 mol 17.755mil = 14.8 MJ mol–1 15 1Pa.m3 * = ⑥ Smo 2. The standard molar entropy of but-1-ene, C H , is 307 J K –1 mol–1. What is the change in 4 8 the molar Gibbs energy of but-1-ene when it is heated from 293 K to 303 K? AT THERMODYNANIC 1 A T G Vdb SdT +3.07 kJ mol–1 - = SQUARE DENION 4 SAT –3.07 J K–1 mol–1 Y = - - - GIBBS OF n b 3.07k5 mot' o–3.07 kJ mol–1 = - +3.07 kJ K–1 mol–1 MOLAR GIBBS - 3. Calculate the change in the molar Gibbs energy of a perfect gas when it is compressed isothermally at a temperature of 298 K from a pressure of 1 MPa to 5 MPa./p> (PERFECTGAS, 335 J mol–1 # Gm p Tm(p+ = (pi) CONSTANT, 1.73 kJ mol–1 P = 8.31755 motk- VARYINGPS 3.99 kJ mol–1 o 28.6 kJ mol–1 4. The vapour pressure p of liquid sulfur trioxide, SO3, may be calculated for different temperatures T using the expression log(a) A - where, for the temperature range 24 to = 48 °C, A = 9.147 and B = 1771 K. What is the vapour pressure of liquid sulfur trioxide at 32 °C? log(pa) A = - B 28.2 kPa 3.34 kPa 0.524 kPa log (a) 9.17-177 0 273.15 = + 32K 2.20 MPa a 2207.432 = e =2204.432997k8a x1000Pa z20UUs2 = Da 5. https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-04?previousFilter=tag_focus-04 4/18/24, 8:43 PM Page 1 of 5  5. The vapour pressure of toluene is 6.811 kPa at 310 K and 24.15 kPa at 340 K. Assuming that the variation of the vapour pressure p with temperature T may be described by the expression What are the values of A and B? 1 A = 7.059, B = 1930 K A = 5.307, B = 1930 K A = 6.971, B = 7780 K A = 5.307, B = 7780 K & - 6. The vapour pressure of liquid carbon disulfide, CS2, is 23.5 kPa at 280 K and 51.3 kPa at 300 K. Calculate the enthalpy of vaporization of carbon disulfide at 290 K. A p(e x,X Arapl(# I) CLAUSMS - - 17.9 kJ mol–1 p = = CAPERON o 27.3 kJ mol–1 In(P2) 1n(P1) Arapl(#1 F) = + 97.2 kJ mol–1 - (Giken 2p, It, and 94.1 kJ mol–1 me in between AvapH 27.3KJmot = 7. The standard enthalpy of vaporization of water, H2O, DvapHʅ = 40.7 kJ mol–1 at 373 K. Assuming this value to remain constant at temperatures close to 373 K, use the Clausius– CLAUSIS Clapeyron equation to estimate the vapour pressure of liquid water at 80 °C. CAPEYRON 95.8 kPa in(P2) in(pi) Arapl(#1 F2) 4 + = notgiven, always assume If - 48.2 kPa 101325Da SolnUsed STP (Iatm 101325pa) A or 1 a SATP (Ibar and 290.15K) = In(P2) (n(1x105pa) 97.9 kPa see 40700Emo as PI, butSATPCIbar HISDAS = + is more updated 101 kPa ex.pl becomes (bar or D2 = 77592.759M Paar47.MPa (U8.2kpa at STP) 1x1050d ↓ 8. The triple point of acetylene, C2H2, lies at a temperature of 192.4 K and pressure of 128 kPa. Assuming that the enthalpy of vaporization of acetylene, DvapH = 31.3 kJ mol–1, is CLAUSMS invariant with temperature, calculate the normal boiling temperature. CAPERON In(P2) (n(pi) 0 195 K Avabu (t, F) = + - cassuming PIbar 1x10TPA C 5.14 K 72.6 K In (120000fa) in = (1x105a) 313005mol" (ie.uk ) + - 205 K Tz 194.054 = - 9. On a pressure–temperature phase diagram, the conditions under which a one-component system exists as two phases in equilibrium corresponds to: a point o a line & an area the entire diagram 10. 3 –1 https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-04?previousFilter=tag_focus-04 4/18/24, 8:43 PM Page 2 of 5  10. The partial molar volumes of water, H2O, and methanol, CH3OH, are 17.8 cm3 mol–1 and 38.4 cm3 mol–1 respectively at 25.0 °C for dilute mixtures of methanol in water. Calculate the total volume when 15 cm3 of methanol is added to 250 cm3 of water at this temperature. The 1 density of methanol is 0.791 g cm–3. 265 cm3 269 cm3 261 cm3 250 cm3 & 11. Calculate the change in the chemical potential of a perfect gas when it expands isothermally at a temperature of 20.0 °C so that its volume doubles. I MOF PERFECTGAS RTVYVz Am RTM*p AN () -> –1.69 kJ mol–1 = = –115 J mol–1 since * PandVare = (293.15) (n (0.31755mol) +1.69 kJ mol–1 inversaxproportional =- 1089.773945molar -1.009k5mol- +115 J mol–1 ↑/ 12. Which of the following statements is always true for a liquid mixture of two components A and B in equilibrium with a mixture of their vapours? mA(l) = mB(l) and mA(g) = mB(g) ~ Atequilibrium, I ofcomponentmustbe the mA(l) = mA(g) = mB(l) = mB(g) same in all phases om (l) = m (g) and m (l) = m (g) A A B B mA(l) ≠ mA(g) ≠ mB(l) ≠ mB(g) - &tral moves 13. Calculate the entropy change when 2.0 mol of a perfect gas A and 3.0 mol of a perfect gas B mix spontaneously. ENTROPY OF +195 J K–1 Amixs =-nR (XAIXA XBlnXB t...) + MIXING ((3/5) (n(3/5))] o +28 J K–1 5(0.31455mTK) [(E)In (2/5)) + =- (AmixS) –12 J K–1 = 27.97954-1 –8.34 kJ K–1 ⑤14. A solution is prepared by dissolving 5.32 g of benzene, C6H6, in 93.2 g of toluene, C6H5CH3, at 25 °C. The vapour pressures of pure benzene and pure toluene are 12.7 kPa and RAOUU'S LAN 3.8 kPa at this temperature. Use Raoult's law to determine the partial vapour pressure of each of the two components for the mixture. B = benzene, Ttomene = pB = 0.68 kPa, pT = 1.01 kPa pF xP* PB xp0p* + + = pB = 0.06 kPa, pT = 0.94 kPa = pB = 77.9 kPa, pT = 23.3 kPa =(0000mll =(10ml)S.OP 10795mol 1.0795mol (12.7kPa o pB = 0.80 kPa, pT = 3.56 kPa = 3.56 kPd 0.8kPa = lmalbenzene 0.000 molbenzene 5.32gbenzene x = 15. 78.1lgbenzene 93.29 toluene xmoomene to mol omene 92.14ghmene https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-04?previousFilter=tag_focus-04 4/18/24, 8:43 PM = 1,0795mol real Page 3 of 5  - 15. The vapour pressure of bromobenzene, C6H5Br, above an ideal–dilute aqueous solution of molality 0.092 mol dm–3 is 23.6 kPa. Calculate the Henry's law constant of bromobenzene. HENRY'S LAW O3.90 mol m –3 kPa–1 (B KPB = Iam)3.090 molKDat m 461 mol m–3 kPa–1 0.092madm -3 (k) (23.0kpa) 1x 3.890x10 -3molKpata = = 2.17 mol m–3 kPa–1 = 2.56 mol m–3 kPa–1 -16. Which of the following statements is true for an ideal–dilute solution? The solute and solvent both obey Raoult's law. SOIH-B The solute obeys Raoult's law and the solvent obeys Henry's law. SoNR-A oThe solute obeys Henry's law and the solvent obeys Raoult's law. The solute and solvent both obey Henry's law. - Back to top  1mo/ns 8 = 17. Calculate the minimum partial pressure of nitrogen that is necessary to achieve an aqueous ideal–dilute solution of concentration 1.00 mmol dm–3. The Henry's law constant for nitrogen in water is 6.48 × 10-3 mol m–3 kPa-1. 154 MPa (B kpB = :commolanx(ram)"a (unox 10- olm concentrations. ku(pT)An,g kp 3.29x10" kcI(0.31455mrl" 15') (290K)]loam = e 8.15 ´ 103 = 8.15 ´ 104 3.29 x10"x1x1058a KcT(0.3158am"mol 15') (2a0k)] = 8.15 ´ 106 Ibar 8.15 ´ 108 https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-05?previousFilter=tag_focus-05 4/18/24, 8:45 PM Page 2 of 7  ***since formula is mhulebook no and cheat sheet HOFF VANT 10. The equilibrium constant for the reaction2HCl(g) + O (g) ∏ H O(g) + Cl (g) is 13300 at 2 2 2 400 K and 429 at 500 K. Use the van't Hoff equation to determine the standard enthalpy of EQUATON mItit is inCkz/KY reaction. +57.1 kJ mol–1 = - An –28.6 kJ mol–1 +28.6 kJ mol–1 mn(tz- t) 0.3145mi", –57.1 kJ mol–1 x = - 571055 mol" or - 57.1k5mot ⑤ 11. Calculate the concentration of hydronium, H3O+, ions in a solution of with pH = 4.3. 20 ´ 103 mol dm–3 - pH 10 5.0 ´ 10–5 mol dm–3 14 ´ 10–3 mol dm–3 74 mol dm–3 - 12. The standard Gibbs energy of reaction, DrGo, for the dissociation of phenol is 56.4 kJ mol– 1 at 298 K. Calculate the pK of phenol. a 1 1.3x10 9.88 pla = -10g4a 22.8 where from ArG RTInK = 4.12 |a 1.3x10 = 1 5.24 - So,pka = - 10g(.3410 - 9.88 = 13. The pKa of hypochlorous acid, HClO, is 7.53. What is the value of pKb? 6.47 p(a pkb + 1 = 7.53 5.37 ´ 10–4 2.02 - 14. Calculate the concentration of OH– ions in a solution with pH = 12.1. 10-pon 79.4 mol dm–3 p0U 1.9, COH = = 0.0126 7.94 ´ 10–13 mol dm–3 = 1.90 mol dm–3 0.0126 mol dm–3 ⑤ 15. Calculate the pH of an aqueous solution of hydrocyanic acid of concentration 0.088 mol dm–3. The acidity constant of hydrocyanic acid is 4.9 ´ 10–10. 6.57 - 10 [X1[X] P(a 1n 1[aN 11.ax10 + = = 0.080M taN] [UE G.563502x10-6, pU=-logTON] 5.18 = https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-05?previousFilter=tag_focus-05 4/18/24, 8:45 PM Page 3 of 7  8.82 5.18 7.43 - 16. Calculate the concentration of SO32– ions in an aqueous solution of sulfurous acid, H2SO3. The second acidity constant of sulfurous acid is Ka2 = 1.2 ´ 10–7. ((az [u 1[S022 - + 1.8 ´ 10–7 mol dm–3 = 1.2 ´ 10–7 mol dm–3 INSOz] 1.2x10 7 [x][X] - 6.3 ´ 10–13 mol dm–3 = 7.7 ´ 10–7 mol dm–3 [X] ⑤ x 1.2x10 -7M = 17. Calculate the pH of an aqueous solution of potassium hydrogensulfite, KHSO3. For sulfurous acid, pKa1 = 1.81 and pKa2 = 6.91. 3.82 # 5.10 8.72 4.36 18. In a titration, 2.7 cm3 of 0.100 mol dm–3 sodium hydroxide, NaOH, solution is added to 25.0 cm3 of 0.125 mol dm–3 benzoic acid, C6H5COOH, solution. Calculate the pH of the resulting solution given that the pKa of benzoic acid is 4.19. 3.13 5.25 1.74 6.67 19. Use the data in Table 8.1 to estimate the pH of a buffer solution formed from equal amounts of potassium carbonate, K2CO3, and potassium hydrogen carbonate, KHCO3. 7.6 10.3 6.4 3.9 - 20. Calculate the concentration of chloride ions, Cl–, in an aqueous solution formed by adding lead chloride, PbCl2, to water. The solubility constant of lead chloride is 1.6 ´ 10–5. 0.025 mol dm–3 1.0x15-5 product[P12+][92 = 0.004 mol dm–3 reactant [P34125a) 0.032 mol dm–3 4+3 = 0.1587,50xx2 = 0.31748 50,X = https://learninglink.oup.com/access/content/echem7e-na-stude…rces/self-test-questions-focus-05?previousFilter=tag_focus-05 4/18/24, 8:45 PM Page 4 of 7  0.050 mol dm–3 ⑤ 2F23 30832 + 21. Calculate the ionic strength of a solution of iron (III) carbonate, Fe2(CO3)3 of - concentration 0.020 mol dm–3. 2 [[A]z...] + 0.25 ( = 0.30 –0.10 = II(0.020x2)(3)2 (0.020x3)( 2)"] + - 0.38 = 0.00 daw I (5x7z...) n 0.12 = (from poster 0.03 wave 22. Use the Debye–Hückel limiting law to calculate the mean activity coefficient for the ions 1 in an aqueous solution of potassium sulfate, K2SO4, of molality 0.010 mol kg–1 at 25°C. 10g(2) Az z-r ( (t(0.010)(1) x2] (0.010)( 2) ) = - + = + - 0.943 = 0.51(1)( 2)00s - - 1.074 8 1.58 = 1.023 S 0.932 23. The pKa of hydrocyanic acid, HCN, is 9.218 at 293 K. Calculate the conductivity of an aqueous solution of perchloric acid, of concentration 0.015 mol dm–3, given that the ionic conductivity of H3O+ ions is 35 mS m2 mol–1 and of CN– ions is 82 mS m2 mol–1. 12 mS m2 mol–1 24 ´ 10–2 mS m2 mol–1 8.1 ´ 10–2 mS m2 mol–1 15 ´ 10–2 mS m2 mol–1 ~24. For a galvanic cell, which of the following statements is never true? The potential of the anode is higher than that of the cathode. The electrons flow in the external circuit from the anode to the cathode. Reduction takes place at the cathode. Oxidation takes place at the anode. +G ⑨ O a St 25. The standard reaction Gibbs energy for the electrochemical reaction Cr2O72– Crc0f + 2 Fe + 14 H+ ® 2 Cr3+ + 2 Fe3+ 7 H2Ois –793 kJ mol–1. Calculate the standard cell - potential. ArG" nEaF = - +1.37 V -