Thermo-Chemistry Curve Balls

Thermo-Chemistry Curve Balls

• KNOW thine enemy!!!!!!!!!!!!

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

CH3OH + O2 CO2 + H2O (l)

1. What is the enthalpy of combustion, ΔHo

c, for the complete combustion of methanol to produce carbon dioxide

and liquid water?

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

CH3OH + 1.5O2 CO2 + 2H2O (l)



• Mistake made: water gas instead of liquidAny test any year (basic chemistry mistake)


• ΔHco = [-393.5 + -241.5] – [-210.0] unbalanced

• ΔHco = -425 kJ/mole of methanol

• ΔHco = [-393.5 + 2(-241.5)] – [-210.0] balanced


• 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 kJB2H6 + 3O2 B2O3 + 3H2O (g) ∆H = [-1272 + 3(-241)] – 35

CH3OH + 1.5O2 CO2 + 2H2O (l)



• 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

CH3OH + 1.5O2 CO2 + 2H2O (l)



• Correct answer:


• ΔHco = [-393.5 + 2(-285.8)] – [-210.0]


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• ΔHc

o = [-400 + 2(-300)] – [-200]• ΔHc

o = -1000 - -200• ΔHc

o = 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

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

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 includedCH3OH + 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

CH3OH + 1.5O2 CO2 + 2H2O (l)




• 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

2CH3OH + 3O2 2CO2 + 4H2O (l)



OR the equation could have the heat included2CH3OH + 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

2CH3OH + 3O2 2CO2 + 4H2O (l)




• 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

2CH3OH + 3O2 2CO2 + 4H2O (l)




• Mistake made: incorrect substitution

• “ΔHof “ = [2(-393.5) + 4(-285.8)] – [2(-755.1)]

confused between combustion and formation

• “ΔHof “ = -420 kJ/mole

CH3OH + 1.5O2 CO2 + 2H2O (l)




• Correct answer:

-755.1 = [-393.5 + 2(-285.8)] – [x]


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?

CH3OH + 1.5O2 CO2 + 2H2O (l)



• 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)

CH3OH + 1.5O2 CO2 + 2H2O (l)



Mistake made: incorrect molar substitutionQuestion #3a 1998, 1979

ΔHoc = [-393.5 + 2(-285.8)] – [-584.1]

ΔHoc = -381.0kJ/mol

CH3OH + 1.5O2 CO2 + 2H2O (l)



• 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)] – [ΔHo

f]

• ΔHof = -1549.2kJ/mol

CH3OH + 1.5O2 CO2 + 2H2O (l)



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

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

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?

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)

CH3OH + 1.5O2 CO2 + 2H2O (l)


• 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

CH3OH + 1.5O2 CO2 + 2H2O (l)


• 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

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?

Mistake made: ΔHof ≠ ΔHo

c

• 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.

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


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?

CH3OH + 1.5O2 CO2 + 2H2O (g)


• Mistake made: ΔHof(water gas) ≠ ΔHo

f (water liquid)

• ΔHco = [-393.5 + 2( -285.8)] – [-210.0]


CH3OH + 1.5O2 CO2 + 2H2O (g)


• 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

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

CH3OH + 1.5O2 CO2 + 2H2O (l)


• 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

CH3OH + 1.5O2 CO2 + 2H2O (l)


• 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

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• ΔHo

c = [ 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!)

Chemistry Olympiad Question

• 23/02. Estimate ∆H for this reaction.• H2(g) + Cl2(g) 2HCl(g)

Bond Energies, kJ·mol–1

H–H 436Cl–Cl 243H–Cl 431(A) 1110 kJ (B) 248 kJ*(C) –183 kJ (D) –248 kJΔH = broken - formed ΔH = [H-H + Cl-Cl] – [ 2(H-Cl)]ΔH = [436 +243] – [ 2(431)]

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

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) ]


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


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?

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!

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

CH3OH + 1.5O2 CO2 + 2H2O (l)

10. What is the value of the standard free energy for the combustion of methanol.

CH3OH + 1.5O2 CO2 + 2H2O (l)


• Mistake made: enthalpy - entropy units

• ΔGoc = ΔHo

c – TΔSoc

• ΔGoc =-755.1kJ – 298(-191.5J)

• ΔGoc = 56311.9???? Units

• ΔHoc from enthalpies of formation not bond

energies ( either is OK) =-755.1kJ

CH3OH + 1.5O2 CO2 + 2H2O (l)


• Mistake made: Kelvin units

• ΔGoc = ΔHo

c – TΔSoc

• ΔGoc =-755.1 – 25oC (-0.1915)

• ΔGoc =-750.3

CH3OH + 1.5O2 CO2 + 2H2O (l)


• Correct Way

• ΔGoc = ΔHo

c – TΔSoc

• ΔGoc =-755.1 – 298(-0.1915)

• ΔGoc = -698 kJ

CH3OH + 1.5O2 CO2 + 2H2O (l)

11. What is the value of the free energy for the formation of methanol.

CH3OH + 1.5O2 CO2 + 2H2O (l)


• Mistake made: absolute entropy does not equal entropy of formation

• ΔGof = ΔHo

f – TΔSof

• ΔGof =-201 – 298(0.238)

238J/k x mol from data of absolute entropies

• ΔGof =-271.9 kJ

CH3OH + 1.5O2 CO2 + 2H2O (l)


• ΔGof = ΔHo

f – TΔSof

C + 0.5 O2 + 2H2 CH3OH

• ΔSof = 240 – [6 + 0.5(205) + 2(131)]

• ΔSof = -130.5 J/K

• -201 ΔHof(methanol) from data table

• ΔGof =-201 – 298(-0.1305)

• ΔGof = -162.1 kJ

R = 8.31 - WHAT

• Force = mass x acceleration• Force = 10,000kg x 9.8 m/s2

• Force = 1.0 x 105 kgxm/s2

• Pressure = force / area• Pressure = 1.0 x 105 kgxm/s2/m2

• Units of pressure = kg /s2m• Work = pressure x ΔVol• Work (Joules) = kg /s2m x m3

• Joules = kg xm2 /s2

CH3OH + 1.5O2 CO2 + 2H2O (l)

12. What is the equilibrium constant for the combustion of methanol?

CH3OH + 1.5O2 CO2 + 2H2O (l)


• Mistake made: units of R• ΔGo = -RTlnK• ΔGo =-698kJ• -698kJ = - 8.31J x 298 lnK• K= 1.33

CH3OH + 1.5O2 CO2 + 2H2O (l)


• Mistake made: wrong R• ΔGo = -RTlnK• ΔGo =-698kJ• -698000J = - 0.0821 “J” x 298 lnK• K= error (time killer)

CH3OH + 1.5O2 CO2 + 2H2O (l)


• Mistake made: 2nd ln• ΔGo = -RTlnK• ΔGo =-698kJ• -698000J = - 8.31J x 298 lnK• K= 281.9 has not taken the antilog

CH3OH + 1.5O2 CO2 + 2H2O (l)


• Mistake made: wrong LOG• ΔGo = -RTlnK• ΔGo =-698kJ• -698000J = - 8.31J x 298 log K• K= error ???? ( time killer)

CH3OH + 1.5O2 CO2 + 2H2O (l)


• Correct Way• ΔGo = -RTlnK• ΔGo =-698kJ• -698000J = - 8.31J x 298 lnK• lnK= 281.9 • K= error -- too big ?? x 10100 or more

• Combustion reactions are not the norm for ΔGo = -RTlnK equation weak acid /base, Ksp, Kp are better suited……

Essay

-ΔH +ΔS always spontaneous -ΔG

-ΔH -ΔS decrease temperature more spontaneous -ΔG

+ΔH +ΔS increase temperature more spontaneous - ΔG

+ΔH -ΔS never spontaneous +ΔG

NEVER say “decrease or increase” when referring to a more negative or more positive change

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)


• 26/01. The ∆Ho and ∆So values for a particular reaction are –60.0 kJ and -0.200 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)


• 27/02. For the reaction PCl3(g) + Cl2(g) PCl5(g), ∆Ho = –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


• 24/02. Which reaction occurs with an increase in entropy?

• *(A) 2C(s) + O2(g) 2CO(g) solid to a gas

• (B) 2H2S(g) + SO2(g) 3S(s) + 2H2O(g)

• (C) 4Fe(s) + 3O2(g) 2Fe2O3(s)

• (D) CO(g) + 2H2(g) CH3OH(l)

Thank You For Listening

• I Hope this helps

• Now let’s try our curve ball skills on a modified 1984 AP thermo question

Thermo-Chemistry Curve Balls

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