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Tutorial 5: Question 11
Presented by 12S7F Yu Tian :D
Problem statement >.<
• Aluminium fluoride is made industrially by reacting aluminium oxide with hydrogen fluoride at a high temperature.
Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
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Problem statement >.<
Al2O3 (s) HF (g) AlF3 (s) H2O (g)
ΔHfo / kJ mol-1 -1676 -271 -1504 -242
ΔSfo / J K-1 mol-1 -313 +7.0 -266 -44.4
UNIT: J K-1 mol-1
NOT kJ K-1 mol-1
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Problem statement >.<(a) Use the data given to calculate:
• The standard enthalpy change, ΔHo , of this reaction
• The standard entropy change, ΔSo , of this reaction
Hess’s Law
Energy cycle
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Problem wracking x.x
Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
ΔHo
?
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
ΔHfo of Al2O3 (s)
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Problem wracking x.x
Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
ΔHo
?
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
ΔHfo of HF(s) x 6
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Problem wracking x.x
Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
ΔHo
?
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
ΔHfo of AlF3 (s) x 2
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Problem wracking x.x
Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
ΔHo
?
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
ΔHfo of H2O (g) x 3
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Problem wracking x.x
Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
ΔHo
?
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
ΔHfo of H2O (g)
x 3ΔHfo of Al2O3 (s)
ΔHfo of AlF3 (s) x 2
ΔHfo of HF(s)
x 6
By Hess’s Law, the enthalpy change of a particular reaction is
determined only by the initial and final states of the system regardless of the pathway taken.
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Problem wracking x.x
Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
ΔHo =
?
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
ΔHfo of H2O (g)
x 3ΔHfo of Al2O3 (s)
ΔHfo of AlF3 (s) x 2
ΔHfo of HF(s)
x 6
ΔHfo of Al2O3 (s)
+ΔHf
o of HF(s) x 6 --ΔHf
o of AlF3 (s) x 2 ΔHfo of H2O (g) x 3
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Problem wracking x.x
Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
ΔHo =
?
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
ΔHfo of H2O (g)
x 3ΔHfo of Al2O3 (s)
ΔHfo of AlF3 (s) x 2
ΔHfo of HF(s)
x 6
( - 1676 )+
( -242 )x 6 --( -1504 ) x 2 ( -242 ) x 3
= -432 kJ mol-1
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Problem wracking x.x
Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
ΔSo =
?
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
ΔSfo of H2O (g)
x 3ΔSfo of Al2O3 (s)
ΔSfo of AlF3 (s) x 2
ΔSfo of HF(s)
x 6
ΔSfo of Al2O3 (s)
+ΔSf
o of HF(s) x 6 --ΔSf
o of AlF3 (s) x 2 ΔSfo of H2O (g) x 3
ΔSo
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Problem wracking x.x
Al2O3 (s) + 6HF (g) 2AlF3 (s) + 3H2O (g)
ΔSo =
?
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
ΔSfo of H2O (g)
x 3ΔSfo of Al2O3 (s)
ΔSfo of AlF3 (s) x 2
ΔSfo of HF(s)
x 6
( - 313 )+
( + 7.0 )x 6 --( - 266 ) x 2 ( - 44.4 ) x 3
= - 394 J mol-1 (to 3 s.f. )
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(b) Use the values calculated in (a) to calculate ΔGo .
Standard Gibbs Free energy change
ΔGo is measured at 298 K and 1 atm
ΔGo = ΔHo – T x ΔSo
= (-432) x 103 – (298) x (-394.2) J mol-1
= - 315 J mol-1 (to 3 s.f.)
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Problem statement
(c) How will the value of ΔGo for this reaction change with temperature?
What consequences will this have for the conditions used to make AlF3 industrially?
What is the relationship between T and ΔGo ?
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Problem wracking
(c) How will the value of ΔGo for this reaction change with temperature?
Formula: ΔGo = ΔHo – ΔSo x T GRADIANT OF THE GRAPH
ΔGo
T0
ΔHo
Y- INTERCEPT
POSITIVE GRADIENT
ΔSo is negative
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• Ans:As value of ΔSo is negative, according to
formula Formula: ΔGo = ΔHo – ΔSo x T, the higher the reaction temperature, the more positive the value of ΔGo becomes.
(c) How will the value of ΔGo for this reaction change with temperature?
Problem wracking
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(c) What consequences will this have for the conditions used to make AlF3 industrially?
Problem wracking
Sign of ΔGo
Positive Reaction is not feasible.
0 The system is at equilibrium
Negative Reaction is feasible
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(c) What consequences will this have for the conditions used to make AlF3 industrially?
Problem wracking
From ΔGo = - 315 J mol-1
- Under standard condition, the reaction is not spontaneous/feasible
For the reaction to be feasible, ΔGo > 0
T > 1100 K or T> 823 ℃- High temperature is needed in industrial production of AlF3
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Thank you!
