#### Transcript Q11 Chem Tut 5 - 12S7F-note

```Presented by 12S7F Yu Tian :D
1
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)
2
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
3
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
4
Problem wracking x.x
Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O
(g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
5
Problem wracking x.x
Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
6
Problem wracking x.x
Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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Problem wracking x.x
Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
8
Problem wracking x.x
Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
10
Problem wracking x.x
Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
11
Problem wracking x.x
Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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Problem wracking x.x
Al2O3 (s) + 6HF (g)  2AlF3 (s) + 3H2O (g)
2Al (s) + 3H2 (g) + 3F2 (g) + O2 (g)
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(b) Use the values calculated in (a) to calculate
ΔGo .
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?
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Problem wracking
(c)
How will the value of ΔGo for this reaction
change with temperature?
GRADIANT OF THE GRAPH
Formula: ΔGo = ΔHo – ΔSo x T
ΔGo
0
Y- INTERCEPT
ΔSo is negative
T
ΔHo
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Problem wracking
(c) How will the value of ΔGo for this reaction
change with temperature?
• 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.
17
Problem wracking
(c) What consequences will this have for the
conditions used to make AlF3 industrially?
Sign of ΔGo
Positive
Reaction is
not feasible.
0
The system is
at equilibrium
Negative
Reaction is
feasible
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Problem wracking
(c) What consequences will this have for the
conditions used to make AlF3 industrially?
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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