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Thermo-Chemistry Curve Balls KNOW thine enemy!!!!!!!!!!!!

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Presentation on theme: "Thermo-Chemistry Curve Balls KNOW thine enemy!!!!!!!!!!!!"— Presentation transcript:

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

2 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

3 CH 3 OH + O 2  CO 2 + H 2 O (l) 1. What is the enthalpy of combustion, ΔH o c, for the complete combustion of methanol to produce carbon dioxide and liquid water?

4 CH 3 OH + XO 2  CO 2 + YH 2 O (l) 1. What is the enthalpy of combustion, ΔH o c, for the complete combustion of methanol to produce carbon dioxide and liquid water? Mistake made: unbalance equation. Any test any year (basic chemistry mistake) ΔH c o = products – reactants ΔH c o = [ ] – [-210.0] ΔH c o = kJ/mole of methanol

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6 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 1. What is the enthalpy of combustion, ΔH o c, for the complete combustion of methanol to produce carbon dioxide and liquid water? Mistake made: water gas instead of liquid Any test any year (basic chemistry mistake) ΔH c o = products – reactants ΔH c o = [ ] – [-210.0] unbalanced ΔH c o = -425 kJ/mole of methanol ΔH c o = [ (-241.5)] – [-210.0] balanced ΔH c o = kJ/mole of methanol

7 21/03. Calculate the amount of energy mol-1) released when mol of diborane, B 2 H 6, reacts with oxygen to produce solid B 2 O 3 and steam. ∆ Hf° kJ B 2 H 6(g) 35 B 2 O 3(s) H 2 O (l) -285 H 2 O (g) -241 *(A) 2030 kJ (B) 2160 kJ (C) 3300kJ (D) 3430 kJ B 2 H 6 + 3O 2  B 2 O 3 + 3H 2 O (g) ∆ H = [ (-241)] – 35

8 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 1. What is the enthalpy of combustion, ΔH o c, for the complete combustion of methanol to produce carbon dioxide and liquid water? Mistake made: reactants – products Any test any year (basic chemistry mistake) ΔH c o = reactants - products ΔH c o = [-210.0] – [ ] unbalanced ΔH c o = kJ/mole of methanol ΔH c o = [-210.0] – [ (-285.8)] balanced ΔH c o = kJ/mole of methanol

9 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 1. What is the enthalpy of combustion, ΔH o c, for the complete combustion of methanol to produce carbon dioxide and liquid water? Correct answer: ΔH c o = products – reactants ΔH c o = [ (-285.8)] – [-210.0] ΔH c o = kJ/mole of methanol

10 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! rounds to rounds to rounds to -200 ΔH c o = [ (-300)] – [-200] ΔH c o = ΔH c o = around -800 the correct answer is The answer should be less than ( more positive) because the rounding up of the product’s enthalpies

11 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 2. What is the heat of formation, ΔH o f, of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat? OR the equation could have the heat included CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) kJ

12 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 2. What is the heat of formation, ΔH o f, of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat? OR the equation could have the heat included CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) kJ Mistake made: ignore the sign of the enthalpy Question #3b = [ (-285.8)] – [ΔH o f ] = x ΔH o f = kJ/mole

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14 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 2. What is the heat of formation, ΔH o f, of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat? OR the equation could have the heat included CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) kJ Mistake made: incorrect substitution ( no true understanding between formation and combustion) and sign problem. Question #3b 1998 “ΔH o f “ = [ (-285.8)] – [755.1] confused between combustion and formation ΔH o f = kJ/mole

15 2CH 3 OH + 3O 2  2CO 2 + 4H 2 O (l) 2. What is the heat of formation, ΔH o f, of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat? OR the equation could have the heat included 2CH 3 OH + 3O 2  2CO 2 + 4H 2 O (l) kJ Mistake made: over balancing the equation leading to a stoich mistake. Any test any year = [2(-393.5) + 4(-285.8)] – [2x] doubled everything BUT the molar enthalpy of combustion ΔH o f =

16 2CH 3 OH + 3O 2  2CO 2 + 4H 2 O (l) 2. What is the heat of formation, ΔH o f, of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat? OR the equation could have the heat included 2CH 3 OH + 3O 2  2CO 2 + 4H 2 O (l) 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 ΔH o f = -420

17 2CH 3 OH + 3O 2  2CO 2 + 4H 2 O (l) 2. What is the heat of formation, ΔH o f, of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat? OR the equation could have the heat included 2CH 3 OH + 3O 2  2CO 2 + 4H 2 O (l) kJ Mistake made: incorrect substitution “ΔH o f “ = [2(-393.5) + 4(-285.8)] – [2(-755.1)] confused between combustion and formation “ΔH o f “ = -420 kJ/mole

18 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 2. What is the heat of formation, ΔH o f, of methanol if one mole of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat? OR the equation could have the heat included CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) kJ Correct answer: = [ (-285.8)] – [x] ΔH o f = kJ/mole

19 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 3. What is the heat of combustion, ΔH o c, of methanol if grams of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat?

20 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 3. What is the heat of combustion, ΔH o c, of methanol if grams of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and 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)

21 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 3. What is the heat of combustion, ΔH o c, of methanol if grams of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat? Mistake made: incorrect molar substitution Question #3a 1998, 1979 ΔH o c = [ (-285.8)] – [-584.1] ΔH o c = kJ/mol

22 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 3. What is the heat of combustion, ΔH o c, of methanol if grams of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat? Mistake made: incorrect molar substitution = [ (-285.8)] – [ΔH o f ] ΔH o f = -381kJ/mol OR use the wrong enthalpy with the wrong sign: = [ (-285.8)] – [ΔH o f ] ΔH o f = kJ/mol

23 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 3. What is the heat of combustion, ΔH o c, of methanol if grams of methanol reacts with excess oxygen to produce carbon dioxide, liquid water, and kJ of heat? Correct way 24.75g = 32 g kJ x kJ x = [ (-285.8)] – [ΔH o f ] = [ (-285.8)] – [ΔH o f ] ΔH o f = kJ/mole

24 Chemistry Olympiad Question 25/00. What is the standard enthalpy of formation of MgO(s) if kJ is evolved when 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) kJ·mol–1 (D) kJ·mol– /300.9 = 40/X kJ

25 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 4.If 8.95 grams of methanol was used to heat 1500 mL of water from 25 o C to 58.7 o C What is the molar enthalpy of combustion of methanol?

26 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 4. If 8.95 grams of methanol was used to heat 1500 mL of water from 25 o C to 58.7 o C What is the molar enthalpy of combustion of methanol? Mistake made: q/grams ≠ ΔH/mole Question #3a 1998, #d 1995, 1979,1989b, 2001a q = 1500g x 4.18 J/ g x o C (58.7 o C – 25 o C) q = joules of heat/ 8.95 grams of methanol combusted ( but the question wanted the molar enthalpy)

27 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 4. If 8.95 grams of methanol was used to heat 1500 mL of water from 25 o C to 58.7 o C What is the molar enthalpy of combustion of methanol? Mistake made: increase in temperature = + ΔH o c +q = 1500g x 4.18 J/ g x o C (58.7 o C – 25 o C) +q = joules of heat/ 8.95 grams of methanol combusted 8.95 grams of methanol = 32 grams /mole joules x J / mole  +755 kJ/mole

28 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 4. If 8.95 grams of methanol was used to heat 1500 mL of water from 25 o C to 58.7 o C What is the molar enthalpy of combustion of methanol? Correct answer : q = 1500g x 4.18 J/ g x o C (58.7 o C – 25 o C) q = joules of heat/ 8.95 grams of methanol combusted 8.95 grams of methanol = 32 grams /mole joules x J / mole  -755 kJ/mole

29 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 5. If 8.95 grams of methanol was used to heat 1500 mL of water from 25 o C to 58.7 o C What is the molar enthalpy of formation of methanol?

30 Mistake made: ΔH o f ≠ ΔH o c Wrong answer: q = 1500g x 4.18 J/ g x o C (58.7 o C – 25 o C) q = joules of heat/ 8.95 grams of methanol combusted 8.95 grams of Methanol = 32.0 grams /mole Methanol joules x J / mole  -755 kJ/mole  the student found the heat of combustion correctly but then uses it incorrectly as the heat of formation: ΔH o c = [ (-285.8)] – [ -755]  incorrect substitution ΔH o c =  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.

31 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 5. If 8.95 grams of methanol was used to heat 1500 mL of water from 25 o C to 58.7 o C What is the molar enthalpy of formation of methanol? Correct answer: q = 1500g x 4.18 J/ g x o C (58.7 o C – 25 o C) q = joules of heat/ 8.95 grams of methanol combusted 8.95 grams of methanol = 32.0 grams /mole joules x -755 = [ (-285.8)] – [ΔH o f ]  correct substitution ΔH o f = kJ/mole

32 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (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?

33 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (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: ΔH o f(water gas) ≠ ΔH o f (water liquid) ΔH c o = [ ( )] – [-210.0] ΔH c o = kJ/mole of methanol

34 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (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? ΔH o c = [ (-241.5)] – [ ] = kJ OR CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) H 2 O (l)  2H 2 O (g) (-88.5) ΔH o c = kJ ΔH o v = 2[-285.8] - 2[-241.5] ΔH o v = -88.5

35 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (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-H413 O=O 495 C-O358 O-H463 C=O799

36 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (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 413O=O 495 C-O 358O-H463 C=O 799 Mistake made  formed - broken C=O H-O C-H C-O O-H O=O ΔH o c = [ 2(799) + 4(463) ]- [ 3(413) (495)] ΔH o c =

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38 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (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 413O=O 495 C-O 358O-H633 C=O 799 Mistake made: bonds ≠ coefficients CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) ΔH o c = [ (413) + ( 463) + 1.5(495)] – [ (799) – 2(463)] ΔH o c =

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

40 Chemistry Olympiad Question 23/02. Estimate ∆ H for this reaction. H 2(g) + Cl 2(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 = [ ] – [ 2(431)]

41 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 8. The complete combustion of 10.5 g of methanol will produce carbon dioxide, liquid water and kJ of heat. Calculate the bond enthalpy of a oxygen hydrogen single bond

42 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 8. The complete combustion of 10.5 g of methanol to produce carbon dioxide, liquid water and 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 = [ 3(413) x + 1.5(495)] - [ 2(799) + 4(x) ] ΔH o bond = kJ/mole C-H C-O O-H O=O C=O O-H = [ 3(413) x + 1.5(495)] - [ 2(799) + 4(x) ] ΔH o bond = kJ/mole

43 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 8. The complete combustion of 10.5 g of methanol to produce carbon dioxide, liquid water and kJ of heat. Calculate the bond enthalpy of a oxygen hydrogen single bond g = 32 g/mol kJ = xkJX= C-H C-O O-H O=O C=O O-H = [ 3(413) x + 1.5(495)] - [ 2(799) + 4(x) ] = [ x ] – [ x] = x ΔH o bond = kJ/mole

44 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (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?

45 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (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 ΔS o = [214+2(70)] – [ (205)] ΔS o = J/mole Some will even try kJ Correct math – just a foot shot on units!

46 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (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 ΔS o = [214+2(70)] – [ (205)] ΔS o = J/K x mole Must have Kelvin

47 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 10. What is the value of the standard free energy for the combustion of methanol.

48 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 10. What is the value of the standard free energy for the combustion of methanol. Mistake made: enthalpy - entropy units ΔG o c = ΔH o c – TΔS o c ΔG o c =-755.1kJ – 298(-191.5J) ΔG o c = ???? Units ΔH o c from enthalpies of formation not bond energies ( either is OK) =-755.1kJ

49 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 10. What is the value of the standard free energy for the combustion of methanol. Mistake made: Kelvin units ΔG o c = ΔH o c – TΔS o c ΔG o c = – 25 o C ( ) ΔG o c =-750.3

50 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 10. What is the value of the standard free energy for the combustion of methanol. Correct Way ΔG o c = ΔH o c – TΔS o c ΔG o c = – 298( ) ΔG o c = -698 kJ

51 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 11. What is the value of the free energy for the formation of methanol.

52 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 11. What is the value of the free energy for the formation of methanol. Mistake made: absolute entropy does not equal entropy of formation ΔG o f = ΔH o f – TΔS o f ΔG o f =-201 – 298(0.238) 238J/k x mol from data of absolute entropies ΔG o f = kJ

53 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 11. What is the value of the free energy for the formation of methanol. ΔG o f = ΔH o f – TΔS o f C O 2 + 2H 2  CH 3 OH ΔS o f = 240 – [ (205) + 2(131)] ΔS o f = J/K -201 ΔH o f(methanol) from data table ΔG o f =-201 – 298( ) ΔG o f = kJ

54 R = WHAT Force = mass x acceleration Force = 10,000kg x 9.8 m/s 2 Force = 1.0 x 10 5 kgxm/s 2 Pressure = force / area Pressure = 1.0 x 10 5 kgxm/s 2 /m 2 Units of pressure = kg /s 2 m Work = pressure x ΔVol Work (Joules) = kg /s 2 m x m 3 Joules = kg xm 2 /s 2

55 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 12. What is the equilibrium constant for the combustion of methanol?

56 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 12. What is the equilibrium constant for the combustion of methanol? Mistake made: units of R ΔG o = -RTlnK ΔG o =-698kJ -698kJ = J x 298 lnK K= 1.33

57 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 12. What is the equilibrium constant for the combustion of methanol? Mistake made: wrong R ΔG o = -RTlnK ΔG o =-698kJ J = “J” x 298 lnK K= error (time killer)

58 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 12. What is the equilibrium constant for the combustion of methanol? Mistake made: 2 nd ln ΔG o = -RTlnK ΔG o =-698kJ J = J x 298 lnK K=  has not taken the antilog

59 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 12. What is the equilibrium constant for the combustion of methanol? Mistake made: wrong LOG ΔG o = -RTlnK ΔG o =-698kJ J = J x 298 log K K= error ???? ( time killer)

60 CH 3 OH + 1.5O 2  CO 2 + 2H 2 O (l) 12. What is the equilibrium constant for the combustion of methanol? Correct Way ΔG o = -RTlnK ΔG o =-698kJ J = J x 298 lnK lnK= K= error -- too big ?? x or more Combustion reactions are not the norm for ΔG o = -RTlnK equation  weak acid /base, Ksp, Kp are better suited……

61 Essay -ΔH+ΔSalways spontaneous -ΔG -ΔH-ΔSdecrease temperature more spontaneous -ΔG +ΔH+ΔSincrease temperature more spontaneous - ΔG +ΔH-ΔSnever spontaneous +ΔG NEVER say “decrease or increase” when referring to a more negative or more positive change

62 Chemical Olympiad Question 28/00. What are the signs of ∆ H˚ and ∆ S˚ for a reaction that is spontaneous at all temperatures? ΔH˚ ΔS˚ (A) + +spont at + (higher temp) (B) + –never spont *(C) – +always spont (D) – –spont at – (lower temp)

63 Chemical Olympiad Question 26/01. The ∆ H o and ∆ S o values for a particular reaction are –60.0 kJ and kJ·K –1 respectively. Under what conditions is this reaction spontaneous? (A) all conditions *(B) T < 300 K (C) T = 300 K (D) T > 300 K – –spont at – (lower temp)

64 Chemical Olympiad Question 27/02. For the reaction PCl 3(g) + Cl 2(g)  PCl 5(g), ∆ H o = –86 kJ. Under what temperatures is this reaction expected to be spontaneous? (A) no temperatures (B) high temperatures only (C) all temperatures *(D) low temperatures only - ∆ H and - ∆ S = more spont at –(low) temp

65 Chemical Olympiad Question 24/02. Which reaction occurs with an increase in entropy? *(A) 2C (s) + O 2(g)  2CO (g) solid to a gas (B) 2H 2 S (g) + SO 2(g)  3S (s) + 2H 2 O (g) (C) 4Fe (s) + 3O 2(g)  2Fe 2 O 3(s) (D) CO (g) + 2H 2(g)  CH 3 OH (l)

66 Thank You For Listening I Hope this helps Now let’s try our curve ball skills on a modified 1984 AP thermo question


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