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Thermo-Chemistry Curve Balls

Thermo-Chemistry Curve Balls. KNOW thine enemy!!!!!!!!!!!!. Equations. q = m C ΔT q = m Hx q = ΔH o ΔH o = products – reactants ΔH o = broken – formed (not on equation sheet) ΔS o = products – reactants ΔG o = products – reactants ΔG o = ΔH o - T ΔS o ΔG o =-RTlnK.

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Thermo-Chemistry Curve Balls

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  1. Thermo-Chemistry Curve Balls • KNOW thine enemy!!!!!!!!!!!!

  2. Equations • q = m C ΔT • q = m Hx • q = ΔHo • ΔHo = products – reactants • ΔHo = broken – formed (not on equation sheet) • ΔSo = products – reactants • ΔGo = products – reactants • ΔGo = ΔHo - T ΔSo • ΔGo =-RTlnK

  3. CH3OH + O2 CO2 + H2O (l)1. What is the enthalpy of combustion, ΔHoc, for the complete combustion of methanol to produce carbon dioxide and liquid water?

  4. CH3OH + XO2 CO2 + YH2O (l) • 1. What is the enthalpy of combustion, ΔHoc, for the complete combustion of methanol to produce carbon dioxide and liquid water? • Mistake made: unbalance equation.Any test any year (basic chemistry mistake) • ΔHco = products – reactants • ΔHco = [-393.5 + -285.8] – [-210.0] • ΔHco = -469.3 kJ/mole of methanol

  5. CH3OH + 1.5O2 CO2 + 2H2O (l) • 1. What is the enthalpy of combustion, ΔHoc, for the complete combustion of methanol to produce carbon dioxide and liquid water? • Mistake made: water gas instead of liquidAny test any year (basic chemistry mistake) • ΔHco = products – reactants • ΔHco = [-393.5 + -241.5] – [-210.0] unbalanced • ΔHco = -425 kJ/mole of methanol • ΔHco = [-393.5 + 2(-241.5)] – [-210.0] balanced • ΔHco = -666.5 kJ/mole of methanol

  6. 21/03. Calculate the amount of energy mol-1) • released when 0.100 mol of • diborane, B2H6, reacts with • oxygen to produce solid B2O3 • and steam. • ∆Hf° kJ • B2H6(g) 35 • B2O3(s)-1272 • H2O(l)-285 • H2O(g)-241 *(A) 2030 kJ (B) 2160 kJ (C) 3300kJ (D) 3430 kJ B2H6 + 3O2 B2O3 + 3H2O (g) ∆H = [-1272 + 3(-241)] – 35

  7. CH3OH + 1.5O2 CO2 + 2H2O (l) • 1. What is the enthalpy of combustion, ΔHoc, for the complete combustion of methanol to produce carbon dioxide and liquid water? • Mistake made: reactants – productsAny test any year (basic chemistry mistake) • ΔHco = reactants - products • ΔHco = [-210.0] – [-393.5 + -285.8] unbalanced ΔHco = 469.3 kJ/mole of methanol • ΔHco = [-210.0] – [-393.5 + 2(-285.8)] balanced ΔHco = 755.1 kJ/mole of methanol

  8. CH3OH + 1.5O2 CO2 + 2H2O (l) • 1. What is the enthalpy of combustion, ΔHoc, for the complete combustion of methanol to produce carbon dioxide and liquid water? • Correct answer: • ΔHco = products – reactants • ΔHco = [-393.5 + 2(-285.8)] – [-210.0] • ΔHco = -755.1 kJ/mole of methanol

  9. Correct way • This type of enthalpy problem (just a products – reactants ) is more often found on the multiple choice test and usually involves a little easier number to add and subtract. • In a multiple choice problem a student should round – DO NOT get bogged down in mathematics! • -393.5 rounds to -400 • -285.8 rounds to -300 • -210 rounds to -200 • ΔHco = [-400 + 2(-300)] – [-200] • ΔHco = -1000 - -200 • ΔHco = around -800 the correct answer is -755.1 • The answer should be less than ( more positive) -800 because the rounding up of the product’s enthalpies

  10. CH3OH + 1.5O2 CO2 + 2H2O (l) 2. What is the heat of formation, ΔHof , of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 755.1 kJ of heat? OR the equation could have the heat included CH3OH + 1.5O2 CO2 + 2H2O (l) + 755.1 kJ

  11. CH3OH + 1.5O2 CO2 + 2H2O (l) 2. What is the heat of formation, ΔHof , of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 755.1 kJ of heat? OR the equation could have the heat included CH3OH + 1.5O2 CO2 + 2H2O (l) + 755.1 kJ • Mistake made: ignore the sign of the enthalpyQuestion #3b 1998 +755.1 = [-393.5 + 2(-285.8)] – [ΔHof] +755.1 = -965.1 -x • ΔHof = -1720.2 kJ/mole

  12. CH3OH + 1.5O2 CO2 + 2H2O (l) 2. What is the heat of formation, ΔHof , of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 755.1 kJ of heat? OR the equation could have the heat included CH3OH + 1.5O2 CO2 + 2H2O (l) + 755.1 kJ • Mistake made: incorrect substitution ( no true understanding between formation and combustion) and sign problem.Question #3b 1998 “ΔHof “ = [-393.5 + 2(-285.8)] – [755.1] confused between combustion and formation ΔHof = -1720.2 kJ/mole

  13. 2CH3OH + 3O22CO2 + 4H2O (l) 2. What is the heat of formation, ΔHof , of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 755.1 kJ of heat? OR the equation could have the heat included 2CH3OH + 3O2 2CO2 + 4H2O (l) + 755.1 kJ • Mistake made: over balancing the equation leading to a stoich mistake.Any test any year • -755.1 = [2(-393.5) + 4(-285.8)] – [2x] doubled everything BUT the molar enthalpy of combustion • ΔHof= -587.6

  14. 2CH3OH + 3O22CO2 + 4H2O (l) 2. What is the heat of formation, ΔHof , of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 755.1 kJ of heat? OR the equation could have the heat included 2CH3OH + 3O2 2CO2 + 4H2O (l) + 755.1 kJ • Mistake made: over balancing the equation leading to a stoich mistake.Any test any year 2(-755.1) = [2(-393.5) + 4(-285.8)] – [x] doubled everything BUT the molar enthalpy of formation • ΔHof=-420

  15. 2CH3OH + 3O22CO2 + 4H2O (l) 2. What is the heat of formation, ΔHof , of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 755.1 kJ of heat? OR the equation could have the heat included 2CH3OH + 3O2 2CO2 + 4H2O (l) + 755.1 kJ • Mistake made: incorrect substitution • “ΔHof “ = [2(-393.5) + 4(-285.8)] – [2(-755.1)] confused between combustion and formation • “ΔHof “ = -420 kJ/mole

  16. CH3OH + 1.5O2 CO2 + 2H2O (l) 2. What is the heat of formation, ΔHof , of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 755.1 kJ of heat? OR the equation could have the heat included CH3OH + 1.5O2 CO2 + 2H2O (l) + 755.1 kJ • Correct answer: -755.1 = [-393.5 + 2(-285.8)] – [x] • ΔHof = -210.0 kJ/mole

  17. CH3OH + 1.5O2 CO2 + 2H2O (l) 3. What is the heat of combustion , ΔHoc , of methanol if 24.75 grams of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 584.1 kJ of heat?

  18. CH3OH + 1.5O2 CO2 + 2H2O (l) 3. What is the heat of combustion , ΔHoc , of methanol if 24.75 grams of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 584.1 kJ of heat? • A common student mistake is to underestimate the test writer and believe the answer is as easy as picking it out of the question . (-584.1kJ)

  19. CH3OH + 1.5O2 CO2 + 2H2O (l) 3. What is the heat of combustion , ΔHoc , of methanol if 24.75 grams of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 584.1 kJ of heat? Mistake made: incorrect molar substitutionQuestion #3a 1998, 1979 ΔHoc = [-393.5 + 2(-285.8)] – [-584.1] ΔHoc = -381.0kJ/mol

  20. CH3OH + 1.5O2 CO2 + 2H2O (l) 3. What is the heat of combustion , ΔHoc , of methanol if 24.75 grams of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 584.1 kJ of heat? • Mistake made: incorrect molar substitution • -584.1 = [-393.5 + 2(-285.8)] – [ΔHof] • ΔHof = -381kJ/mol • OR use the wrong enthalpy with the wrong sign: 584.1 = [-393.5 + 2(-285.8)] – [ΔHof] • ΔHof = -1549.2kJ/mol

  21. CH3OH + 1.5O2 CO2 + 2H2O (l) 3. What is the heat of combustion , ΔHoc , of methanol if 24.75 grams of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 584.1 kJ of heat? Correct way 24.75g = 32 g -584.1kJ x kJ x = [-393.5 + 2(-285.8)] – [ΔHof] -755.1 = [-393.5 + 2(-285.8)] – [ΔHof] ΔHof = -210.0 kJ/mole

  22. Chemistry Olympiad Question • 25/00. What is the standard enthalpy of formation of MgO(s) if 300.9 kJ is evolved when 20.15 g of MgO(s) is formed by the combustion of magnesium under standard conditions? *(A) –601.8 kJ·mol–1 (B) –300.9 kJ·mol–1 (C) +300.9 kJ·mol–1 (D) +601.8 kJ·mol–1 20.15/300.9 = 40/X kJ

  23. CH3OH + 1.5O2 CO2 + 2H2O (l) • If 8.95 grams of methanol was used to heat 1500 mL of water from 25oC to 58.7 oC What is the molar enthalpy of combustion of methanol?

  24. CH3OH + 1.5O2 CO2 + 2H2O (l) 4. If 8.95 grams of methanol was used to heat 1500 mL of water from 25oC to 58.7 oC What is the molar enthalpy of combustion of methanol? • Mistake made: q/grams ≠ ΔH/moleQuestion #3a 1998, #d 1995, 1979,1989b, 2001a • q = 1500g x 4.18 J/ g x oC (58.7 oC – 25 oC) q = 211299 joules of heat/ 8.95 grams of methanol combusted • ( but the question wanted the molar enthalpy)

  25. CH3OH + 1.5O2 CO2 + 2H2O (l) 4. If 8.95 grams of methanol was used to heat 1500 mL of water from 25oC to 58.7 oC What is the molar enthalpy of combustion of methanol? • Mistake made: increase in temperature = + ΔHoc • +q = 1500g x 4.18 J/ g x oC (58.7 oC – 25 oC) • +q = +211299 joules of heat/ 8.95 grams of methanol combusted • 8.95 grams of methanol = 32 grams /mole • +211299 joules x • 755482J / mole +755 kJ/mole

  26. CH3OH + 1.5O2 CO2 + 2H2O (l) 4. If 8.95 grams of methanol was used to heat 1500 mL of water from 25oC to 58.7 oC What is the molar enthalpy of combustion of methanol? • Correct answer : • q = 1500g x 4.18 J/ g x oC (58.7 oC – 25 oC) • q = 211299 joules of heat/ 8.95 grams of methanol combusted • 8.95 grams of methanol = 32 grams /mole • - 211299 joules x • -755482J / mole -755 kJ/mole

  27. CH3OH + 1.5O2 CO2 + 2H2O (l) 5. If 8.95 grams of methanol was used to heat 1500 mL of water from 25oC to 58.7 oC What is the molar enthalpy of formation of methanol?

  28. Mistake made: ΔHof ≠ ΔHoc • Wrong answer: • q = 1500g x 4.18 J/ g x oC (58.7 oC – 25 oC) • q = 211299 joules of heat/ 8.95 grams of methanol combusted • 8.95 grams of Methanol = 32.0 grams /mole Methanol • - 211299 joules x • -755482J / mole  -755 kJ/mole  the student found the heat of combustion correctly but then uses it incorrectly as the heat of formation: • ΔHoc = [-393.5 + 2(-285.8)] – [ -755]  incorrect substitution • ΔHoc = -210.1  correct answer found the wrong way • This error IF caught should send a red flag to the teacher that this student does not know the difference between enthalpy of reaction and enthalpy of formation.

  29. CH3OH + 1.5O2 CO2 + 2H2O (l) 5. If 8.95 grams of methanol was used to heat 1500 mL of water from 25oC to 58.7 oC What is the molar enthalpy of formation of methanol? • Correct answer: • q = 1500g x 4.18 J/ g x oC (58.7 oC – 25 oC) • q = 211299 joules of heat/ 8.95 grams of methanol combusted • 8.95 grams of methanol = 32.0 grams /mole • - 211299 joules x • -755 = [-393.5 + 2(-285.8)] – [ΔHof]  correct substitution • ΔHof = -210.1 kJ/mole

  30. CH3OH + 1.5O2 CO2 + 2H2O (g) 6. Calculate the enthalpy of reaction for the combustion of one mole of methanol with excess oxygen to produce carbon dioxide and gaseous water?

  31. CH3OH + 1.5O2 CO2 + 2H2O (g) 6. Calculate the enthalpy of reaction for the combustion of one mole of methanol with excess oxygen to produce carbon dioxide and gaseous water? • Mistake made: ΔHof(water gas) ≠ ΔHof (water liquid) • ΔHco = [-393.5 + 2( -285.8)] – [-210.0] • ΔHco = -755.1 kJ/mole of methanol

  32. CH3OH + 1.5O2 CO2 + 2H2O (g) 6. Calculate the enthalpy of reaction for the combustion of one mole of methanol with excess oxygen to produce carbon dioxide and gaseous water? • ΔHoc = [-393.5 + 2(-241.5)] – [ -210.0] = -666.5kJ • OR • CH3OH + 1.5O2 CO2 + 2H2O (l) -755 • 2H2O(l) 2H2O(g)(-88.5) • ΔHoc = -666.5kJ • ΔHov= 2[-285.8] - 2[-241.5] • ΔHov = -88.5

  33. CH3OH + 1.5O2 CO2 + 2H2O (l) 7. Using the table bond enthalpies calculate the molar enthalpy for the combustion of methanol with excess oxygen to produce carbon dioxide and liquid water. • C-H 413 • O=O 495 • C-O 358 • O-H 463 • C=O 799

  34. CH3OH + 1.5O2 CO2 + 2H2O (l) 7. Using the table bond enthalpies calculate the molar enthalpy for the combustion of methanol with excess oxygen to produce carbon dioxide and liquid water. • C-H 413 O=O 495 • C-O 358 O-H 463 • C=O 799 Mistake made formed - broken C=O H-O C-H C-O O-H O=O • ΔHoc = [ 2(799) + 4(463) ]- [ 3(413) + 358 + 463 + 1.5(495)] • 3450 - 2802.5 • ΔHoc = +647.5

  35. CH3OH + 1.5O2 CO2 + 2H2O (l) 7. Using the table bond enthalpies calculate the molar enthalpy for the combustion of methanol with excess oxygen to produce carbon dioxide and liquid water. • C-H 413 O=O 495 • C-O 358 O-H 633 • C=O 799 • Mistake made: bonds ≠ coefficients • CH3OH + 1.5O2 CO2 + 2H2O (l) • ΔHoc = [ (413) + ( 463) + 1.5(495)] – [ (799) – 2(463)] • ΔHoc = -106.5

  36. Correct Way • Correct answer: Must have structural formula • H • | • H—C –O-H + 1.5 O=O  O=C=O + H-O-H • | H-O-H • H • C-H C-O O-H O=O C=O H-O • ΔHoc = [ 3(413) + 358 + 463 + 1.5(495)] - [ 2(799) + 4(463) ] • 2802.5 - 3450 • ΔHoc = - 647.5kJ/mole • Bond enthalpies give different enthalpy of combustion because they don’t take into account intermolecular forces. (good essay question!)

  37. Chemistry Olympiad Question • 23/02. Estimate ∆H for this reaction. • H2(g)+ Cl2(g) 2HCl(g) Bond Energies, kJ·mol–1 H–H 436 Cl–Cl 243 H–Cl 431 (A) 1110 kJ (B) 248 kJ *(C) –183 kJ (D) –248 kJ ΔH = broken - formed ΔH = [H-H + Cl-Cl] – [ 2(H-Cl)] ΔH = [436 +243] – [ 2(431)]

  38. CH3OH + 1.5O2 CO2 + 2H2O (l) 8. The complete combustion of 10.5 g of methanol will produce carbon dioxide, liquid water and 217.7 kJ of heat. Calculate the bond enthalpy of a oxygen hydrogen single bond

  39. CH3OH + 1.5O2 CO2 + 2H2O (l) 8. The complete combustion of 10.5 g of methanol to produce carbon dioxide, liquid water and 217.7 kJ of heat. Calculate the bond enthalpy of a oxygen hydrogen single bond. Mistake made: sign mistake or incorrect substitution or both! C-H C-O O-H O=O C=O O-H • -217.7 = [ 3(413) + 358 + x + 1.5(495)] - [ 2(799) + 4(x) ] • ΔHobond = 319.7 kJ/mole • C-H C-O O-H O=O C=O O-H • 217.7 = [ 3(413) + 358 + x + 1.5(495)] - [ 2(799) + 4(x) ] • ΔHobond = 174.6 kJ/mole

  40. CH3OH + 1.5O2 CO2 + 2H2O (l) 8. The complete combustion of 10.5 g of methanol to produce carbon dioxide, liquid water and 217.7 kJ of heat. Calculate the bond enthalpy of a oxygen hydrogen single bond. 10.5 g = 32 g/mol -217.7kJ = xkJ X= -663.5 C-H C-O O-H O=O C=O O-H -663.5 = [ 3(413) + 358 + x + 1.5(495)] - [ 2(799) + 4(x) ] -663.5 = [2339.5 +x ] – [1598 -4x] -663.5 = 741.5 -3x • ΔHobond = 468.3 kJ/mole

  41. CH3OH + 1.5O2 CO2 + 2H2O (l) 9. Calculate the entropy of reaction for the combustion of one mole of methanol with excess oxygen to produce carbon dioxide and liquid water?

  42. CH3OH + 1.5O2 CO2 + 2H2O (l) • 9. Calculate the entropy of reaction for the combustion of one mole of methanol with excess oxygen to produce carbon dioxide and liquid water? • Mistake made: UNITS • ΔSo = [214+2(70)] – [238 + 1.5(205)] • 354 - 545.5 • ΔSo = -191.5 J/mole • Some will even try kJ • Correct math – just a foot shot on units!

  43. CH3OH + 1.5O2 CO2 + 2H2O (l) • 9. Calculate the entropy of reaction for the combustion of one mole of methanol with excess oxygen to produce carbon dioxide and liquid water? • Correct Way • ΔSo = [214+2(70)] – [238 + 1.5(205)] • 354 - 545.5 • ΔSo = -191.5 J/K x mole • Must have Kelvin

  44. CH3OH + 1.5O2 CO2 + 2H2O (l) 10. What is the value of the standard free energy for the combustion of methanol.

  45. CH3OH + 1.5O2 CO2 + 2H2O (l) 10. What is the value of the standard free energy for the combustion of methanol. • Mistake made: enthalpy - entropy units • ΔGoc=ΔHoc – TΔSoc • ΔGoc=-755.1kJ – 298(-191.5J) • ΔGoc= 56311.9???? Units • ΔHoc from enthalpies of formation not bond energies ( either is OK) =-755.1kJ

  46. CH3OH + 1.5O2 CO2 + 2H2O (l) 10. What is the value of the standard free energy for the combustion of methanol. • Mistake made: Kelvin units • ΔGoc=ΔHoc – TΔSoc • ΔGoc=-755.1 – 25oC (-0.1915) • ΔGoc=-750.3

  47. CH3OH + 1.5O2 CO2 + 2H2O (l) 10. What is the value of the standard free energy for the combustion of methanol. • Correct Way • ΔGoc=ΔHoc – TΔSoc • ΔGoc=-755.1 – 298(-0.1915) • ΔGoc= -698 kJ

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