1. No category

Thermo-Chemistry Curve Balls
• KNOW thine enemy!!!!!!!!!!!!
Equations
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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,
ΔHoc, 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)
• 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 liquid
Any 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
• 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
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 – products
Any 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)
• 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
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
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 included
CH3OH + 1.5O2  CO2 + 2H2O (l) + 755.1 kJ
• Mistake made: ignore the sign of the enthalpy
Question #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)
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
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
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
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
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
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)
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)
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
Question #3a 1998, 1979
ΔHoc = [-393.5 + 2(-285.8)] – [-584.1]
ΔHoc = -381.0kJ/mol
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
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
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
+300.9 kJ·mol–1 (D) +601.8 kJ·mol–1
20.15/300.9 = 40/X kJ
(C)
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/mole
Question #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)
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
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
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 ≠
•
•
•
•
•
•
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.
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
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)
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
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
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)
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 -
• ΔHoc = +647.5
2802.5
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 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
• 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!)
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)]
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) ]
• ΔHobond = 174.6 kJ/mole
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
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)
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
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
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
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)
11. What is the value of the free energy for the
formation of methanol.
• Mistake made: absolute entropy does not equal
entropy of formation
• ΔGof = ΔHof – 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)
11. What is the value of the free energy for the
formation of methanol.
• ΔGof = ΔHof – 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)
12. What is the equilibrium constant for the
combustion of methanol?
•
•
•
•
•
Mistake made: units of R
ΔGo = -RTlnK
ΔGo =-698kJ
-698kJ = - 8.31J x 298 lnK
K= 1.33
CH3OH + 1.5O2  CO2 + 2H2O (l)
12. What is the equilibrium constant for the
combustion of methanol?
•
•
•
•
•
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)
12. What is the equilibrium constant for the
combustion of methanol?
•
•
•
•
•
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)
12. What is the equilibrium constant for the
combustion of methanol?
•
•
•
•
•
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)
12. What is the equilibrium constant for the
combustion of methanol?
•
•
•
•
•
•
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
-ΔH
+ΔH
+ΔH
+ΔS always spontaneous -ΔG
-ΔS decrease temperature more
spontaneous -ΔG
+ΔS increase temperature more
spontaneous - ΔG
-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)
Chemical Olympiad Question
• 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)
Chemical Olympiad Question
• 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
Chemical Olympiad Question
• 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